ML20085E981
| ML20085E981 | |
| Person / Time | |
|---|---|
| Site: | Arkansas Nuclear  |
| Issue date: | 01/20/1992 |
| From: | Dib T ENTERGY OPERATIONS, INC. |
| To: | |
| Shared Package | |
| ML20085E940 | List: |
| References | |
| 91-E-0094-03, 91-E-0094-03-R00, 91-E-94-3, 91-E-94-3-R, NUDOCS 9506190144 | |
| Download: ML20085E981 (54) | |
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l ATTACHMENT 2
_T_Q ICAN039402 ENGINEERING CALCULATION 91-E-0094-03 1
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Sape.l ARKANSAS NUCLEAR ONE
i TABLE OF CONTENTS l 1
Page Table of Contents 1 Purpose 2 Design Approach 2-3 References 3 l l
Notes and Assumptions 4 Calculation 5 - 32 Conclusion 33 i
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i 4'- opuscons eni, CALCULATION NUMBER 91-E-0094-03 ARKANSAS 0 ///##/9/ M #ed NUCLEAR PAGE REV. DATE BY CHK'D ONE NUMBER / OF 33
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PURPOSE:
To evaluate and perform shipping cask drop analysis for the concrete floor area above the ANO-1 Control Room, with and without hexagonal honeycomb, using the following data:
- Shipping cask diameter = D = 70"
- Shipping cask weight = 100 Ton, use 200 kips
- maximum drop = 9."
DESIGN APPROACH Using Bechtel Topical Report BC-TOP-9A, Rev. 2, an i analysis will be performed utilizing Kinetic Energy Formulas to determine the true Kinetic Energy that the slab can absorb. Steps in this analysis are:
- 1) Establish four different loading cases taken for this analysis, page 5 to page 6.
- 2) Determine the combined moment of inertia for the structure, page 7 to page 8.
- 3) Calculate the maximum moment Mu capacity of the slab for all four (4) cases, page 9.
- 4) Calculate the ultimate load (Pu) based on Mu for each of the four cases, page 10 to page 15.
- 5) Calculate the deflection (Du) based on Pu for each of the four cases, page 10 to page 15.
- 6) Calculate the ultimate shear capacity of the slab, Vmax. based on concrete / reinforcement strength, page 16 to page 17.
- 7) Calculate the maximum load Ps to cause Vmax. for each of the four (4) cases, page 17 to page 18.
- 8) Calculate the deflection due to Ps and compare with deflections due to Pu and dead weight of slab, for each of the four (4) cases. page 19.
- 9) Check bond stress at the face of the concrete wall, page 20.
i
- 10) Plot the deflection VR load curve to calculate the kinetic energy for each of the four (4) cases, page 21 to page 24.
CALCULATION NUMBER AsEs
" op 91-E-0094-03 r ARKANSAS l 6 ///M'/41 d AC# NUCLEAR PAGE gV l DATE BY CHK'D ONE NUMBER I OF U l
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. DESIGN APPROACH (cont.)
- 11) Calculation of maximum drop height of cask that the slab can take based on equivalency of strain energy to kinetic energy, page 25 to page 26.
- 12) Calculation of the thickness of energy absorption material required for a 9" cask drop, page 27.
- 13) Calculation of the safety factors using 3" energy
.' absorption material with 9" cask drop, page 28 to page 32.
REFERENCES:
- 1) ACI 318-77
- 2) ACI 318-63
- 3) Topical Report BC-TOP-9A Rev. 2
- 4) Civil Calculation No. 88 Book 26, Pg. J23-J26 &
J70-J78
- 5) Civil Calculation No. 11406-130
- 6) Reinforced Concrete Design by Wang & Salmon, Third Ed
- 7) Drawing No. C-206 Rev. 16
- 8) Drawing No. C-212 Rev. 9
- 9) AISC, Eighth Edition
- 13) Design of Concrete Structures by Winter & Nilson
- 11) Calculation NO. 83-D-2200-11 Pg. 124 of 555
- 12) Specification 6600-C-302
$ent,y CALCULATION NUMBER Operations 91-E-0094-03 ARKANSAS O N/24/9/ 799 Acd NUCLEAR PAGE BY CHK'D ONE NUMBER 2 OF 33 REV. DATE
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NOTES AND ASSUMPTIONS:
! 1) f'c = 4850 psi 0 28 days (Pour No. 517) (Ref. no. 11) f'c = 6230 psi 0 90 days (Pour No. 517) (Ref. no. 11) f'c = 6230 X 1.1 = 6850 psi 9 4 years (Ref. 10)
= 6500 psi 9 4 years (Conservative)
Use Fy = f'E8300 psi per mill test report (Ref. no. 5)
- 2) This calculation is to qualify the slabs between Column Lines B, C, and the north face of the Cask Pit which is located 3 feet north of Column Line D, and Column Lines 4 and 5. The span length between B and C is 27 feet. The span between C and'the north face of the Cask Pit is 24 feet. Four cases were considered in this calculation:
Case # 1 : Span length = 24 feet and the cask drop location is at a distance equal to the effected depth of the slab from the face of the wall.
