ML19296D752

Rept on Design Criteria for Masonry Walls

ML19296D752
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Site:	Trojan File:Portland General Electric icon.png
Issue date:	02/13/1980
From:	Colville J MARYLAND, UNIV. OF, COLLEGE PARK, MD
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ML19296D739	List: ML19296D738 ML19296D743 ML19296D746 ML19296D752
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TAC-12369, NUDOCS 8003130111
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Text

REPORT ON DESIGN CRITERIA FOR MASONRY WALLS IN nlE TROJAN POWER PLAhT Submitted to:

Mr. Kenneth Herring Nuclear Regulatory Commission Division of Operating Reactors By Dr. James Colville, P.E.

8003180l

TABLE OF CONTENTS Page A.

In-Plane Behavior 1.

Introduction.

1 2.

Flexure Failure 1

3.

Sliding Failure 3

4 Diagonal Tension Failure.

4 5.

Verification of Proposed Equations..

6 a) California Test Results.......

6 b)

Bechtel Test Results.

7 6.

Application to Trojan Plant 11 a)

Introduction 11 b) Type A Walls 11 c) Type B Walls 13 15 B.

Out-of-Plane Behavior 1.

Introduction 15 2.

Shear Bond Strength of Masonry Collar Joint 15 3.

Metal Tie Capacity for Shear Transfer 18 4.

Support Conditions at Wall Ends 19 5.

Tensile Bond Strength of Masonry Bed Joints 19 6.

Induced Loauing Due to Seismic Effects.

20 7.

Out-of-Plane Arching 20 21 S.

Cracked Moment of Inertia of Wall 9.

Design Check of Wall No.46 (Auxiliary Building)

Appendix A - Development of Flexure Failure Equations 23 Appendix B - Mortared Double Wythe Wall Calculations Design Check of Wall No.46 (Auxiliary Building)

e Appendix C - References.

3-C ' u c :59 y

1 A.

In-Plane Behavior 1.

Introduction For walls with in-plane lateral loads, three distinct failure modes may be identified as follows:

(a) flexure failure, (b) sliding failure, (c) diagonal tension failure.

Equations for each of these failure mechanisms are developed below and are applied to the test data presented in Refs. (1,2) and subsequently to conditions existing at the Trojan nuclear power plant.

2.

Flexure Failure Determination of the lateral load producing flexural failure in the grouted cavity walls of the Trojan plant is complicated by the influence of the steel frame embedded in the concrete core of the walls.

The procedure developed by Bechtel in which a typical segme-t of the wall bounded by two adjacent columns and two adjacent f'c:r levels is examined is considered a reasonable model for estimating the flexural capacity of the wall system.

This observation is suggested by the reported presence of vertical cracking in the field of the masonry along vertical column lines.

Bechtel's ap-preach in developing failure equations for both single and double curvature response of a typical wall is also reasonable althou;h the single curvature mode of behavior requires substantiation cf the possibility of displacement compatibility along element inter-fac e s.

2 Equations for predicting the nominal shearing stress in the masonry wall units at ultimate failure in flexure are presented in Appendix A.

These equations are based on the following assump-tions:

(a)

Failure strain in concrete masonry in compression equal 1.5 fE/Em.

Evidence in Ref. (3) indicates that the compression failure stress of masonry in flexure is around 1.5 times the nominal wall com-pressive strength,fy. Using the relation that E =1000 fE, yields m

a compressive failure strain of 1.5(10-3) m/in.

This is half of the value used for concrete and is, therefore, considered a con-servative estimate of the failure strain.

(b) Maximum tensile strain in the extreme tension reinforcing steel is 4.0 times the yield strain.

This is also a conservative esti-mate of the tensile steel strain.

(c) The vertical reinforcing steel in the wall is uniformly distribut-ed over the wall length.

(d) Wall height to length ratio = 0.5 The resulting equation for double curvature is:

f = 1.55c f 1.62:

(1)

+

v y y a

for single curvature, the equation is:

'V f = 1.4 ch.v

.'v ^

• I#

()

v o

tt w

In these equations vf = equivalent nominal shearing stress at flexure failure c. = percentage of vertical wall reinforcing (i.e. core plus masonry) h = percent ge f horizontal reinforcing in the masonry wythes c

(i.e.

neglecting discontinuous core reinforcing)

3 o = compressive stress in wall due to external gravity g

loads t = length of wall panel in inches y

t = total thickness of wall f = yield strength of reinforcing steel y

V = shear force capacity of the steel frame beam to column connection These two equations are essentially in basic agreement with those presente/ by Bechtel. There is no philosophical difference between Eqs. (1,2) and the equations presented by Bechtel.

Differences exist, however, in the coefficients applied to each of the various terms in Eqs. (1,2).

3.

Sliding Failure In presenting an estimation of the shearing stress correspond-ing to sliding failure, the occurrence of a horizontal crack at the wall-floor interface is presumed and the shear friction developed along such a failure plane is determined.

The assumptions used are listed below:

(1) The resistance to sliding offered' by the steel frame columr.s equals the sum of the shear resistances of the columns. The shear yield strength of the steel is taken as f //5.

This y

value is suggested in Ref. (4).

