ML031250607
| ML031250607 | |
| Person / Time | |
|---|---|
| Site: | Diablo Canyon  |
| Issue date: | 03/03/2003 |
| From: | Tumminelli S Enercon Services |
| To: | Office of Nuclear Reactor Regulation |
| References | |
| +sispmjr200505, -RFPFR, DIL-03-003, PGE-009, TAC L23399 PGE-009-CALC-007, Rev 0 | |
| Download: ML031250607 (161) | |
Text
ATTACHMENT 3
Calc. No.
PGE-009-CALC-007 ENGINEERING CALCULATION Rev.
C E
E
~COVER SHEET ENERCON SERVICES, INC.
Sheet 1 of 160
Title:
ISFSI Cask Storage Pad Steel Reinforcement Client:
PG&E Job No.
PGE-009 Purpose Of Calculation:
The purpose of this calculation is to compute the size and spacing of the steel reinforcement necessary to accommodate the forces and moments within the storage pad for the temperatures resulting from the heat of hydration during the curing process, from the shrinkage of the concrete and from the seismic demand. The ISFSI Facility will contain (7) pads, which will support (20) HI-Storm Storage Casks per pad. Some results from this Calculation will be used in Calculation No. PGE-009-CALC-00 1 to evaluate the anchor plate embedment capacity.
NOTE: This Calculation is furnished as part of PG&E Contract No. 4600010841, Change Order No. 001 Scope Of Revision:
Initial issue.
Revision Impact On Results:
N/A D9C Safety Related Preliminary Calculation LI Non-Safety Related Z
Final Approvals (Print Name and Sign)_
Originator S.C. TUMMINELLI Date March 11, 2003 Reviewer Verification Engineer K.L. WHITMORE Date March 11, 2003 Approver R.F. EVERS Date March 11, 2003
ENGINEERING CALCULATION REVISION STATUS SHEET ENGINEERING CALCULATION REVISION
SUMMARY
REVISION NO.
DATE DESCRIPTION 0
3/11/03 Initial Issue CALCULATION SHEET REVISION STATUS SHEET NO.
REVISION NO.
SHEET NO.
REVISION NO.
All 0
N/A N/A APPENDIX AND ATTACHMENT REVISION STATUS APPENDIX NUMBER ISSUE REV.
REISSUE DATE DATE DATE APPENDIX NUMBER ISSUE REV.
REISSUE DATE DATE DATE
SHEET 3
OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCO ENERCON SERVICES, INC.
Method of Review:
The calculation has been independently reviewed in accordance with the requirements of ENERCON Corporate Standard Procedure 3.01. The independent verification of the calculation was performed by a detailed review and check of the entire calculation. This included verification of inputs, methodology, results and conclusions as well as a check of the mathematical accuracy of the computations.
Results:
The calculation has been independently verified to be mathematically correct and to be performed in accordance with license and design basis requirements and applicable codes. Inputs are appropriate and are obtained from verified source documents. The calculation is sufficiently documented and detailed to permit independent verification. No assumptions are made other than conservative simplifying assumptions, which are identified and do not require confirmation.
The methodology used to evaluate the Cask Storage Pad Steel Reinforcement has been independently verified to be appropriate and consistent with design and code requirements. The design requirements specified in the reference documents have been adequately addressed and the code requirements have been appropriately considered. All applicable provisions of the design codes have been addressed and all design requirements have been met. The results are clearly identified and the conclusions are supported by the results of the calculation.
The steel reinforcement is capable of resisting all applied loads including those due to curing stresses and seismic loads. Furthermore, the steel reinforcing has been shown to be adequate to limit cracks during curing to widths that are within acceptable limits. Forces have been developed for use in design of the embedment structure.
Thus, the analysis has been independently verified to be technically correct and to be consistent with license and design basis requirements. The results and conclusions accurately reflect the findings of the calculation.
SHEET 4
OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Table of Contents Sheet Introduction 5
References 6
Section 1 - Assessments of Design Requirements Design Requirements per PG&E Specification 10012-N-NPG Section 6.4 Design Requirements per Holtec Design Criteria Document, Reference 2 Design Requirements per ANSI/ANS-57.2 1992, Reference 10 Design Requirements per NUREG 1536, Reference 7 Design Requirements per NUREG 1567, Reference 8 Load Combinations per ACI 349-77, Reference II Assessment of NRC DG 1098, Reference 9 Reduction of the Load Combinations Section 2 - Sizing of Reinforcement for Thermal and Shrinkage Demand 18 Thermal Demand due to Curing Temperatures and Shrinkage Initial Sizing of Reinforcement Approximate Allowable Thermal and Shrinkage Stresses Detailed Evaluation for Thermal and Shrinkage Demands - Constrained Model Detailed Evaluation for Thermal Demands - Unconstrained Model Section 3 - Evaluation for Seismic Loads 66 Basic Data for Strength Method North-South Section Concrete Capacity 67 East-West Section Concrete Capacity 100 North-South and East-West Section Evaluations 133 Shear Evaluation 154 Development Length and Lap Splice Requirements 154 Seismic Forces in Reinforcement 157 Summary and Conclusions 160
SHEET 5
OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Introduction The purpose of this calculation is to size the reinforcement for the ISFSI storage pads. This calculation will therefore refer to the governing document for the design, which is the PG&E Specification, 10012-N-NPG (Reference 1) including its subordinate specifications and referenced regulatory requirements.
The PG&E Specification is written to procure an entire Dry Cask Storage System which includes the casks, cask transfer equipment etc. as well as related structures. This calculation will only draw upon those requirements of the Specification that relate specifically to the ISFSI Storage Pad. Those requirements are specified in Section 6.4, Proposed ISFSI Facility, of Reference 1. The various requirements of Section 6.4 will be addressed as well as the various subordinate specifications and regulatory documents.
The hierarchy of the design requirement documents begins with the governing document for the work, which is the PG&E specification, Reference 1. This requires the cask supplier to develop a design criteria document, which is the Holtec Design Criteria, Reference 2. The Holtec document, in turn, invokes some regulatory and industry documents, as does the PG&E specification. These subordinate documents are NUREG 1536, NUREG 1567, USNRC DG 1098, ANSI 57.9 and ACI 349 (References 7, 8, 9, 10 and 11 respectively.) In some areas, the requirements overlap and are slightly different from one another. Further, the documentation that supports the design and demonstrates that the design complies with all the various criteria does not all reside in one complete document. Therefore, as much as possible, this calculation will compile all the criteria and identify where specific requirements have been met.
The implementing documents that demonstrate compliance with the criteria are the Holtec Cask analysis report (Reference 3) and the ENERCON calculations for the Embedment, Seismic Analysis of the pad/cask configuration, Thermal/Shrinkage Analysis of the pad (References 4, 5 and 6) and this calculation. The Holtec Cask analysis provides input loads for the Embedment and Pad Seismic Analyses. Further, the Holtec Cask analysis provides some "filtering" for the controlling loads. In addition, PG&E has had some analyses performed by others to fulfill the various aspects of the design criteria. Where this occurred, it is noted.
This calculation is organized into three main sections. The first is an assessment of the requirements of the PG&E Specification and its subordinate and referenced documents. This assessment provides a reference to the location of the documentation that demonstrates compliance to each requirement where that is known. It also sorts and assesses the load structural requirements, particularly, the load combinations specified by the design requirement documentation. The second is a calculation for the size of the reinforcement for the thermal and shrinkage demand that will be placed on the pad during construction. This reinforcement is then evaluated for the seismic demand in the third section.
SHEET 6
OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 References 1
PG&E Specification, 10012-N-NPG.
2 HOLTEC Report HI-20025 11, Rev.2, Design Criteria Document for the ISFSI Pad for Anchored HI-Storm 100 Deployment at the Diablo Canyon Power Plant, dated Julyl2, 2001.
3 HOLTEC Report HI-2012618, Rev. 5, Analysis of Anchored HI-Storm Casks at the Diablo Canyon ISFSI, dated December 11, 2001.
4 ENERCON Calculation PGE-009-CALC-001, Embedment Support Structure, latest revision.
5 ENERCON Calculation PGE-009-CALC-003, ISFSI Cask Storage Pad Seismic Analysis, latest revision.
6 ENERCON Calculation PGE-009-CALC-006, ISFSI Cask Storage Pad Concrete Shrinkage and Thermal Stresses, latest revision.
7 USNRC, NUREG 1536,: Standard Review Plan for Dry Cask Storage Systems 8
USNRC, NUREG 1567, Standard Review Plan for Spent Dry Fuel Storage Facilities 9
USNRC Draft Regulatory Guide DG-1098, Safety-Related Concrete Structures for Nuclear Power Plants (Other Than Reactor Vessels and Containments), dated August 2000.
10 American Nuclear Society, ANSI/ANS-57.9-1992, design criteria for an independent spent fuel storage installation (dry type) 11 ACI 349-97, Code Requirements for Nuclear Safety Related Concrete Structures 12 ACI 207.1R-96, Mass Concrete 13 ACI 207.R2-95, Effect of Restraint, Volume Change, and Reinforcement on Cracking of Mass Concrete.
Section 1 - Assessments of the Design Requirements Design ReQuirements per PG&E Specification 10012-N-NPG Section 6.4 The various design requirements from the PG&E specification are presented below. All the requirements are listed for completeness. Those that are addressed elsewhere or those with no structural significance are noted. Those that are carried forward, to be addressed within this calculation, will also be noted. The requirements are listed in bullet form with the specific paragraph numbers noted in parentheses:
(6.4.1) The supplier shall provide the design criteria for the pad. The supplier, in this case, is the cask storage supplier, Holtec. The design criteria document is Holtec Report HI-20025 11, Reference
- 2. The requirements of the Holtec Design Criteria are discussed below.
(6.4.2) Site Conditions (6.4.2.1) Location is already addressed on the site drawings.
(6.4.2.2) Topography is already addressed on the site drawings.
(6.4.2.3) Air Temperature and Marine Environment. The requirements are addressed in the thermal data calculation provided by PG&E. The PG&E calculation is Reference 2 of the ENERCON thermal calculation, Reference 6 herein. The finished concrete pad will be coated as required to provide protection from environmental conditions.
SHEET 7
OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 (6.4.2.4) Precipitation. This has no structural significance.
(6.4.2.5) Isolation. These requirements are really for cask cooling. There are no effects on the pad.
(6.4.2.6) Geology. No requirements for the pad design.
(6.4.2.7) Ground Water. No requirements for the pad design.
(6.4.2.8) Liquefaction. No requirements for the pad design.
(6.4.2.9) Slope Stability. This requirement is being addressed elsewhere.
(6.4.3) ISFSI Site-Specific Design Requirements (6.4.3.1) Soil-Rock Stiffness. The soil stiffness properties have been addressed in the seismic analysis calculation (Reference 5). The engineered backfill will be a concrete pad. Thus, there is no impact on the seismic analyses.
(6.4.3.2) Seismic Design Criteria. Refers to (6.2.5) for criteria. Those portions of 6.2.5 that are applicable to the storage pad are:
(6.2.5.1 IV) IOCFR72 Criteria. Qualify the pad for the four (4) seismic events provided in Appendix A. These are 1) the Design Earthquake (DE), Figures A.1.1 and A.1.2; 2) Double Design Earthquake (DDE), Figure A.1.3; 3) Hosgri Earthquake (HE), Figures A.1.4, A.1.5 and A.1.6; and 4) the Long-Term Seismic Program earthquake (LTSP), Figures A.1.7 and A.1.8. The load combinations and allowables shall be per NUREG-1536, NUREG-1567 and other applicable requirements. These requirements are discussed below.
(6.2.5.3 I) Damping Ratios shall be per the included Table. This is not applicable for the pad design since the seismic forces from the casks will be supplied by Holtec and the inertia force applied to the pad itself is the ZPA of the response spectrum.
(6.2.5.4) Static Analysis. Applicable for structures with fundamental frequencies above 33 Hz.
Use the ZPA, this is addressed in Reference 5.
(6.2.5.5 V) Methods of Directional Combinations of Loads.
(A) Use the SRSS or 100-40-40 rule for HE and LTSP. Addressed in Reference 5.
(B) Use the 2D-ABSUM for DE and DDE. Not required to be addressed, see the discussion in the Reduction of the Load Combinations section.
(6.2.5.6 VI) Use the maximum interface loads between the cask and pad to design the pad.
Addressed in Reference 5. The implementing document, Reference 3 provides a Table of applied loads.
(6.4.3.2) Requires an evaluation if the supplied spectra could be invalidated by the specifics of the pad design. Addressed in Reference 2, see below.
(6.4.3.3) Lateral Soil Pressure (H). The lateral soil pressure is negligible since the pad is massive and a solid. Using the specified weight of 130 pcf and the at rest coefficient of pressure of 0.5, the maximum soil pressure at a height of 8 feet is (0.5)(130)(8) = 520 psf = 3.6 psi, neglect.
(6.4.3.4) Interface Friction Coefficient. This is the value to use for the friction between the bottom of the pad and the top of the rock. This issue is addressed by others for the seismic analyses, and it is addressed in the calculations supporting the data in Table 3 herein.
(6.4.3.5) Bearing Pressure. Evaluated in Reference 5.
(6.4.3.6) Soil Elastic Settlement Loads (Sc). Not applicable since there is no soil beneath the pads.
(6.4.3.7) Rock Displacement Loads. No expected displacement beneath the pads.
" W, ENERCON SERVICES, INC.
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 8
OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0
- (6.4.4) ISFSI Layout and Design Criteria
- (6.4.4.1) General Requirements. No structural requirements
- (6.4.4.2 I D) Sequencing addressed in Reference 5.
- (6.4.4.2 II) Structural Criteria for the Pad
- (6.4.4.2 II A) Required for thickness of pad and reinforcement. The pad thickness is determined in Reference 5 that demonstrates that the thickness is adequate to resist all the specified applied loads.
(6.4.4.2 II B) Required fill beneath the pad. A concrete construction pad will be used, see Reference 6.
(6.4.4.2 II C i) Loads and calculations for drop and tip-over. Addressed in Reference 2, see below.
- (6.4.4.2 II C ii) Loads and calculations for stability. Addressed in Reference 2, see below.
- (6.4.4.2.11 C iii) Loads and calculations for rigging. Not applicable to the pad
- (6.4.4.2 II C iv) Loads and calculations on pad due to sequencing. Addressed in Reference 5
- (6.4.4.2.11 D) Detailed design and material specifications for the anchor system. Design is provided in Reference 4.
- (6.4.4.2 II E) Utility requirements. No structural requirements
- (6.4.5) Instrumentation and Control Requirements. No structural requirements
- (6.4.6) Electrical Requirements. No structural requirements Though not specifically referenced in Section 6.4, the pad is also included in the requirements of Section 6.1, DCPP Requirements:
- (6.1.4 (f)) The Storage Pad is specifically included as part of the Storage System.
(6.1.9.1) Loads, as applicable, for the Storage System (including the pad) shall be evaluated in accordance with NUREG-1536, NUREG-1567, ANSI 57.9 and other applicable requirements.
- Loads are:
I.
Dead Loads II.
Live Loads III.
Thermal Loads IV.
V.
VI.
VII.
VIII.
Ix.
X.
XI.
- Normal operational
- Off-normal
- Accidental or Abnormal (due to limitation or loss of cooling air for an extended period of time)
Pressure loads Earthquake Loads Wind Loads Tornado Loads Cask/MPC Handling and Onsite Transport Loads External Man-induced events Blast and Tip-Over Events Soil Loads at the ISFSI Site:
Lateral soil pressure
SHEET 9
OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0
- Soil reaction
- Burial under debris The applicable pressure loads and handling loads are addressed by the Holtec design documents. The other loads are addressed in the Holtec Design Criteria (Reference 2) and subsequently in the Holtec Cask Analysis (Reference 3), as well as the ENERCON analyses, References 4, 5 and 6. Sizing of reinforcement for the applicable design load is provided below.
+ (6.1.9.2) The 10CFR72 load combinations shall be as specified in NUREG-1536, NUREG-1567 and other applicable requirements. These will be addressed below.
- (6.1.10) Requires detailed reports
- (6.2) DCPP Site-Specific Design Requirements. Specifies the magnitude of loads and specific methodology to be used for the Storage System. The loads are:
- (6.2.1) Wind. The pad itself is not subject to wind loads since it is below grade. The applied wind load on the storage casks is insignificant and bounded by the seismic loads, see Reference 3, Section 9.3.
- (6.2.2) Tornado and tornado missiles. The pad itself is not subject to tornado loads since it is below grade. The seismic loads bound the tornado loads (tornado missile plus tornado wind), see Reference 3, Section 9.3.
- (6.2.3) Tsunami. The pad is not subject to tsunami.
- (6.2.4) Flood. The pad is not subject to flood.
- (6.2.5) Earthquake. See presentation above.
The net result of the requirements of the specification as they relate to the structural evaluations performed here are 1) a design criteria document from the cask supplier, 2) loads and load combinations shall be per NUREG-1536, NUREG-1567 and other documents as applicable, 3) seismic loads per Appendix A of the Specification.
In response to the requirement of 6.4.1 of 10012-N-NPG, Holtec, the cask supplier, provided HI-2002511, Rev. 2, Design Criteria Document for the ISFSI Pad for Anchored HI-Storm 100 Deployment at the Diablo Canyon Power Plant, dated Julyl2, 2001, Reference 2.
Design Requirements per Holtec Design Criteria Document. Reference 2 This document is written in a narrative form. The requirements will be listed as they appear in the document. Further they will be related to PG&E specification 10012-N-NPG (Reference 1) as applicable.
o (Section 2.0) Background. Tip over analysis not required since casks are anchored. Thus "soft pad" requirements are not applicable. Ref 1, (6.4.4.2 II C i).
o (Section 2.0) Background. Pad must be "...secured against excessive uplift (> 1/8" nominal) during extreme environmental events (viz., tornado missile, earthquakes, etc.". Ref. 1 (6.4.3.2). This is
SHEET 10 OF 160 JOB. NO.
PGE-009 DATE March 11,_2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 interpreted to mean that that as long as the pad does not deflect more than 1/8 inch beneath the casks, the spectra used to analyze the storage system remain valid. The maximum pad displacements for any of the seismic events are as follows:
Hosgri/LTSP seismic events, soft or hard rock, 0.101 inches anywhere, -0.045 inches at a cask. See Table 2, Reference 5.
Hosgri/LTSP seismic events, soft rock, reduced concrete density, 0.112 inches anywhere, 0.0359 at a cask. See Table 13, Reference 5.
Hosgri seismic events, soft rock, sequencing, all bounded by the basic load cases. See tables 2 and 14, Reference 5.
Hosgri seismic events, soft rock, cask extraction, all bounded by the basic load cases. See Tables 2 and 15, Reference 5.
Hosgri/LTSP seismic events, soft rock, reduced Poisson's ratio, 0.1023 inches anywhere, 0.0272 inches at a cask. See Table 16, Reference 5.
All displacements are computed to be less than the 1/8 inch; thus the spectra used are valid.
o The two key design objectives for the pad are (Reference 2):
- a. "The pad is sufficiently robust to withstand the overturning moments exerted on it by the cantilevered HI-STORMs due to destabilizing loadings such as the earthquake or a tornado missile. The overturning moments on the surface of the pad will act to cause local pad rotation that may need to be resisted by appropriately anchoring the pad to the subterranean half-space."
- b. "The anchored construction calls for a thick heavily reinforced pad, which means that the pad presents a stiff target to any object colliding with it. Because the maximum g-load that the MPC is permitted to sustain due to an impact event is limited by the HI-STROM FSAR, the handling device used to move and position the loaded cask on the pad shall be Single-Failure Proof...."
o (Section 4.1) Design and Construction Criteria. The pad is categorized as a Category B Important-To-Safety Structure.
o (Section 4.1) Design and Construction Criteria. The pad does not need to be designed to accommodate cask tip-over since the casks are anchored. Ref. 1, (6.4.4.2 II C i) and (6.4.4.2 II C ii).
o (Section 4.1) Design and Construction Criteria. There is no need to perform a drop analysis and establish a lift height since the cask handling equipment meets NUREG-0612 criteria.
o (Section 4.2) Applicable Codes and Load Combinations. Factored load combinations for the pad design are to be from ACI-349-97, NRC DG 1098 and NUREG-1536.
o (Section 4.2 a) Holtec will supply earthquake loads to the pad from the casks. Pad designed to include pad self weight inertia terms. This issue is addressed in Reference 5.
o (Section 4.2.b) Load Combinations for the Concrete Pad. Notation and acceptance criteria of NUREG-1536 apply (see discussion of ANSI/ANS 57.9 below for definitions):
ENERCON SERVICES, INC.
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 11 OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0 Normal Events Uc > 1.4D + 1.7L Uc > 1.4D + 1.7 (L+H)
Off-Normal Events Uc > 1.05D+1.275 (L+H+T)
Uc > 1.05D +1.275(L+H+T+W)
Accident-Level Events Uc > D+L+H+T+(E or F)
Uc> D+L+H+Ta Uc > D+L+H+T+Wt Uc > D+L+H+T+A These load combinations are addressed below.
The Design Criteria goes on to discuss overturning and sliding issues, which are others.
being addressed by o (Section 4.2.c) Load Combinations for the Anchor Studs. Addressed by Holtec o
(Section 4.2.d) Load Combinations for Embedment Structure. Addressed in Reference 4.
o (Section 4.3) Limiting Design Parameters. Embedment ductility. Addressed in Reference 4.
o (Section 4.3) Lift height limitation is not applicable since lifting devices are single-failure-proof.
o (Section 4.4) Additional Requirements.
o (Section 4.4 i) Interface. The details for the pad design and embedment structure are not in the HI-STORM FSAR o (Section 4.4 ii) Applicable Code. The applicable code for the pad is ACI 349-97 or as specified by owner. Since 10012-N-NPG does not define a Code, the Code is ACI-349-97. Addressed below.
o (Section 4.4 iii) Grounding. No structural requirements o
(Section 4.4) There are no requirements in the CoC pertaining to cask load sequencing. Thus cask load sequence issues are left to the pad designer to evaluate. Sequencing is considered in Ref. 5.
o (Section 4.5) Provides general data for the design.
o (Section 4.6) Maximum Permissible Tornado Missile and Wind Load. Requires the cask designer to provide values but expects the seismic to control. Seismic bounds Wind and Tornado, including tornado missile, see Reference 3, Section 9.3.
o (Section 4.7) DCPP Cask Transporter Input Data. Provides dimensions and weights.
o (Section 5.1) General Comments. Requires the licensee to determine that the seismic and tornado loads are applicable for the site. These are specified in reference 1.
SHEET 12 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI b
SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 o
(Section 5.2) General Requirements for the ISFSI Pad. These requirements are:
- 1) Steel embedment to comply with ACI 349, and the 10/01/2000 Draft Appendix B. Addressed in Reference 4.
- 2) Pad to comply with all structural requirements of NUREG-1567 and ACI 349. Addressed below.
- 3) Pad/foundation qualified so that sliding will not have a significant impact on the cask qualification. This is being addressed elsewhere.
- 4) Compression block/coupling in the embedment able to resist the maximum stud load provided by the cask designer. Addressed in Reference 4.
Design Reguirements ner ANSI/ANS-57.9 1992, Reference 10 ANSI/ANS-57.9 does not prescribe any detailed structural design requirements. Section 6.17 of ANSI 57.9 provides the design requirements. This section references ACI 349 and AISC N690 for load combinations but it does not specifically enforce them. The load definitions are:
Normal Operating Loads D - dead weight structure and attachments. Should be varied by +/- 5% to simulate the most severe response.
L - snow, rain, transient loads of equipment T-thermal loads during operation, can be neglected if shown that the resultant stresses are secondary and self-limiting, and the response of the structure is ductile H-lateral soil pressure
- Natural Phenomena Loads E - Seismic, equivalent to the Safe Shutdown Earthquake W - Design Basis Wind at a 100 year reoccurrence level Wt - Design tornado - including wind pressure, differential pressure and missiles Off-Normal Operating and Accident Loads Ta - Temperatures due to loss of cooling A - Loads due to heavy load drop as described in ASCE Manual No. 58 Design Reguirements per NUREG 1536. Reference 7 The structural design requirements per NUREG 1536 that are enforced by the PG&E design documents are the load combinations and allowable forces/stresses. These are provided in NUREG 1536 Table 3-1, Loads and load combinations:
Reinforced Concrete Structures
SHEET 13 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Normal Events and Conditions Uc > 1.4D+1.7L Uc > 1.4D+1.7(L+H)
Off-Normal Events and Conditions Uc > 1.05D+1.275(L+H+T)
Uc > 1.05D+1.275(L+H+T+W)
Accidents and Conditions Uc > D+L+H+T+(E or F)
Uc > D+L+H+T+A Uc > D+L+H+Ta Uc > D+L+H+T+Wt Where Uc is the reinforced concrete available strength as computed from ACI 349 and the loads D, L etc are defined by ANSI 57.9 and F is the design basis flood. These load combinations are the same as specified in the Holtec Design Criteria Document and are addressed below.
Design Requirements per NUREG 1567, Reference 8 The structural design requirements per NUREG 1567 that are enforced by the PG&E design documents are the load combinations and allowable forces/stresses. These are provided in NUREG 1567 Table 7-1, Loads and load combinations Steel and Reinforced Concrete Structures Normal Events and Conditions Uc > 1.4D+1.7L Uc > 1.4D+1.7(L+H)
Off-normal Events and Conditions Uc > 1.05D+1.275(L+H+T)
Uc > 1.05D+1.275(L+H+T+W)
Accident-Level Events and Conditions Missile concurrent with wind - same for 1536 Uc > D+L+H+T+E Uc > D+L+H+T+A Uc > D+L+H+Ta Uc > D+L+H+T+Wt Uc > D+L+H+T+F
kSHEET 14 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 The load combinations from NUREG 1536 and NUREG 1567 are the same. These load combinations are addressed below.