Case # 2 : Span length = 27 feet and the cask drop location is at a distance equal to the effected depth of the slab from the face of the wall.
Case # 3 : Span length = 24 feet and the cask drop location is at Mid Span.
Case # 4 : Span length = 27 feet and the cask drop location is at Mid Span.
These four (4) loading cases will provide worst condition stress in the slab for each load drop.
CALCULATION NUMBER
$w%s
~- opw 91-E-0094-03 ARKANSAS O #/24/4 / 727 Rcn NUCLEAR PAGE REY. DATE BY CHK'D ONE NUMBER 4 OF 33
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CASE i 1 & 2 X = \ ( DIA. OF CASK) = \ (70") = 35" = 28-11" (Ref. Pg. 2) t = Slab Thickness = 3'-6" = 42" (Ref. # 7 & 8) d = Effective Depth = t - 4" = 42" - 4" = 38" (Ref. # 7 & 8)
Reinforcing Rebars:
This slab has Top and Bottom reinforcing rebars. To be conservative, ignore the Top rebars. Bottom rebars are 3 # 11 per foot.
=
(Ref. # 7 & 8)
- 5 stirrups @ 3'-10" in E-W (Ref. # 7 & 8)
- 5 stirrups @ l'-0" in N-S (Ref. # 7 & 8)
For this case, the critical location for moment and shear is @ a distance d from the face of the wall and the load is located at a distance (d') from the center of the wall. d' = X+d+ wall thickness d'= 2'-11" + 3'-2" + 18-6" = 7'-7" Used distance d' = 7d-6" = 7.5' from the center of the wall.
L = 24' for Case # 1 L = 27' for Case # 2 9 ans, CALCULATION NUMBER
~ operations ARKANSAS
@l-E-OGAN-03 o v/z/,/ 4/ -rd RcN NUCLEAR PAGE REY. DATE BY CHK'D ONE NUMBER .5' or 33
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CASE # 3 & 4 X = \ ( DIA. OF CASK) = (70") = 35" = 2'-11" (Ref. Pg. 2) t = Slab Thickness = 3'-6" = 42" (Ref. # 7 & 8) d = Effective Depth = t - 4" = 42" - 4" = 38" (Ref. # 7 & 8)
Reinforcing Rebars:
This slab has Top and Bottom reinforcing rebars. To be conservative, ignore the Top rebars. Bottom rebars are 3 # 11 per foot. 1 (Ref. # 7 & 8) l
- 5 stirrups @ 3'-10" .in E-W (Ref. # 'i & 8)
- 5 stirrups @ l'-0" in N-S (Ref. # 7 & 8)
For this case, the critical location for moment is-@ mid span and the critical location for shear is 9 a distance d from the face of the wall or at a distance (d') from the center of the wall.
d' = d + \ wall thickness d'= 3'-2" + 1'-6" = 4'-8" (Ref. # 1, Sec. 11.1.3.1)
L = 24' for Case # 3 L = 27' for Case # 4 CALCULATION NUMBER
$~~eniEs ope ARKANSAS 9 l- s - o09+- 01 O ///24/f/ '1BY
/ RcN NUCLEAR PAGE f REY. DATE CHK'D ONE NUMBER // 07 A7 l
CALCULATION:
Calculate the beam properties:
I. b = Effective width of the beam b = T + 2d where T = Diameter of the cask = 70" = 5'-10" d = Distance from extreme compression fiber to the centroid of the tension reinforcements d = 3'-2" (see page 5 - 6 ) I Therefore, b = (5'-10) + 2 (3'-2") = 12'-2" = 12.167 feet
= 146 inches Using Topical Report BC-TOP-9A (Ref. # 3, Formula 4-2)
I, = (I, + I, )
bt3 I,=\( + Fbd3 )
12 i
where:
I,. = Average Moment of Inertia I, = Moment of Inertia of Gross Concrete I, = Moment of Inertia of Cracked Concrete t = Thickness of the slab beam = 42" = 3.5' F = Coefficient for Moment of Inertia of Cracked section with tension reinforcements r=
bd where:
A, = Area of Tensile reinforcing steel per foot
=3 # 11 bars = 3
- 1.56 = 4.68 in8 4 .68 in8 .
P= = .010263 12"
- 38" 4 an, CALCULATION NUMBER ope br s 91-E-0094-03 ARKANSAS O ///F4/11 TA Red NUCLEAR PAGE REY. DATE BY CIIK'D ONE NUMBER 7 OF M
1 . e i -
CALCULATION: (cont.)