(2) The coefficient of friction for the normal force developed as a result of yielding of the vertical wall reinforcement is. 4 This value is applicable for concrete cast monolithicall:- (see Ref. (5)) and since all of the reinforcing steel is in high strength grout concrete a value of 1.4 is considered appli-cable.

4

(.5) The coefficient of friction for the normal force developed from axial compression in the wall is taken as a weighted average based on the relative areas of bearing composed of nortar bed joints and grout or concrete.

The coefficient of friction for the mortar bed joints is taken to be 0.75 which is a lower bound value used by Hat:inikolas (Ref. 6).

The basic equation for sliding shear is given below.

A f

N g = l.75 AB 1.4(A ) },9 g

s g3) 1,4c g v

y

)

in which A

= n t ar a f m rtar bed joints B

A

= area of concrete grout y

A,.

=A

+A B

wall compressive stress due to gravity load

=

o,

= percentage of vertical reinforcing steel in wall f,

= yield strength of wall steel A

= cross-sectional aren of a steel colura ge f,s = yield strength of column steel t,,

= length of wall panel t

= thickness of wall panel N

= total number of colu as available for sliding c

resistance 4

Diagonal Tension Failure For wall segments with a height to length ratio of 0.5 the usual beam shear stress conditions are not applicable since under lateral loading the beam span / depth ratio is sna11 (i.e. 0.5).

For

5 deep beams the AC1 code permits an increase in the usual concrete shear capacity. The concrete shear strength may be taken as v = (3.5-2.5

) 2@

(4)

For an aspect ratio of 0.5, Eq. (4) gives v = 2.25(2) F = 4.5 F (5) e e

e The compressive strength of the grouted masonry walls is taken as a weighted average of the masonry and grout. Thds the shear strength of the wall is

-, + /y Ac A,, }

A f l (6) v = 4.5 {/f' A

c m

in which f' = compressive strength of the masonry wythes f' = compressive strength of the grout concrete A = area f masonry block wythes B

A = area of grout concrete g

A

=A

+A w

B w

xial compression stress in the wall will increase the shearing strength of concrete.

Following Ref ( 6 ), which is based on a Mohr's circle analysis, the shearing strength with axial compres-sion is

= ' (v +h)2, (

)2 (7) v c c

in which v is obtained from Eq. (6).

c A further contribution to diagonal tension strength is provid-ed by both the hori ontal and vertical wall steel.

Since test

6 results (1.2) indicate cracking at a 45* angle for walls with a height to length ratio of 0.5 the horizontal and vertical steel are considered to be equally effective as shear reinforcement.

Thus the final equation for diagonal tension failure is A

+A g = $ {v* + 0.5( s

)ff (7) v t

y w

= & { v* + 0. 5(p f)f}

(8) e y

y in which v = grouted masonry shear strength obtained from Eq. (6A)

= capacity reduction factor for shear = 0.85 p = percentage of vertical reinforcement y

c = Percentage of horizontal reinforcement (based on vertical h

wall area) f = yield strength of reinforcing steel y

5.

Verification of Proposed Equations a.

California Test Results The results of a series of tests performed on hollow concrete masonry wall elements are presented in Ref. (1).

These walls had a height to length ratio of 0.5 and the 7-5/8" thick concrete masonry units were grouted and reinforced with 3 #7 vertical re-inforcing bars.

The major variable in these tests was the amount of horizontal reinforcing steel.

The governing equations for these tests are obtained from Eqs.

(1), (3), and (8). After substitution of appropriate values (using f =50 ksi) the relations become y

~

7 (a) flexure f = 229 + 1.62 o (9) v g

(b) sliding f = 206 + 1.075 o (10) v g

(c) shear f = 0. 85 { v* + 74 +

f}

(11) o y

From these equations it is apparent that Eq. (9) will always yield larger values than Eq. (10) so that the failure mechanism will be either sliding on diagonal tension. Substitution of the ap-propriate test values in Eqs. (10) and (11) yields the results gisca in Table 1.

The theoretical equations consistently predict a diagonal tension failure.

In fact, Ref. (1) states that specimen 1,2,3 c.nd 5 did in fact fail in diagonal tension.

Although the final failure of specimen 4 and 6 consisted of a combined diagonal tension and sliding failure, the initial cracking of these two specimen in-dicated diagonal tension behavior. The predicted failure stresses are in good agreement with measured stresses with an average error of only around 6*6.

This is taken as evidence that Eq. (8) is a reasonable predic-tor of diagonal tension failure, b.

Bechtel Test Results A series of 23 tests performed on concrete masonry walls with grouted cores and grouted cavities are also available for checking the above equations.

8 Table 1 - Comparison of Theory with California Test Results Wall Capacity Test Normal Horizontal Test Sliding Diagonal Error No.

Stress Steel Ratio Result Shear Tension psi ph Psi psi psi 1

194 0

310 415 335*

+0.1 2

200 0.00102 330 422 347*

+5.2 3

243 0.00203 398 468 371*

-6.8 1

4 212 0.00305 344 4 34 373*

+S.4 5

215 0.00407 361 43S 384*

+6.4 6

234 0.00577 413 455 408*

-1.2 Average 6*o Note:

• denotes theoretical failure capacity

9 Details of the test specimen will not be repeated herein, although the concrete masonry had an average compressive strength of 2500 psi and the grout concrete had an f[=5500 psi.