The Holtec Design Criteria document invokes Load Combinations of ACI 349-97, as modified by NRC DG 1098 (Reference 2, Section 4.2).
Load Combinations ver ACI 349-97, Reference 11 The load combinations are specified in Chapter 9 of ACI-349:
- 1. U > 1.4D+1.4F+1.7L+1.7H+1.7Ro
- 2. U > 1.4D+1.4F+1.7L+1.7H+1.7Eo+1.7Ro
- 3. U > 1.4D+1.4F+1.7L+1.7H+1.7W+1.7Ro
- 4. U > D+F+L+H+To+Ro+Ess
- 5. U > D+F+L+H+To+Ro+Wt
- 6. U > D+F+L+H+Ta+Ra+1.25Pa
- 7. U > D+F+L+H+Ta+Ra+1.1 5Pa+1.O(Yr+Yj+Ym)+1.1 5Eo
- 8. U > D+F+L+H+Ta+Ra+l.OPa+1.O(Yr+Yj+Ym)+l.OEss
- 9. U > 1.05D+1.05F+1.3L+1.3H+1.05To+1.3Ro
- 10. U > 1.05D+1.05F+1.3L+1.3H+1.3Eo+1.05To+1.3Ro
- 11. U > 1.05D+1.05F+1.3L+1.3H+1.3W+1.O5To+1.3Ro The load definitions are provided in ACI 349. They are consistent with ANSI 57.9 where they overlap, however, ACI includes more loads since it is a general code and not developed specifically for ISFSI applications. These applicable load combinations are addressed below.
ACI also specifies a few additional requirements:
In Section 9.2.2, when the effects of differential settlement, creep or shrinkage may be significant, include them with D in combinations 4-11. The code requires a realistic assessment of such effects occurring in service. It is, therefore, not intended that the designer consider a "bounding" assessment of these effects. Thermal stresses and shrinkage are addressed below.
In Section 9.2.3, where a load reduces effects of other loads in any combination the coefficient shall be 0.9 if it can be shown the load is always there, otherwise use 0. Thus, the dead load must be reduced by 10% if it assists, rather than the 5% specified by ANSI 57.9. This is addressed in Reference 5 where the mass of the pad is reduced by 10%. The reduction applied to both the horizontal seismic acceleration loads on the pad as well as the vertical accelerations holding the pad down. The cask loads were not reduced. The displacements of the pad and casks due to a reduced concrete density increased slightly, however, the applied stresses decreased as shown in Reference 5. The displacements are acceptable, see above, and the internal forces from the basic analyses using normal weight concrete bound those using the reduced density concrete.
SHEET 15 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI f
i SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 The reduced mass of the pad also needs to be evaluated relative to the tornado wind plus missile loads.
The applied tornado wind plus missile load is 29,376 in kips (Reference 3, Section 9.3) Vs the maximum seismic load of 61,000 in kips (Reference 3, Table 3). Thus, since the pad is acceptable for the seismic load with the 10% reduction in mass, it is also acceptable for the tornado wind plus missile load since the seismic applied load is more than twice the tornado wind plus missile load.
The Holtec Design Criteria document invokes NRC DG 1098 (Reference 2, Section 4.2).
Assessment of NRC DG 1098. Reference 9 The draft regulatory guide accepts 349-97 with 15 additional regulatory positions. The only position that could effect the design is position 6, which modifies some of the load factors in the ACI 349 load combinations presented above. These modifications are:
Load Combinations 9, 10 and 11, use 1.2To rather than 1.05To Load Combination 6, use 1.5Pa rather than 1.25Pa Load Combination 7, use 1.25Pa rather than 1.1 5Pa The ACI load combinations are repeated below with the DG 1098 changes reflected in BOLD.
- 1. U > 1.4D+1.4F+1.7L+1.7H+1.7Ro
- 2. U > 1.4D+1.4F+1.7L+1.7H+1.7Eo+1.7Ro
- 3. U > 1.4D+1.4F+1.7L+1.7H+1.7W+1.7Ro
- 4. U > D+F+L+H+To+Ro+Ess
- 5. U > D+F+L+H+To+Ro+Wt
- 6. U > D+F+L+H+Ta+Ra+1.5Pa
- 7. U > D+F+L+H+Ta+Ra+1.25Pa+1.0(Yr+Yj+Ym)+1.1 5Eo
- 8. U > D+F+L+H+Ta+Ra+l.OPa+l.0(Yr+Yj+Ym)+l.OEss
- 9. U > 1.05D+1.05F+1.3L+1.3H+1.2To+1.3Ro
- 10. U > 1.05D+1.05F+1.3L+1.3H+1.3Eo+1.2To+1.3Ro
- 11. U > 1.05D+1.05F+1.3L+1.3H+1.3W+1.2To+1.3Ro Reduction of the Load Combinations The load combinations specified by the Holtec Design Criteria, NUREG 1536 and NUREG 1567 are all the same. The ACI Code load combinations are significantly more complex. The ACI load combinations will be reduced first, followed by the NUREG 1567 Code load combinations.
The load combinations are assessed in light of the specification of the loads:
F = 0 Flood (and all other potential fluid loads) are zero. See Reference 1, (6.2.4).
SHEET 16 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 H = 0 Soil pressures are not applicable for these analyses, see discussion of PG&E specification section (6.4.3.3).
Ro and Ra are normal operating and accident equipment reactions other than the casks. The transporter is the only equipment load, and since its load is always down, i.e., it can not apply any moment to the pad, it does not have any effect on the pad design. The casks are treated as live loads.
W is insignificant, see Reference 3, Section 9.3.
To and Ta are normal operating and accident thermal forces. The bounding transient concrete temperature is specified in the Holtec Design Criteria (Reference 2, Section 4.5) to be 150 degrees F. This is considered to be below the threshold where explicit analyses for structural response are required. Further, the temperature of 150 degrees is the long-term general area limit, and is well below the short-term limit of 350 degrees per ACI 349 (see Reference 11, A.4.1 and A.4.2). Thus no material degradation or reduction in strength will occur.
Pa is an accident pressure load and is not applicable for this analysis Yr, Yj and Ym are all associated with pipe break and are not applicable for this analysis.
Given the above definitions and applicability of loads, the ACI load combinations reduce to:
- 1. U > 1.4D+1.7L
- 2. U > 1.4D+1.7L+1.7Eo
- 3. U > 1.4D+1.7L
- 4. U > D+L+Ess
- 5. U > D+L+Wt
- 6. U> D+L
- 7. U> D+L+l.l5Eo
- 8. U > D+L+1.OEss
- 9. U > 1.05D+1.3L
- 10. U > 1.O5D+1.3L+1.3Eo
- 11. U > 1.05D+1.3L The Eo is the Design Earthquake and Ess is the most severe of the Double Design Earthquake (DDE),
Hosgri (HE) or the Long Term Seismic Program (LTSP) earthquake. The HE, 7% damped curve, has a ZPA of 0.75G and a peak of approximately 1.8G (Reference 1, Figures A.1.4 and A.1.5). The damping values used in this discussion are from the Table in Section 6.2.5.3.1, of the PG&E Specification, Reference 1. The envelope of the DE, 5% damped curves, has a ZPA of 0.20G and a peak of approximately 0.76G (Reference 1, Figure A.1.3). Thus the DDE, 5% damped curve, has a ZPA of 0.40G and a peak of 1.52G. Comparing both the numerical values and the shape of the curves, it is clear that the HE, 7% damped curve, bounds the DDE curve, and therefore bounds the DE by more than a factor of 2.0.
SHEET 17 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Considering the load factors for the Eo load combinations, and the results of the reduced concrete density analyses described above, the use of 1.4D will result in a less demanding pad response than using just the dead weight. The same is true for the 1.7 factor on cask weight, L.
Further, Wt is bounded by the seismic responses, Reference 3, Section 9.3. Given the above the 11 equations further reduce to:
- 1. U > 1.4D+1.7L
- 4. U > D+L+Ess Finally, given that equation 1 has no horizontal force component associated with it, it is acceptable by inspection, including the effect of the transporter. Hence the ACI 349 load combinations reduce to:
- 4. U > D+L+Ess Where Ess is the most demanding of either the Hosgri or LTSP seismic events. An evaluation of the Hosgri curves and LTSP curves did not result in a clear governing seismic event. Therefore, both seismic events were explicitly evaluated, see References 3 and 5.
Now, using the above evaluation, the NUREG 1536 and 1567 load combinations also reduce to:
Steel and Reinforced Concrete Structures Normal Events and Conditions Uc > 1.4D+1.7L Uc > 1.4D+1.7(L)
Off-normal Events and Conditions Uc > 1.05D+1.275(L)
Uc > 1.05D+1.275(L)
Accident-Level Events and Conditions (Tornado missile concurrent with tornado wind)
Uc > D+L+E Uc>D+L Uc>D+L Uc > D+L+Wt Uc > D+L And, applying the same arguments as above regarding D+L combinations and the Wt being bounded by seismic, the NUREG 1567 equations further reduce to ACI equation 4, where E is Ess.
SHEET 18 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Therefore, all the relevant structural criteria for the applied loads remaining to be satisfied are embodied in equation 4 above. The seismic demand is provided in Table 11, Reference 5. These are to be factored up by 1.15 to demonstrate compliance with the Design Codes and Specifications, see Reference 5, Summary and Conclusions.
Section 2 - Sizing of Reinforcement for Thermal and Shrinkage Demand Thermal Demand due to Curing Temperatures and Shrinkage In addition to the seismic evaluation required above, the only other significant structural demands on the pad are the thermal stresses/forces due to heat up during the curing process and shrinkage stresses/forces. The requirements are provided in ACI 349 Section 7.12. Since the pad is more than 72 inches thick, the reinforcement is proportioned using the provisions of ACI 207, (References 12 and 13),
see ACI 349 Section 7.12.4. Also, the minimum reinforcement ratio shall be 0.0018 unless the area of reinforcement is 1/3 greater than that required by analysis, see ACI 349 Section 7.12.5. Rather than compute the thermal demand using the approach prescribed by ACI 207, the demands on the pad due to temperature and shrinkage were computed using a finite element model similar to that used to compute the seismic demand, Reference 6.
A review of the seismic analysis and the thermal/shrinkage analysis shows that the thermal/shrinkage demand is highly likely to control the design of the reinforcement. Thus, the design of the reinforcement is based on the thermal/shrinkage demand, using the results from Reference 6 and additional guidance as applicable from ACI 207. Once sized, the reinforcement is evaluated for the seismic demand in accordance with ACI 349.
Relevant requirements of ACI 349-97:
3.3.2 Maximum aggregate must be < 3/4 distance between bars 7.6.1 Minimum clear spacing between bars is db, but not less than 1 inch 7.6.2 Where parallel bars are used in two or more layers, the upper bars shall be placed directly above the lower bars with a clear distance not less than 1 inch.
7.6.4 The clear distance requirements apply to clear distance between contact splices and adjacent bars or splices.
7.7.1 Cover requirements: 3 inches for concrete cast against earth, 2 inches for concrete exposed to weather.
7.12.4 Concrete sections thicker than 72 inches should be designed using ACI 207 7.12.5 Where tension steel required, the minimum reinforcement ratio is 0.0018 unless the area of reinforcement provided is 1/3 greater than required by analysis.
SHEET 19 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 21.2.2.3 Members below base shall also comply with Chapter 21 21.2.5.1 Reinforcement shall be ASTM 706 or ASTM 615 with caveats 21.2.6.1 Reinforcement may be welded. Alternate bars. May also be mechanically connected per 12.14.3.3 or 12.14.3.4 Initial Sizing of Reinforcement The layout of the embedment structures will only permit a spacing of the top steel of 9 inches. Further, the geometry of the embedment structures and their proximity on the pad will not allow a reasonable lap splice arrangement for the reinforcement. Hence, the design will be developed using mechanical connectors. The connectors used in the design are Bar Lock "L-series" couplers; however, equivalent connectors can be used providing they meet the strength and geometric requirements. The connectors are 11.5 inches long; therefore at 9 inch spacing the traverse bars need to be far enough apart that the couplers can fit between them. Figures 1 and 2 show a layout for two layers of #10 bars with the couplers.
The layout provides for the specified cover to the surface of the coupler rather than the reinforcement.
However, should this be relaxed and the coupler is allowed to be closer to the surface than the reinforcement cover, these calculations will still be valid.
Addressing some of the basic code requirements:
3.3.2 The maximum aggregate is 11/2 inches. The minimum clear distance between parallel reinforcement is 2.9 inches. And, 2.9 > 4/3xl l/2. Therefore OK.
7.6.1 The 2.9 inches is > db (1.27) and > 1 inch. Therefore OK.
7.6.2 The top bars are placed directly above the lower bars. Therefore OK.
7.6.4 The location of the connectors will be staggered along the lengths of the bars. The clear distance between a connector and a bar is 1.27 + (2.90-1.27)/2 = 2.085 inches. And, 2.085 inches >
4/3xl 1/2. Therefore OK.
7.12.6 Minimum reinforcement is 0.0018 x 9 x 96 = 1.555 square inches per 9" width. Conservative since full depth of pad is used, rather than d. Area steel of 2 #10 is 2x1.27 = 2.54 >1.55.
Therefore OK without enforcing the 1/3 greater requirement.
FI
..I ENERCON SERVICES, INC.
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 20 OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0 Top of Concrete 2.9 inches OD Coupler 1.27 inches #10 bar 2.9 inches OD coupler 1.27 inches #10 bar Figure 1 - Section of Top Steel - Looking N or S
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 21 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 1.27 inches #10 bar 2.9 inches OD coupler 8.93 inches.
Provides 2 inches
'minimum clearance between the top of the anchor plate and the coupler of the second
-layer of NS steel.
1.27 inches #10 bar 2.9 inches OD Coupler 2 inches cover Bottom of Concrete Figure 2 - Section of Bottom Steel - Looking N or S
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR ENERCON REVIEWER K. L. Whitmore APPROVED SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 22 OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0 Approximate Allowable Thermal and Shrinkage Stresses The guidance from ACI 207.2, Reference 13, Chapter 5 discusses the issue of crack width vs. calculated steel stress. The following are approximate calculations to determine if the reinforcement arrangement and criteria are reasonable. The suggested acceptable crack width is 0.013 inches for exterior exposure.
The relation between crack width and steel stress, is equation 5.1 from ACI 207:
.0 b w =0.076Vj7Of, 10-3 CG Bars Where:
w is the crack width at surface dc is the cover to the center of the bar L
A is the average effective concrete area around a bar
= 2yb/N where N is the number of bars - see figure B distance from neutral axis to the tensile face divided by the distance from neutral axis to the steel f, is the calculated steel stress For the NS top steel:
Now, allow w to equal 0.013 inches d, from the surface to the CG of the closest NS bar is 2+2.9/2 = 3.45 inches A is 2 x (2+2.9+1.27/2) x 9/2 = 49.815 sq. in. per bar (see Figure 1 for dimensions)
Let 13 equal approximately (96/2)/(96/2-(2+2.9+1.27/2)) = 1.13 Recasting the equation for with f. as the unknown:
w 0.076VK Af1 o-0.013
, = 27.2 ksi for NS steel 0.076l 3.45x49.815 x 1.13x10-3 For the EW top steel:
d, from the surface to the CG of the closest EW bar is 2+2.9+1.27/2 = 5.535 inches A is 2 x (2+2.9+1.27+2.9/2) x 9/2 = 68.58 sq. in. per bar (see Figure 1 for dimensions)
Let B equal approximately (96/2)/(96/2-(2+2.9+1.27+2.9/2)) = 1.19
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 23 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 w
5 0.076 vd7A 1 0-3 For the NS bottom steel:
0.013 0.0764 5.535 x 68.58 x 1.19x10-3 19.9 ksi for EW steel Consider only the bottom bar, the second bar is too high into the concrete to consider.
Now, allow w to equal 0.013 inches dc from the surface to the CG of the closest NS bar is 2+2.9/2 = 3.45 inches A is 2 x (2+2.9/2) x 9 = 62.1 sq. in. per bar (see Figure 2 for dimensions)
Let B equal approximately (96/2)/(96/2-(2+2.9/2)) = 1.08 5 =
7 aw 0.076 VdJI P10-3 0.013
=.3
= 26.5 ksi for NS steel 0.076VA3.45x62.1 x 1.08 x10-3 For the EW bottom steel:
d, from the surface to the CG of the closest EW bar is 2+2.9+1.27/2 = 5.535 inches A is 2 x (2+2.9+1.27/2) x 9 = 99.63 sq. in. per bar (see Figure 2 for dimensions)
Let B equal approximately (96/2)/(96/2-(2+2.9+1.27/2)) = 1.1 3
=
w 0.013
- 18.5 ksi for EW steel 5 0.076 Ap0-3 0.076 V5.535 x 99.63 x 1. 13x10-3 Shrinkage forces from Reference 6, Table 14: (for the top 12 inches):
NS direction, sign convention: (-) indicates tension, see sign convention in Reference 5, forces adjusted from 17 feet (204 inches) to 9 inch bar spacing (where 2.54 is the area of 2 #10 bars):
-. 1510.694 x 9/204 =26.2 ksi 2.54 EW direction:
fe
- 1437.67 x 9/204 25.0 ksi The shrinkage forces are conservatively calculated because they do not include the stresses in the concrete at the bottom of the 12-inch layer. Thus it appears that the reinforcement arrangement and criteria are reasonable and warrant a more detailed assessment. The allowable stresses will be recomputed using more detailed information.
SHEET 24 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Detailed Evaluation for Thermal and Shrinkaue Demands - Constrained Model Analyses This bar arrangement will now be evaluated for moments and forces from the more detailed analysis results presented in Reference 6. First, the equations of equilibrium will be presented for the section.
Since the reinforcement is not symmetric two sets of equilibrium equations will be developed: one where the sense of the moment produces tension on the bottom of the pad and one where the sense of the moment produces compression on the bottom of the pad. The equations are developed with the applied force and moment located at the mid-height of the pad since this is where the applied forces from the ANSYS analysis are computed, see Figures 3 and 4.
The thermal demand on the pad is computed using two bounds. The first analysis assumes that pad is constrained from sliding in any horizontal direction by the underlying rock. This is referred to as the "constrained model" in Reference 6. This is the most reasonable approximation to the actual expected behavior of the pad and it is the condition that is considered in the literature, see Reference 13.
However, the potential that the pad may slide, locally, in some areas can not be denied. Thus, a second thermal model was analyzed in Reference 6 where the pad was allowed, analytically, to slide freely in any horizontal direction. This is referred to as the "unconstrained model" in Reference 6. The force results for the constrained model are provided in Table 9 of Reference 6, while the stress results for unconstrained model are provided in Table 11 and Figures 27 to 33 of Reference 6.
Since these assessments are focused on crack width estimation and to gain a sense for the actual pad stresses, the equations are developed using working stress methodology.
SHEET 25 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC. CALCULATION NO.
PGE-009-CALC-007 REVISION 0 X or Z Strip Evaluation - moment produces tension on the bottom of the Pad:
Sc
- _ d A d2A C2 0 Ccc x
M h
4k C
h 2
T 2 A
h-x 4-T IV St =h-x C
Figure 3 - Equilibrium for moment that produces tension on the bottom of the vad
SHEET 26 OF 160 March 11, 2003 JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 CALCULATION NO.
PGE-009-CALC-007 Equilibrium equations:
The applied internal forces C and M are at h/2. Width of section is b.
_F=O C, = 2 xbECEV C2 C 2 = AsEs (x-d2
( h -- d3 )
=AE h-xd3)(h-x C = AEs hxd3
) c And similarly, T2 =AsEs h-xd4>c x
Substituting:
C=bE gcx+ASE~(X
)£c+AEs, x
)
(
2 c
Cx
)
h-x-d3 )>
AsEsx)d4 Gathering terms:
C=6.{bEcx2+ASE,[x-d, +x-d 2 - (h-x-d3)-(h-x-d4 Canceling terms:
SHEET 27 OF 160 DATE March 11, 2003 JOB. NO.
PGE-009 PROJECT SUBJECT CLIENT DCPP ISFSI ISFSI Cask Storage Pad Steel Reinforcement PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 APPROVED R. F. Evers REVISION 0 C=. -{ib-ECX2 +AE 8[4x-2h-di +d 3 -d 2 +d If x
2T r
Therefore:
°=K {b ECx +AsEsj4x-2h+dnit]}-C Where:
dnet =-di + d3 -d 2 + d4 Multiply by x
- 0, and expanding the terms:
0 = Sc {2EcX2 + 4AEsx-2AEh + ASESdet }- Cx Now, bring the Cx term into the coefficient for x, and recognize that e. *0, o
= £b {-ECX2 + 04AEs-C x - 2ASESh + AsEsdn, }
Therefore the { } must equal 0, since sc
- 0, 0 = 2 X2 + t4ASES -
x-2AsEsh + ASESdnet }
Now divide by E,
- 0, 02= {(x2+
4A E, _
E, E,
Let a = E, and recognize that Ece = a,, the concrete stress:
0= {kx2
(
G4As
)o
-C x-2ASah+Asadet}
SHEET 28 OF 160 March 11, 2003 7
JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR ENERCON REVIEWER K. L. Whitmore APPROVED SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 S. C. Tumminelli R. F. Evers REVISION 0 This is a quadratic for x, therefore:
l
- (4Asa -- C +/-.4ASa-C
-4 2(-2Asah+Asadnet) x b
Only the + sign for the radical makes physical sense, since the - sign will always lead to a negative value for x, since the radical will always be larger than the first term in the numerator. This is true because the sum of the terms 2h+dnet is always positive. (dnet 2 -2h). Therefore:
l
- 4Asa -- C+ J4Aa
)
+2bAsoa(2h-dnet) x b
Equation (1)
- Also, EM=0 Therefore:
M Equatn (2( ))+C2(h d2)+TI(-d)+T2(h-d4)
Equation (2)
SHEET 29 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
X or Z Strip Evaluation - moment produces compression on the bottom of the pad E =h-x E tx c
4-0T 2
M h
C
_0 C2 cc Sc Fieure 4 - Equilibrium for moment that produces compression on the bottom of the pad
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 30 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 Equilibrium equations:
The applied internal forces C and M are at h/2. Width of section is b.
JF=O
.-. C=CC+C 1 +C2 -T. -T 2 Cc = IxbEc~c C2 C =AE, x - d3 )
(
x
)
TAs st hh-x
) '
(h-x-d, Y(h -=xA, h-xd-1d>
As h -x
)tx
£c=Ass And similarly, T 2 =A(E, h-x-d C
Substituting:
C=-ECcx+AAEsI X
c +ASEsI X
-A-E, I
SC -A
)
('XC)
Gathering terms:
C=E-{kb ECX2 +AE.[x -d3 +x-d 4 -(h -x -d)-
(h -x-d2 Canceling terms:
JOB. NO.
PROJECT SUBJECT CLIENT REVIEWER CALCULATII SHEET 31 OF 160 PGE-009 DATE March 11, 2003 DCPP ISFSI ISFSI Cask Storage Pad Steel Reinforcement PG&E-DCPP ORIGINATOR S. C. Tumminelli K. L. Whitmore APPROVED R. F. Evers ON NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
C= x{kEcx +AE,[4x-2h-d 3 -d 4 +d +dJ]}
Therefore:
°=. X{2 E0 X + AsEs [4X - 2h + d]}-C Where:
d,,m, =-d 3 -d 4 +di +d2 Multiply by x
- 0, and expanding the terms:
0 = EC {2ECX2 + 4AEx - 2AEh + ASESdS }- Cx Now, bring the Cx term into the coefficient for x, and recognize that EC : 0,
° =
bsc{Ecx
+ 4AsEs -
x - 2AsEsh + ASEsdsn Therefore the { } must equal 0, since -,
- 0, 0 = j 2
+ (4ASES -Cx
- 2AEsh + AsEsdsm }
Now divide by E,
- 0, 0=
x2+
4A5 E
)x2As. El h+A, E-d,.
E, Let a =
, and recognize that E.c = aO, the concrete stress:
0{bx2+
4Asax'C x -2ASaxh +Asadsum
SHEET 32 OF 160 March 11, 2003 JOB. NO.
PROJECT SUBJECT PGE-009 DATE DCPP ISFSI ISFSI Cask Storage Pad Steel Reinforcement PG&E-DCPP ORIGINATOR K. L. Whitmore APPROVED f
CLIENT ENERCON REVIEWER SERVICES, INC. CALCULATI, S. C. Tumminelli R. F. Evers REVISION 0 ON NO.
PGE-009-CALC-007 This is a quadratic for x, therefore:
4Asa - 6Cj j(4Asa --
)j -4 2 (- 2Asah + Aad. )
x-=
b Only the + sign for the radical makes physical sense, since the - sign will always lead to a negative value for x, since the radical will always be larger than the first term in the numerator. This is true because the sum of the terms -2Asah+Asadsum is always negative. (d,. Ž -2h). Therefore:
- 4Asa -yC j+4As - C
+2bASa(2h-d U) b Equation (3)
- Also, YM=0 Therefore:
M =C h -x+Cl(h-d3)+C (h d
q)+T(h d )+T(h-d )
Equation (4)
SHEE T 33 OF 160 E March 11, 2003 JOB. NO.
PGE-009 DATI PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATO ENERCON REVIEWER K. L. Whitmore APPROVED SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007
'R S. C. Tumminelli R. F. Evers REVISION 0 Now, developing the equations for M and solving explicitly with the equations for C using x and a,, as the unknowns will result in cubic equations for x. This becomes too complicated for an explicit solution.
Therefore, the preferred method of solution is to just use the equation for C (Equation (1) or (3)) using an assumed value for ac to compute x. The values for x and the assumed value for a;" are then used in Equations (2) or (4) to compute the internal moment. This is then compared to the applied moment.
When the two values are close, the solution is considered to have converged. The equations were programmed using Excel. Only the final results are provided here.