E , = w ,L5
- 33 (f',)% (Ref # 1, 8.5.1) where:
3 w = 150 lb/ft (Ref # 5) f', = 6500 psi (see pg. 3)
E, = (150)1.5
- 33 (6500)* = 4,88,7,733 psi ,
use, E, = 4888 Ksi E, = 29,000,000 psi for steel (Ref # 5)
E, 29,000,000 psi n= = =6 E, 4,887,733 psi From FIG 4-1 of BC-TOP-9A (Ref # 3)
(n = 6, F = .010263) ---> F = .045~
Therefore, 146 * (42)3 I, = ( + .045
- 146
- 383 )
12 I, = \ ( 901404 + 360509 )
I, = 63 0956 in' Calculate the maximum moment capacity of the slab (M o) using tension reinforcement only:
M, = $ [ A,
- f y * (d- a))
where:
$ = 0.9 (Ref # 1)
A, = Area of tension steel = 4.68 in' fy = 48.3 Ksi A,
- f y
. 85 f ',
- b a
4.68 in8/ft
- 12.167 ft
- 48.3 Ksi
=
= 3.41" u
.85
- 6.5 Ksi
- 12.167 ft
- 12 in 1
CALCULATION NUMBER
&"oaMs 91-E-0094-03 ARKANSAS O ///,24/f/ '76 kc # NUCLEAR PAGE REV. DATE BY CIIK'D ONE NUMBER 8 OF N l
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, CALCULATION: (cont.)
M, = 0 . 9 [4.68 in'/ft *12.167 ft *48.3 Ksi *(38"-\ 3.41)]*1'/12" M,= 7486.5 Kip-ft Using beam formulas, calculate the maximum load P that the slab can take, utilizing the. maximum moment of the beam,M,.
Consider a sin 1 supported beam, find the maximum P in addition to the dead we g t D.W. of the beam in all four case,s addressed on I page 4 of this calculation. These calculations are on page 10 l thru page 15. Also, deflections O point of load due to the maximum I load p, (ultimate) were calculated. These deflections will be used later in this calculation to find the Kinetic Energy that this I. slab could absorbe in each of the four cases.
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i CALCULATION NUMBER
$~~e,a opw% s 91-E-0094-03 ARKANSAS O N/34/T/ 7"23 Rcr/ NUCLEAR PAGE N
REY. DATE BY CIIK'D ONE NUMBER OF
.I.-
CALCULATION: (cont.)
For Case # 1 & 2, calculate the maximum load P in addition to D.W. of the beam, O distance d from the face oY ,the wall.
Loadina # 1 P,
- a
- c
.Mu = (Ref # 9, Pg. 2-116) f M,
- L
- _> p -
a*c Loadinc 4 2 w, = weight of concrete beam w,= b
- H
- 0.15 (use wt. of concrete = .15 k/ft )3 w, = 12.167 f t
- 3. 5 f t
- 0.15 k/ f t 3= 6.388 Kip /ft P = Equivalent concentrated load which has the same impact on the 9 beam as the D.W.
w,
- a a*c M, = ( L - a) = P,q (Ref # 9 Pg 2-114) 2 L M,
- L w,
- a L Pg ( L - a) ( )
a*c 2 a*c P, = P3 - P Mu *L w,
- a L P, = -
( L - a) ( )
a*c 2 a*c .
L Pu *
,, [Mu "" IWo *a* (L - a)) ------EQ. #1 CALCULATION NUMBER
&~~ eni op M ons 91-E-0094-03 ARKANSAS O ///J.J/4f/ % Red NUCLEAR PAGE REV. DATE BY CHK'D ONE NUMBER /8 OF M
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CALCULATION: (cont.)
CASE # 1 ( Ref. page 4)
L = 24 ft a = 16.5 ft = (L-d') = 24'- 7.5' = 16.5 ft c = d' = 7.5' I l
Using EQ. # 1 (PG. 10) l l
L P, =
[M u- \ w,
- a * (L - a))
24' Pu =
,, ,7486.5
[ K-ft - 6.388 K/ft
- 16.5'* (24'-16.5')]
Pu = 1452 Kips - 76.6 Kips = 1375 Kips Pu = 1375 Kips
- Calculate the deflection Du due to Pu and D.W.
Py a8 c8 Do = + D, (Ref # 9 3 E, I, L ,
where D, = deflection due D.W., Py = 76.6 Kips i .
w *c D, = (L3- 2Lc' +c3 ) ---EQ # 2 (ref #9,Pg 2-114) 24 E,I, 6.388 K/ft
- 7.5 ft 3 D, =
24
- 4 888 Ksi
- 630956 in'[24 -2 (24 ') (7. 5 ') 8 +7. 5 '3) (12"/ f D, = . 012 9 in 1375 K (7.5' X 16.5')2 (12"/ft)3 + 0.0129" = .177" D" =
3
- 4888 Ksi
- 630956 in'
- 24 '
Do = 0.177 in CALCULATION NUMBER I
& ani %rs
" ope 91-E-0094-03 ARKANSAS O ///Sk/f/ '727 Acd NUCLE /.R PAGE REV. DATE BY CHK'D ONE NUMBER // OF I
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CALCULATION: (cont.)
CASE # 2 ( Ref. page 4) 1 L = 27 ft j a = 19.5 ft = (L-d') = 27'- 7.I'= 19.5 ft c = d' = 7.5' Using EQ. #1 (PG. 10)
- L Pu = [ M, - ), w,
- a * (L - a)]
l a*c 27 P, =
, ,7486.5
[ K-ft - % 6.388 K/ft
- 19.5'* (27*- 19.5'))
Pu = 1382.12 Kips - 86.2 Kips = 1296 Kips P, = 1296 Kips
- Calculate the deflection D, due to P and u 0.W.