Since the test specimen were constructed between two relative-ly large reinforced concrete beams, the walls, which had a height to length ratio of 0.5 are considered to have been bent in double curvature.

A comparison of the test failure stresses, and failure stresses predicted for each of the three failure modes discussed above are pre-sented in Table 2 These results indicate sliding shear failures should have occur-red in all tests except A4, F1, J1, L1 and L2, which fail in diagonal tension,and El and E2 which fail in flexure. The Bechtel report dif-fers somewhat in that diagonal tension failures are also reported in tests F2, H1 and H2.

However, figures of failed specimen H1 and H2 show indications of sliding failures, with large diagonal cracks in specimen Hl.

No figure of specimen F2 is available for review.

Thus the above equations predict the correct failure made in at least 20 and possibly all 23 specimen.

Errors in the predicted failure stresses range from 1.5% to 31.5%.

However, except for specimen El, E2, F1 and G2, the errors are less than 15%, with an average error of around 8%.

Both specimens El and E2 have an aspect ratio of 1.0 and are predicted to fail in fl exure.

The significant errors in the predicted stresses indicate that the flexure failure equations may be too conservative. This is partly a result of the assumption that the maximum steel tension

Table 2 - Comparison of Theory with llechtel Test Results Specimen I:ailure St ress Sliding Shear Diagonal Tension Flexure Failure Error No.

from Test I:ailure St ress Iailure St ress Stress psi psi psi psi

?.

Al/A3 268 272*

359 334

-1. 5 A2/A5 325 360*

384 456

+10.8 A4 457 479 416*

>uo

-9.0 til /ii3 239 209*

329 265

-12.6 112 /111 310 297*

354 386

-4.2 11 5 141 120*

304 143

-14.9 D1 200 206*

324 250

+3.0 til 264 318 364 197*

-25.4 li2 183 221 342 133*

-27.3 1:1 308 426 405*

505

+31.5 1:2 318 338*

376 383

+6.3 GI 192 162*

350 275

-15.6 G2 105 81*

317 163

-22.9 111 312 292*

378 348

-4.8 l12 232 366*

398 436

+13.3

.I l 475 517 456*

584

-4.0 K1 165 156*

320 184

-5.5 1.1 428 560 377*

422

-11.9 1.2 367 579 375*

466

+2.2 Note:

• denot es theoret ical failure capacity 5

11 strain is limited to four times the yield strain. The error for F1 stems from the fact that this specimen failed - t a relatively low stress in comparison to member F2 which was subjected to a lower precompression stress. The error in specimen G2 tends to be somewhat magnified since this specimen failed at a relatively low stress.

Overall the comparisons between the theoretical predictions and the Bechtel test results are considered reasonable especially considering the variability of the materials involved.

6 Application to Trojan Plant a.

Introduction All of the various conditions prevaling in walls in place are not known.

It will be assumed that a typical wall is three bays wide and four levels high. Story heights are 16'-0" and column spacings are 31'-0".

General assumptions for Type A walls are listed below.

Also for a typical wall panel in the field the lower value for flexure stress from Eqs. (1),(2) will be used.

b.

Tvoe A Walls 0.0021 v =

y

.0019 v =

f 50 ksi

=

y shear capacity of the beam-column connection V

=

350 kips

=

t = wall thickness, ranging from 23" to 48", with ext erior masonry wythes of 7-5/8" f' = 6600 psi (core concrete strength) f' = 2700 psi (light block in the structure)

Steel columns - W14x142

( Area =41.8 in )

12 Flexure Failure:

Based on the assumed values listed above the nominal shear stress at failure in flexure from Eq. (1) is v = 163 + 1.62 o (12) f g

For single curvature f = 254 +.81c (48" thick walls)

(13) v g

v = 296 +.81c (23" thick walls)

(14) f g

Thus as long as the axial compression stress o is less g

than 112 psi, Eq. (12) governs and the double curvature failure mode is critical.

Sliding Failure:

For sliding failure, Eq. (3) gives f = 212 + 1.30 c (48" thick walls)

(14) v g

f = 282 + 1.18 o (23" thick walls)

(15) v g

Diagonal Tension:

In diagonal tension, if the effect of the axial compression stress is ignored a conservative estimate of the failure stress from Eq. (8) is f = 338 psi (23" thick walls)

(16) v f = 338 psi (48" thick walls)

(17) v Summary:

In reviewing Eqs. (12) through (17), the following estimates of noninal failure shearing stress are obtained:

f = 338 psi maximum (from Eq. 16)

(18) v If the axial compressive stress is below 100 psi compres-sion either a flexural on sliding shear failure will occur and

13 g = 163 + 1,62 c (flexure failure)

(19) v or g = 212 + 1. 30 o (sliding failure)

(20) v g

However, an examination of Eqs. (19) and (20) indicates that g _100 psi.

Thus depend-Eq. (20) will not become cdtical for o ing on the value of o, the failure stress, vg will vary from g

163 psi to 338 psi and a general equation for vf becomes v = 163 + 1.62 e < 338 psi (21) g Thus when o 1108 psi, flexural failure occurs, with c > 105 psi, g

g diagonal tension failure results.

For walls in the field a slid-ing failure is not likely.