Below is the evaluation of the section for the forces and moments due to curing from 1.125 days to 3.125 days. The sign convention established in Reference 5 and the signs used here are not the same.
The sense of the moment as depicted in Figures 3 and 4 is the sense observed. The applied positive moment for the Z strips in Reference 5 produces tensile stresses on the bottom of the pad. However, the applied moments for the X strips in Reference 5 are positive when they produce compressive stresses on the bottom of the pad. This sign change results from the change in process where the transition is made from the finite element model, observing the right hand rule, to the concrete evaluation process.
Thus the sign conventions from Reference 5 will be interpreted using the sense of the applied moment shown on Figures 3 and 4. When the applied moment produces tensile stresses on the bottom of the pad Equations (1) and (2) will be utilized, and when the sense of the applied moment produces compressive stresses on the bottom of the pad Equations (3) and (4) will be used. The results are then tabulated in Table 1.
SHEET 34 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Numerical Evaluation:
Time is 1.125 days NS (Z) Strip b = 9 inches; h = 96 inches; d, = 3.45 inches; d2 = 7.63 inches; d3 = 3.45 inches; d4 16.55 inches A, = 1.27 sq. in.
- 10 bar d8 m = -8.92 f, = 1.054 ksi Table 4, Reference 6:
Ec = 1964 ksi; Es = 29000 ksi; a = 29000 -14.77 1964-0.574
= 0.574 ksi from off-line calculations; Fc -
964 = 0.000292 in/in From Table 9 (4/11), Reference 6:
C = 4817.00 kip/17 foot section; C = 212.51 kip/9 inch section M =121000 in-kip/17 foot section; M = 5338.24 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
C
~212.51 04Aa --
= 4 x 1.27 x 14.77 -
= -295.22 0
5
)0.574 2bAa(2h - d,) = 2 x 9 x 1.27 x 14.77 x (2 x 96 + 8.92) = 67819.71
- (4Aca --
)
+ j4Asa l
+ 2bA,a(2h - d,)
2bA(2hd) 295.22 +
295.222 +67819.71 x=
b=76.54 inches b
9
- Now,
=x bbEs
= xbc;0 76.54x9x0.574 =197.71kip C
2
= 2 29000
=
2 7
2 0
10.28 kip C, = AEs, X-3 FeC 1.27 x 29000(
76.54 -34) x 0.000292 = 10.28 kip
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 35 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 C2 =ASE (x 4 ) 6C = 1.27 x 29000 (7654
) x 0.000292 = 8.44 kip A hx d 2 2 o j 6 76.54 ~ 3 4 T, = AEt h 1 )
=1.27 x 29000 96 -76.54
) x 0.000292 =2.25 kip
( h x - d )9 6-76.54 - 7 3 \\
T2 = AE, (
x 2 ), = 1.27 x 29000(
7654-
) x 0.000292 = 1.66 kip Check on equilibrium F = 0
..C=Cc +C, +C2 -T1 -T2 212.51 = 197.71+10.28+8.44-2.25-1.66 = 212.52 OK Now the internal moment is:
M =C h-x +C (h-d3)+C (h-d )+T(h-d
)+T2(h-d )
Equation (4)
M=197.71
_ 76.54) +(10.28) 2 3.45) + (8.44)96-16.55)+
(2.25) 9
- 3.45) + (1.66) 296 - 7.631 = 5336.31 _ 5338.24 OK Concrete compressive stress of 0.574 ksi is 54% of f, and the maximum tensile steel stress is 2.25 = 1.77 ksi, therefore OK.
1.27
SHEET 36 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
Numerical Evaluation:
Time is 1.125 days EW (X) Strip b = 9 inches; h = 96 inches; d, = 5.535 inches; d2 = 9.705 inches; d3 = 5.535 inches; d4 = 18.635 inches A, = 1.27 sq. in.
- 10 bar d5,, = -8.93 fc = 1.054 ksi Table 4, Reference 6:
E, = 1964 ksi; Es = 29000 ksi; a=29000 =14.77 1964-qc = 0.582 ksi from off-line calculations; g-= 1964=0.000296in/in From Table 9 (4/11), Reference 6:
C = 4319.608 kip/17 foot section; M = 123000 in-kip/17 foot section; C = 190.57 kip/9 inch section M = 5426.47 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
4A5a --
= 4 x 1.27 x 14.77 - 0 58 252.43 (190.57
=
-252243 2bA~ct(2h -d.)= 2 x 9 x 1.27 x 14.77 x (2 x 96 +8.93) = 67823.08
- (4Asa --
) + j(4As C)
+ 2bAsa(2h - d,
)
b
- Now, 1
xba 68.35 x 9 x 0.582 2 =-xbEc
=
2 2
252.43 + ;- 252.432 + 67823.08 = 68.35 inches 9
(x-d3
=1 685 35)6 C1 = AS C= 1.27 x29000(
68.35 -
- 3) x 0.000296 =10.03 kip
SHEET 37 OF 160 DATE March 11, 2003 JOB. NO.
PGE-009 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel I CLIENT PG&E-DCPP REVIEWER K. L. Whitmore CALCULATION NO.
PGE-009-CALC-007 Reinforcement ORIGINATOR S. C. Tumminelli ENERCON SERVICES, INC.
APPROVED R. F. Evers REVISION 0 C2 = AE (
d4
(
5 1
= 18-27 x 29000 6868).635 x 0.000296 = 7.94 kip C(s~ f d4)J 6 1 2 7 x 2 0 6 8.3 5
)
TI= AE (h x-
)
=1.27x2900(
'96 - 68.35 - 5535 x 0.000296 = 3.53 kip 68.35
)
T2 =AsE h
(
d2
,C = 1.27 x 2900k 68.-9.705 x 0.000296 = 2.87 kip Check on equilibrium YF = 0
.-. C = Cc + C1 +C2 -T 1 -T2 190.57 = 179.00+10.03+7.94-3.53-2.87 = 190.57 OK Now the internal moment is:
M=C c(hx)+C,(-d3)+C2 (
d4)
-2 )T(2
)
Equation (4)
M = 179.00 96 2
6 8.3 5 j 3
+ (10.03)96 _5.535) + (7.94) 96 _ 18.6352+...
(3-3) 6-5.535~ + (2.87) 296
-_9.705j=
5432.77 -=5426.5 OK
)( 2
)
( 2)
Concrete compressive stress of 0.582 ksi is 55% of fc and the maximum tensile steel stress is
= 2.78 ksi, therefore OK.
1.27 Numerical Evaluation:
Time is 1.125 days Shear NS (Z) and EW (X) Strips Max Fy is 21.692 kip, Table 9 (4/11), Reference 6, largest of NS or EW values.
21.692 c
204 x (96 -9.705)
= 0.00123 ksi = 1.23 psi UK by inspection.
SHEET 38 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Numerical Evaluation:
Time is 1.625 days NS (Z) Strip b = 9 inches; h = 96 inches; d, = 3.45 inches; d2 = 7.63 inches; d3 = 3.45 inches; d4 = 16.55 inches A =1.27 sq. in.
- 10 bar dsm= -8.92 f, = 1.523 ksi Table 4, Reference 6:
Ec = 2358 ksi; Es = 29000 ksi;
_ 29000 - 12.30 2358 ac = 0.724 ksi from off-line calculations; C
724 =0.000307 in/in Fc 2
-3 5 8 From Table 9 (5/11), Reference 6:
C = 5390.02 kip/17 foot section; C = 237.79 kip/9 inch section M =151000 in-kip/17 foot section; M =6661.77 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
C 237.79 t4Asa _C
= 4 x 1.27 x 12.30-
= -265.969 x 9c 0.724 2bAsa(2h -d5,,) = 2 x 9 x 1.27 x 12.30 x (2 x 96 + 8.92) = 56487.66
- (4Asa -
+ j4A a - C b
+ 2bAsa(2h -d,)
265.969 + I4-265.9692 + 56487.66 -69.18 inches 9
x
- Now, 1
CC~ = -xbE,, 6, 2
xbac 69.18x9x0.724 225.40 kip 2
2 z
-3 3) C =1.27 x 29000 698
) x0.000307=10.74 kip
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 39 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 C.,
-d4)," =1.27x29000 69 18 x 0.000307= 8.60kip xK 69.18)
T 1= AE (-
x -(
d C = 1.27 x 290001(
6918
) x 0.000307 =3.82 kip K
69.18
)
(h-x-d 2 )..
96 -69.18-7.63 T2 =ASE s
()
1c
=1.27 x 29000(
9
)x 0.000307 =3.14 kip Check on equilibrium XF = 0
.-. C=CC+C +C2-T -T2 237.79 = 225.40+10.74+8.60-3.82-3.14 = 237.78 OK Now the internal moment is:
M=C
=h-x)+C,(h-d3)+C2(h d4)
(2
)
(2
)
Equation (4)
M = 225.4096 69.18 (10 74) 3.45 + (8.60)96 1655 +
(3.82)( 2 -3.45)+(3.14)( 2 _7.63=6667.19=6661.77OK Concrete compressive stress of 0.724 ksi is 48% of fc and the maximum tensile steel stress is
- 3.
= 3.01 ksi, therefore OK.
1.27
SHEET 40 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
Numerical Evaluation:
Time is 1.625 days EW (X) Strip b = 9 inches; h = 96 inches; d, = 5.535 inches; d2 = 9.705 inches; d3 = 5.535 inches; d4 = 18.635 inches A, = 1.27 sq. in.
- 10 bar dsm, = -8.93 fc = 1.523 ksi Table 4, Reference 6:
EC=2358 ksi; E, = 29000 ksi; a = 29000 = 12.30 2358 a, = 0.728 ksi from off-line calculations; C = 2358 = 0.000309 in/in 2358 From Table 9 (5/11), Reference 6:
C = 4877.105 kip/l7 foot section; C = 215.17 kip/9 inch section M = 150000 in-kip/17 foot section; M =6617.65 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
C 215.17 4Asa --
= 4 x 1.27 x 12.30 -
= -233.08 sc J 0.728 2bAsa(2h - d
) = 2 x 9 x 1.27 x 12.30 x (2 x 96 + 8.93) = 56490.47
- ~4A~oa -
+ j4A~a -
+ 2bAsa(2h -
,)
b 233.08 + J 233.082 + 56490.47
=
9=
62.89 inches 9
- Now, cc = 2 xbG=, = 62.89 x 9 x 0.728 C 2-x s = 2 2
= 0.1kp C, = ASE (x d 3 ) c = 127 x 29000 62.89 -5.535 x 0.000309 = 10.37 kip S S C62.89
)
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 41 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0
C2 =AsEs,(
= 1.27 x 29000 (
62.89
) x 0.000309 = 8.00 kip T. = AE, h
) gc =1.27x29000(
62.89
) x 0.000309 = 4.99 kip T2 =AE, hxd2
)s,
= 1.27 x 2900( 96-62.89-9.705) x 0.000309 = 4.23 kip Check on equilibrium F = 0 C = Cc
+CQ +C 2 -T 1 -T 2 215.17 = 206.01+10.37+8.00-4.99-4.23 = 215.16 OK Now the internal moment is:
M=C
=(h-x)+Cl(-d3)+C2(h d4) T(-
)
(2
)
Equation (4)
M = 206-01 96 _ 62.89) + (10.37) 96 _5.535 + (8.0) 96 _18635)+
(4.99) (
- 5.535) + (4.23)
- 9.705) = 6619.39 _ 6617.65 OK Concrete compressive stress of 0.728 ksi is 48% of fc and the maximum tensile steel stress is
= 3.93 ksi, therefore OK.
1.27 Numerical Evaluation:
Time is 1.625 days Shear NS (Z) and EW (X) Strips Max Fy is 22.181 kip, Table 9 (5/11), Reference 6, largest of NS or EW values.
=
22.181
= 0.00126ksi =1.26psi OK by inspection.
204 x (96 - 9.705)
SHEET 42 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Numerical Evaluation:
Time is 2.125 days NS (Z) Strip b = 9 inches; h = 96 inches; d, = 3.45 inches; d2 = 7.63 inches; d3 = 3.45 inches; d4 = 16.55 inches A, = 1.27 sq. in.
- 10 bar dslm = -8.92 fe = 1.961 ksi Table 4, Reference 6:
Ec = 2674 ksi; Es = 29000 ksi; a = 29000 = 10.85 2674-ac = 0.794 ksi from off-line calculations; c = 2674 = 0.000297 in/in c2674 From Table 9 (6/11), Reference 6:
C = 5574.946 kip/17 foot section; C = 245.95 kip/9 inch section M = 163000 in-kip/1 7 foot section; M = 7191.18 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
C 245.95 54Aa --
= 4 x 1.27 x 10.85 -
= -254.672
)=
90.794 2bA,a(2h -d,.
= 2 x 9 x 1.27 x 10.85 x (2 x 96 +8.92) = 49812.23 4Asj -
+ %4Ascr
+ 2bAsa(2h - ds,,)
b 254.672 + I/- 254.6722 + 49812.23 65.92 incheE 9
- Now, 1
xb=c= 65.92 x 9 x 0.794 cc= -xbECE6
=
22
= 235.54 kip C, = AsEs (
-3 F)c = 1.27 x 29000 65.92
) x 0.000297 = 10.36 kip
SHEET 43 OF 160 March 11, 2003 JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers
_-REVISION 0
,(x-d4 65.92 -16.55 C2 = ASEs
) _c = 1.27 x 29000(
65.92
) x 0.000297 = 8.19 kip T1 =AEs X
)£¢ = 1.27x29000(9 65.92 -345 x 0.000297 4.42 kip (hX-9665.92-76 T2 = AsEs (
2 )ec =1.27 x 29000(
65.92 -
) x 0.000297 =3.72 kip Check on equilibrium E F = 0 C.- C=C +Cl+C2-T1-T2 245.95 = 235.54+10.36+8.19-4.42-3.72 = 245.95 OK Now the internal moment is:
(2
- 3) 2 (2
2 2
Equation (4)
M 235.5496 - 65 92)+ (10.36) 96 _3.45 + (8.19) 96 -16.55 +...
(4.42)( 2 _ 345) + (372)( 2 _ 7.63) 7196.5 7 7191.18OK Concrete compressive stress of 0.794 ksi is 40% of fc and the maximum tensile steel stress is 4427 = 3.48 ksi, therefore OK.
1.27
SHEET 44 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC. CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Numerical Evaluation:
Time is 2.125 days EW (X) Strip b = 9 inches; h = 96 inches; d1 = 5.535 inches; d2 = 9.705 inches; d3 = 5.535 inches; d4 = 18.635 inches A, = 1.27 sq. in.
- 10bar d]=-8.93 f, = 1.961 ksi Table 4, Reference 6:
Ec = 2674 ksi; Es = 29000 ksi; a = 29000 = 10.85 2674-ac = 0.793 ksi from off-line calculations; 0.793 C = 2674 =0.000297 in/in From Table 9 (6/11), Reference 6:
C = 5070.757 kip/l7 foot section; C = 223.71 kip/9 inch section M =160000 in-kip/17 foot section; M =7058.82 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
4Asa -)-
= 4 x 1.27 x 10.85 -
= -227.01 (Ye 20.793 2bAsa42h -ds.= 2 x 9 x 1.27 x 10.85 x (2 x 96 +8.92) = 49814.71 4Asc-Cc
+
4AS -
C
+ 2bASa(2h - d( )
227.01+4-I227.012 +49814.71 = 60.60 inches b
9
- Now, 1
xbo 60.60 x 9 x 0.793 Cc = - xbE&Cc = 2
=
2=
216.24 ip C, = AsE(
x 3 ),
= 1.27 x 29000 60.60
) x 0.000297 = 9.92 kip
(
K 60.60
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 45 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 C2= AE, (
4 )
= 1.27 x 29000 (.60-18.635 x 0.000297 = 7.56 kip T= AsEsX
')x-dl e = 1.27 x 29000(
60.60 5) x 0.000297 5.38 kip 0
x }
60.60
)
T2 AE, h Ixsd
=1.27x2900(
6060 x 0.000297=4.63 kip Kx 60.60
)
Check on equilibrium F = 0
.C=Cc +C +C2 -T 1 -T 2 223.71 = 216.24+9.92+7.56-5.38-4.63 = 223.71 OK Now the internal moment is:
M (2
)
(2 2
2 (2
Equation (4)
M = 216.24 C96
_ 60.60)
(2 3 )
+ (9.92) (926 -5535
+ (7.56)(26 -18.635)+
( 2
)
( 2 (5.38 96 5.55 + (4.63) (96 _9.705~
7061.24 =_7058.82 OK K
22)
Concrete compressive stress of 0.793 ksi is 40% of fc and the maximum tensile steel stress is 5 = 4.24 ksi, therefore OK.
1.27 Numerical Evaluation:
Time is 2.125 days Shear NS (Z) and EW (X) Strips Max Fy is 27.83 kip, Table 9 (6/11), Reference 6, largest of NS or EW values.
Vcx=-27.83
=0.00158 ksi =1.58 psi OK by inspection.
204 x (96 - 9.705)
SHEET 46 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
Numerical Evaluation:
Time is 2.375 days NS (Z) Strip b = 9 inches; h = 96 inches; d1 = 3.45 inches; d2 = 7.63 inches; d3 = 3.45 inches; d4 = 16.55 inches As = 1.27 sq. in.
- 10 bar d,,,m = -8.92 fc = 2.133 ksi Table 4, Reference 6:
E0= 2788 ksi; E, = 29000 ksi; a =
-= 10.40 2788-ac = 0.813 ksi from off-line calculations;
=
8
= 0.000292in/in C
2788 From Table 9 (7/11), Reference 6:
C = 5581.821 kip/17 foot section; M = 166000 in-kip/1 7 foot section; C = 246.26 kip/9 inch section M = 7323.53 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
C246.26=25.8 b4Aa
-C
= 4 x 1.27 x 10.40 -
2 9
=
4250.058 0' J 0.813 2bAsa(2h - dr,.B) = 2 x 9 x 1.27 x 10.40 x (2 x 96 + 8.92) = 47775.43 4Asa - C + 4t4Asa - C b
+ 2bA~ct(2h - d,,,)
250.058 + - 250.0582 + 47775.43
= 64.69 inches 9
- Now, c =1x
= xbc, = 64.69 x 9 x 0.813 2
2 2
=236.66khp C, = ASE( x 3 ) SC = 1-27 x 29000 (6469
) x 0.000292 = 10. 17 kip x d
.2 9 0 64.69-4
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 47 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 C2 =AEs (
- 4) 8 c = 1.27 x 29000 (64C69
) x 0.000292 = 7.99 kip 2(
h-x d
19664.69 -.
5 T= AE, hx
)
Ec =1.27x2900( 9
) x 0.000292= 4.63 kip
( h x -d 6 64.69 7. 3 T2 =AAsEsh
- 2) £C = 1.27 x 29000(
64.69 -
x 0.000292 = 3.93 kip Check on equilibrium XF = 0
- -C=C +C 1+C2 -T 1 -T 2 246.26 = 236.66+10.17+7.99-4.63-3.93 = 246.26 OK Now the internal moment is:
M=C
=(h-x)+Cl(-d3)+C2(h d4)
-d )
(2
)
Equation (4)
M236.66 C (10.17)
_3.45+(7.99)96_-16.55 +...
(4.63) 3.45 (393)
- 7.63) =7325.79 _ 7323.53OK Concrete compressive stress of 0.809 ksi is 38% of f. and the maximum tensile steel stress is 4
= 3.64 ksi, therefore OK.
1.27
SHEET 48 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
Numerical Evaluation:
Time is 2.375 days EW (X) Strip b = 9 inches; h = 96 inches; d, = 5.535 inches; d2 = 9.705 inches; d3 = 5.535 inches; d4 = 18.635 inches A, = 1.27 sq. in.
- 10 bar d... = -8.93 fC = 2.133 ksi Table 4, Reference 6:
EC = 2788 ksi; E, = 29000 ksi; a = 29000 = 10.40 2788-Cc = 0.814 ksi from off-line calculations;
.C = 0288 = 0.000292 in/in From Table 9 (7/11), Reference 6:
C = 5119.114 kip/17 foot section; C = 225.84 kip/9 inch section M = 163000 in-kip/17 foot section; M = 7191.18 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
C 225.84 4ASa --
= 4 x 1.27 x 10.40 -
= -224.608 Iy,
0.8 14 2bA,a(2h -d,.)= 2 x 9 x 1.27 x 10.40 x (2 x 96 + 8.93) = 47777.81
- 54Aja --
) +.l/4Aca - -
+ 2bAa(2h - d,
)
l 224.608 + /- 224.6082 +47777.81 -59.78 inches x
b 9
- Now, C1 =-xbE£ = xbo C
x2 0
x s
2 59.78 x 9 x 0.814 = 218.97 kip C, = AsE ( -3 ) SC = 1.27 x 29000 ( 59.78
) x 0.000292 = 9.76 kip
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 49 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 C2 =AsEs
-4
) Co = 1.27 x 29000 (759 78
) x 0.000292 = 7.40 kip
= (
)x--=.7 2900(
59.78-5.53.>
T1 = AESL h -
d
, = 1.27 x2096-59.78 -
9.3 x 0.000292 =5.52 kip T2 = AsE, (hxd2 )>C = 1.27 x 29000( 96 - 59.78 -9)705 x 0.000292 = 4.77 kip 59.78 Check on equilibrium E F = 0
.-. C= C, +C +C2 -T -T2 225.84 = 218.97+9.76+7.40-5.52-4.77 = 225.84 OK Now the internal moment is:
M Ch 3)
'(2 d3)+C2 (
d4) +T=(-d )+T2(h-d2)
Equation (4)
M=218.97 96 59.78 )
(9.76) 2 _5.535+ (7.40) 96_18635 +
(5.51)( 96 _ 5.535) + (4.77)( 96 -_9705) 7196.05 _- 7191.18 OK Concrete compressive stress of 0.814 ksi is 38% of fc and the maximum tensile steel stress is 5 = 4.35 ksi, therefore OK.
1.27 Numerical Evaluation:
Time is 2.375 days Shear NS (Z) and EW (X) Strips Max Fy is 35.96 kip, Table 9 (7/11), Reference 6 largest of NS or EW values.
204=x35.96
= 0.00204 ksi = 2.04 psi OK by inspection.
204 x (96 -9.705)
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 50 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 Numerical Evaluation:
Time is 3.125 days NS (Z) Strip b = 9 inches; h = 96 inches; d, = 3.45 inches; d2 = 7.63 inches; d3 = 3.45 inches; d4 = 16.55 inches A, = 1.27 sq. in.
- 10bar d,
=-8.92 f, = 2.604 ksi Table 4, Reference 6:
Ec = 3080 ksi; Es = 29000 ksi;
_ 29000 = 9.42 3080 a, = 0.815 ksi from off-line calculations;
£ = °.
= 0.000265 in/in c 3080 From Table 9 (8/11), Reference 6:
C = 5456.216 kip/17 foot section; C = 240.72 kip/9 inch section M
164000 in-kip/17 foot section; M = 7235.29 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
C-240.72 l4Asa-
= 4 x 1.27 x 9.42 -
= -247.525 2bAsa(2h -d
) = 2 x 9 x 1.27 x9.42 x (2 x 96 + 8.92) = 43246.07
- (4ASa - Cj +
4Asot -
C
+ 2bAsa(2h - d.)
247.525 + /-247.5252 + 43246.07 = 63.42 inches b
9
- Now, 1
xb,
= 63.42x9x0.815 2
2 2
p C, = AE, (
3 )c
= 1.27 x 29000 E
- 63.
42 45 ) x0.000265 = 9.22 kip
JOB. NO.
PGE-009 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforc JCLIENT PG&E-DCPP ORIGI ENERCON REVIEWER K. L. Whitmore APPR(
SERVICES, INC. CALCULATION NO.
PGE-009-CALC-007 SHEET 51 OF 160 DATE March 11, 2003 cement NATOR OVED S. C. Tumminelli R. F. Evers REVISION 0 C2 = AsEs(x -d4)C = 1.27 x 29000 (
6
.4
) x 0.000265 = 7.20 kip T 1 = AsE h-x d
£c = 1.27 x 29000(
T2 = AE h-x-d2>EC =1.27x29000(
96 - 63.42 - 345 x 0.000265 = 4.48 kip 63.42
)
'96 - 63.42 - 763 x 0.000265 = 3.83 kip 63.42
)
Check on equilibrium _ F = 0
.,.C =C, +C 1+C2 -T 1 -T 2 240.72 = 232.61+9.22+7.20-4.48-3.83 = 240.72 OK Now the internal moment is:
2 3
C1(h -d 3) +C2 (h-d4 )+T1 h-d.
+T2(h -d2)
Equation (4)
M=232.61 296 63.42)
(9.22)( 2 -3445)+(7.20)(96
_-1655)+
(2 3 )
2
)
2)
(4.48)
- 3.45) + (3.83) 96 - 7.63) = 7238.72 _ 7235.29 OK Concrete compressive stress of 0.815 ksi is 31% of fc and the maximum tensile steel stress is 4 = 3.52 ksi, therefore OK.
1.27
SHEET 52 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC. CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Numerical Evaluation:
Time is 3.125 days EW (X) Strip b = 9 inches; h = 96 inches; d, = 5.535 inches; d2 = 9.705 inches; d3 = 5.535 inches; d4 = 18.635 inches A, = 1.27 sq. in.
- 10 bar d,,,,, = -8.93 fo = 2.604 ksi Table 4, Reference 6:
E, = 3080 ksi; Es = 29000 ksi; Q = 29000 = 9.42 3080 qc = 0.811 ksi from off-line calculations; C = 3080 =0.000263in/in From Table 9 (8/11), Reference 6:
C = 4996.156 kip/l7 foot section; C = 220.42 kip/9 inch section M = 160000 in-kip/17 foot section; M =7058.82 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
C'
~220.42 4Ara
=
'4 x 1.27 x 9.42-
= -223.96 oC 0.811 2bAsa(2h -d,, ) = 2 x 9 x 1.27 x 9.42 x (2 x 96 + 8.93) = 43248.23 4Aa -
+ J4AS
-C b
+ 2bA,a(2h - d,.)