P, a' c' D, = + D, (Ref # 9) 3 E, I, L where D, = deflection due D.W. Py = 86.2 Kips w *c D, = (L3- 2Lc' + c 3) (ref #9,Pg 2-114) 24 E,I, 6.388 K/ft
- 7.5 ft )
D, =
24
- 4888 Ksi
- 630956 in'[273 -2 (27 ') (7. 5 ' ) 8 +7. 5 '3] (12"/f t) 3 D, = . 019 in 1296 K (7.5'X 19.5')8 (12"/ft)3 D" = + 0.019" = .211" 3
- 4888 Ksi
- 630956 in'
- 27 '
D, = 0. 211 in CALCULATION NUMBER
$~- emes op 91-E-0094-03 ARKANSAS 6 N/24/4/ T# Rc # NUCLEAR PAGE REY. DATE BY CHK'D ONE NUMBER /1 OF N
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- l CALCULATION: (cont.) !
For Case # 3 & 4, calculate the maximum load P, , in addition to D.W. of the beam, e mid span of the beam. ,
I Loadina # 1 t
P,
- L Mu = (Ref # 9, Pg. 2-116) 4 M,
- 4
> P 3 =
L Loadina # 2 w, = weight of concrete beam ,
3 w, = b
- H
- 0.15 (use wt. of concrete = .15 k/ft )
w,= 12.167 ft
- 3.5 ft
- O.15 k/ft3 = 6.388 Kip /ft P = Equivalent concentrated load which has the same impact on the beam as the D.W.
w,
- L2 L M= =P y (Ref # 9 Pg 2-114) 8 4 4 w'
- L2 p" =
8L P, =P i -P y 2
M u *4 4W o *L P, = -
L 8L 2
4 w *L P, = [ M, - ] ------EQ. #3 i
L 8 I & CALCULATION NUMBER
" h%s 91-E-0094-03 ARKANSAS i O N/s4/9/ 2:Zi Rc" NUCLEAR PAGE REV. DATE BY CHK'D ONE NUMBER /3 OF 33 l
CALCULATION: (cont.)
CASE # 3 ( Ref. page 4)
L = 24 ft Using EQ. # 3 (PG. 13) 4 w,
- L2 P, =
L
[4- 8
]
4 P, =
[ 7486.5 K-ft - 1/8 ( 6.388 K/ft
- 24 2 ))
Pu = 1247.77 Kips - 76.6 Kips = 1171.17 Kips P, = 1171.17 Kips
- Calculate the deflection D, due to P and u D.W.
Po L 3 D, = + D, (Ref # 9) 48 E,I, where D, = deflection due D.W.
5 w, L' D, = (ref #9,Pg 2-114) 38'4 E,I,
=
5
- 6.388 K/ft * (24 ft)' * (12"/ f t) 3 = .0154" D
384
- 4888 Ksi
- 630956 in' D, = 0. 0154 in i
1171.17 K (24')3 (12"/ f t) 3
+ .0154" = 0.204" D" =
48
- 4888 Ksi
- 630956 in' i
Du = 0.204 in 1
$ ant ,,y CALCULATION NUMBER opwations
- 91-E-0094-03 ARKANSAS 0 ///24/ f/ 74 RC /4 NUCLEAR PAGE
/Y REY. DATE BY CHK'D ONE NUMBER OF IT
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CALCULATION: (cont.)
CASE #4 ( Ref. page 4)
L = 27 ft l Using EQ. #3 (PG. 13) 2 4 w *L P, =
L
[4- 8
)
4 P, =
[ 7486.5 K-ft - 1/8 ( 6.388 K/ft
- 27'2 )) '
27' P, = 1109.1 Kips - 86.2 Kips = 1022.9 Kips P, = 1022.9 Kips
- Calculate the deflection Du due to Pu and D.W.
P, L3 Do = + D, (Ref # 9) 48 E,I, where D, = deflection due D.W.
5 w, L' D, = (ref #9,Pg 2-114) 384 E,I, 5
- 6.388 K/ft * (27 f t)' * (12"/ft)3 = .0246" D = 4 384
- 4888 Ksi
- 630956 in' )
l D, = .0246 in 1022.9 K (27')3 (12"/ft)3 + 0.0246" = .260" D" =
48
- 4888 Ksi
- 630956 in' Du = 0.260 in CALCULATION NUMBER "oaMs 91-E-0094-03 ARKANSAS 0 ///24/ 4 / 'Ta Pr Al NUCLEAR PAGE I D, 33 REY. DATE BY CHK'D ONE NUMBER OF
CALCULATION: (cont.)