In summary, it is recommended that Eq. (21) be used for Type A grouted double wythe wc11s.

c.

Type B Walls All assumptions for Type A walls apply for Type B walls ex-cept as follows :

c

=.0012 y

c

=.0013 h

Flexure Failure:

The nominal shear stress at failure in double curvature is vf = 93 + 1. 62 O (22) g For single curvature wg = 148 +.S o (48" thick walls)

(23) g v = 190 +. Sic (23" thick walls)

(24) g g

14 Thus as long as the axial compression strese o, is less g

than 68 psi, Eq. (22) governs and the double curvature failure mode is critical.

Sliding Failure:

For sliding failure, Eq. (3) gives f = 149 + 1.30 o (48" thick walls)

(25) v g

f = 219 + 1.18 o (23" thick walls)

(26) v g

Diagonal Tension:

In diagonal tension, if the effect of the axial compression stress is ignored a conservative estimate of the failure stress from Eq. (8) is f = 308 psi (23" thick walls)

(27) v v = 332 psi (48" thick walls)

(28) f Summary:

In reviewing Eqs. (23),(24),(25), and (26) it is apparent that sliding failure will not occur, since single curvature flexural failure will always precede a sliding failure.

Thus the following general equations may be stated:

For 23" walls f = 93 + 1.62 c for 0<c 1 120 psi (29) v g

f =190 +

.81 o i 308 psi for o > 120 psi (30) v g

g _

For 48" walls f = 93 + 1.62 o for 0<o 1 6S psi (31) v g

g g

9 _ 6S psi (32) g =14S +

.81 o 1 332 psi for 7 >

t It is reco:mnended that Eqs. (29) through (32) be used for Type B grouted double wythe walls.

15 B.

Out-of-Plane Behavior 1.

Introduction For double wythe mortared walls with out-of-plane lateral loads induced by seismic effects, the response of the walls is significant-ly affected by the interaction of the two wythes.

In particular the following items are of importance in determining the ability of such walls to withstand weak axis bending:

(a) Shear bond capacity of the collar joint between the two wythes (b)

Capacity of metal ties to transfer shear between the two wythes (c)

Support conditions at the wall ends (d) Tensile bond strength of the masonry bed joints (e)

Induced loading due to seisnic effects (f)

Possibility of 1 way or 2 way out-of-plane arching of the wall due to confinement of bounding supporting elements (g) Cracked moment of inertia of wall Each of these items are reviewed be.1 for walls consisting of two wythes of 7-5/8" thick concrete masonry, constructed with type M mortar.

2.

Shear Bond Strength of Masonry Collar Joint The collar joint shear bond strength is a major factor in the be-havior of multi-wythe masonry construction, particularly with respect to weak axis bending. A widely stated position is that for composite construction the collar joint must be completely filled with mortar.

However, even if this joint is filled, there must be a transfer of

16 shearing stress across this joint without sigaificant slip in order for full composite interaction of the multiple wythes to be realized. Since the cracking strength, moment of inertia, and ultimate flexural strength, of the wall cross section are significantly influenced by the interaction of multiple wythes, it is crucial to establish the collar joint shear bond strength.

Unfortunately, no existing published data on the shear bond strength of collar joints has been located.

Data is available from several sources on the strength of mortar bed joints.

However, these results produce a scatter of values due to the influence of different test procedures, masonry units, mortars, precompression across the joint, etc.

Also the bond strength in a bed joint can be expected to exceed that in a collar or head joint because the weighting action of the masonry units will tend to promote better bonding in a horizontal mortar joint. Also the bed joint will be under precompression from other gravity leading which will increase the joint capacity in shear.

In Ref. ( 7 ), a brief review of test results from several sources on the strength of concrete masonry joints is presented.

Ref. (-)

also indicates, based on the results of tests of 2 specimen, that the shear bond strength of a head joint is only around 15 psi.

Since the shear bond strength of cellar joints in concrete masonry construction is not presently known, reasonable values for use in design will be deduced below using a combination of existing test data. As stated above a problem in determining joint strengths stems from the test methods employed.

Recently

17 a method of evaluation of tensile bond and shear bond by means of centrifugal force has been suggested ( 6 ).

This test procedure is considered to be an improvement over present methods since joint precompression is essentially eliminated as a result of the test-ing procedure.

Test results using this procedure on concrete masonry specimen with type M mortar indicate the following strength values:

Tensile bond strength = 74 psi (mean of 30 specimen)

Shear bond strength

= 47 psi (mean of 20 specimen)

Although there was a considerable scatter of results, these values are considered reasonable estimates of strength.

However, the main use of these results in this report is to establish the relation between shear bond and tensile bond strength.

The values indicate that shear bond strength is around 60% cf the tensile bond strength.

From a limited series of field tests performed at the Trojan plant ( 8 ), the tensile bond strength of the masonry collar points is estimated to be 45 psi, based on the net area of mortar encounter-ed in the test specimen. Actual test values for 4 specimen were 29 psi, 40 psi, 160 psi, and 64 psi.

However, in estaelishing the streng-th value for design use, the 160 psi value has been discarded since it is significantly in excess of the other three values.

It should also be noted that these values are based on estimates of the mortar in the collar joints that range from 15% to 85%, with less than 50%

mcrtar in 3 of the 4 tests.