223.96 + J-223.962 + 43248.23
=9
= 58.84 inches
- Now, 1
xbcF_ 58.84x9x0.811 Cc = - xbEsC =
2 2
214.74kip C2 2
C, = AE, 3F
= 1.27 x29000 5884 -535xO0.000263 =8.79 kip
SHEET 53 OF 160 DATE March 11, 2003 JOB. NO.
PGE-009 PROJECT SUBJECT CLIENT DCPP ISFSI ISFSI Cask Storage Pad Steel Reinforcement PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 APPROVED R. F. Evers REVISION 0 C2 =AEs x 4 c)s, = 1.27 x 29000 58.84
) x 0.000263 = 6.63 kip T1 = AsES thx
)dic==1.27x29000( '96 - 58.84 -5535 x 0.000263 = 5.21 kip 58.84
)
T2 = AsEs(
) ec = 1.27 x 29000 9 5 8.8 4
) x 0.000263 = 4.52 kip Check on equilibrium E F = 0
.-. C=
C,
+ C1 +C2 -T1 -T2 220.42 = 214.74+8.79+6.63-5.21-4.52 = 220.43 OK Now the internal moment is:
2 3
2 2
2 2
Equation (4)
M=214.74 96 (2
- 58.84) 3)
+ (8.79)( 96 _.53 (6.63) 96 -18.635) +...
(5.2 1) 96 _5.535~ + (4.52i 96 _ 9.705J 7057.97 =-7058.820OK
(
( 2
)
( 2)
Concrete compressive stress of 0.811 ksi is 31% of f, and the maximum tensile steel stress is 5 21 = 4. 10 ksi,therefore OK.
1.27 Numerical Evaluation:
Time is 3.125 days Shear NS (Z) and EW (X) Strips Max Fy is 49.868 kip, Table 9 (8/11), Reference 6 largest of NS or EW values.
VC = 204 x(9689.705) = 0.00283 ksi = 2.83 psi OK by inspection.
The largest shear in Table 9 is 98.386 kip, see Table 9 (11/11). Results in vc of 5.59 psi, also OK by inspection.
SHEET 54 OF 160 E March 11, 2003 JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore SERVICES, INC. CALCULATION NO.
PGE-009-CALC-007 APPROVED R. F. Evers REVISION 0 Numerical Evaluation:
Shrinkage NS (Z) Strip b = 9 inches; h = 96 inches; d, = 3.45 inches; d2 = 7.63 inches; d3 = 3.45 inches; d4 = 16.55 inches A, = 1.27 sq. in.
- 10 bar d,. = -8.92 fc = 5.000 ksi Ec = 2844 ksi (see Concrete Properties, Ref.6);
E, = 29000 ksi; a = 29000 = 10.20 2844-or = 0.459 ksi from off-line calculations; C = 0.459 = 0.000161in/in
-2844 Ollni From Table 13, Reference 6:
C = 39.019 kip/17 foot section; M = 85400 in-kip/17 foot section; C = 1.72 kip/9 inch section M = 3767.65 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
C4Asa -
C 1.72
= 4 x 1.27 x 10.20 -
= 48.05 ac )
0.459 2bAsa(2h - d5,,,l,) = 2 x 9 x 1.27 x 10.20 x (2 x 96 +8.92) = 46834.71 4Asa -C
+ j4AsU -
+ 2bAa(2h - d,)
- 48.05+ J48.052 +46834.71 = 19.29 inches 9
A-b
- Now, c2 = I xbEE,
- xbo, 19.29 x 9 x 0.459
=
= 39.85 kip 2
21 C, = AE, (
3 )
= 1.27 x 29000 (199
) x 0. 000161 = 4.88 kip
SHEE'T 55 OF 160 March 11, 2003 JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 APPROVED R. F. Evers REVISION 0 C2 = Ass( x-d4 ) c = 1.27 x 29000 19.29
) x 0.000161 = 0.85 kip T 1= AES hxd)
Fc = 1.27 x 29000(96 19.29
) x 0.000161 = 22.57 kip T 2 = AsE hxd2c
=1.27x 2900
- 19. 29 -763 x 0.000161 = 21.28kip Check on equilibrium E F = 0
.-. C=C,+Cl+C2-T,-T2 1.72 = 39.85+4.88+0.85-22.57-21.28 = 1.73 OK Now the internal moment is:
M=C
=
d3)+C2(2 d4)
)
(2
) Equation (4)
M = 39.85 (96 -19.29)+
(4.88) (92 K.
2 3 )(
- 3.45) + (0.85) 96 _ 16.55) +
(22.57) 96 - 3.45) + (21.28) (9
- 7.63) = 3765.23 _ 3767.65 OK Concrete compressive stress of 0.459 ksi is 9% of fc and the maximum tensile steel stress is 2257 = 17.78 ksi, therefore OK.
1.27
SHEET 56 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
Numerical Evaluation:
Shrinkage EW (X) Strip b = 9 inches; h = 96 inches; d, = 5.535 inches; d2 = 9.705 inches; d3 = 5.535 inches; d4 = 18.635 inches As = 1.27 sq. in.
- 10 bar dsu. = -8.93 fe = 5.000 ksi Ec = 2844 ksi (see Concrete Properties, Ref.6);
Es = 29000 ksi; a = 29000 = 10.20 2844-a, = 0.485 ksi from off-line calculations; 6 = 0.485 = 0.000171 in/in c
2844 From Table 13, Reference 6:
C 489.971 kip/17 foot section; M = 84000 in-kip/1 7 foot section; C = 21.62 kip/9 inch section M = 3705.88 in-kip/9 inch section Moment produces compression on the bottom of the pad, therefore use Equations (3) and (4).
CT 0
21.62=72 24A(a - -
= 4 x 1.27 x 10.20 -
= 7.23 2bAsa(2h -d5,,,,) = 2 x 9 x 1.27 x 10.20 x (2 x 96 + 8.93) = 46837.04
- (4Asa - C + j(4Asa -
+ 2bAsa(2h - d,
)
b
- Now, 1
- xba, 23.26 x 9 x 0.485 C
2=
1 XbEcs0 =
2 2
=50.76kip
- 7.23 + V7.232 + 46837.04 = 23.26 inches 9
C1 = ASE (
) EC = 1.27 x 29000 2326
) x 0.000171 = 4.79 kip x
23.26
)
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR ENERCON REVIEWER K. L. Whitmore APPROVED SERVICES, INC. CALCULATION NO.
PGE-009-CALC-007 57 OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0 C2 = AsEs d 4 )Ec = 1.27 x 29000 23.26
) x 0.000171 = 1.25 kip T= ASESh-x -d
, = 1.27 x 29000( 9 6 2 326
)
5.535 x 0.000171 = 18.15 kip (hx-d 623.269.5 T2 =AE,
) Ec = 1.27 x 29000(
23.26
) x 0.000171 = 17.02 kip Check on equilibrium E F = 0
.C=CC
+C1 +C2 -TI -T2 21.62 = 50.76+4.79+1.25-18.15-17.02 = 21.63 OK Now the internal moment is:
M =C h-_
+C, h d3)+C2(h -d4)+T (h -d1)+T2(h d2)
Equation (4)
M=50.76 9(6 23.26) + (4.79) (
_ 5-535 + (1.25) ( 2_ 18.635 +...
C 963024~30.8O (18.15)(
-5.535 + (17.02) 6 -9.705
=3705.48_3705.88OK Concrete compressive stress of 0.485 ksi is 10% of fc and the maximum tensile steel stress is 18.15= 14.29 ksi, therefore OK.
1.27 Numerical Evaluation:
Shrinkage Shear NS (Z) and EW (X) Strips Max Fy is 63.084 kip, Table 13, Reference 6, largest of NS or EW values.
63.084 204 x (96 -9.705)
).00358 ksi =3.58 psi OK by inspection.
SHEET 58 OF 160 DATE March 11, 2003 SO=
JOB. NO.
PGE-009 PROJECT DCPP IS]
SUBJECT ISFSI Ca CLIENT PG&E-D, ENERCON REVIEWER K. L. Wh SERVICES, INC.
CALCULATION NO.
PSI sk Storage Pad Steel Reinforcement CPP ORIGINATOR APPROVED itmore S. C. Tumminelli R. F. Evers PGE-009-CALC-007 REVISION 0 Detailed Evaluation of Allowable Stresses The calculations above are tabulated in Table 1. This data shows that the percentage of applied concrete stress decreases over time while the steel stress peaks at 2.375 days.
Table 1 - Tabulation of Thermal/Shrinkaze Stresses Constrained Model Analyses Time X (inches) f c c (ksi)
% f' a, (ksi)
(days)
NS/EW (ksi)
NS/EW NS/EW NS/EW 1.125 76.54/68.35 1.054 0.574/0.582 54/55 1.77/2.78 1.625 69.18/62.89 1.523 0.724/0.728 47/48 3.01/3.93 2.125 65.92/60.60 1.961 0.799/0.793 40/40 3.48/4.24 2.375 64.69/59.78 2.133 0.813/0.814 38/38 3.64/4.35 3.125 63.42/58.84 2.604 0.815/0.811 31/31 3.52/4.10 Shrinkage 19.29/23.26 5.000 0.459/0.485 9/10 17.77/14.29 Using the previously presented equations, the only difference between the approximate evaluation process and the detailed process is the assessment of P which is a function of the distance from the tensile face to the neutral axis. All of the evaluations above are for the applied moment producing compression on the bottom of the pad, hence tension on the top. Thus all the steel stresses are for the top steel. The equation for f, shows that it is inversely proportional to j3, which is inversely proportional to X, the distance to the neutral axis. Thus fs is proportional to X. The lowest value for the f, for the thermal conditions is from analysis for 3.125 days where NS X is 63.42 inches and EW X is 58.84 inches, see Table 1.
Thermal Conditions For the NS top steel:
Allowing w to equal 0.013 inches dc from the surface to the CG of the closest NS bar is 2+2.9/2 = 3.45 inches A is 2 x (2+2.9+1.27/2) x 9/2 = 49.815 sq. in. per bar (see Figure 1 for dimensions)
B equals (63.42)/(63.42-(2+2.9+1.27/2)) = 1.10 fs w
0.013 0.076V3.45x49.815 x l.xO10-3 28.0 ksi for NS steel 0.076 7K10-
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 59 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 For the EW top steel:
d, from the surface to the CG of the closest EW bar is 2+2.9+1.27/2 = 5.535 inches A is 2 x (2+2.9+1.27+2.9/2) x 9/2 = 68.58 sq. in. per bar (see Figure 1 for dimensions)
B equals (58.84)/(58.84-(2+2.9+1.27+2.9/2)) = 1.15 5 =7 Cw 0.076 Va7I 01-0.013 0.076V55.535x 68.58 x 1.15x10-3 20.5 ksi for EW steel Shrinkage Conditions For the NS top steel:
Allowing w to equal 0.013 inches
- d. from the surface to the CG of the closest NS bar is 2+2.9/2 = 3.45 inches A is 2 x (2+2.9+1.27/2) x 9/2 = 49.815 sq. in. per bar (see Figure 1 for dimensions)
B equals (19.29)/(19.29-(2+2.9+1.27/2)) = 1.40 0.7 vw 80.076V~j~A 010-'
0.013 0.0764V3.45x49.815 x 1.40x10-3 = 22.0 ksi for NS steel For the EW top steel:
d, from the surface to the CG of the closest EW bar is 2+2.9+1.27/2 = 5.535 inches A is 2 x (2+2.9+1.27+2.9/2) x 9/2 = 68.58 sq. in. per bar (see Figure 1 for dimensions) 13 equals (23.26)/(23.26 -(2+2.9+1.27+2.9/2)) = 1.49 w
5 0.076 W j~10-3 0.013 0.076 85.535 x 68.58 x 1.49x10-3 -= 15.9 ksi for EW steel Since the applied stresses are below the acceptable values, all crack widths are expected to be below the 0.013 inches which is considered acceptable, per ACI 207. Further, since all the steel stresses are well below yield, the cracks, if they occur, will all close.
SHEET 60 OF 160 DATE March 11, 2003 Kill JOB. NO.
PGE-009 PROJECT DCPP IS]
SUBJECT ISFSI Cai CLIENT PG&E-Di ENERCON REVIEWER K. L. Wh SERVICES, INC. CALCULATION NO.
PSI sk Storage Pad Steel Reinforcement CPP ORIGINATOR APPROVED S. C. Tumminelli itmore R. F. Evers PGE-009-CALC-007 REVISION 0 Detailed Evaluation for Thermal Demands - Unconstrained Model Analysis The evaluations above use the forces/moments from the constrained model presented in Reference 6.
This modeling technique maximized the constraint of the rock on the pad, resulting in the highest values for the net internal compressive force. However, one can not guarantee that this compressive force will actually develop throughout the pad/rock interface in the field. Local high strain early life concrete creep and/or local shear failure at the rockl"mud" pad interface or at the "mud" pad/pad interface may all combine in some form to reduce the compressive forces resulting from the analytical constraint applied to the pad in the model. Therefore, it is prudent to consider a reduction in the compressive force in the concrete along with a reduction in the applied moments due to the possible slip between the pad and the rock. The unconstrained model analysis presented in Reference 6, bounds this condition in a very conservative manner. This analysis results in pad stress fields that are in equilibrium internally, since there are no external forces acting upon the pad. (The stress field results from Internal Restraint due to the non-uniform temperature distribution within the pad, see Reference 13, Section 4.4.)
The compressive stresses for all the time steps taken from Table 11 of Reference 6 show that ox Min -
Z Min -3 Min and all three stresses are a fraction of fe.
Table 2 - Pad Concrete Compressive Stresses (psi) Vs Time Unconstrained Model Time Load X Min OZM(n 03 Mn E
(R)
(days)
Casef 0.25 1
-7
-7
-11 234 4.7 0.50 2
-67
-67
-68 469 14 0.625 3
-82
-82
-82 586 14 1.125 4
-190
-190
-190 1054 18 1.625 5
-261
-261
-262 1523 17 2.125 6
-284
-285
-285 1961 15 2.375 7
-283
-285
-285 2133 13 3.125 8
-256
-258
-259 2604 10 4.125 9
-218
-218
-219 1 2939 7.5 6.125 10
-154
-147
-160 3448 4.7 7.875 11
-220
-220
-232 3699 6.3
SHEET DATE JOB. NO.
PGE-009 61 OF 160 March 11, 2003 PROJECT DCPP ]
SUBJECT ISFSI (
CLIENT PG&E-REVIEWER K. L. V CALCULATION NO.
ISFSI Cask Storage Pad Steel Reinforcement ORIGINATOR
-DCPP ai7tmore ORIGINATOR APPROVED S. C. Tumminelli ENERCON SERVICES, INC.
R. F. Evers PGE-009-CALC-007 REVISION 0 The evaluations below consider the top reinforcement and then the bottom.
Top Reinforcement-NS and EW Steel After examining the path stress plots in Reference 6, the most demanding stress field for the top reinforcement occurs at time 1.625 days and is shown in Figure 28 of Reference 6. The stresses occur at the center (X and Z are both 0.0) of the pad. A portion of Figure 28 along with some construction added is shown in Figure 5 below. A conservative means to compute the force demand on the reinforcements is to compute the moment demand of the stress field at point "O" only from the tension side, and require the steel to be able to react the moment.
343. 863 272.0 0.0 28XD4.0____
224.212 164.384 _lI-104.556.
\\
44.728 ------ e
-15.099..
15--i
-74.927. a
-134.755
-194.583.
A
-'>qA All
-C,Jw -J X
=
s-0 19.2 38.4 9.6 28.8 nTo Figure 5 - Path plot for Sx and Sz at the ton of the pad With additional construction for analysis
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 62 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 Thus, the applied moment of the tension stress field about point 0 is:
M (2)(3 4 3.9 - 272.0)(11.5(26.9 - 13 )+
(272.OXl 1.5 2 6.9-f l Ž5+( )(272.0X15.4(3.(15.4) in - pounds For the NS steel, the center of gravity of the top steel is 2
= 5.535 inches (see Figure 1) from the concrete surface. Thus the necessary force to react the moment is 874,764
= 40.94 kip. And the bar stress is 40.94
=16.12 ksi. And, for the EW steel, the center of gravity of the top steel is s s(2~)(1.27~)
5.535 + 9.705 2 5=
7.62 inches (see Figure 1) from the concrete surface. Thus the necessary force to react 2
the moment is (2
'762) = 45.37 kip. And the bar stress is 4537) = 17.86 ksi.
26.9 -7.62)
(2X1.27)
Bottom Reinforcement-NS Steel The governing path for Sz occurs at 2.125 days and is shown in Figure 6, which is a portion of Figure 33 from Reference 6. Again as for the top steel, the sum of the moments about 0 is:
410.853S,-.. -
341.232.[
271.4614 1- -'
201.996&*---.-
132.378. ---
62.760E_.-. --
-6.857
-76. 475.
20.2
........ 1.- - -...Ali
-146.093.__.
-21.711..........._
-285.329L.
76.8 96 86.4 0
2 Figure 6 - Path plot for Sz at bottom of Dad with additional construction for analysis
ENERCON SERVICES, INC.
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 63 OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0 Mo = (-(410.9X20.2{
32)(20.2)(9) = 502,991 in pounds And the long (Z) steel (using only the bottom layer) is located 3.45 inches from the bottom of the pad. Thus the force in this bar is:
502.991 30.03
= 30.03 kip. The bar stress is therefore,
= 23.7 ksi 20.2 -3.45 1.27 Bottom Reinforcement-EW Steel The governing path for Sx occurs at 2.125 days and is shown in Figure 7, which is a portion of Figure 32 from Reference 6. Again as for the top steel, the sum of the moments about 0 is:
4 01. 339--.
333.042 264.743.1-196.444.
128.145
- 59. 846]1-
-8. 452j._-
-76.751
.0.
- 76.
8 9
.2 86.4
-145.0501
-213.349}....
-281.648 _
0 6
Figure 7 - Path iplot for Sx at bottom of Dad with additional construction for analysis Mo = 1(2!401.3)(20.7( 32)(20.7X9)= 515,859 in pounds And the short (X) steel (using only the bottom layer) is located 5.535 inches from the bottom of the pad. Thus the force in this bar is:
5.59 = 34.02 kip. The bar stress is therefore,
- = 26.8 ksi 20.7 -5.535 1.27
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 64 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 Evaluation of the Unconstrained Model Stresses Using the previous equations for crack width:
For the NS top steel:
dc from the surface to the CG of the closest NS bar is 2+2.9/2 = 3.45 inches A is 2 x (2+2.9+1.27/2) x 9/2 = 49.815 sq. in. per bar (see Figure 1 for dimensions)
B equals (26.9/(26.9-(2+2.9+1.27/2)) = 1.26 w = 0.07634-Fpf,10-3 = 0.076 (33.45 x 49.815)1.2 6 x 16.12 x10-3 = 0.0086 inches For the EW top steel:
dc from the surface to the CG of the closest EW bar is 2+2.9+1.27/2 = 5.535 inches A is 2 x (2+2.9+1.27+2.9/2) x 9/2 = 68.58 sq. in. per bar (see Figure 1 for dimensions)
B equals (26.9)/(26.9-(2+2.9+1.27+2.9/2)) = 1.40 w =0.076 f T f,1 l 0-3 = 0.076 (35.535 x 68.58)1.40 x 17.86 x10-3 = 0.0138 inches For the NS bottom steel:
Consider only the bottom bar, the second bar is too high into the concrete to consider.
dc from the surface to the CG of the closest NS bar is 2+2.9/2 = 3.45 inches A is 2 x (2+2.9/2) x 9 = 62.1 sq. in. per bar (see Figure 2 for dimensions)
B equals (20.2)/(20.2-(2+2.9/2)) = 1.21 w =0.076F XOf,10-3 = 0.076 (3 3.45 x 62.1)1.21 x 23.7 x10-3 =0.0130inches For the EW bottom steel:
d, from the surface to the CG of the closest EW bar is 2+2.9+1.27/2 = 5.535 inches A is 2 x (2+2.9+1.27/2) x 9 = 99.63 sq. in. per bar (see Figure 2 for dimensions)
B equals (20.7)/(20.7-(2+2.9+1.27/2)) = 1.36 w = 0.076 3VKA Pfl 0-3 = 0.076 (3 5.535 x 99.63)1.36 x 26.8 x10-3 = 0.0227 inches
SHEET DATE 65 OF 160 March 11, 2003 JOB. NO.
PROJECT SUBJECT PGE-009 DCPP ISFSI ISFSI Cask Storage Pad Steel Reinforcement ENERCON SERVICES, INC.
CLIENT PG&E-REVIEWER K. L. M CALCULATION NO.
DCPP Ihitmore ORIGINATOR APPROVED S. C. Tumminelli R. F. Evers PGE-009-CALC-007 REVISION 0 Table 3 - Tabulation of Thermal Crack Widths Unconstrained Model Analyses Time Location/
Applied Crack Width (days)
Direction Stress (ksi)
(mils) 1.625 Top NS 16.12 9
1.625 Top EW 17.86 14 2.125 BotNS 23.7 13 2.125 Bot EW 26.8 23 The crack widths of 9 to 14 mils are acceptable since they are very small. The 23 mil crack width is also acceptable since: 1) the analysis that produced the applied forces is very conservative, 2) the mud mat protects the steel from water intrusion (also the water table is a considerable distance below the bottom of the pad) and 3) since the applied stress is well below yield, cracks, if they occur, will close in a matter of days.
The thermal and shrinkage evaluations presented above demonstrate that the #IO bar arrangement shown in Figures 1 and 2 is acceptable and not overly conservative. The expected cracking due to temperature and shrinkage is within the ACI recommended values for external exposure. Lastly, should a bond break occur during construction either between the rock and the "mud" pad, or the mud pad and the pad, the pad will not suffer any detrimental consequences.
SHEET 66 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Section 3 - Evaluation for Seismic Loads The following calculation presents an evaluation of the concrete section with the #10 bar configuration shown in Figures 1 and 2 for the seismic loads. Actually, the evaluation simply evaluates ACI equation 4, U>D+L+Ess. As demonstrated above, this is the only equation that needs to be evaluated. The moments/forces from the seismic analysis, Reference 3, are the D+L+Ess terms. Thus, the following evaluation will demonstrate that the section capacity U bounds all of the moment/ force combinations provided in Reference 5.
The section capacities are computed for the entire range of anticipated loads. Thus the full moment/force interaction diagram for each strip direction, i.e., North-South (Z) and East-West (X) are computed and plotted. Since the steel arrangement is not symmetrical, the section capacities must be computed for the two senses of the applied moment, i.e., moment that produces tension on the bottom and moment that produces compression on the bottom. Figures 8 through 15 provide the graphical information needed to follow the calculations for the North-South section capacities. As, before, these are computed for a 9 inch width of pad.
The thermal calculations presented above assess the pad for the thickest the pad can be, which is 8 feet (96 inches) at its center. This is the basis for the calculation of the heat of hydration, the subsequent thermal demand computed in Reference 6 and the sizing of the reinforcement computed above. The seismic evaluation assesses the pad for the thinnest the pad can be, which is 7 feet 6 inches (90 inches) at its edges. This value is the thickness used in the computation of the seismic demand in Reference 5.
This value will be used to evaluate the section for the seismic demand.
Basic data for Strength Method Concrete strength:
fl = 5000 psi Whitney stress:
0.85f0 = 4250 psi= 4.25 ksi Depth of stress block: PBX wherep, is 0.80 for 5000psi concrete and X is the distance from the compression fiber to the NA Concrete strain:
sc = 0.003 in/in Steel yield stress:
fy = 60000 psi = 60.0 ksi 60 Steel yield strain:
FY =
=0.002069 in/in 29E3 Net section depth for evaluation is 90 inches (7'-6")
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR ENERCON REVIEWER K. L. Whitmore APPROVED SERVICES, INC. CALCULATION NO.
PGE-009-CALC-007 67 OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0 North-South Section Concrete Capacity
- 3.45
-7.62 k s 0.003 P.i
/,r 41.55
/-
/7 37.38
`
45.00 86.55 28.45 4.25 ksi C,2 4-4-
C 4-C.- C
-11
<- A C
I I
I
-I
- 41.55
>0.00206 F,
-4
- 16.55
- 3.45 Figure 8 - North-South (Z strip) Section - Net Axial Compression No Moment
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 68 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0
- 3.45
-7.62 0.003 A
L SC2 41.55
/z
/
_' i' C 37.38 45.00 86.55
.- 28.45 x
a 4.25 ksi 4-C C2 4-C T2 86.55-X ST2 -
41.55 T Ir Is 0.002069
- 16.55
- 3.45 Fieure 9 - North-South (Z strip) Section - Balanced Condition Tension on the Bottom This is +Mx. see Reference S
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 69 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0
- 3.45 4762 0.003 A
L F
.-/
41.55 37.38
-FC2 7
4.25 ksi
-0 SC, 4-C2 C
2
<4-
, 0 a
x Cc
-t-45.00 K..
FL 86.55 28.45 N 41.55
\\-
86.55-X AL
- T 2
_ Ly-
- 16.55 STi
- 3.45 Figure 10 - North-South (Z strip) Section - Compression Controls Tension on the Bottom This is +Mx, see Reference 5
JOB. NO.
PGE-009 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforc CLIENT PG&E-DCPP ORIGI REVIEWER K. L. Whitmore APPR(
CALCULATION NO.
PGE-009-CALC-007 SHEE DATE cement T
70 OF 160 B March 11, 2003 ENERCON SERVICES, INC.
NATOR S. C. Tumminelli OVED R. F. Evers REVISION 0
-3.45 7 762 K-k I A
I,--
i 41.55 SC2 4.25 ksi 41-I, A
I z
x a
_' 14A
-. 37.38 45.00 86.55
- 28.45 86.55-X s i 41.55 s-16.55
- 3.45 STI>Sy Figure 11 - North-South (Z strip) Section - Tension Controls Tension on the Bottom This is +Mx. see Reference 5
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 71 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0
- 3.45 7 62 0.002069 A
F F -a T
_-il T2 41.55
/7 CT2 37.38 86.55-X e,..