Calculate the maximum shear capacity of the slab l II. Maximum shear Capacity:
a) Shear carried by concrete: j Vd V,=$ (1.9 (f ',)% + 2500 r ) bd (Ref # 1 Eq. 11-6) ,
m l Vd where =1 ( Ref # 1, Section 11) m
$ = 0.85 for shear ( Ref # 1, Section 11)
V, = 0 . 8 5 ( 1.9 (6500 psi)* + 2500 * . 010263) (146"
- 38")
V, = 84 3,37 5 lb < 3. 5 * (f',)% bd
< 3.5 * (6500 psi)%
- 146"
- 38" = 565,529 #
Therefore, V, = 8 4 3 . 4 Kips b) Shear carried by web reinforcement:
- 5 stirrup 9 3'-10" in the E-W = 46" (Ref # 7 & 8)
- 5 stirrup 9 l'-0" in the N-S = 12" (Ref # 7 & 8)
Ay = 0. 31 in' (consider one leg)
A y f, V ', = $ bd (Ref # 6 , 5.10.7) b, s 0.31 in8 48300 psi V',= .85
- 146"
- 38" = 127,916 lb V', = 128 Kips V,,, = V, + V ' ,
V,,, = 84 3. 4 kips + 128 kips = 971. 4 Kips CALCULATION NUMBER "oanMs 91-E-0094-03 ARKANSAS O /M24/4 / M A e ^/ NUCLEAR PAGE REY. DATE BY CHK'D ONE NUMBER /d OF 87 >
d P
. CALCULATION: (cont.)
For all four cases, the maximum shear will occur at a distance "d" from the face of the wall ( see page 5 - 6, Ref.# 1 Sec. 11.1.3.1)
Calculate the maximum load "P'" in addition to the D.W. of the slab, that will cause a shear of 971.4 Kips @ a distance "d" from the face of the wall ( d" = 1'-6" + 38") = 56" =4.7 feet )
1 Case # 1 & 2 V,,,, = Shear due to P, + Shear due to D.W.. w, V,,,, = 971.4 Kips (see Page 16)
Shear due to P, =
Pg
- a/L (Ref. # 9, 2-116)
Shear due to D.W. = %
- L
- w, - d '
- w, (Re f . # 9, 2-114) w,= 6.388 Kip /ft (see Page 10) I V, = P,
- a/L + %
- L
- 6.388 K/ft - dH
- 6.388 K/ft -EQ # 4 CASE #1 ( Ref. page 4)
L = 24 ft a = 16.5 ft = (L-d') = 24'- 7.5' = 16.5 ft using EQ # 4 - !
16.5' 971. 4 Kip = P, * ,
+
- 24'
- 6.388 K/ft - 4.7'* 6.388 K/ft
---> P, = 13 5 4 Kips CASE # 2 ( Ref. page 4)
L = 27 ft a = 21 ft = (L-d') = 27'- 7.5'= 19.5 ft using EQ # 4 19.5' 971.4 Kip = P, * + \
- 27'
- 6.388 K/ft - 4.7'* 6.388 K/ft
---> : P, = 12 67 . 2 Kips i
$ eni, CALCULATION NUMBER
~~ operatkms 91-E-0094-03 ARKANSAS O ///SG/f/ TD Rc.// NUCLEAR PAGE REY. DATE BY CHK'D ONE NUMBER /7 OF M ]
f' CALCULATION: (cont.)
Case # 3 & 4:
Calculate the shear at distance d' when the load is located at mid span of the beam.
V ,, = Shear due to T, + Shear due to D.W. w, V.,, = 971.4 Kips (see Page 16)
Shear due to P, = \ P, Shear due to D.W. = w, ( L - d')
V,,, = 971. 4 Kip = % P, + w, ( L - d') --- EQ # 5 CASE # 3 ( Ref. page 4)
L = 24 ft using EQ # 5 971. 4 Kip = \ P, + 6. 388 k/ f t (
- 24' - 4.7')
---> P, = 1849.5 Kips > P, due to Ultimate Moment = 1171.17 K CASE #4 ( Ref. page 4)
L = 27 ft using EQ # 5
< 971.4 Kip'= P, + 6. 3 88 k/ f t ( \
- 27' - 4.7')
---> P, = 18 3 0 Kips > P, due to Ultimate Moment = 1022 K 4
$ ento,,y CALCULATION NUMBER
~ operatx>ns 91-E-0094-03 ARKANSAS 9 ///M/9/ -TD RCM NUCLEAR PAGE REV. DATE BY CIIK'D ONE NUMBER 8 OF 33
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CALCULATION: (cont.)
Calculate deflections "D" in all four Cases Case # 1 : ( Ref. page 4)
For P, = 1375 Kips -----> D = .177 in (Ref Pg. 11) 1354 k P, = 13 54 Kips -----> D= * .177 = .174 in Pot = 76.6 Kips -----> D = .0129 in (Ref Pg. 11)
Case # 2 : ( Ref. page 4) l For P, = 129 6 Kips -----> D = 0.211 in (Ref Pg. 12) 1267 k P, = 12 67 Kips -----> D=
- 0.211 = 0.206 in Pot = 86.2 Kips -----> D = .019 in (Ref Pg. 12)
Case # 3 : ( Ref. page 4)
For P, = 1171. Kips -----> D = 0.204 in (Ref Pg. 14)
Pot = 76.6 Kips -----> D = 0.0154 in (Ref Pg. 14) i Case # 4 : ( Ref. page 4) !