18 The shear bond strength of the collar joints in the Trojan plant is now estimated as follows Shear bond strength = 0.60.x tensile bond strength

= 0.60 x 45 psi

= 27 psi in which the factor 0.60 is based on Ref. ( 6 ) data.

Since Bechtel has indicated that the collar joints are 80%

filled in the field the shear bond strength on the gross area becomes 0.8(27) = 21.6 psi.

However, due to the scatter of test data on mortar joints, and the crucial nature of this joint on the behavior of the Trojan walls, a strength reduction factor of 0.6 is recommended for design purposes.

Thus the reliable design strength capacity of the mortared collar joints is estimated to be 0. 6 (21. 6) = 12 p si.

The head joint test value of 15 psi, if reduced to account for partial (80%) filling of the collar joints, also yields a design value of 0.S()

= 12 psi.

In summary, it is recommended that the shearing bond strength of the mortared collar joints be taken as 12 psi. This is not con-sidered a conservative value since discussions with individuals ex-perienced in masonry research indicate that they would not rely on e

the collar joint for any transfer of shearing stress.

It is recom-mended that in-situ testing be performed to confirm this design strength value.

3.

5 fetal Tie Canacity for Shear _ Transfer Tests of multiple wythe masonry wall units included in a Eechtel submission ( 9 ) and other tests described in Ref. (10), indi:ste a

19 4

high degree of composite action. However,in the tests in Ref. (10) the collar joint was grouted rather than mortared. Also in both series of tests the masonry wythes were tied with 9 gage continuous metal joint reinforcement 16" on center.

Since the masonry wythes in the Trojan plant are tied with in-dividual metal ties spaced 4' apart, the contribution of such ties to the development of composite action is minimal.

Thus it is recommend-ed that the metal tie capacity for shear transfer be neglected.

4 Support Conditions at Wall Ends The degree of restraint afforded to the masonry walls at the wall-floor interface is difficult to establish.

Factors influencing this restraint include relative rigidity of wall and floor members, which in turn depends on the extent of wall and floor cracking.

Also if the walls vibrate in the fundamental mode the rigidity of the wall-floor joint will not eliminate near pinned end conditions of support.

Consideration is given below in the wall capacity com-putations of both fixed and pinned support conditions.

However, it is recommended that a pinned support condition be used to more closely represent the wall support condition.

5.

Tensile Bond Strength of Masonry Bed Joints In the absence of data to indicate the tensile bond strength of the type M masonry bed joints it is recommended that the value of 74 psi, obtained from Ref. ( 6 ), reduced by a : factor of 0.6, be used in design calculations.

It is further recommended that in-situ tests be performed to confirm this value.

20 6.

Induced Loading Due to Seismic Effects In computing the induced lateral loading due to seismic effects, Bechtel has employed the following equations:

(1) fixed-fixed supports

. 1 ",

K I33) natural frequency, f =

2 2L (2) pinned-pinned supports all (34 )

natural frequency, f =

' AY 2L'[5 in which L = wall height, in inches E = modulus of elasticity of masonry wall material I = weak axis moment of inertia A = cross sectional area of wall, in inches y = unit weight of masonry g = acceleration constant Once the natural frequency is obtained from the appropriate equation, the spectral acceleration is determined from response spectrum curves with 5'. damping. The inertia load then equals the wall mass times the spectral acceleration.

This procedure is considered appropriate and will also be followed herein.

7.

Out-of-Plane Arching It is generally accepted that significant out-of-plane strength can be developed in masonry walls due to compressive reactions that develop in boundary supporting elements. Arching can develop in one or two directions and depends on the details of the structure enclosing the wall panel.

1 21 8.

Cracked hbment of Inertia of Wall After tensile cracking of a laterally loaded masonry wall element, the effective moment of inertia is reduced.

Bechtel calculations use the following relation for the effective moment of inertia, I,,

in the post cracking condition.

I = 3- (1

+ Ier)

(35) e 2

g in which 1

gross uncracked moment of inertia 8

I = m ment f inertia f a cracked section r

The AC1 Code (5 ), user the following value for reinforced concrete flexural members M

cr)3 (36)

I = (I -1er) ( M

+I e

g cr Ma = maximum moment in the beam span M = moment causing cracking er 7

=f 1

ty 7 f

tensile cracking stress

=

v = distance from uncracked section centroidal axis

't to extreme fiber in tension From Eq. (36), a conservative value for the effective moment of inertia would be to use the cracked section moment of inertia.

In Ref. (10), it is suggested that EI

= 0.2 EI (3-)

e g

22 In the absence of information to substantiate the relation used by Bechtel (Eq. (35)) it is recommended that the more con-servative value from either Eq. (36) or (37) be used in design computations.

The influence of the value used for I on the wall inertia e

loading is discussed in Appendix B.

9.

Design Check of Wall No.46 (Auxiliary Building)

Wall No.46 is assumed to be a nominal 16" thick wall composed of two wythes of 7-5/8" block with a 3/8" collar joint.

Detailed design calculations are presented in Appendix B.

These calculations indicate that the collar joint shear strength will be insufficient to maintain composite behavior. As a result the wythes will separate and act as individual segments.

The everall conclusion is that this wall will fail above E1.77' under its own inertia londing unless two way action can be developed.