45.00 86.55 I
28.45 I-41.55 x
a
.I+-
Cc
< r I
I 16.55 4-Cl 4.25 ksi 0.003
- 3.45 Fieure 12 - North-South (Z strip) Section - Balanced Condition Compression on the Bottom This is -Mx. see Reference S
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 72 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0
'1
- 3.45 4 762 A
_-o Tj T2 41.55
/
37.38 ET2 45.00 86.55 86.55-X L
pS C2 a
a x
k I
28.45 N 41.55 4-4-
4.25 ksi
-Ccc
-l C2 C1
- 16.55
- 3.45 0.003 Figure 13 - North-South (Z strip) Section - Compression Controls Compression on the Bottom This is -Mx, see Reference 5
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 73 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 8 T1>SY
_--Ti
_-O T2
- 41.55
- 37.38
- 45.00 86.55 86.55-X u L 71-28.45 x
a
.1
.1
.1
- z-
.1 41.55
+Cc 4-C2 4-C1 4.25 ksi n t
- 16.55 3.45 0.003 Figure 14 - North-South (Z strin) Section - Tension Controls Compression on the Bottom This is -Mx. see Reference 5
SHEET 74 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
I
- 3.45 7.62 L
41.55
`
37.38
`
45.00 81 F,-I 1\\82 8 ~y
-__17-1-C F
T T2 PT
_* T3 T4 86.55 28.45
'N1 I
Fk N
41.55
- 16.55
- 3.45 Figure 15 - North-South (Z strip) Section - Net Axial Tension No Moment
SHEET 75 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
Compute the axial force/moment interaction diagram.
North-South Net Axial Compression - No Moment - Figure 8 Since the reinforcement is not symmetrical, in order to achieve a net axial moment at the mid-height of the section, the concrete stress must vary slightly over the height of the section so that the unbalanced moment produced by the steel at a stress of Fy is offset by the concrete. Thus, the unbalanced moment is AM = AsFy(Net moment arm difference) = 1.27 x 60 x (37.38-28.45) 680.5 in kips. Therefore the A concrete stress (Aa) is computed from the equilibrium equation AM As x S where S is the section 9X90 2 A
8.
modulus. S =
6
12150inches 3. And, AG
= 680.5 =0.056 ksi. Thus, the concrete stress 6
S5 25 varies linearly from 4.25 ksi at the top to 4.25-(2)(0.056) = 4.138 at the bottom of the section. And, P. = (4.138 + ( 2j) (2) 0.0563 (9) (90) + (60) (4) (1.27) = 3701.9 kips Check on the moment:
M = (4.25-4.138)(0.5)(9)(90) ((2/3)(90)-45))-76.2(37.38-28.45) = -0.1 OK North-South Balanced Condition - Tension on the Bottom - Figure 9 Compute X (distance to NA):
X
(
0.003 K0.003 + 0.002069
- ) 86.55 = 51.22 inches 6T1 =£Y = 0.002069 T2 35.33+3.45 16-5 0.002069=0.001302 35.33 6 C2
_ (
.22-7.62 0.003 = 0.002554 > 0.002069 and ecc > 0.0 0 206 9 also.
Compute individual internal forces:
Depth of stress block: a = 0.80 x 51.22 = 40.98 inches
ENERCON SERVICES, INC.
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 76 OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0 C= 0.85fcab = 4.25 x 40.98 x 9 = 1567.5 kips C, = A, (f - 0.85fj)= 1.27 x (60 - 4.25) = 70.8 kips C2 =70.8 kips T2 = EeAs = 29E3 x 0.001302 x 1.27 = 47.9 kips T1 =A fA =1.27 x 60 = 76.2 kips Net nominal compression is:
Cn =Cc +C 1 +C 2 -T. -T 2 = 1567.5 +70.8+70.8 - 76.2 - 47.9 =1585.0kips Net nominal moment is:
Mn = Cc(45-j+(C+/-+TI)41.55+(C2 )37.38+(T2)28.45 Mn= 1567.5 (45 -
29)+ (70.8 + 76.2)41.55 + (70.8)37.38 + (47.9) 28.45 = 48,536.5 in - kips
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 77 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 N-S Nominal Fnrce/Moment Interaction - Tension on Bottom - Comoression Controls - Figure 10 Set 0.002069 0.001035, therefore:
2 2
X =
0.003 86.55 = 64.35 inches (0.003 + 0.001035 6T2 = (
86.5 5 6435 0.001035= 0.000424 SC2 = (64.35-7.62) 0.003 = 0.002645 > 0.002069 and 8cc > 0.002069 also.
64.35
)
Compute individual internal forces:
Depth of stress block: a = 0.80 x 64.35 = 51.48 inches C= 0.85fcab = 4.25 x 51.48 x 9 = 1969.1 kips C, = As (fy - 0.85f )= 1.27 x (60 -4.25) = 70.8 kips C2 =70.8 kips T= EeAs = 29E3 x 0.001035 x 1.27 = 38.1 kips T2 =EEA =29E3 x 0.000424 x 1.27 =15.6 kips Net nominal compression is:
Cn = Cc + C1 + C2 -T -T2 = 1969.1 + 70.8 + 70.8-38.1-15.6 =2057.0 kips Net nominal moment is:
Mn =Cc 45 - a)+ (Cl + T 1
)41.55 + (C2 )37.38 + (T2) 28.45 M = 1969.1 (45 - 5148) + (70.8 + 38.1) 41.55 + (70.8) 37.38 + (15.6) 28.45 = 45,540.0 in - kips
ENERCON SERVICES, INC.
SHEET 78 OF 160 JOB. NO.
PGE-009 DATE March 11,2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 N-S Nominal Force/Moment Interaction - Tension on Bottom - Compression Controls - Figure 10 SetsTI = 0.0, therefore:
X = 86.55 inches T2 3.45 -16.55 0.003 = - 0.000454
(-) indicates compression TI 86.55
)
(86.55-7.62 0.003 = 0.002736 > 0.002069 and ec, > 0.002069 also.
12 86.55
)
Compute individual internal forces:
Depth of stress block: a = 0.80 x 86.55 = 69.24 inches Cc = 0.85fcab = 4.25 x 69.24 x 9 = 2648.4 kips C, = A, (f, - 0.85f,)= 1.27 x (60 - 4.25) = 70.8 kips C2 = 70.8 kips T, =0.Okips T 2 = ErA, = 29E3 x - 0.000454 x 1.27 =-16.7 kips Net nominal compression is:
C' = C,
+ C1 + C2 - T, - T2 = 2648.4 + 70.8 + 70.8 - (-16.7) =2806.7 kips Net nominal moment is:
Mn =CC,45 -aj)+(C' +Tl)41.55+(C2 )37.38+(T2)28.45 Mn = 2648.4 (45 -
- 24) + (70.8) 41.55 + (70.8) 37.38 -(16.7)28.45 = 32,603.5 in - kips
SHEET 79 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
N-S Nominal Force/Moment Interaction - Tension on Bottom - Tension Controls - Fieure 11 Set ET2 = Sy = 0.002069, (Note the change from ETI to ET2), therefore:
X = 1-0.003 (86.55 + 3.45 -16.55) = 43.47 inches (0.003 + 0.002069) gT1 = (
86.55 - 43.47
)0.002069 = 0.002973 >
T 86.55 -43.47 + 3.45-16.55) y
&c2 = (43.47-7.62 0.003 = 0.002474 > 0.002069 and scl > 0.002069 also.
C2 43,47
)
Compute individual internal forces:
Depth of stress block: a = 0.80 x 43.47 = 34.78 inches C, = 0.85fcab = 4.25 x 34.78 x 9 = 1330.2 kips C, = A, (fy - 0.85f)= 1.27 x (60-4.25) = 70.8 kips C2 =70.8 kips T1 =Asfy =1.27 x 60 =76.2 kips T2 =Asfy = 1.27 x 60 =76.2 kips Net nominal compression is:
Cn =C, +C1 +C 2 -T1 -T2 =1330.2+70.8+70.8-76.2-76.2=1319.4kips Net nominal moment is:
Mn= Cc(45 - )+(C, + T 1)41.55+(C 2)37.38+(T 2)28.45 Mn = 1330.2 (45 -
24.7)
+ (70.8 + 76.2) 41.55 + (70.8) 37.38 + (76.2)28.45 = 47,649.1 in - kips
SHEET 80 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISPSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 N-S Nominal Force/Moment Interaction - Tension on Bottom - Tension Controls - Figure 11 Set s=
2.0y = 2.0 x 0.002069 = 0.004138, therefore:
X
(
0.003 86.55 = 36.38 inches 0.003 + 0.004138 (86.55 -36.38 +3.45 -16.55 "
£T2 =
86.55 - 36.38
)0.004138 = 0.003058 > Ey 6 c2 = (36.38-7.62 0.003 = 0.002372 > 0.002069 and Ec, > 0.002069 also.
c'~36.38
)
Compute individual internal forces:
Depth of stress block: a = 0.80 x 36.38 = 29.10 inches C, =0.85fcab =4.25x29.10x9=1113.1kips C, = A,(f, -0.85fc )=1.27 x (60-4.25) = 70.8kips C2 =70.8 kips T= A, fy = 1.27 x 60 =76.2 kips T2 = Afy = 1.27 x 60 =76.2 kips Net nominal compression is:
C. =Cc +C1 +C 2 -T1 -T 2 =l113.1+70.8+70.8-76.2-76.2=1102.3kips Net nominal moment is:
Mn =C,(45-a) +(C' +TI)41.55+(C 2 )37.38+(T 2 )28.45 M" =1113.1 (45-2910) +(70.8+ 76.2) 41.55 + (70.8) 37.38 + (76.2) 28.45 =44,816.1 in - kips
SHEET 81 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
N-S Nominal Force/Moment Interaction - Tension on Bottom - Tension Controls - Fieure 11 Set 6TI = 4.06Y = 4.0 x 0.002069 = 0.008276, therefore:
X =
0.003 186.55 = 23.03 inches p0.003 + 0.008276)
( 86.55 -23.03 +3.45 - 1 655 6T2 = I 86.55 - 23103 J 0.008276 = 0.006569 > £y 86.55 -23.03
)
C(23.03-7.62 0.003 = 0.002007 < 0.002069
-~23.03
)
c =23.03 3.45 0.003 = 0.002551 > 0.002069 Compute individual internal forces:
Depth of stress block: a = 0.80 x 23.03 = 18.42 inches Co = 0.85fcab = 4.25 x 18.42 x 9 = 704.6 kips C, = A, (fy - 0.85fc )= 1.27 x (60 - 4.25) = 70.8 kips C2 = (EsC2 - 4.25)1.27 = (29E3 x 0.002007 -4.25)1.27 = 68.5 kips T1 =Afy =1.27x60=76.2kips T2 =Asfy = 1.27 x 60 =76.2 kips Net nominal compression is:
Cn = Cc +C] +/-C 2 -T1 -T 2 = 704.6 + 70.8 + 68.5-76.2-76.2 = 691.5 kips Net nominal moment is:
Mn =C4I_45-
+(C,+T,)41.55+(C2 )37.38+(T2)28.45 Mn = 704.6 45 - 1842) +(70.8 + 76.2) 41.55 + (68.5) 37.38 + (76.2) 28.45 = 36,053.9 in - kips
SHEET 82 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 N-S Nominal Force/Moment Interaction - - Tension on Bottom - Tension Controls - Figure 11 Set cTl =8.0£y =8.0x0.002069=0.016552, therefore:
X=
0.003 886.55 =13.28 inches 0.003 + 0.016552)
CT2 = (
86.5 -13 28
) 0.016552 = 0.013593 > £y 8613.28-73.28 FC2 = ( 13.28 -7.62 0.003 = 0.001279 < 0.002069
-~13.28
)
Fc 0 =(13.28345 D 0.003 = 0.002221 > 0.002069 Compute individual internal forces:
Depth of stress block: a =0.80x 13.28=10.62inches C, = 0.85fcab = 4.25 x 10.62 x 9 = 406.2 kips C, = A, (fy - 0.85f ) = 1.27 x (60 - 4.25) = 70.8 kips C2 = A, (EC0 2 - 0.85fj ) = 1.27 x (29E3 x 0.001279 - 4.25) =41.7 kips T. =Arfy = 1.27 x 60 =76.2 kips T2 A fy = 1.27 x 60 =76.2 kips Net nominal compression is:
Cn = C +C1 +C 2 -1T7 -T2 =406.2+70.8+41.7-76.2-76.2=366.3kips Net nominal moment is:
Mn = C4(45 - 2) +(Ci+T,)41.55+(C 2)37.38+(T2 )28.45 Mn =406.2( 4 5 -
+(70.8+76.2) 41.55+(41.7) 37.38+(76.2)28.45=25,956.6in-kips
SHEET 83 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 N-S Nominal Force/Moment Interaction - Tension on Bottom - Tension Controls - Figure 11 Set STi = 166y = 16 x 0.002069 = 0.033104, therefore:
X
(
0.003 86.55 = 7.19 inches (0.003 + 0.033104)
(86.55 - 7.19 + 3.45 -16.5502
£T2 = K 86.55-7.
0.0331047= 0.027640>£y 8 C2 = 7 119-7.62 ) 0.003 = -0.000179 (-) indicates bar in tension 71-(
0.003 = 0.001561 < 0.002069 Compute individual internal forces:
Depth of stress block: a = 0.80 x 7.19 = 5.75 inches Cc = 0.85fcab = 4.25 x 5.75 x9 = 219.9 kips C, = A, (Ecc, - 0.85fc) = 1.27 x (29E3 x 0.001561-4.25) = 52.1 kips C2 = ASEEC2 = 1.27 x 29E3 x - 0.000179 = -6.60 kips (bar in tension)
T= Afy =1.27 x 60 =76.2 kips T2 =Afy =1.27 x 60 =76.2 kips Net nominal compression is:
Cn = C, +C1 +C 2 -T-T 2 = 219.9+52.1-6.60-76.2-76.2 = 113.0kips Net nominal moment is:
Mn =Cc(45-a)+ (Cl + T)41.55+(C2 )37.38+(T2)28.45 Mn = 219.9(45 -
275)+ (52.1 + 76.2) 41.55 + (-6.60) 37.38+ (76.2)28.45 =16,515.3 in -kips
SHEET 84 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 N-S Nominal Force/Moment Interaction - Tension on Bottom - Tension Controls - Flizure 11 Set cT1 = 32cy = 32 x 0.002069 = 0.066208, therefore:
X
(
0.003 86.55 = 3.75 inches k0.003 + 0.066208)
T 86.55 - 3.75 + 3.45 - 16.550 CT2 = K 86.55 - 375 0.066208=0.055733>
6 C2 (3.75-7.62 0.003 =-0.003096> 8y (-) indicates bar in tension c
375 -
)3450.003
= 0.000240 < 0.002069 Compute individual internal forces:
Depth of stress block: a = 0.80 x 3.75 = 3.00 inches Co = 0.85fcab = 4.25 x 3.00 x 9 = 114.8 kips C, = A, (E6,1 - 0.85f ) = 1.27 x (29E3 x 0.000240 - 4.25) = 3.4 kips C 2 = Afy = 1.27 x (-60) = -76.2 kips (bar in tension)
T, = Af = 1.27 x 60 =76.2 kips T2 =Afy = 1.27 x 60 =76.2 kips Net nominal compression is:
C, = Cc + CI +C 2 -T1 -T 2 = 114.8 + 3.4-76.2-76.2-76.2 =-1 10.4 kips (-)indicates tension Net nominal moment is:
Mn= Cc(45 - a) + (Cl + T1)41.55 + (C2)37.38 + (T2)28.45 Mn = 114.8 (45 -320)
+ (3.4 + 76.2) 41.55 + (- 76.2) 37.38 + (76.2) 28.45 = 7620.7 in - kips
SHEET 85 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 North-South Net Axial Tension - No Moment - Figure 15 As with the net compression case, the strain across the section must vary slightly in order to achieve zero moment at mid-height.
Set &3 = 6y = 0.002069 and therefore T3 = T4 = 76.2 kips Moment equilibrium requires:
M = 0 = (TI) 41.55 + (T2) 37.38 - (T3) 28.45 - (T4 ) 41.55 Use the strain:
C 7.62 -3.45 A
' 41.55 + 28.45 2=1 + 0.000123-0.05957E F2 =0.9404F1 +0.000123 Now, use the stress-strain relation and the bar areas:
T1 = AEc =1.27 x 29E3 x F, = 36,830Fl and T2 =AE62 =1.27x 29E3x(0.9404c, +0.000123)=34,634.9el + 4.53 Substitute in to equation for M 0 = (36,830o
)41.55 + (34,634Fl + 4.53) 37.38 -(76.2) 28.45- (76.2) 41.55 0 = 2,824,905E] -5,164.7 5167 = 0.001828 2,824,905 2 = (0.9404)0.001828 + 0.000123 = 0.001842 And, Ti = 1.27 x 29E3 x 0.001828 = 67.3 kip
SHEET 86 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
And, T2 = 1.27 x 29E3 x 0.001842 = 67.8 kip Thus, T = Pt = 67.3+67.8+76.2+76.2=287.5 kip Check on the moment:
0 = (67.3)41.55 + (67.8)37.38 - (76.2)28.45 - (76.2)41.55 =-3.3 OK North-South Calculation of i2 per ACI - Tension on Bottom p = 0.90 for tension plus flexure p = 0.70 for compression plus flexure 0.1ftAg =0.1 x5.00x9x90=405kip>pPb=((1585.0) 0.90 0.90 C,,
366.3kips, (I 2C
=
2363
= 0.76 1 + 2n 1+2x363 fcAg 5x9x90 C
0.90 0.90
=08 Cn =113.0kips, (
2C
=
2113
.8 1+C 1+5xx90 fLA9 5x9x90
SHEET 87 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 North-South Balanced Condition - Compression on the Bottom - Fizure 12 Compute X (distance to NA):
X =~.0 0.003 86.5 5 = 5 1.22 inches (0.003+ 0.002069)
STI Ey = 0.002069
£T2 (35.33 +33.45 62 0.002069 =0.001825 C2 51.22 -16.55 0.003 = 0.002031 < 0.002069 scl (51.22
)0.003=0.002798>0.002069 Compute individual internal forces:
Depth of stress block: a = 0.80 x 51.22 = 40.98 inches Cc = 0.85fcab = 4.25 x 40.98 x 9 = 1567.5 kips C, = A, (fy - 0.85f ) = 1.27 x (60 - 4.25) = 70.8 kips C2 = A, (Es - 0.85fo ) = 1.27 x ((29E3X0.002031) - 4.25) = 69.4 kips T2= EeA, = 29E3 x 0.001825 x 1.27 = 67.2 kips T.
A fy =1.27 x 60 = 76.2 kips Net nominal compression is:
Cn = Cc + C + C2 -T
-T 2 = 1567.5 + 70.8 + 69.4-76.2-67.2 = 1564.3 kips
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 88 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 Net nominal moment is:
Mn =C4 45-a
.+(C 1 +T1 )41.55+(C2 )28.45+(T2 )37.38 M n = 1567.5 (45 -
2
)+ (70.8 + 76.2) 41.55 + (69.4) 28.45 + (67.2)37.38 = 49,013.6 in - kips
SHEET 89 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 N-S Nominal Force/Moment Interaction - Compression on Bottom - Comp Controls - Figure 13 S = =0.002069 = 0.001035, therefore:
SetCFTI =-
2 X= (
0.003
) 86.55 = 64.35 inches 0.003 + 0.001035 (86.55 - 64.35 + 3.45-7-62 0 001035 = 0.000840 86.55 -64.35 FC 2 = (64.35 -16.55 0.003 = 0.002228 > 0.002069 and Scc > 0.002069 also.
64.35 Compute individual internal forces:
Depth of stress block: a = 0.80 x 64.35 = 51.48 inches Cc =0.85fcab =4.25x51.48x9 =1969.1kips C, = A, (fy - 0.85fj )= 1.27 x (60 - 4.25) = 70.8 kips C2 = 70.8 kips T1 = EAS =29E3 x 0.001035 x 1.27 =38.1 kips T2 = EeA, = 29E3 x 0.000840 x 1.27 = 30.9 kips Net nominal compression is:
Cn =C, +C 1 +C 2 -T -T2 = 1969.1+70.8+70.8-38.1-30.9 =2041.7kips Net nominal moment is:
Mn =Cc 45-a) + (Cl + T 1
) 41.55 + (C2 ) 28.45 + (T2 ) 37.38 Mn =1969.1(45-51.48)+ (70.8 + 38.1) 41.55 + (70.8) 28.45 + (30.9)37.38 = 45,619.0 in - kips
SHEET 90 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 N-S Nominal Force/Moment Interaction - Compression on Bottom - Como Controls - Figure 13 Set sT1 = 0.0, therefore:
X = 86.55 inches ET2 =(3.45 -
)7.62 0.003=-0.000145
(-)indicatescompression (86.55
)
C2 = (6.55 -16.55) 0.003 = 0.002426 > 0.002069 and ec, > 0.002069 also.