For P, = 102 2 Kips -----> D= .260 in (Ref Pg. 15) l Po , = 86.2 Kips -----> D = .0246 in (Ref Pg. 15) i l
~
l l
l CALCULATION NUMBER
@ ent brs
~ ope 91-E-0094-03 i ARKANSAS I O ///S/f/ @ Rca NUCLEAR PAGE REY. DATE BY CIIK'D ONE NUMBER / 4 OF33
CALCULATION: (cont.)
Check Bond Stress @ the Face Of Column: l V 9.5 ( f ',)
- f V, = $ $ 800 psi (REf # 6, Sec. 6.3.2) ,
j (N r db)J d db where:
V = 971.4 + DW d = distance from the face of the wall N = No. of bars = 3 bars /ft
- b effective width of beam db = Diameter of bar J = 1 - K/3 (REf # 6)'
1 1
=
K=
1 +(f y)/nf', 1+ (48.3 ksi)/(6
- 6.5 ksi)
K = 0.44"
> J = 1 - 0.44/3 = .852 in
[971.4 Kip + 6.388 K/ft
- 3.2']
- 1000 lb/ kip V"
(3 # 1.41 in'
- 12.167')
- 0.852"
- 38" 9.5 (f',)% _ 9.5 (6500 psi)% = 543.2 psi V, = 189.5 psi $
db 1 41 ID*
I; 1
CALCULATION NUMBER
$ emes
~~
op 91-E-0094-03 ARKANSAS 0 gg4/ f/ rp Rc d NUCLEAR PAGE DATE BY CHK'D ONE NUMBER M OF M REV.
1 i .
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~ e,n op. % s 4)- E - #O94-#3 ARKANSAS 8 ///JE///T/ 73 Re N NUCLEAR PAGE REV. DATE BY CHK'D ONE NUMBER A1 or 33 l
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l
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CALCULATION NUMBER
$ eni%s opm 611- E - 00 %- o3 ARKANSAS 6 ///14/1/ # Red NUCLEAR PAGE M
REV. DATE BY CHK'D ONE NUMBER OF 33
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CALCULATION NUMBER ARKANSAS 6ll-6 -00 94 d] ,
C 11/74/61 -TD Rc 4 NUCLEAR PAGE i REV. DATE BY CHK'D ONE NUMBER 13 or 3 3 l l
CAS& $4 (d4 G G 9.)
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$ sni,5ons CALCULATION NUMBER operat ARKANSAS 91 o09+-r3
'O ///2h/4) -rd .
Re ^/ NUCLEAR PAGE REY. DATE BY CIIK'D ONE NUMBER M or S3
CALCULATION: (cont.) II - Allowable Drop Height: V,8 =2gh (Ref # 3, Eq. 3-8 ) where: V, = cask striking velocity' h = Maximum height 200 Kip M, = = Mass of Cask (Ref. pg. 2) 9 M, = Average Effective Mass oc*T M, = ( D, + 2T ) (Dy+ 2T )~(Ref # 3, Eq. 3-15) Since the cask is circle, Dia. = 70" = 5.833' (see Pg. 2) oc*T r (D + 2T)*
----> M* =
g 4 where oc = Weight per unit volume of concrete = .15 k/ft 3 T = Thickness of the floor
=
0.15
- 3.5 r (5.833 + 7.0)8 M*
g 4 67.9
=
M, 9 E, = Strain Energy = Z Kinetic Energy
= Shaded area Pg. 21-24 M,8
- V,8 E, =
(Ref # 3, Eq. 3-8 ) 2 (M, + M,) l CALCULATION NUMBER O~~ ant opw %ns : 91-E-0094-03 ARKANSAS 0 ###f/ 5# Rc N NUCLEAR PAGE REY. DATE BY CHK'D ONE NUMBER SI OF 3
. CALCULATION: (cont.)
{ Case # 1 : ( Ref, page 4) i (200/g)8 *2gh , E, = = \ (1354 K - 76.6 K) (.174" - .0129") 2 (200/g + 67.9/g )
; ------> 149.3 h = 102.9 h = 0.69 in Case # 2 : ( Ref. page 4)
(200/g)8 *2gh E, = = \ (1267. 2 K - 86. 2 K) ( . 206"- .019") 2 (200/g + 67.9/g)
------> 149.3 h = 110.4 h = 0.74 in Case # 3 : ( Ref. page 4)
(200/g)8 *2gh
~
2 (200/g + 67.9/g )
------> 149.3 h = 103.27 8 h = 0.69 in Case # 4 : ( Ref. page 4)
(200/g)8 *2gh E, = = (1022 K - 86.2 K) (.260" - .0246") 2 (200/g + 67.9/g )
------> 149.3 h = 110.14 h = 0.74 in The max. drop height without any energy absorption material allowed is .69". Therefore, Honeycomb Energy Obsorption Material is required for any drop.