23 APPENDIX A Development of Flexure Failure Eauations t

1.5f

= 1. 5 (10~) in/in Assume:

Masonry failure strain c,=

Em

)

= 8.276(10~) in/in Max. steel strain = 4 x c

=

y 29 0 The strain diagram at failure is shown below x

k 1

S.276(10-#)

'Y

.63751 1.5(10~3) i t

I 1

Assuming the tension steel is uniformly distributed along the wall length L.

I'

( S.2 76+1. a- ) t = 0.15; k=

1 x = 7(1

.15L) = 0.2125; For a linear stress variation in the compression :one the moment capacity of the steel and the axial compression force, N, assumed to g

act at the wall mid-length is

9 24 APPENDIX A: (Cont'd)

~

y y [ 63751 I2

/

\\

9

'k) + 125 Af X.21251

~

M=$

.6375A f

+ x+

2 a

2 v y\\ 3 2

1 k

+ -- k

+N a

o2 a

'125

=$

.6375A f (.6311) + '"2 A f (.2421) +.45 N 1, yy vy o

= 4 (0.428 Avf E +.45 N 1) y o

V 1

h i

l 1

1 I

For single curvature Vh=M M

V =5 M

" " hit

/.428 A f

.45 N i

. v = 9 *, t9 (.478 A f I +.45 N 1) =.9 Y+

nu vy o

nt ht i

h=f or since

/.S56 A f

.95 v = 0. 9 - Et at

+

\\

9

25 APPENDIX A: (Cont'd)

=.9(.8560 f

.9c )

+

vy o

0.77p f

.81c

+

=

vy o

For double curvature v = 2(.77p f +.81a ) = 1.54p f + 1.62c vy o

yy o

The above relations indicate the contribution of the vertical steel and the axial compression to the moment capacity of the wall element.

However, for singic curvature vertical shearing forces are developed between adjacent panels due to the horizontal steel and the beam-column shear connection. Thus for single curvature an additional moment capacity is obtained as follows:

M = 1.4A Hy c

M Again since v = ht*

The total nominal shearing stress for single curvature failure is 1.4A f V

gy c

v=. 77p,f

.81c

+

+ gp

+

y g

This becomes

+{'V 1.4c f v=

.77p f +.81c

+

y y g

gy t

Summary:

The following relations indicate the nominal shearing stress corresponding to flexure failure:

.g = 1.540,f. + 1.62c (double curvature) s g

,y (singic curvature)

'b

= 1.4

^

.g Hy vy o

t

26 APPENDIX B hfortared Double Wythe Wall Calculations Design Check of Wall No.46 (Auxiliary Building)

Assumed Data t

= 15.625" f ' = 2000 psi (light weight block) m 6

E

= 1000 f' = 2.0(10 ) psi m

m y

= 138 lbs/cf 2

g

= 386.4 in/sec (1)

Full Composite Action

(

625)

I=

j

= 3815 in Ends pinned 7.12 / EIg g,

2L /5 ) AY 6

~.12 2(10 1386.4(1728)(3815) 2(16x12) r'5 s

(12)(15.625)(138) 19.2 cps

=

Ends fixed f = 19.2/5 = 42.9 cps From Fig. 15 - Floor Response Spectrum Curves (Auxiliary Building) at E1.61' Spectral Acc. = 0.44 for either of above frequencies Bending load =.44(138)(15.625)

.,9.06 ps-

=

For cinned ends 2

"' 2

.\\1

= 2530(12) = 30360""

max 8

51 30360f6) s ensile Stress

- =

,- = 62 psi 12(15.625)~

27 APPENDIX B:

(Cont'd)

Tensile Bond Strength = 0.6(74) = 44 psi

==

Conclusion:==

Cracking Occurs For Fixed Ends 2

WE y' max 12 s

62(8) = 41.3 psi < 44 psi tensile stress

=

12

==

Conclusion:==

No Cracking Occurs Final Conclusion For a 4 level wall with lateral load in the same direction in each story and assuming a pinned base the maximum moment (from moment distribution) indicates ul' giving a tensile stress of 49.6 psi > 44 psi M

max 10 Thus cracking will be assumed irrespective of end support conditions.

(2)

Full Composite Action With Cracking E s t = 15.625" modular ratio = p - = 15 m

d = 11.8125" uncracked depth = c Determine Uncracked Depth 6c' = 15 A (11.8125-c) s Note:

Assuming #6 G24" in each wythe A = 0.22 in~/ ft s

thus c=2.2S9 in 4

12 (~' ' ~'8 91

+ 15 (. 22) (11. S125-2. 259)

= 547 in I

=

cr a

5S15 in I

=

S

28 APPENDIX B: (Cont'd)

Determine I e 1

4 Eq. (35)

I = -(I +Ier) = 2081 in e

g I

=I

= 347 in e

er 4

Eq. (36)

I = 0.2 I = 763 in e

g Determine Frecuency, f For each of the three values of I, given above the following frequency and acceleration values are obtained at Elev. 61' Pinned Ends Fixed Ends I =2081 in f=16.65 cps, acc.=.6 cps f= 37. 2 cps, acc. =.44 cps I,=

347 in f= 6.8 cps, ace.=1.2 cps f=15.2 cps, acc.=.68 cps I,=

763 in f=10.1 eps, ace.=1.4 cps f=22.6 cps, acc.=.44 cps Determine Loading u The above acceleration values give the following inertia loadings Pinned Ends Fixed Ends a=

108 lbs/ft 79 lbs/ft x =

u=

216 lbs/ft u=

122 lbs/ft u= 252 lbs/ft u=

79 lbs/ft Determine Shear Bond Stress on Collar Joint For the pinned end condition, it will be assumed that the wall section is uncracked at the supports since the moment at the supports is zero.