12=~86.55 Compute individual internal forces:
Depth of stress block: a = 0.80 x 86.55 = 69.24 inches Cc= 0.85fcab = 4.25 x 69.24 x 9 = 2648.4 kips C1 = A,(fy -0.85f )=1.27 x (60-4.25) = 70.8 kips C2 =70.8 kips T1 =O.Okips T2 = EeA, = 29E3 x - 0.000145 x 1.27 =-5.3 kips Net nominal compression is:
C, = Cc + Cl + C 2 - T-T 2 = 2648.4 + 70.8 + 70.8 -(-5.3) = 2795.3 kips Net nominal moment is:
M, = Cc(45-2) + (C, +T,)41.55 + (C2)28.45 + (T2)37.38 Mn = 2648.4 (45 -
2
+ (70.8) 41.55 + (70.8) 28.45 - (5.3)37.38 = 32,248.3 in - kips
SHEET 91 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 N-S Nominal Force/Moment Interaction - Compression on Bottom - Tension Controls - Figure 14 Set FTI = 1.5 £y = 1.5 x 0.002069 = 0.003104, therefore:
X (0
0.003
) (86.55) = 42.54 inches (0.003 + 0.003104)
F 86.5 5 -42.54+/-+3.45 -762 0030
.08 ET2 86.55-42.54
)0.003104=0.002811>y c2 (42.54-6.55 0.003 = 0.001833 < 0.002069 l =(42.54 -3.45 0.003 = 0.002757 > 0.002069 Compute individual internal forces:
Depth of stress block: a = 0.80 x 42.54 = 34.03 inches Cc = 0.85fcab = 4.25 x 34.03 x 9 = 1301.6 kips C, = A, (fy -0.85fc )= 1.27 x (60-4.25) = 70.8 kips C2
= A, (Es - 0.85fc) = 1.27 x ((29E3X0.001833) - 4.25) 62.1 kips T= Afy =1.27 x 60 = 76.2 kips T2 =A fy = 1.27 x 60 = 76.2 kips Net nominal compression is:
C. = Cc + C1 + C2 -T -T2 = 1301.6 + 70.8 + 62.1-76.2-76.2 = 1282.1 kips Net nominal moment is:
Mn = Cc(45 - a) + (Cl + Tl) 41.55 + (C2 )28.45 + (T2 ) 37.38 M = 1301.6 45 -
2403) + (70.8 + 76.2) 41.55 + (62.1) 28.45 + (76.2)37.38 = 47,148.2 in - kips
SHEET 92 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 N-S Nominal Force/Moment Interaction - Commression on Bottom - Tension Controls - Figure 14 Set 8 T1 = 2.0y = 2.0 x 0.002069 = 0.004138, therefore:
X=
0.003 86.55 = 36.38inches (0.003 + 0.004138) 6T2 86.55 - 36.38 + 3 45 7.62 0.004138 = 0.003794>s T2 86.55 -36.38 J
9c2 = ( 36.38 -16.55 0.003 = 0.001635 < 0.002069
= 36.38 -3.45 0.003 = 0.002716 > 0.002069 c'
36.38 Compute individual internal forces:
Depth of stress block: a = 0.80 x 36.38 = 29.10 inches C, =0.85fcab =4.25x29.10x9 =1113.1kips C, = A, (fy - 0.85f, )= 1.27 x (60 -4.25) = 70.8 kips C2 = A, (Es - 0.85fc) = 1.27 x ((29E3)(0.001635) - 4.25) = 54.8 kips T. = Afy = 1.27 x 60 = 76.2 kips T2 = Asfy = 1.27 x 60 = 76.2 kips Net nominal compression is:
Cn = C + C1 + C2 -T -T2 = 1113.1+ 70.8 + 54.8-76.2-76.2= 1086.3 kips Net nominal moment is:
Mn = Cc(45 - 2)+(C+TI)41.55+(C 2)28.45+(T 2)37.38 Mn =1113.1 45-2
+ (70.8 + 76.2) 41.55 + (54.8) 28.45 + (76.2)37.38 = 44,409.2 in - kips
SHEET 93 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 N-S Nominal Force/Moment Interaction - Compression on Bottom - Tension Controls - Figure 14 Set CTI = 4.Oy = 4.0 x 0.002069 = 0.008276, therefore:
X 0
0.003 86.55 = 23.03 inches (0.003 + 0.008276J T2 (86.55 - 23.03 + 3.45 -
) 0.008276 = 0.007733 > c (23.086.55-23.03
£C2 = (
) 0.003 = 0.000844 < 0.002069 k2 t23.03 cl = (
23.03 -
) 0.003 = 0.002551 > 0.002069 Compute individual internal forces:
Depth of stress block: a = 0.80 x 23.03 = 18.42 inches Cc = 0.85fab = 4.25 x 18.42 x 9 = 704.6 kips C, = A, (fy - 0.85f )= 1.27 x (60-4.25) = 70.8 kips C2 = (EsC2 - 4.25)1.27 = (29E3 x 0.000844 - 4.25)1.27 = 25.7 kips T, = Asfy = 1.27 x 60 =76.2 kips T2 = Asfy = 1.27 x 60 =76.2 kips Net nominal compression is:
C, = C + C1 + C2 -T -T2 = 704.6 + 70.8 + 25.7-76.2-76.2 = 648.7 kips Net nominal moment is:
Mn = Cc(45 -2) +(C, +T 1)41.55+(C 2 )28.45+(T 2)37.38 M" = 704.6 (45 - 18.42) + (70.8 + 76.2) 41.55 + (25.7) 28.45 + (76.2) 37.38 = 34,905.0 in - kips
SHEET 94 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
N-S Nominal Force/Moment Interaction - Compression on Bottom - Tension Controls - Fieure 14 Set FTl = 8.0y = 8.0 x 0.002069 = 0.016552, therefore:
X (
0.003 186.55 = 13.28 inches K0.003 + 0.016552)
(T2 86.55 -13.28 +3.45 -7.62 00165
.0 610>F 86.55 -13.28
)
(C2
=13.28-16.55 0.003 = -0.000739 > -0.002069 (-) indicates bar in tension (13 1328-3 5
.28-3.45 0.003 = 0.002221 > 0.002069 cl 13.28 Compute individual internal forces:
Depth of stress block: a =0.80 x 13.28=10.62inches Cc = 0.85fcab = 4.25 x 10.62 x 9 = 406.2 kips C, = A, (fy - 0.85fc ) = 1.27 x (60 - 4.25) = 70.8 kips C2 = A, (EsC2) = 1.27 x (29E3 x(- 0.000739)) = -27.2 kips T1 =Afy =1.27 x 60 =76.2 kips T2 = Afy =1.27 x 60 =76.2 kips Net nominal compression is:
Cn = Cc + C1 + C 2 -T
-T 2 = 406.2 + 70.8 - 27.2-76.2-76.2 =297.4 kips Net nominal moment is:
Mn = Cc 45-a)+ (Cl +TI)41.55+(C2 )28.45+(T2 )37.38 Mn = 406.2 45 -
.6 2 + (70.8 + 76.2) 41.55 + (- 27.2) 28.45 + (76.2) 37.38 = 24,304.4 in - kips
SHEET 95 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 N-S Nominal Force/Moment Interaction - Compression on Bottom - Tension Controls - Fiaure 14 Set ETI = 16&y = 16 x 0.002069 = 0.033104, therefore:
X 0.003 186.5 5 = 7.19 inches (0.003 + 0.0331048 T2 6.5 -7.19+3.45-7.62 0.033104 = 0.031365 > Ey 8T2
=
86.55-7.19
)
1c2 9
6
)50.003
= -0.003905 < -0.002069 (-) indicates bar in tension cl= (7.197 -3.45) 0.003 = 0.001561 < 0.002069 Compute individual internal forces:
Depth of stress block: a =0.80 x 7.19 = 5.75 inches Co = 0.85fcab = 4.25 x 5.75 x 9 = 219.9 kips C, = A, (Es%1 - 0.85f, ) = 1.27 x (29E3 x 0.001561-4.25) = 52.1 kips C2 = -Afy = -1.27 x 60 = -76.2 kips (bar in tension)
T =Afy =1.27x60=76.2kips T2 =A sfy =1.27 x 60 =76.2 kips Net nominal compression is:
Cn =C, +C 1 +CC2 -T1 -T 2 =219.9+52.1-76.2-76.2-76.2=43.4kips Net nominal moment is:
Mn = Cc(45 -- )+ (C, +TI)41.55+ (C2 )28.45 + (T2)37.38 M.= 219.9 (45-5.75 + (52.1+ 76.2) 41.55 + (- 76.2) 28.45 + (76.2)37.38 =15,274.6 in - kips
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR ENERCON REVIEWER K. L. Whitmore APPROVED SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 96 OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0 N-S Nominal Force/Moment Interaction - Compression on Bottom - Tension Controls - Figure 14 Set sT1 = 326y = 32 x 0.002069 = 0.066208, therefore:
x
(
0.003 0.003 + 0.066208,)86.55 = 3.75 inches (86.55 - 3.75 + 3.45 - 7.62 0060
.684>F 6T2 =
86.55 - 3.75
)
EC2 =(3753716.5 ) 0.003=-0.01024<-Fy (-)indicatesbarintension scl 3=75 - 3.45 )0.003 = 0.000240 < 0.002069 Compute individual internal forces:
Depth of stress block: a = 0.80 x 3.75 = 3.00 inches C, = 0.85fcab = 4.25 x 3.00 x 9 = 114.8 kips C, = A, (Ec, - 0.85fc ) = 1.27 x (29E3 x 0.000240 - 4.25) = 3.4 kips C 2 = Asfy = 1.27 x (-60) = -76.2 kips (bar in tension)
T= Afy = 1.27 x 60 =76.2 kips T2 =Asfy = 1.27 x 60 =76.2 kips Net nominal compression is:
Cn = C + C + C2 - T - T2 = 114.8 + 3.4 - 76.2 - 76.2 - 76.2 = -110.4 kips (-)indicates tension Net nominal moment is:
M" = C4(45 -
)+(C, +T,)41.55+(C 2)28.45+(T 2)37.38 Mn = 114.8(45 -
2-00 + (3.4 + 76.2) 41.55 + (-76.2) 28.45 + (76.2)37.38 =8981.6 in -kips
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 97 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 North-South Calculation of !2 per ACI - Compresion on Bottom p = 0.90 for tension plus flexure p = 0.70 for compression plus flexure 0.lf"Ag =0.1 x5.OOx9x9O=405kip>qPb=9(1585.0)
Cn297.4 kips,
.009 0.78 l+ 2 1+ 2 x 297.4 f 0Ag 5x9x90 C~ 43.4 kips, (p 029 0.2434
= 0.88 l+,C 1+ 2 x 434 fcAg 5x9x90 North-South Maximum Desien Axial Load Strength Per Reference 11, 10.3.5.2, the maximum design axial load strength shall not be greater than 0.80(pP0.
Therefore, pC is limited to 0.80 x 0.70 x 3703.2 = 2073.8 in Table 4 below.
Table 4 below presents the North-South (Z Strip) section capacity data computed above. The signs of the moment are adjusted to conform to the sign convention established in Reference 5. Therefore, the moments that produce tension on the bottom are (+) and those that produce compression on the bottom are (-).
SHEET DATE 98 OF 160 March 11, 2003 JOB. NO.
PGE-009 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinfort CLIENT PG&E-DCPP ORIGI ENERCON REVIEWER K. L. Whitmore APPR(
SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 cement INATOR 3VED S. C. Tumminelli R. F. Evers REVISION 0 Table 4 - North-South Section Capacitv Data Moment Sign (+/-) and M.
C,,
-D
( Mn (p C,,
e Condition in-kips kips in-kips kips inches Compression no moment 0
3701.9 0.7 0
2591.3*
0 ACI Code Maximum 18,848.8 2073.8 9.089
(+) Compression controls 32,603.5 2806.7 0.7 22,822.5 1964.7 11.616
(+) Compression controls 45,540.0 2057.0 0.7 31,878.0 1439.9 22.139
(+) Balanced Condition 48,536.5 1585.0 0.7 33,975.6 1109.5 30.622
(+) Tension controls 47,649.1 1319.4 0.7 33,354.4 923.6 36.113
(+) Tension controls 44,816.1 1102.3 0.7 31,371.3 771.6 40.657
(+) Tension controls 36,053.9 691.5 0.7 25,237.7 484.1 52.133
(+) Tension controls 25,956.6 366.3 0.76 19,727.0 278.4 70.858
(+) Tension controls 16,515.3 113.0 0.85 14,038.0 96.1 146.077
(+) Tension controls 7620.7
-110.4 0.9 6858.6
-99.4
-69.000 Tension no moment 0
-287.6 0.9 0
-258.8 0
ACI Code Maximum
-18,408.4 2073.8
-8.877
(-) Compression controls
-32,248.3 2795.3 0.7
-22,573.8 1956.7
-11.537
(-) Compression controls
-45,619.0 2041.7 0.7
-31933.3 1429.2
-22.343
(-) Balanced Condition
-49,013.6 1564.3 0.7
-34,309.5 1095.0
-31.333
(-) Tension controls
-47,148.2 1282.1 0.7
-33,003.7 897.5
-36.773
(-) Tension controls
-44,409.2 1086.3 0.7
-31,086.4 760.4
-40.882
(-) Tension controls
-34,905.0 648.7 0.7
-24,433.5 454.1
-53.806
(-) Tension controls
-24,304.4 297.4 0.78
-18,957.4 232.0
-81.713
(-) Tension controls
-15,274.6 43.4 0.88
-13,441.6 38.2
-351.874
(-) Tension controls
-8981.6
-110.4 0.9
-8083.4
-99.4 81.322
ENERCON SERVICES, INC.
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 99 OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0
- This value is used only for linear interpolation for the allowable moment associated with the Code maximum compressive load, see notes ** and *** below.
- Linearly interpolated from the values above and below using 2591.3 = 0.70 x 3701.9 for the (p Cn with the zero moment.
- Linearly interpolated as above.
SHEE T 100 OF 160 E March 11, 2003 R
JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATO ENERCON REVIEWER K. L. Whitmore APPROVED SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 R S. C. Tumminelli R. F. Evers REVISION 0 East-West Section Concrete Capacity The calculations below present the section capacity data for the EW section. Figures 16 to 23 provide the graphical information needed to follow the calculations.
0.003 i4N_
39.465 35.295 45.00 4.25 4-1 C
ksi Cl C2 84.465 PO 26.365 39.465
>0.00206
. C3 18.635 5.535 Fizure 16 - East-West (X strip) Section - Net Axial Compression No Moment
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 101 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0
- 5.535 F 9.705Kf'f 5
0.003
- K 39.465 F-C2 4.25 ksi 4-C, 4-C
=1
' ff "I_1 I0
/
/
-,17-,,
35.295 45.00 84.465 26.365 a
x Cc
\\N 84.465-X
£T2 -
I F
'N-39.465 I
F I 4-,
0.002069
- 18.635
- 5.535 Fizure 17-East-West (X strip) Section - Balanced Condition Tension on the Bottom This is -Mz, see Reference 5
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 102 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 7 5.535 9.705 0.003 I
'A-I 39.465
-C2 4.25 ksi 4-Cl 4-C 2 4-'
/
x a
/
L II 35.295 45.00 84.465 I
26.365 39.465 84.465-X F I F 18.635
- 5.535 ST1>8Y Filure 18-East-West (X striv) Section - Compression Controls Tension on the Bottom This is -Mz. see Reference 5
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 103 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 A
- 5.535 9.705 i
L
-A7 39.465 8-C2 4.25 ksi n-C1 4-C 2
~11
.E x
a cc 35.295
-A-7 'A F
`
45.00 84.465 I
26.365
'N 39.465 84.465-X
\\
-*T 2
I E18.635
- 5.535 8T1>8y Figure 19 - East-West (X strip) Section - Tension Controls Tension on the Bottom This is -Mz, see Reference 5
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 104 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0
_-o Tj l
-- P T2 0.002069 39.465 ST2 35.295 84.465-X 45.00 84.465 x
26.365 Fc2 a
14Cc 4--C 2
39.465 Y-- I 18.635 5.535 0.003 4.25 ksi Figure 20 - East-West (X strip) Section - Balanced Condition Compression on the Bottom This is Mz. see Reference 5
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 105 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0
- 5.535
-9705 P171 F
-L T2 T2 39.465 84.465-X
/-
A
- 35.295
£T2 45.00 84.465 26.365 I
39.465 4-x a
.L
< 4--
K-4.25 ksi cc C2 AzL I!
C,
_- 18.635
- 5.535 0.003 Figure 21 - East-West (X strip) Section - Compression Controls Compression on the Bottom This is Mz, see Reference 5
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 106 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 A2
- 5.535 9.705 k
L
_-oTj
_-O T, 39.465 35.295 84.465-X 45.00 84.465
\\
26.365 IF 39.465 18.635
- 5.535 x
a F.
4Cc 4-C2 4-C 1 Ie 0.003 4.25 ksi Figure 22 - East-West (X strin) Section - Tension Controls Compression on the Bottom This is Mz. see Reference 5
SHEET 107 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
- 5.535 F 9.705 A
I A L
L El 39.465 35.295
--+-i 45.00 84.465 I
26.365 39.465 Sy
_-0 T, T2 PT T3
-IA rI
- 18.635
- 5.535 Figure 23 - East-West (Z strip) Section - Net Axial Tension No Moment
SHEET 108 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Compute the axial force/moment interaction diagram.
East-West Net Axial Compression - No Moment - Figure 16 Since the reinforcement is not symmetrical, in order to achieve a net axial moment at the mid-height of the section, the concrete stress must vary slightly over the height of the section so that the unbalanced moment produced by the steel at a stress of Fy is offset by the concrete. Thus, the unbalanced moment is AM = AsFy(Net moment arm difference) = 1.27 x 60 x (35.295-26.365) = 680.5 in kips. Therefore the A concrete stress (Aa) is computed from the equilibrium equation AM = Aa x S where S is the section 9X90 2 AM 680.5_
modulus. S =
= 12150inches3. And, Au
=
=
=0.056 ksi. Thus, the concrete stress 6
S 12150 varies linearly from 4.25 ksi at the top to 4.25-(2)(0.056) = 4.138 at the bottom of the section. And, Po = (4.138 + (I)(2)0.0561(9)(90)+ 60 x 4 x 1.27 = 3701.9 kips Check on the moment:
M = (4.25-4.138)(0.5)(9)(90) ((2/3)(90)-45))-76.2(37.38-28.45)
-0.1 OK East-West Balanced Condition - Tension on Bottom - Figure 17 Compute X (distance to NA):
X= 0 0.003 84.465 = 49.989 inches
( 0.003 +0.002069)
STI = 6y = 0.002069 (T2
=
34.476 + 5.535-18-635 0002069 = 0.001283 (34.6 534.4765 EC2 =
4
)0.003 = 0.002418 > 0.002069 and eC! > 0.002069 also.
K t 49.989 Compute individual internal forces:
Depth of stress block: a = 0.80 x 49.989 = 39.991 inches C, = 0.85fcab = 4.25 x 39.991 x 9 = 1529.7 kips
SHEET 109 OF 160 March 11, 2003 JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 C1 =A(f - 0.85f, )= 1.27 x (60-4.25) = 70.8 kips C2 =70.8 kips T2= EAs= 29E3 x 0.001283 x 1.27 = 47.3 kips T 1 =A fy =1.27x60=76.2kips Net nominal compression is:
Cn = Cc + Cl + C 2 -T 1 -T 2 = 1529.7 + 70.8 + 70.8-76.2-47.3 = 1547.8 kips Net nominal moment is:
Mn = Cc(45-a)+(C1 +TI)39.465+(C 2 )35.295+(T2 )26.365 M. = 1529.7 45 - 392991 + (70.8 + 76.2)39.465 + (70.8)35.295 + (47.3)26.365 = 47,796.7 in - kips
SHEET 110 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 E-W Nominal Force/Moment Interaction - Tension on Bottom - Comy Controls - Figure 18 Sets..
=
0.002069.001035, therefore:
2 2
X =(
) 84.465 = 62.799 inches t0.003+ 0.001035)
(84.465-62.799+5.535-18.635 1
=
8 T2
=~0.001035
=0.000410 T2 84.465-62.799 EC2 = (62.7
) 0.003 = 0.002536 > 0.002069 and Scl > 0.002069 also.
Compute individual internal forces:
Depth of stress block: a = 0.80 x 62.799 = 50.239 inches C, = 0.85fcab = 4.25 x 50.239 x 9 = 1921.6 kips C, = A, (fy - 0.85f, )= 1.27 x (60-4.25) = 70.8 kips C2 = 70.8 kips T1 = EeA, = 29E3 x 0.001035 x 1.27 = 38.1 kips T2 = EeAs = 29E3 x 0.000410x 1.27 = 15.1 kips Net nominal compression is:
C. =C, +Cl +C 2 -T1 -T 2 =1921.6+70.8+70.8-38.1-15.1 =2010.0kips Net nominal moment is:
Mn =Cc 45-2)+(Cl + T1 ) 39.465 + (C 2) 35.295 + (T2) 26.365 M = 1921.6(45 -
+ (70.8 + 38.1) 39.465 + (70.8) 35.295 + (15.1)26.365 =45,397.1 in -kips
SHEET 111 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
E-W Nominal Force/Moment Interaction - Tension on Bottom - Comrn Controls - Figure 18 Set 6TI = 0-0 therefore:
X = 84.465 inches 6T2 =(s35844165
) 0.003= - 0.000465
(-) indicates compression C2 = 84.465 - 9.705 0.003 = 0.002655 > 0.002069 and Ec, > 0.002069 also.
EC2= ~84.465 Compute individual internal forces:
Depth of stress block: a =0.80 X 84.465 =67.572 inches Cc = 0.85fcab = 4.25 x 67.572 x 9 = 2584.6 kips C, = A, (f, - 0.85f )= 1.27 x (60- 4.25) = 70.8 kips C2 = 70.8 kips T. =0.O kips T2 = EeA, = 29E3 x - 0.000465 x 1.27 =17.1 kips Net nominal compression is:
C, = Cc + C1 + C 2 - T1 - T2 = 2584.6 + 70.8 + 70.8 - (-17.1) =2743.3 kips Net nominal moment is:
Mn =C, 45-a)+(C1 +T I)39.465+ (C 2 )35.295+(T2)26.365 M = 2584.6 45 -
+ (70.8) 39.465 + (70.8) 35.295 - (17.1)26.365 =33,825.9 in - kips
SHEET 112 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
E-W Nominal Force/Moment Interaction - Tension on Bottom - Tension Controls - Fiaure 19 Set sT2 = Fly = 0.002069, (Note the change from eTi to CT2), therefore:
X =
0.003 N
(84.465 + 5.535 - 18.635) = 42.473 inches K0.00 3 + 0.002069J
(
84.465 - 42.473 0.002069 = 0.003007 > c 84.465 - 42.473 + 5.535 - 18.635 Y
C2 (42.473-9.705 )0.003=0.002315>0.002069 and ec] >0.002069 also.
(C 42.473 Compute individual internal forces:
Depth of stress block: a = 0.80 x 42.473 = 33.978 inches Cc = 0.85fcab = 4.25 x 33.978 x 9 = 1299.7 kips C, = A,(fy - 0.85f1) = 1.27 x (60 - 4.25) = 70.8 kips C2 =70.8 kips T= A fy = 1.27 x 60 =76.2 kips T2 =Asfy = 1.27 x 60 =76.2 kips Net nominal compression is:
C. =Cc +CI +C 2 -T1 -T 2 =1299.7+70.8+70.8-76.2-76.2=1288.9kips Net nominal moment is:
Mn =C,(45-a))+ (Cl + Tj)39.465 + (C 2 )35.295 + (T 2)26.365 Mn = 1299.7 (45-32
+(70.8 + 76.2) 39.465 + (70.8) 35.295 + (76.2)26.365 =46,715.2 in - kipE
SHEET 113 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 E-W Nominal Force/Moment Interaction - Tension on Bottom - Tension Controls - Figure 19 Set CT[ = 2.0py = 2.0 x 0.002069 = 0.004138, therefore:
X 0003 84.465 = 35.499 inches
( 0.003 + 0.00413 8J (84.465-35.499 + 5.535-18.635 0.004138 = 0.003031 > £ T2 =
84.465-35.499 6C2 =
35.49 0 ) 0.003 = 0.002180 > 0.002069 and Scl > 0.002069 also.
Compute individual internal forces:
Depth of stress block: a = 0.80 x 35.499 = 28.399 inches Cc = 0.85fcab = 4.25 x 28.399 x 9 = 1086.3 kips C, = A,(f, -0.85fC)= 1.27 x (60-4.25) = 70.8kips C2 =70.8 kips T1 =Afy =1.27 x 60 =76.2 kips T2 =Asfy = 1.27 x 60 =76.2 kips Net nominal compression is:
Cn = Cc + Cl + C2 -T-,
=1086.3 + 70.8 + 70.8-76.2-76.2 =1075.5 kips Net nominal moment is:
Mn =C, 45-a) + (C, + T1) 39.465 + (C2) 35.295 + (T2) 26.365 Mn= 1086. 345 -
+ (70.8 + 76.2) 39.465 + (70.8) 35.295+ (76.2)26.365 = 43,767.8 in - kips
SHEET 114 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 E-W Nominal Force/Moment Interaction - Tension on Bottom Tension Controls - Fieure 19 Set 6T1 = 4.Ocy = 4.0 x 0.002069 = 0.008276, therefore:
X=
000 3
84.465 = 22.472 inches V 0.003 +0.008276)
( 84.465 - 22.472 + 5.535 -18.635 0.008276 = 0.006527 > &Y T2 =
84.465-22.472 CC2 = 22.472 - 9.705 0.003 = 0.001704 < 0.002069 (22.472 555 cl = (
0.003 = 0.002261 > 0.002069 22.472 Compute individual internal forces:
Depth of stress block: a = 0.80 x 22.472 = 17.978 inches C, = 0.85fcab = 4.25 x 17.978 x 9 = 687.7 kips C, = As (fy - 0.85fj )= 1.27 x (60-4.25) = 70.8 kips C2 = (EEC2 - 4.25)1.27 = (29E3 x 0.001704 - 4.25)1.27 = 57.4 kips T1 = Asfy = 1.27 x 60 =76.2 kips T2 = Afy = 1.27 x 60 =76.2 kips Net nominal compression is:
C, =Cc+C1 +C 2 -T 1 -T 2 =687.7+70.8+57.4-76.2-76.2=663.5kips Net nominal moment is:
Mn =Cc (45-2)+(C1 + T) 39.465 + (C2) 35.295 + (T2 ) 26.365 M. = 687.7(45 -
- 29)
+ (70.8 + 76.2) 39.465 + (57.4) 35.295 + (76.2) 26.365 = 34,601.1 in - kips
SHEET 115 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 E-W Nominal Force/Moment Interaction - Tension on Bottom - Tension Controls - Filure 19 Set 8TI = 8.0ey = 8.0 x 0.002069 = 0.016552, therefore:
X 0=003 84.465 =12.960 inches (0.003 + 0.016552) f (84.465-12.960+5.535 -18.635E 0016552=0.013520 CT2 84.465-12.960 Pc (12.960-9.705 0.003 = 0.000753 < 0.002069 c'
12.960
, =12.960 9 5.5)350.003 = 0.001719 < 0.002069 Compute individual internal forces:
Depth of stress block: a =0.80x 12.960=10.368inches C, = 0.85fcab = 4.25 x 10.368 x 9 = 396.6 kips C, = A,(Esc, - 0.85fe) = 1.27 x (29E3 x 0.001719 - 4.25) = 57.9 kips C2 = A, (EsC2 - 0.85f) =1.27 x (29E3 x 0.000753-4.25) = 22.3 kips T.
Afy = 1.27 x 60 =76.2 kips T2 =Afy = 1.27 x 60 =76.2 kips Net nominal compression is:
Cn = C + C1 + C 2 -T -T 2 = 396.6 + 57.9 + 22.3-76.2-76.2 = 324.4 kips Net nominal moment is:
Mn = Cc(45-2)+ (Cl +T,)39.465+ (C2)35.295+ (T2)26.365 M = 396.6 (45 -
2
) + (57.9 + 76.2) 39.465 + (22.3) 35.295 + (76.2) 26.365 = 23,879.4 in - kips
SHEET 116 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
E-W Nominal Force/Moment Interaction - Tension on Bottom - Tension Controls - Figure 19 Set gTI = 16ey = 16 x 0.002069 = 0.033104, therefore:
X t(
0.003 O 84.465 = 7.018 inches
= 0.003 + 0.033104) 84.465 - 7.018 + 5.535 - 18.6350 84.465-7.018 y
6C2 = (7.018 -9.705 0.003 = -0.001149 (-) indicates bar in tension (7.018 cl = (7 018-5535 ) 0.003 = 0.000634 < 0.002069 Compute individual internal forces:
Depth of stress block: a = 0.80 x 7.018 = 5.614 inches Cc = 0.85fcab = 4.25 x 5.614 x 9 = 214.7 kips C, = A (Erc, - 0.85fc) = 1.27 x (29E3 x 0.000634 - 4.25) = 18.0 kips C2 =AsEPU = 1.27 x (29E3 x (-0.001149)) = -42.3 kips (bar in tension)
T1 = Afy = 1.27 x 60 =76.2 kips T2 =Asfy = 1.27 x 60 =76.2 kips Net nominal compression is:
Cn = Cc +C1 +C 2 -T -T2 = 214.7+18.0-42.2-76.2-76.2 = 38.1kips Net nominal moment is:
Mn =Cc(45- )+(C, +T1)39.465+(C 2)35.295+(T 2)26.365 Mn = 214.7 (45 -
56142
+ (18.0 + 76.2) 39.465 + (-42.2) 35.295 + (76.2)26.365 = 13,296.0 in - kips
SHEET 117 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC. CALCULATION NO.