i Ae s CALCULATION NUMBER l i 91-E-0094-03 ARKANSAS 0 ///74/4 ) 7D Red NUCLEAR PAGE DATE BY CHK'D ONE NUMBER A /S OF 32 REV. , 1 l
i . CALCULATION: (cont.) i III. Calculate the thickness of honeycomb material required for 9" drop Per Attachement # 1, Design Data for the Preliminary Selection of Honeycom' Energy Absorption System TSB122 K.E. = f er A (.7) t, (ref. Attach. # 1 pg. 13) where : f e,= Honeycomb Crush Strength = 260 psi t, = Honeycomb thickness A = Impact Area
----> t, = K . E . / (.7 fe ,A)
K.E. = (M, V,8 ) / 2 (ref. # 3, EQ 3-5) where: V,8 =2gh V, = Cask striking Velocity h = Maximam height 200 Kip a M, = = Mass of Cask 9 is* (200000 lb /g) *2*g*9"
------>t*=
260 psi (70") 8
- fr/4 * .7 t, = 2.6 in Therefore, 2.6" of 260 psi Hexcell Honeycomb Material is required for 9" Drop of 100 Ton Cask.
The rest of this. calculation is to calculate the safety i factors using 3" of 260 psi Hexcell Honeycomb Material at ! the bottom of the cask i 1 CALCULATION NUMBER
$ mms o
91-E-0094-03 ARKANSAS
# /// MAT / 78 BC /4 NUCLEAR PAGE REY. DATE BY CIIK'D ONE NUMBER l'/ OF M
CALCULATION: (cont.) 'i Summary of Stresses for 100 Ton Cask 9" Drop with 3" thick 260 psi Hexcell Honeycomb Material at the bottom of the cask K.E. =f,A e (.7) t, where : f,,= Honeycomb Crush Strength = 260 psi t, = Honeycomb thickness = 3" A = Impact Area = (70")** r/4
----> f e,= K.E. / ( . 7 t, A)
K.E. = (M, V,8 ) / 2 (ref. # 3, EQ 3-5) where. V,8 =2gh V, = cask striking Velocity h = Maximum height = 9" 200 K4 0 M, = -
= Mass of Cask C '1 K.E. = \ (200000 lb / g)
- 2
- g
- 9" K.E. = 1,800,000 K-in 1,800,000 K-in
----> f,=
e
= 222.7 psi < 260 psi OK (70") 8
- r/4 *- .7
- 3" P= 222.7 PSI
- A
= 222.7 psi * (70")8* r/4 = 857,050 lb P = 857 Kips
$ ento,,Yons CALCULATION NUMBER operat 91-E-0094-03 ARKANSAS
/) ///%4/ f/ 7D Rc N NUCLEAR PAGE REY. DATE BY CliK'D ONE NUMBER Pf OF 3 3
l1 CALCULATION: (cont.) 1
. CASE # 1 ( Ref. page 4) l L = 24 ft !
r a = 16.5 ft = (L-d') = 24'- 7.5' = 16.5 ft
- Maximum Shear Force and Safety Factor:
, Use Eq. 4 on page No. 17 V ,, = P
- a/L +
- L
- 6.388 K/ft - d"
- 6.388 K/ft 16.5' V,,, = 8 5 7 k *
+ \
- 24'
- 6. 3 8 8 K/ f t - 4. 7 '
- 6. 3 8 8 K/ f t V ,, = 63 6 Kips 971.4 Kips Safety Factor = = 1.53
- Maximum Moment and Safety Factor:
P*a*b M ,, = + \ 6.388 K/ft
- L'* (L'- a')
L (Ref # 9, Pg. 2-116) 857 k
- 16.5'* 7.5' N ,, = + \ 6.388 K/ft
- 16.5'* (24'-16.5')
M ,, = 4814.2 Kip-ft a 7486.5 Kip-ft Safety Factor = = 1.56 4814.2 Kip-ft
- Maximum Bond stress and Safety Factor:
V V = (N r d b)J d 857 Kip + (24'
- 6.388 K/ft) - 1.5'
- 6.388 K/ft V =
(3 x 1.41 in'
- 12.167')
- 0.852"
- 38" V = .177 ksi = 177 psi 543.2 psi Safety Factor = = 3.08 177 psi - - -
l
$ an,1s CALCULATION NUMBER op I 91-E-0094-03 ARKANSAS
# N/MN/ 7/) Nd NUCLEAR PAGE REY. DATE BY CIIK'D ONE NUMBER Ai OF T T
k'. l
', CALCULATION: (cont.)