For an uncracked rectangular section v = b3b = 5bb Ib 2A in which V = wi/2 A = 12(15.623) in

29 APPENDIX B:

(Cont'd)

The coller joints shear stresses are computed as follows:

u= 108 lbs/ft v=

6.9 psi u = 216 lbs/ft v = 13.8 psi u = 252 lbs/ft v = 16.1 psi For the fixed end condition, it will be assumed that the wall section is cracked at the supports since the maximum moment occurs at the ends. The shear stress across the collar joint is difficult to calculate at a crack-ed section.

Using the ACI Code approach of a nominal average stress v

V/bd gives o=

79 lbs/ft v = 4.46 psi o = 122 lbs/ft v = 6.88 psi e=

79 lbs/ft v = 4.46 psi Conclusion The effective moment of inertia used by Bechtel is too large anc thus the collar jcint shear bond stress is between 13.8 psi and 16.1 psi for the pinned end case.

Corresponding values for fixed ends are 4.5 psi to 6.9 psi.

Also the wall ends range between pinned and fixed. Using average values for the pinned and fixed cases indicates around 10 psi stress on the collar joint irrespective of whether I =I or 0.2 I.

Since e er g

the capacity of the joint is estimated to be on 12 psi and interstory drift and bending from pipe support loads have not been considered in the computation of the collar joint stress, it is concluded that these effe: s added to the inertia loading will cause separatien of t he wyt he s.

30 APPENDIX B:

(Cont 'd)

(3) Noncomposite Action of a Single Wvthe As a result of the separation of the two concrete masonry wythes, the following calculations are based on the response of a single 7-5/8" thick masonry wythe.

Determine Inertia Loading on Single Wythe

, 12(7.625)

= 443.3 in#

7 g 12 f(pinned ends) = 9.35 cps acc. = 1.375 f(fixed ends) = 20.91 cps acc. = 0.44 Thus w (pinned ends)= 1.

5( 8)(7.625) = 120.6 psf (fixed tnds) = 0.44(1 (7.625) 3S.6 psf w

=

Using the smaller 1 cad of 38.6 psf, the end moment is:

M = f0 = 988.16 ft/lbs 988.16(12)6 tensile stress =

= 102 psi > 44 psi 12(7.625)'

Thus even assuming fixed end conditions, cracking will occur.

Since the single wythe will also be cracked under its inertia loading and the wall effective moment of inertia will be greatly reduced, it is considered reasonable to assume pinned supports in the re: aining calculations.

31 APPENDIX B:

(Cont'd)

Determine Cracked Deoth of Wall (c) 6c" = nA (d-c) 3 15 (. 22) (3. 8123-c)

=

from which c = 1.2" che,ck ductility T = 20(.22) = 4.4 kips 2000 2(4400) = 611 ps 2T f

t<

=

=

m 12(1.2) 12(1.2) 3 OK cracked I = 12(1)# + 15(.22)(3.8125-1.2)~

~

7

= 29.43 in#

Use 4

= 29.43 in or I effective

= 0.2 I

0. 2(443. 3) = 88.7 in effective S

The corresponding frequencies are 2.4 cps and 4.18 cps, respectively.

From the response spectrum curves for the auxiliary building in the N-S direction with 3'e damping,

the following spectral ace'.lerations are obtained:

f = 2.4 cps f = 4.18 cp1 El 61' O.66 0.92 E1 --'

1.25 1.69 El 93' 1.10 1.63 (Note: Since the frequencies fall below the maximum response range,using the smaller I value gives Icwer accelerations),

e

32 APPENDIX B:

(Cont'd)

Using the lower frequency of 2.4 cps, the wc11 loads and moments may be obtainer!

Load Moment (e2 /8)

El. 61'

.66(138)(7.625)/12= 57.9 psf M=1852.8 ft.lbs/ft.

El. 77' 1.25(138)(7.625)/12=109.6 psf M=3507.2 ft. lbs/ft.

El. 93' 1.10(138)(7.625)/12= 96.5 ps f M=3088 ft. lbs/ f t.

(Note: Wall obviously remain cracked, so that the pinned end assump-tion is considered satisfactory)

Compute wall ultimate moment capacity using a straight line, stress dis-bribution and f = 1.5 f' (Ref. ( 3))

f = 50 ksi y

Steel force at ultimate = 50(.22) = 11 kips Depth of compression block at failure = c Thus f (1.5)(2)(12)c = 11000 los or c = 0.61 ins.

Thus M = 0.9 { 50(. 22)(3. 8125 '

)

0}

u

= 2978 ft.lbs/ft.