PGE-009-CALC-007 REVISION 0 E-W Nominal Force/Moment Interaction - Tension on Bottom - Tension Controls - Fieure 19 Set 8T1 = 326y = 32 x 0.002069 = 0.066208, therefore:
X 0.003 84.465 =3.661 inches y0.003+ 0.066208)
T2 84.465-3.661 + 5.535 -
0.066208 =0.055474 > y 8T2 K
~84.465 - 3.661 1.3)0060=.544 C2= (3.661
- 9.705 )0.003 = -0.004953
> cy (-) indicates bar in tension 61-5 53.661 cl= 3.661-5.53 )0.003 = -0.001536<0.002069 (-)indicates bar in tension K 3.661 Compute individual internal forces:
Depth of stress block: a = 0.80 x 3.661 = 2.929 inches Cc = 0.85fcab = 4.25 x 2.929 x 9 = 112.0 kips C, = AEc, = 1.27 x (29E3 x - 0.001536) = -56.6 kips (bar in tension)
C 2 = Afy = 1.27 x (-60) = -76.2 kips (bar in tension)
Tl = Afy =1.27 x 60 =76.2 kips T2 = Afy 1.27 x 60 =76.2 kips Net nominal compression is:
C. = Cc + C1 + C 2 - Tl T2 = 112.0 - 56.6 - 76.2 - 76.2 - 76.2 = -173.2 kips (-)indicates tension Net nominal moment is:
Mn =C,(45-2a)+(CI +T I )39.465+(C 2 )35.295+(T2 )26.365 K, = 112.0 (45
- 2.929) + (- 56.6 + 76.2) 39.465 + (- 76.2) 35.295+ (76.2)26.365 = 4969.0in-kips.
SHEET 118 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0
East-West Net Axial Tension - No Moment - Figure 23 As with the net compression case, the strain across the section must vary slightly in order to achieve zero moment at mid-height.
Set 63 = Fy = 0.002069 and therefore T3 = T4 = 76.2 kips Moment equilibrium requires:
M = 0 = (TI) 39.465 +(T 2) 35.295 - (T3) 26.365 - (T4) 39.465 Use the strain:
( 9.705-5.535 639.465 + 2 6.3 6 5 ) Y 62 =&s +0.000131-0.063345s, 62 =0.93665561 + 0.000131 Now, use the stress-strain relation and the bar areas:
T, = AEE =1.27 x 29E3 x E] = 36,830s, and T2 = AIE&2 =1.27 x 29E3 x (0.936655s, + 0.000131) = 34,497.06l + 4.825 Substitute in to equation for M 0 = (36,830,)39.465 + (34,4976l + 4.825) 35.295 -(76.2) 26.365 - (76.2) 39.465 0 = 2,671,06861 -4,845.9 4845.9 61 =
= 0.001814 2,671,068 62 = (0.93665)0.001814+0.000131=0.0018303 And, T, = 1.27 x 29E3 x 0.001814 = 66.8 kip
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 119 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 And, T2 = 1.27 x 29E3 x 0.001830 = 67.4 kip Thus, T = Pt = 66.8+67.4+76.2+76.2=286.6 kip Check on the moment:
0 = (66.8)39.465 + (67.4)35.295 - (76.2)26.365 - (76.2)39.465
-1.1 OK East-West Calculation of 2 ner ACI - Tension on Bottom (p = 0.90 for tension plus flexure p = 0.70 for compression plus flexure
- 0. 1fAg = 0.1 x 5.00x9x90=405kip C= 324.4 kips,
° 0290 0 90234 0.78 1+ i 1+2x344 fcAg 5 x 9 x 90 Cn = 38.1 kips, 09=
0C0 2381 =0.88 1+
1+
fL~g 5x9x90
SHEET 120 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
East-West Balanced Condition - Compression on Bottom - FiEure 20 Compute X (distance to NA):
X =
0.003 84.465 = 49.989 inches (0.003 + 0.002069)
&TI =
= 0.002069 F T2 = (
) 0.002069 = 0.001819 34.476
)
Sc2 (49.989 i8.635) 0.003 = 0.001882 < 0.002069 (49.989-535 cl= (49.89 -
0.003 = 0.002668 > 0.002069 K49.989 Compute individual internal forces:
Depth of stress block: a = 0.80 x 49.989 = 39.991 inches C, = 0.85fcab = 4.25 x 39.991 x 9 = 1529.7 kips C, = A, (fy -0.85f)= 1.27 x (60-4.25) = 70.8 kips C2 = As (Es - 0.85fj ) = 1.27 x ((29E3)(0.001882) - 4.25) = 63.9 kips T2 =EeAs =29E3 x 0.001819 x 1.27 = 67.0 kips T= A fy =1.27 x 60 = 76.2 kips Net nominal compression is:
Cn = C, + C1 + C2 -T -T2 = 1529.7 + 70.8 + 63.9-76.2-67.0 = 1521.2 kips Net nominal moment is:
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 121 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 M =Cc (45
- 2) + (C, + T1) 39.465 + (C2) 26.365 + (T2) 35.295 Mn = 1529.7 (45 -
- 29)
+ (70.8 + 76.2) 39.465 + (63.9) 26.365 + (67.0) 35.295 = 48,100.2 in - kips
SHEET 122 OF 160 March 11, 2003 JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 E-W Nominal Force/Moment Interaction - Compression on Bottom - Comy Controls - Fieure 21 Set T1
_= 2 = 0.002069 = 0.001035, therefore:
2ec l -
2 x
0.003 84.465 = 62.799 inches
=0.003+0.001035) 6 84.465 - 62.799 + 5.535 - 9.705 00 gT2
=(
84562790.001035
= 0.000836 84.465 - 62.799 SC2 (62.799 -18635 ) 0.003 = 0.002110 > 0.002069 and eci > 0.002069 also.
62.799)
Compute individual internal forces:
Depth of stress block: a = 0.80 x 62.799 = 50.239 inches Cc = 0.85fcab = 4.25 x 50.239 x 9 = 1921.6 kips C, = A, (fy - 0.85f,)= 1.27 x (60 - 4.25) = 70.8 kips C2 = 70.8 kips T, = EcA, = 29E3 x 0.001035 x 1.27 = 38.1 kips T2 = EeA, = 29E3 x 0.000836 x 1.27 = 30.8 kips Net nominal compression is:
Cn = Cc + C] + C 2 - T-T 2 = 1921.6 + 70.8 + 70.8-38.1-30.8 =1994.3 kips Net nominal moment is:
Mn =Cc 45-2) + (Cl + T 1) 39.465 + (C2) 26.365 + (T2) 35.295 Mn =1921.6 45-52
+(70.8+38.1) 39.465+(70.8)26.365+(30.8) 35.295=45,453.8in-kips
SHEET 123 OF 160 March 11, 2003 JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 E-W Nominal Force/Moment Interaction - Compresison on Bottom - Comip Controls - Figure 21 Set STI = 0.0, therefore:
X = 84.465 inches T2 (5.535 - 9.705 0.003= -0.000148
(-) indicates compression
-I 84.465 c2 84.465 - 18.63 0.003 = 0.002338 > 0.002069 and Fc- > 0.002069 also.
C2 84.465 Compute individual internal forces:
Depth of stress block: a = 0.80 x 84.465 = 67.572 inches Cc = 0.85fab = 4.25 x 67.572 x 9 = 2584.6 kips C, = As (f, - 0.85f )= 1.27 x (60 -4.25) = 70.8 kips C2 =70.8 kips T1 =0.0 kips T2 =EeA, = 29E3 x - 0.000148 x 1.27 =-5.5 kips Net nominal compression is:
Cn = C,
+ C 1 + C 2 - T1 - T2 = 2584.6 + 70.8 + 70.8 - (-5.5) = 2731.7 kips Net nominal moment is:
Mn =Cc(45-a) )+(C + T 1)39.465+(C 2 )26.365+(T 2 )35.295 M. = 2584.6 (45 -
2 ) + (70.8) 39.465 + (70.8) 26.365 + (- 5.5) 35.295 = 33,450.3 in - kips
fg..
t; ENERCON SERVICES, INC.
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 124 OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0 E-W Nominal Force/Moment Interaction - Compression on Bottom - Tension Controls - FiLure 22 Set cT1 = 1.5cy = 1.5 x 0.002069 = 0.003104, therefore:
x
(
0.003 t0.003+ 0.003104
)84.465 = 41.513inches (84.465 -41.513+ 5.535 -9.705 0.003104 = 0.002803 > E T K 84.465-41.513 J
C2 = (41.513 -18635 ) 0.003 = 0.001653< 0.002069 C2 t 41.513 cl =(41.513-5.535 ) 0.003 = 0.002600 > 0.002069 Compute individual internal forces:
Depth of stress block: a =0.80 x 41.513 = 33.210inches C, = 0.85fcab = 4.25 x 33.210 x 9 = 1270.3 kips C, = A(fy -0.85f)
= 1.27 x (60-4.25) = 70.8 kips C2 =A, (Es - 0.85fc)= 1.27 x ((29E3)(0.001653) - 4.25) = 55.5 kips T.
Asfy = 1.27 x 60 =76.2 kips T2 =Asfy = 1.27 x 60 = 76.2 kips Net nominal compression is:
C. = Cc + C + C2 -T -T2 = 1270.3 + 70.8 + 55.5-76.2-76.2 = 1244.2 kips Net nominal moment is:
Mn =Cc 45-2) + (C, + T.) 39.465 + (C2) 26.365 + (T2) 35.295 M= 1270.3 45 -
2
+ (70.8 + 76.2) 39.465 + (55.5 )26.365 + (76.2) 35.295 = 46,024.3 in - kip:
SHEET 125 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 E-W Nominal Force/Moment Interaction - Compression on Bottom - Tension Controls - Figure 22 Set 6TI = 2.0sy = 2.0 x 0.002069 = 0.004138, therefore:
x 0.00 3 84.465 = 35.499 inches
= (0.003 +0.004138)
(84.465 - 35.499 + 5.535 -9705
£T2 84.4-9.70110.004138 = 0.003786>E 84.465 - 35.499 EC2 = (.499
-18.635 0.003 = 0.001425 < 0.002069 3 5.499 5C2 = (
35499 535 0.003 = 0.002532 > 0.002069 Compute individual internal forces:
Depth of stress block: a = 0.80 x 35.499 = 28.399 inches Cc = 0.85fcab = 4.25 x 28.399 x 9 = 1086.3 kips C, = A, (fy - 0.85fc )= 1.27 x (60-4.25) = 70.8 kips C2 = A, (Es - 0.85fji) = 1.27 x ((29E3)(0.001425) - 4.25) = 47.1 kips T.
Afy = 1.27 x 60 =76.2 kips T2 -Afy = 1.27 x 60 = 76.2 kips Net nominal compression is:
Cn = Cc + C1 + C2-T,-T2= 1086.3 + 70.8 + 47.1-76.2-76.2 = 1051.8 kips Net nominal moment is:
Mn =Cc 45-a)+ (C + T,)39.465 + (C2 )26.365+ (T2)35.295 Mn = 1086.3 45 - 28399 + (70.8 + 76.2) 39.465 + (47.1)26.365 + (76.2) 35.295 =43,191.2 in - kips N-
SHEET 126 OF 160 S
JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 E-W Nominal Force/Moment Interaction - Compression on Bottom - Tension Controls - Fizure 22 Set ETI = 4.0y = 4.0 x 0.002069 = 0.008276, therefore:
X=
.003 84.465 = 22.472 inches
( 0.003 + 0.008276)
(84.465 - 22.472 + 5.535 - 9.7050 84.465 - 22.472 0
c2 = (22.472 -18.635 ) 0.003 = 0.000512 < 0.002069
-~22.472
)
c 22.472-5.535 ) 0.003 = 0.002261 > 0.002069 Compute individual internal forces:
Depth of stress block: a = 0.80 x 22.472 = 17.978 inches Cc = 0.85fab = 4.25 x 17.978 x 9 = 687.7 kips C, = A (fy -0.85f)=
1.27 x (60-4.25) = 70.8 kips C 2 = (EFC2 -4.25)1.27
= ((29E3)(0.000512)-4.25)1.27=13.5 kips T. =Asfy = 1.27 x 60 =76.2 kips T2 -Afy
= 1.27 x 60 =76.2 kips Net nominal compression is:
C. =C, ++/-C +C 2 -T 1
-T 2
=687.7+70.8+13.5-76.2-76.2=619.6kips Net nominal moment is:
M. =Cc 45 --
) + (C1 + T1 )39.465 + (C2 )26.365 + (T 2 ) 35.295 M
= 687.7 (45 -
29)+
(70.8 + 76.2) 39.465 + (13.5) 26.365 + (76.2) 35.295 = 33,611.5 in - kips
SHEET 127 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
E-W Nominal Force/Moment Interaction - Compression on Bottom - Tension Controls - Fivure 22 Set FT1 = 8.0cy = 8.0 x 0.002069 = 0.016552, therefore:
X =
0003 84.465 = 12.960 inches
( 0.003 + 0.016552)
( 84.465 - 12.960 + 5.535 - 9.705 0.016552 = 0.015587 >
84.465 -12.960
)
(12.960 - 18.635 8 C2 =
12960
) 0.003 = -0.001314 > -0.002069 (-) indicates bar in tension
=(12.960 -
)5.535 0.003 = 0.001719 < 0.002069 cly
~12.~960 Compute individual internal forces:
Depth of stress block: a=0.80x 12.960=10.368inches Cc = 0.85fcab = 4.25 x 10.368 x 9 = 396.6 kips C, = A,(Egc, - 0.85fc) = 1.27 x (29E3 x 0.001719 - 4.25) = 57.9 kips C2 = A, (E6, 2 ) = 1.27 x ((29E3)(- 0.001314)) = 48.4 kips T. =Afy =1.27x60=76.2kips T2 =Afy = 1.27 x 60 =76.2 kips Net nominal compression is:
C. = Cc + C + C2 -T -T2 = 396.6 + 57.9 - 48.4-76.2-76.2 = 253.7 kips Net nominal moment is:
Mn =Cc 45-a))+(C1 +T 1 )39.465+(C2 )26.365+(T 2 )35.295 Mn = 396.6 (45 - 10.368) + (57.9 + 76.2) 39.465 - (48.4) 26.365 + (76.2) 35.295 = 22,496.7 in - kips
SHEET 128 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 E-W Nominal Force/Moment Interaction - Compression on Bottom - Tension Controls - Fi2ure 22 Set 6TI = 16sy = 16 x 0.002069 = 0.033104, therefore:
X
(
0.003 4
84.465 = 7.018 inches 0.003 + 0.033104)
(8 6-7.018+5.535-9.705')
£T2 84.465-7.018.
0.033104 = 0.031322 > e£ 84.465 - 7.018 8 C2 (7.0 18 18.635 0.003 = -0.004966 < -0.002069 (-) indicates bar in tension 12=
7.018
=(7.018 5.535) 0.003=0.000634<0.002069 Compute individual internal forces:
Depth of stress block: a = 0.80 x 7.018 = 5.614 inches Cc = 0.85fcab = 4.25 x 5.614 x 9 = 214.7 kips C, = A, (EE,1 - 0.85f) = 1.27 x (29E3 x 0.000634 - 4.25) = 18.0 kips C2 = -Afy = -1.27 x 60 = -76.2 kips (bar in tension)
T1 =Afy = 1.27 x 60 =76.2 kips T2 =Ajfy = 1.27 x 60 =76.2 kips Net nominal compression is:
Cn, =Cc +C 1 +C 2 -T1 -T2 =214.7+18.0-76.2-76.2-76.2=4.1kips Net nominal moment is:
Mn =Cc (45-
) + (Cl + T1) 39.465 + (C2) 26.365 + (T2) 35.295 M. = 214.7(45-2 5 6 4 )+(18.0+ 76.2) 39.465 - (76.2)26.365 + (76.2) 35.295 =13,456.9 in - kips
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 129 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 E-W Nominal Force/Moment Interaction - Compression on Bottom - Tension Controls - Fiwure 22 Set FTs = 32&y = 32 x 0.002069 = 0.066208, therefore:
X
(
0.003
= 0.003 + 0.066208;') 84.465 = 3.661inches (84.465 - 3.661 + 5.535-9.705 N0066208
= 0.062791 >F tT2 K
84.465 - 3.661 0
0 8C2 = (3.661 - 18.635)0.003 = -0.012270< -£y (-) indicates bar in tension K 3.661
)
cl (3.661 -5.350.003 = -0.001536> -0.002069 (-) indicates bar in tension Compute individual internal forces:
Depth of stress block: a = 0.80 x 3.661 = 2.929 inches Cc = 0.85fcab = 4.25 x 2.929 x 9 = 112.0 kips C, = AsEscl = 1.27 x (29E3 x -0.001536) = -56.6 kips (bar in tension)
C2 =Asfy =1.27 x (-60) = -76.2 kips (bar in tension)
T1 = Af, =1.27 x 60 =76.2 kips T2 =Af = 1.27 x 60 =76.2 kips Net nominal compression is:
Cn= Cc +C 1 +C2 -T 1 -T 2 = 112.0-56.6-76.2-76.2-76.2 =-173.2kips(-)indicatestension Net nominal moment is:
Mn =Cc(45 - a) + (Cl + T 1)39.465+ (C 2 )26.365+ (T2) 35.295 M. =112.0 (45-2
)
9 29+(- 56.6+76.2) 39.465 -(76.2)26.365+ (76.2) 35.295 = 6330.0 in-kips
SHEET 130 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
East-West Calculation of q per ACI - Compression on Bottom (p 0.90 for tension plus flexure p
0.70 for compression plus flexure 0.lfcAg = 0.1 x 5.00x9x90=405kip Cn 253.7kips,
(
2C 2x2537 0.80 1+,n 1+
fcAg 5x9x90 C,,=4.1 kips, 9 0.900241
= 0.90 1+ i 1+2x.
fLg 5 x9 x90 East-West Maximum Desitn Axial Load Strength Per Reference 11, 10.3.5.2, the maximum design axial load strength shall not be greater than 0.80(pP0.
Therefore, (pC is limited to 0.80 x 0.70 x 3701.9 = 2073.1 in Table 5 below.
Table 5 below presents the East-West (X Strip) section capacity data computed above. The signs of the moment are adjusted to conform to the sign convention established in Reference 5. Therefore, the moments that produce tension on the bottom are (-) and those that produce compression on the bottom are (+).
SHEET 131 OF 160 DATE March 11, 2003 JOB. NO.
PROJECT SUBJECT CLIENT REVIEWER PGE-009 DCPP ISFSI ISFSI Cask Storage Pad Steel Reinforcement PG&E-DCPP ORIGINATOR APPROVED S. C. Tumminelli ENERCON K. L. Whitmore R. F. Evers SERVICES, INC. CALCULATION NO.
PGE-009-CALC-007 REVISION 0 Table 5 - East-West Section Capacity Data Moment Sign (+/-) and Mn Cn P
(P M.
(P C, e
Condition in-kips in-kips kips inches Compression no moment 0
3701.9 0.7 0
2591.3*
0 ACI Code Maximum
-18,283.4 2073.1
-8.886
(-) Compression controls
-33,825.9 2743.3 0.7
-23,678.1 1920.2
-12.331
(-) Compression controls
-45,397.1 2010.0 0.7
-31,778.0 1407.0
-22.586
(-) Balanced Condition
-47,796.7 1547.8 0.7
-33,457.7 1083.5
-30.879
(-) Tension controls
-46,715.2 1288.9 0.7
-32,700.6 902.2
-36.245
(-) Tension controls
-43,767.8 1075.5 0.7
-30,637.5 752.9
-40.693
(-) Tension controls
-34,601.1 663.5 0.7
-24,220.8 464.5
-52.144
(-) Tension controls
-23,879.4 324.4 0.78
-18,625.9 253.0
-73.620
(-) Tension controls
-13,296.0 38.1 0.88
-11,700.5 33.5
-349.269
(-) Tension controls
-4969.0
-173.2 0.9
-4472.1
-155.9 28.686 Tension no moment 0
-286.6 0.9 0
-257.9 0
ACI Code Maximu 17,867.4 2073.1 8.619
(+) Compression controls 33,450.3 2731.7 0.7 23,415.2 1912.2 12.245
(+) Compression controls 45,453.8 1994.3 0.7 31,817.7 1396.0 22.792
(+) Balanced Condition 48,100.2 1521.2 0.7 33,670.1 1064.8 31.621
(+) Tension controls 46,024.3 1244.2 0.7 32,217.0 870.9 36.993
(+) Tension controls 43,191.2 1051.8 0.7 30,233.8 736.3 41.067
(+) Tension controls 33,611.5 619.6 0.7 23,528.1 433.7 54.250
(+) Tension controls 22,496.7 253.7 0.8 17,997.4 203.0 88.657
(+) Tension controls 13,456.9 4.1 0.9 12,111.2 3.7 3273.3
(+) Tension controls 6330
-173.2 0.9 5697
-155.9
-36.543
SHEET 132 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0
- This value is used only for linear interpolation for the allowable moment associated with the Code maximum compressive load, see notes ** and *** below.
- Linearly interpolated from the values above and below using 2591.3 - 0.70 x 3701.9 for the (p C, with the zero moment.
- Linearly interpolated as above.
SHEET 133 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli ENERCON REVIEWER K. L. Whitmore APPROVED R. F. Evers SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 REVISION 0 North-South and East-West Section Evaluations The section capacity data from Tables 4 and 5 is plotted in Figures 24 and 29 below. The applied seismic forces necessary to qualify the concrete and reinforcement are taken from Table 11 (1/2) for the N-S (Z) sections and Table 11 (2/2) for the E-W (X) sections, see Reference 5, and are shown in Tables 6 (1/2) and (2/2). This data is referred to as "Selected Forces" since they were selected to be the values from the entire data set to bound all the points in the data set. Thus they are the only points that require evaluation to qualify the design.
The forces from the seismic calculation are in pounds (Ibs) and inch-pounds (in-lbs) for a 17-foot (204 inches) wide section (strip) of the pad, and are presented in the 2nd and 3rd columns in Tables 6 (1/2) and (2/2) below. This reinforcement calculation uses kips and in-kips and evaluates a 9 inch wide concrete section. Thus the forces from the seismic calculation are converted. Further, the seismic calculation identified a factor of 1.15 to be applied to all the forces to provide for the variability in Young's modulus and Poisson's ratio. Thus the values in the 2nd and 3rd columns are factored by:
Factor=
9 1 )1.15 = 5.074E-5 204 )1000 Therefore, the values in the 4 th and 5t columns in Tables 6 (1/2) and (2/2) are the values in the 2nd and 3rd columns factored by 5.074E-5. These are plotted in Figures 25 and 30 below. Presenting this data graphically, the selected forces are plotted on the section capacity curves in Figures 26 and 31. This data graphically presented shows that the applied forces are well within the allowable ACI factored capacity envelope. The graphical presentation of the data is then presented using all the applied forces from the seismic analysis calculation, Reference 5. Thus, Figures 27 and 32 present all the applied force data points factored by 5.074E-5, and Figures 28 and 33 present these same data points plotted with the ACI factored capacity envelope as before. Again, the graphical presentation shows that all the calculated applied forces are well with the envelope.
SHEET 134 OF 160 DATE March 11, 2003 JOB. NO.
PGE-009 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel I CLIENT PG&E-DCPP REVIEWER K. L. Whitmore CALCULATION NO.
PGE-009-CALC-007 Reinforcement ENERCON SERVICES, INC.
ORIGINATOR S. C. Tumminelli APPROVED R. F. Evers REVISION 0 Table 6 (1/2) - N-S (Z) Strips - Selected Forces for Desien Quadrant Mx (17 feet)
Fz (17 feet)
Mx (9 inches)
Fz (9 inches) e in-lbs lbs in-kips kips in (1)
(2)
(3)
(4)
(5)
(6)
+Mx Ten. On Bot.
0.0449E8 0.971E6 227.8 49.3 4.621 0.184E8 0.638E6 933.5 32.4 28.812
+Fz Camp. on Sect.
0.336E8 0.224E6 1704.7 11.4 149.535 0.345E8 0.00812E6 1750.4 0.4 4376.00
-Mx Comp. On Bot.
-0.443E8 0.984E6
-2247.6 49.9
-45.042
+Fz Comp. on Sect.
-Mx Comp. On Bot.
-0.00187E8
-0.224E6
-9.5
-11.4 0.833
-Fz Ten. On Sect.
+Mx Ten. on Bot.
0.333E8
-0.097E6 1689.5
-4.9
-344.796
-Fz Ten. on Sect.
0.272E8
-0.606E6 1380.0
-30.7
-44.951 Shear Force Fy
-731078 Fy
-37.1 NA Table 6 (2/2) - E-W (X1 Strips - Selected Forces for Design Quadrant Mz (17 feet)
Fx (17 feet)
Mz (9 inches)
Fx (9 inches) e in-lbs in-lbs in-kips kips in (1)
(2)
(3)
(4)
(5)
(6)
+Mz Comp. on Bot.
0.4165E8 0.8925E6 2113.1 45.3 46.647
+Fx Comp. on Sect.
-Mz Ten. On Bot.
-0.00231E8 0.7565E6
-11.7 38.4
-0.305
+Fx Comp. on Sect.
-0.238E8 0.5355E6
-1207.5 27.2
-44.393
-0.476E8 0.177E6
-2415.0 9.0
-268.333
-0.481E8 0.171E6
-2440.4 8.7
-280.506
-Mz Ten. On Bot.
-0.370E8
-0.08415E6
-1877.2
-4.3 436.558
-Fx Ten. On Sect.
-0.2516E8
-0.5185E6
-1276.5
-26.3 48.536
-0.2482E8
-0.5372E6
-1259.3
-27.3 46.128
+Mz Comp. on Bot.
0.00255E8
-0.2397E6 12.9
-12.2
-1.057
-Fx Ten. On Sect.
Shear Force Fy
-653720 Fy
-33.2 NA
SHEET 135 OF 160 DATE March 11, 2003 JOB. NO.
PROJECT SUBJECT PGE-009 DCPP ISFSI ISFSI Cask Storage Pad Steel Reinforcement ENERCON SERVICES, INC.
CLIENT PG&E-REVIEWER K. L. V CALCULATION NO.
- DCPP VWhitmore PGE-009-CALC-007 ORIGINATOR S. C. Tumminelli APPROVED R. F. Evers REVISION 0 I
a0 N-S Sectlon Capacity 4000 Moe* (-t O
I
{
s
))00
-40000
-20000 i
t20000 4000:0 Mont Qn~dp)
Figure 24 - North-South Section Canacities From Table 4 l+
Nominal Capacity
-U-Factored Capacity I oo0
SHEET 136 OF 160 DATE March 11, 2003 JOB. NO.
PGE-04 PROJECT DCPP]
SUBJECT ISFSI (
CLIENT PG&E-REVIEWER K. L. M CALCULATION NO.
109
[SFSI Cask Storage Pad Steel Reinforcement
-DCPP ORIGINATOR APPROVED S. C. Tumminelli ENERCON SERVICES, INC.
Nldtmore R. F. Evers PGE-009-CALC-007 REVISION 0 N-S Selected Forces i
30 II 30 I
40 A
33 10 A
00
-2000
-1500
-1000
-500 500 1000 1500 A 2C 00
-20
-3E)
A
--40 Moment (in-kips)
Fieure 25 - North-South Selected Forces From Table 6 (1/2)
Compare to Fizure 33. Reference 5 IA Applied Forces
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 137 OF 160 March 11, 2003 S. C. Tumminelli ENERCON SERVICES, INC.
R. F. Evers REVISION 0 Selected N-S Forces I
-60
/0 30000
\\200 200.000 6
Moment (In-Icip)
Fl/ure 26-North-South Section Selected Applied Forces From Table 6 (1I2)
Comp~are to Fieure 33. Reference 5 Nominal Capacity
-U-Factored Capacity Applied Forces Co -Z-
SHEET 138 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
NSAI Fmos I
0U-A AledFaat.
Figure 27 - North-South All Aiplied Forces Compare to Figure 33, Reference 5 3/4A--
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 139 OF 160 March 11, 2003 S. C. Tumminelli R. F. Evers REVISION 0 ENERCON SERVICES, INC.
All N-S Forces I
'A 2009
\\4
/
\\
11000\\
Se 1~000
) 00
-40000
-20000 20000 40000 601 Momlent (anzdpe)
--- Nominal Capacity U
Factored Capacity
-*-All Applied Forces 00
-40 Fiigure 28 - North-South Section All ADpDiied Forces Commnare to Figure 33. Reference 5 CD3
SHEET 140 OF 160 DATE March 11, 2003 JOB. NO.
PGE-009 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel I CLIENT PG&E-DCPP REVIEWER K. L. Whitmore CALCULATION NO.
PGE-009-CALC-007 w
Reinforcement ORIGINATOR APPROVED S. C. Tumminelil ENERCON SERVICES, INC.
R. F. Evers REVISION 0 E-W Section Capacity 4000 Of
{~
2000 00 40000
-20000 20000 40000 60 F Moment (n-Ipd)
Figure 29 - East-West Section Capacities From Table 5
-4Nominal Capacity I Factored Capacity I 00
-6c
ENERCON SERVICES, INC.
SHEET 141 OF 160 DATE March 11, 2003 JOB. NO.
PGE-01 PROJECT DCPPl SUBJECT ISFSI (
CLIENT PG&E-REVIEWER K. L. M CALCULATION NO.
[F9 ISFgSI Cask Storage Pad Steel Reinforcement DCPP Nbitmore ORIGINATOR APPROVED S. C. Tumminelli R. F. Evers PGE-009-CALC-007 REVISION 0 E-WSelected Forces a
-3'
+
.*~-4 40 a0 20* k-N s I*AO WFie Fa Figure 30 - East-West Selected Forces From Table 6 (2/2)
Compare to Fligure 34. Reference S
SHEET 142 OF 160 DATE March 11, 2003 JOB. NO.
PGE-04 PROJECT DCPP ]
SUBJECT ISFSI (
CLIENT PG&E-REVIEWER K. L. V CALCULATION NO.
tD9 ISFSI Cask Storage Pad Steel Reinforcement
-DCPP VWhitmore ORIGINATOR APPROVED S. C. Tumminelli ENERCON SERVICES, INC.
R. F. Evers PGE-009-CALC-007 REVISION 0 Selected E-W Forces I
-60 0-400
-2000 20000 40000 6
C S 1~000f Moment (In-hpe)
Fieure 31-East-West Section Selected ADDlied Forces From Table 6 (2/2)
Comnare to Rigure 34. Reference 5
--+-Nominal Capacity N Factored Capacity
-A-Applied Forces 00
ENERCON SERVICES, INC.
SHEET DATE 143 OF 160 March 11, 2003 JOB. NO.
PGE-Oi PROJECT DCPP]
SUBJECT ISFSI (
CLIENT PG&E.
REVIEWER K. L. V CALCULATION NO.
[F9 ISFSI Cask Storage Pad Steel Reinforcement
-DCPP ORIGINATOR APPROVED S. C. Tumminelli Vhitmore R. F. Evers PGE-009-CALC-007 REVISION 0 E-W All Forces
- I 0
U.
40*
30 AU Ao
,ce to 3_eeec 00
-200w
-1000 1000 2000 3
4 30
-40 Moment (In-kips)
Figure 32 - East-West All ADplied F~orces Comn~are to Ffrure 34. Reference 5
-AII Applied Forces DO
SHEET 144 OF 160 DATE March 11, 2003 JOB. NO.
PROJECT SUBJECT PGE-009 DCPP ISFSI ISFSI Cask Storage Pad Steel Reinforcement ENERCON SERVICES, INC.
CLIENT PG&E-REVIEWER K. L. V CALCULATION NO.
- DCPP Vhitmore PGE-009-CALC-007 ORIGINATOR APPROVED S. C. Tumminelli R. F. Evers REVISION 0 All E-W Forces 40F 0
EatWs Section
/
2~600\\
-400
_200 2.000 400 All Applied Forces Compare to Fieure 34. Reference 5 Nominal Capacity Factored Capaciy
-- AIl Applied Forces
.40 CO0&
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 145 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 The applied seismic forces (Selected Data) are evaluated numerically relative to the ACI factored capacity data by extending the vector from the origin to the applied force data point (Ms,Ps) to the ACI factored capacity curve holding the eccentricity e constant. The process does not hit the capacity curve exactly since the intersection is for two straight lines. Since the capacity curve is always concave inward, the straight line approximation will always yield conservative results.
M2,P2 Force M. P.
-,/ -
M1.PI e
Ms.Ps Moment Figure 34 - Intersection of Applied Force Vector with the ACI Factored Capacity Curve.
Developing the equation for a numerical evaluation, the following definitions are used (see Figure 20):
The applied moment/force is Ms, Ps The ACI factored moment/force values are Ml, P1 and M2, P2 The intersection of the applied force line with the line on the ACI factored line (linearly interpolated between the two closest points is) Mi, Pi.
And, the capacity beyond the ACI Code allowable is computed as either Mi/Ms or Pi/Ps. Both will yield the same numerical value.
SHEET 146 OF 160 March 11, 2003 JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 The algebra to solve for Mi and Pi is as follows:
In Figure 34, the equation for the line passing through Ms, Ps is:
P =-M Ms Also, in Figure 34, the equation for the line passing through MI, P1 and M2, P2 is:
P=P1+( P2-P1 (M-M1)
M2 -M1)
Now, at the intersection of the two lines P = P, therefore, PsM=P+
P(M-Ml)=Pl+2P MM(
Pl Ml Ms M2-M1 M2-M1 M2-Ml Therefore, CPs P2-PI M=PI-( P2-P1 Ml Ms M2-M1)
M2-M1)
- Thus, M =Mi= =0PI_
P2-P1 A4.[
Ps _ P2-P1 1 L
M2-M1)
J LMs M2-Mlj
- And, P1i =-Mi Ms Further, the ratios of the allowable Code values divided by the applied values is:
R = Mi = Pi Ms Ps
SHEET 147 OF 160 March 11, 2003 JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 The values of R are computed for all the forces in Tables 6 (1/2) and (2/2). The values of factored capacity used are those where the eccentricity of the seismic loads e, is between the values of the capacity eccentricities e:. As an example, for the N-S applied seismic forces (Table 6 (1/2), Ps of 32.4 kips and Ms of 933.5 in-kips, the e, is 28.812 inches. Therefore, the values for the calculation of Mi and Pi are Ml = 31,878.0 in-kips and P1 = 1439.9 kips (ec = 22.139 inches) and M2 = 33,975.6 in-kips and P2 = 1109.5 kips (e, = 30.622 inches), see Table 4. Thus:
M=Mi P
_(
)2P Ml +
s P-P M2 - Ml)
[Ms Mi2 - M1
[
1109.5-1439.9 ir[ 32.4 L33,975.6-31,878.0) 933.5 1109.5 -1439.9 1
33,975.6 - 31,878.0 =
- And, 33,612.9 R =
5=
36.0 933.5 All of the values of Mi, Pi and the corresponding R are provided in Tables 7 (1/2) for the N-S (Z) sections and in Table 7(2/2) for the E-W (X) sections.
SHEET 148 OF 160 DATE March 11, 2003 JOB. NO.
PGE-009 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel I CLIENT PG&E-DCPP REVIEWER K. L. Whitmore CALCULATION NO.
PGE-009-CALC-007 Reinforcement ENERCON SERVICES, INC.
ORIGINATOR S. C. Tumminelli APPROVED R. F. Evers REVISION 0 Table 7 (1/2) - N-S (Z) Sections Numerical Evaluation of the Selected Forces for Desian*
Ms/Ps MI/Pi M2/P2 Mi/Pi R
e, e,
eeI from Table 6 (1/2) from Table 4 from Table 4 227.8 / 49.3 0/2073.8 18,848.8 / 2073.8 9582.4 / 2073.8 42.1 4.621 0
9.089 933.5 / 32.4 31,878.0 / 1439.9 33,975.6 / 1109.5 33,612.9 / 1166.6 36.0 28.812 22.139 30.622 1704.7 / 11.4 14,038.0 / 96.1 6858.6 / -99.4 13,929.8 / 93.2 8.2 149.535 146.077
-69.000 1750.5 /0.4 14,038.0 / 96.1 6858.6 / -99.4 10,597.8 / 2.4 6.1 4376.0 146.077
-69.000
-2247.6 / 49.9
-31,086.4 / 760.4
-24,433.5 / 454.1
-28,140.1 / 624.8 12.5
-45.042
-40.882
-53.806
-9.5 / -11.4
-8083.4 / -99.4 0 / -258.8
-212.2 / -254.6 22.3 0.833 81.322 0
1689.5 / -4.9 14,038.0 / 96.1 6858.6 / -99.4 9497.0/ -27.5 5.6
-344.796 146.077
-69.000 1380.0 / -30.7 6858.6 / -99.4 0 / -258.8 5689.5 / -126.6 4.1
-44.951
-69.000 0
- Units are: in-kips for Ms, MI, M2 and Mi, kips for Ps, P1, P2 and Pi, and inches for e, and ec.
SHEET 149 OF 160 DATE March 11,2003 JOB. NO.
PGE-O(
PROJECT DCPP I SUBJECT ISFSI (
CLIENT PG&E-REVIEWER K. L. M CALCULATION NO.
[9 ISFSI Cask Storage Pad Steel Reinforcement DCPP lhitmore ORIGINATOR APPROVED S. C. Tumminelli R. F. Evers ENERCON SERVICES, INC.
PGE-009-CALC-007 REVISION 0 Table 7 (2/2) - E-W (X1 Sections Numerical Evaluation of the Selected Forces for Desian*
Ms/Ps MI/PI M2/P2 Mi/Pi R
- e.
e, e.
from Table 6 (2/2) from Table 5 from Table 5 2113.1 /45.3 30,233.8/736.3 23,528.1 /433.7 26,508.5/568.3 12.5 46.647 41.067 54.250
-11.7 / 38.4 0/2073.1
-23,678.1 / 1920.2
-631.6 /2073.1 54.0
-0.305 0
-12.331
-1207.5 / 27.2
-30,637.5 / 752.9
-24,220.8 / 464.5
-27,838.0 /627.1 23.1
-44.393
-40.693
-52.144
-2415.0 / 9.0
-18,625.9 / 253.0
-11,700.5 / 33.5
-12,061.8 /45.0 5.0
-268.333
-73.620
-349.269
-2440.4 / 8.7
-18,625.9 / 253.0
-11,700.5 / 33.5
-11,992.4 /42.8 4.9
-280.506
-73.620
-349.269
-1877.2 / -4.3
-11,700.5 / 33.5
-4472.1 / -155.9
-9584.1 / -22.0 5.1 436.558
-349.269 28.686
-1276.5 / -26.3
-11,700.5 /33.5
-4472.1 / -155.9
-5834.3 / -120.2 4.6 48.536
-349.269 28.686
-1259.3 / -27.3
-11,700.5 / 33.5
-4472.1 / -155.9
-5703.3 / -123.6 4.5 46.128
-349.269 28.686 12.9 / -12.2 5697.0 / -155.9 0 / -257.9 267.3 / -253.1 20.7
-1.057
-36.543 0
- Units are: in-kips for Ms, MI, M2 and Mi, kips for Ps, PI, P2 and Pi, and inches for e. and e,.
The N-S assessment is shown graphically in Figures 35 and 36; the E-W data is shown in Figures 37 and 38.
SHEET 150 OF 160 DATE March 11, 2003 JOB. NO.
PROJECT SUBJECT PGE-009 DCPP ISFSI ISFSI Cask Storage Pad Steel Reinforcement ENERCON SERVICES, INC.
I
/
I CLIENT PG&E-REVIEWER K. L. V CALCULATION NO.
-DCPP Vhitmore ORIGINATOR APPROVED S. C. Tummineli R. F. Evers PGE-009-CALC-007 REVISION 0 N-S Section Evaluation
+ Nominal Capacity
-UFactored Capacity
-k-Series3 Series4 N Series5
+ Series6 Serles7 Series8
-Series9
-+- Seriesl0 Moment (In.dpa)
Figure 35 - Graphical Presentation of the N-S Numerical Evaluation data from Table 7 (1/2) col
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 151 OF 160 March 11, 2003 S. C. Tumminelli ENERCON SERVICES, INC.
R. F. Evers REVISION 0 I
1
-20000 20000 4
_~
-900 Moment (in-kips)
Fiaure 36 - Graphical Presentation of the N-S Numerical Evaluation data from Table 7 (1I2) zoom at the orifn to show detail COQ9
JOB. NO.
PGE-009 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Stor CLIENT PG&E-DCPP REVIEWER K. L. Whitmore CALCULATION NO.
PGE-0i SHEET 152 OF 160 DATE March 11, 2003 age Pad Steel Reinforcement ORIGINATOR APPROVED S. C. Tumminelli ENERCON SERVICES, INC.
R. F. Evers REVISION 0 019-CALC-007 E-W Section Evaluation i" tt 000
-+-Nominal Capacity
-U-Factored Capacity Serles3
-*-Series4 Se-es5
+ Series6 I Series7
-Series8
-Series9 Sedes10 Series1 I
-80 oo40000
-2000 oo* o 20000 40000 69 o0 GOO '
Moment (In-dps)
Figure 37 - Granhical Presentation of the E-W Numerical Evaluation data from Table 7 (2/2)
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 153 OF 160 March 11, 2003 S. C. Tummineili ENERCON SERVICES, INC.
R. F. Evers REVISION 0
.4 nn I
I I II II I
-ffi.
K*1_
K" I %J W J
- 10
-U El I
-U*
1, I
n o'4
-20000 Q
20000 Moment (in-kips)
Figure 38 - Graphical Presentation of the E-W Numerical Evaluation data from Table 7 (2/2) zoom at the oriain to show detail CIO
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 154 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 Shear Evaluation Evaluation of the applied shears per ACI.
The applied shears are:
V=
37.1 kips see Table 6 (1/2)
Vew = 33.2 kips see Table 6 (2/2)
Calculate the effective depths (since the bending capacities of both the NS and EW sections are greater than twice the applied loads, the inner layers of reinforcement do not need to be included in the calculation for d):
dns= 90.00 - 3.45 = 86.55 inches dew = 90.00 - 5.535 = 84.465 inches and, calculate the Code strength and compare to the applied forces:
pVc = 2(pqf cbd see Reference 11, 11.3.1.1 Therefore, (pVcns = 2 x 0.85 x -0 x 9 x 86.55 = 93,636.1 lbs = 93.6 kips > 37.1 kips OK and
- pVc,
=2x0.85xV5 6 0x9x84.465 = 91,380.4lbs = 91.4kips> 33.2kipsOK Therefore, the applied shears are well within ACI Code limits.
Development Length and Lay Splice Reguirements All reinforcement in the horizontal plane will be spliced using mechanical connectors. However, at the edges, the bars will be bent to the vertical and lap spliced.
Compute the basic development length id per ACI (Reference 11, 12.2) for the #10 bars.
The basic length is computed from either 12.2.2 or 12.2.3. Compute spacing values to determine the appropriate equation for Id. The clear distance is nominally 9-1.27 = 7.73 inches. If bars are moved say
SHEET 155 OF 160 March 11, 2003 L
JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR ENERCON REVIEWER K. L. Whitmore APPROVED SERVICES, INC.
CALCULATION NO.
PGE-009-CALC-007 S. C. Tumminelli R. F. Evers REVISION 0 1/2 inch and lap spliced in a horizontal plane, then the clear distance becomes 8.5 - 2 x 1.27 = 5.96 inches. Now, 5.96 inches is greater than 2db = 2.54 inches. Also, the clear cover will be at least 2 inches (exposed to weather), (See Reference 11, 7.7.1). Since 2 inches is greater than db = 1.27 inches, the equation to use for ld from 12.2.2.is:
Id fyan db 20/f,
- Now, fy = 60,000 psi fc =5000 psi V- =70.7psi<100psi (seel2.1.2) x = 1.0 (not a top bar)
P = 1.0 bars are uncoated ld fyaP 60,000 x 1 xl 424 db 20~f -
20 00 and lb =42.4 x db = 42.4 x 1.27 = 53.9 inches from 12.2.2 The other ACI method is Eq. (12-1)
Id = 3 fy apBy db 40f C c+Ku, db
)
where newly introduced terms are:
7= 1.0 size factor c = spacing or cover dimension (smallest of center to surface or l/2 center to center)
Ktr transverse reinforcement index = 0 (conservative) and the term:
c + Kt must be less than 2.5 db
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 156 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 Smallest c (yields largest id) is 2+1.27/2 = 2.635.
- And, c +K tr =2.635 + ° = 2.075 less than 2.5 db 1.27 Therefore, Id
_ 3 fy apy 3 60,0001x1x 1 3 0.7 db 40f7 (c+K, 40 V
2.075 db) id=30.7xdb=30.7x1.27=39.0inches fromI12.2.3 Therefore the basic development length is 39 inches 3'-3".
Class B splice, factor by 1.3, 1.3 x 39.0 = 51 inches 4'-3" If, in the development of the details, bundled bars result, the splice lengths shall be increased by 20% for 3 bar bundles to 1.2 x 51 = 61 inches and by 33% for 4 bar bundles to 1.33 x 51 = 68 inches.
SHEET 157 OF 160 March 11, 2003 JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 Seismic Forces in Reinforcement The design of the reinforcement that is located just above the embedment structure anchor plates has additional requirement beyond the bending response to the thermal, shrinkage and seismic. These are addressed in the embedment structure calculation, see Reference 4. An input to that calculation is the set of forces in the reinforcement due to the seismic demand. There will be no appreciable forces in these bars due to the thermal and shrinkage demand since considerable time will have elapsed between the construction of the pad and the placement of the fuel casks.
As noted in the details below, this calculation uses 90 inches for the depth of the pad while making use of the working stress equations developed for the thermal analyses. Use is made of equations 1 and 2 since the applied moments produce tension on the bottom of the pad.
Forces in Reinforcement:
Bounding Seismic - Tension on Bottom NS (Z) Strip b = 9 inches; h = 90 inches; d, = 3.45 inches; d2 = 7.63 inches; d3 = 3.45 inches; d4 = 16.55 inches A =1.27 sq. in.
- 10 bar dnet=-8.92 fc = 5.000 ksi E, =4030 ksi; Es = 29000 ksi; a= 29000 = 7.196 4030 q, = 0.221 ksi from off-line calculations; 0.22 1 EC = 4
= 5.48E - 5 in/in c4030 Bounding Seismic Force/Moment Combination from Table 6 (1/2).
This is the largest positive Mx combined with the largest tension:
C = -695.867 kip/1 7 foot section; C = -30.7 kip/9 inch section (-) indicates tension M = 39675.73 in-kip/1 7 foot section; M = 1750.4 in-kip/9 inch section Moment produces tension on the bottom of the pad, therefore use Equations (1) and (2).
(4Asa -
) = 4 x 1.27 x 7.196 - 02
= 175.470 2bAsa(2h-dnet)= 2x 9 x 1.27 x 7.196x (2 x 90-8.92) = 28,142.76
SHEET 158 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
4ASa -
+ j(4Asa --
)
+ 2bAsa(2h - dnfl)
- x. =
b
-175.470 + 1175.4702 + 28,142.76 9=
7.477inches
- Now, Cc = IxbE.6. =
2xa C22 7.477 x 9 x 0-221 2
=7.436kip C, = AsEs (xdl )c = 1.27 x 29000 (7477 -.4) x5.48E-5 = 1.088 kip C 2 = AsEs( xd2 c= 1.2 7 x 2 9 0 0 0 7.477
- 63) x 5.4 8E -5= -0.041 kip (-) indicates tension T. = AsES (hxd3c
= 1.27 x 2900 90-7.477-3.45 x 5.48E - 5 = 21.344 kip T.= AE(h-x-d4 1 27 29000-7.477-16.55 x 5.48E-5 =17.808kip Check on equilibrium F = 0
.. C =Cc +C1 +C2 -T. -T2
-30.7 = 7.436+1.088-0.041-21.344-17.808 = -30.67 OK Now the internal moment is:
M C
=)
'(2 d')+C2(h d2)+T'(h-d3 +T2(h-d4 Equation (2)
M = 7.436 (90 7.477 +(1.088)(90 -3.45)+(-0.041) 90 -7.63(+..9 (32 3
-7.6
-6.5 (21.344)90 - 3.45+( 17.808) C90
-16.551 1753.2 =-1750.40K
( 2 (2
)
SHEET 159 OF 160 JOB. NO.
PGE-009 DATE March 11, 2003 PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR S. C. Tumminelli REVIEWER K. L. Whitmore APPROVED R. F. Evers CALCULATION NO.
PGE-009-CALC-007 REVISION 0 ENERCON SERVICES, INC.
Forces in Reinforcement:
Bounding Seismic - Tension on Bottom EW (X) Strip b = 9 inches; h = 90 inches; di = 5.535 inches; d2 = 9.705 inches; d3 = 5.535 inches; d4 = 18.635 inches A, = 1.27 sq. in.
- 10 bar dnet = -8.93 fc = 5.000 ksi Ec=4030 ksi; Es = 29000 ksi; a = 29000 = 7.196 4030 a', = 0.398 ksi from off-line calculations; C = 0.398 = 9.88E - 5 in/in c4030 Bounding Seismic Force/Moment Combination from Table 6 (2/2).
This is the largest negative Mz combined with the largest tension:
C = -618.8 kip/17 foot section; C = -27.3 kip/9 inch section M = 55315.73 in-kip/17 foot section;M = 2440.4 in-kip/9 inch section Moment produces tension on the bottom of the pad, therefore use Equations (1) and (2).
C
-27.3 4Aa--=
= 4 x 1.27 x 7.196 --
= 105.149 cc 20.398 2bAjIt(2h -dnet) = 2 x 9 x 1.27 x 7.196 x (2 x 90 -8.93) = 2814 1.111I
-l4Asa-Cj+
1t04As a--C
+ 2bAsa(2h -dnet)
-105.149+V105.149 2 +28141 111 10.315inches b
9
- Now, 1
xboc 10.315 x 9 x 0.398 C =2xbEcsc=
2 2
18.474 kip C, = AEr xd1
=l 2 7X 2 9 000C 10.315 -5.535x988E-5=1.686kip ss y
)
C
.10.315
)
SHEET JOB. NO.
PGE-009 DATE PROJECT DCPP ISFSI SUBJECT ISFSI Cask Storage Pad Steel Reinforcement CLIENT PG&E-DCPP ORIGINATOR REVIEWER K. L. Whitmore APPROVED CALCULATION NO.
PGE-009-CALC-007 160 OF 160 March 11, 2003 ENERCON SERVICES, INC.
S. C. Tumminelli R. F. Evers REVISION 0 C2 =AsEs, x
)Ec
=1.27x29000(
10315 9
x 9.88E-5 = 0.215 kip TI =ASEsthd
&C =1.27x29000(-
T 2 =ASE (h-xd4>)c
= 1.27 x29000(
- - 5 x 9.88E -5 = 26.158 kip 10.315
)
90-10.315-18.635)x 9.88E -5 = 21.536 kip 10.315 Check on equilibrium F = 0
.-. C=Cc+C1+C2-T
-T2
-27.3 = 18.474+1.686+0.215-26.158-21.536 = -27.32 OK Now the internal moment is:
M C
=)
'(2 l)+C2(h d2)+Tl(h-d3 +T2(hd4)
Equation (2) 90 10.315)
( 686)
-5.535)+(0.215) (9-9.705+...
(2.18 20 535
+ (21.536) 20 -18.635)=24421 1_ 24404 OK Thus, the largest N-S or E-W force in a bar just above the anchor plates (T2) is 21.536 kip.
Summary and Conclusions This calculation has provided the determination of the evaluation required by the governing documents for the design, i.e., the PG&E specification and all of its relevant subordinate documents and the regulatory requirements. Further, the selected reinforcement, #10 bars at 9 inches, Figures 1 and 2, has been shown to limit the expected crack widths due to curing stresses (temperature and shrinkage) to within the recommended values prescribed by ACI 207. The seismic evaluation shows that the pad is compliant with all the required design parameters. Finally, forces in reinforcement are computed for use in the embedment structure design.