Check steel strain at ultimate (3.8125.6101 (3000)

.00786 in/in

=

.610 D

2r10 )

.001724 Steel yield strain

=

29 0 Steel stri.n at ultimate = 4.56 x yield strain

33 APPEND 1X B:

(Cont'd)

Conclusion Walls above El. 77 will fail in flexure due to inertia loading.

Checking High Strength Block Assuming f =4000 psi reduces the modular ratio from 15 to 10.

A summary of calculations for high strength block is given below.

(1)

Full Comnosite Action Calculations indicate that the spectral acceleration will be 0.44 cps for either pinned on fixed ends.

Thus cracking will ocer-even with high strength block.

(2)

Full Composite Action with Cracking The uncracked depth = 1.906" 12(1.90613 4

=

+ 10(. 22) (11. 8125-1. 906) ~ = 24 3.6 in I er a

Loadings at Elev. 61'become Pinned ends Fixed ends 243.6 in w= :19 lbs/ft u= 123,6 lbs/ft I

=

e I = 763 in u=

.80 lbs/ft e=

79 lbs/ft t

Collar joint shear stresses at El. 61 Pirned ends Fixed ends I = 243 in 1o.0 psi 6.97 psi I

= 763 in 1.3 osi 4.46 psi These values indica e a shear bond stress due to inertia load-ing of around S psi to 10 psi. Thus it is concluded that adding interst.ory drift and pipe loading to these walls will cause wythe separation.

34 APPENDIX B:

(Cont'd)

(3) Noncomposite Single Wythe Action Uncracked depth, c=1.013" I

= 21.4 in Cr I

=443.3 in#

S Using I

=I gives the following moments at the various e

er elevations El. 61' f= 2. 9, ace.=0.69 M=1936 ft lbs/ft El. 77' f= 2. 9, acc.=1'56 M=4377 ft.lbs/ft El. 93' f= 2. 9, acc.=1.23 M=3451 ft.lbs/ft The wall moment capacity = 3061 ft.lbs/ft Thus the walls above El. 77' will fail in flexure Ccnsider Yield Line Theory for a Typical Wall Panel x

m Floor Line

,/

A 16' i

I B

l 31' From page 32 Mu = 2978 ft.lbs/ft Assume no moment restraint along boundaries but Muf +)

• M (-) = 2978 f*. lbs/ ft u

35 APPENDIX B:

(Cont'd)

Let x = 10' Segment A 16(2978) = h10)I10) "A I

1. "179 Psf A

3 Segment B 31(2978) = 11(8)(4) WB + 10(8) 3 W A

= 163 psf B

B Let x = 10.25 Segment A 16(2978) =

(10 25)* g A

W =170 psf A

A Segment B 31(2978) = 10.5(8)(4) WB + 10.25(8)IW A

B = 166 psf B

Thus Wf = i.68 psf Assuming a 10'6 reduction for corner effect f =.9(168) = 150 psf W

Maximum load on wall = 109.6 psf Thus 150 Factor Safety = 109.6 Conclusion If the masonry wythes are tied to the vertical columns so as to provide a simple support condition, the walls will resist the inertia loading due to seismic conditions without collapse.

However, using I,=0.2 Ig, the load at E1.77' is 148.2 psf which is essentially equal to the estimated failure load of 150 psf.

s 36 APPENDIX B:

(Cont'd)

It is concluded that 2 way wall action will not provide suf-ficient capacity to prevent failure unless negative moment capacity can be developed along the wall boundaries.

37 APPENDIX C References 1.

Hidalgo, P.A., Mayes, R.L., McNiven, H.D., and Clough, R.W., " Cyclic Loading Tests of Masonry Single Piers - Volume 3 - Height to Width Ratio of 0.5," Report No. UCB/EERC-79/12, College of Engineering, Univ. of California, Berkeley, California, hby 1979.

2.

" Appendix A-Shear Wall Specimen Testing Program," submitted to NRC, DOR, by Bechtel Corp.

3.

Yokel, F.Y.,

and Dikkers, R.D., " Strength of Load-Bearing Masonry Walls," Journal of the Structural Division, ASCE, Vol.97, No.STS, Proc. Paper 8143, iby 1971, pp. 1593-1609.

4 Manual of Steel Construction, 7th " Edition, American Institute of Steel Construction, Inc., New York, N.Y.

5.

Building Code Requirements for Reinforced Concrete (AC1 318-77),

American Concrete Institute, Detroit, Michigan.

6.

Hat:inikolas, M., Longworth, J., and Warwaruk, J., " Evaluation of Tensile Bond and Shear Bond of thsonry by Means of Centrifugal Force,"

Alberta Masonry Institute, Edmonton, Alberta.

7 Hegemier, G. A., et al., "On the Behavior of Joints in Concrete Masonry,"

Proceedings of the North American bbsonry Conference, Univ. of Colorado, Bouder, Colorado, August, 1978.

S.

Results of Pullout Tests on Mortared Collar Joints, submitted to NRC, COR, by Bechtel Corp.

9.

" Compressive and Transverse Strength Tests of Eight-Inch Brick Walls,"

Research Report No.10, Structural Clay Products Research Foundation, Geneva, Illinois, October, 1966.

10.

Fattal, S.G., and Cattaneo, L.E., " Structural Performance of Masonry Walls Under Compression and Flexure," Center for Building Technology, NBS, Washington, D.C., June 1976.