EME#2 ( Ref. page 4) L = 27 ft a = 19.5 ft = (L-d') = 278- 7.5' = 19.5 ft j
- Maximum Shear Force and Safety Factor:
Use Eq. 4 on page No. 17 V,,, = P
- a/L + \
- L
- 6.388 K/ft - d'
- 6.388 K/ft 19.5' V,,, = 8 57 * + \
- 27' * 'i.388 K/ft - 4.7'* 6.388 K/ft 27' V,,,, = 67 5 Kips 971.4 Kips Safety Factor = = 1.43
- Maximum Moment and Safety Factor:
P*a*b M ,, = + \ 6.388 K/ft
- L'* (L'- a')
L (Ref # 9, Pg. 2-116) 857 k
- 19.5'*7.5' M ,, = ,
+ \ 6.388 K/ft
- 19.5'* (27'- 19.5')
M ,, = 5109 Kip-ft 7486.5 Kip-ft Safety Factor = = 1.47 5109 Kip-ft
- Maximum Bond stress and Safety Factor:
V V = (N r d b )J d 857 Kip + (27'
- 6.388 K/ft) - 1.5'
- 6.388 K/ft V =
(3 x 1.41 in'
- 12.167')
- O.852"
- 38" V = .178 ksi = 178 psi 543.2 psi Safety Factor = = 3.05 9 ent CALCULATION NUMBER
~ Oper ns 91-E-0094-03 ARKANSAS
# ///M/ il 'TD k e ^/ NUCLEAR PAGE l REY. DATE BY CHK'D ONE NUMBER 3 9 OF_33 l 4
CALCULATION: (cont.) CASE #3 ( Ref. page 4) L = 24 ft Loaded at mid span
- Maximum Shear Force and Safety Factor:
V,,, = P + \
- L
- 6.388 K/ft - d'
- 6.388 K/ft V,,, = \
- 8 57 kip + \
- 2 4 '
- 6.388 K/ft - 4.7'* 6.388 K/ft i
V,,,, = 47 5 Kips 971.4 Kips Safety Factor = = 2.04
- Maximum Moment and Safety Factor:
2 P*L w *L
+ (Ref # 9, Pg. 2-116) 4 8 857 kips
- 24'
+
6.388 k' * (24')8 M ,, = 4 8 M,,,, = 5 6 0 2 Kip-ft 7486.5 Kip-ft Safety Factor = = 1.34 5602 Kip-ft '
- Maximum Bond stress and Safety Factor:
V V = (N r d 3)J d
\* 857 Kip + \ (24'* 6.388 K/ft) - 1.5'
- 6.388 K/ft V=
(3 fr 1.41 in'
- 12.167')
- 0.852"
- 38" V = .095 ksi = 95 psi Safety Factor = = 5.72 95 psi l
@ enio,,y CALCULATION NUMBER Operat6ons 91-E-0094-03 ARKANSAS O ///24/41 77) Rc N NUCLEAR PAGE REV. DATE BY CIIK'D ONE NUMBER 8/ OF 33 -
'I
. CALCULATION: (cont.)
CASE # 4 ( Ref. page 4) L = 27 ft Loaded at mid span
- Maximum Shear Force and Safety Factor:
V,=%P + %
- L
- 6.388 K/ft - d'
- 6.388 K/ft V ,, = %
- 857 K + \
- 27 '
- 6. 388 K/ f t - 4 . 7 '
- 6. 3 88 K/ f t V ,, = 485 Kips 971.4 Kips Safety Factor = = 2.0
- Maximum Mome.nt and Safety Factor:
P*L w *L2
+ (Ref # 9, Pg. 2-116) 4 8 857 kips
- 27' 6.388 k' * (27')2 M ,, = +
4 8 M ,, = 63 66.7 Kip-f t t 5 7486.5 Kip-ft , Safety Factor = = 1.18
- Maximut Bond stress and Safety Factor:
V V = l i (N r db )J d
\* 857 Kip + \ (27'* 6.388 K/ft) - 1.5'
- 6.388 K/ft V=
(3 x 1.41 in'
- 12.167')
- 0.852"
- 38" 97 psi j V = .097 ksi =
Safety Factor = = 5.6 97 psi -- I
$ entorgy CALCULATION NUMBER Operatlons 91-E-0094-03 ARKANSAS r7 ///d/4 / TM Ac N NUCLEAR PAGE 31 OF M REV. DATE BY CIIK'D <,
ONE NUMBER
l CONCLUSION: Drop evaluation is performed in this calculation for all the four cases addressed on page 3. Case # 1 is found to be the mest critical case for the 100 Ton cask drop. Drcp evaluation summaries for the 100 Ton handling cask (70" diameter) at elevation 404 (Over the U-1 Control Room) are listed below:
~
A drop of 100 Ton is not allowed for any height without hexagonal ! honeycomb. l CONDITION-A - A drop evaluation for 9" travel height with 3" of hexagonal honeycomb. CONDITION-A GOVERNING CASE Shear Capacity 071.4 kips
= --- = 1.44 ; Case # 2 Shear Computed 675.0 kips Moment Capacity 7486.5 kip-ft
= 1.18 ; Case # 4 Moment Computed 6366 kip-ft Bond Stress Capacity 543.20 psi
= = 3.05 ; Case # 2 Bond Stress Computed 178 psi i
i
- The minimum thickness of a 260 psi hexagonal honeycomb required for the 100 Ton cask drop of 9 inches travel = 3".
- Spalling is designed not to occur because the analysis of the above cases that the slab stays in elastic range.
l i
$ en,%s CALCULATION NUMBER opw 91-E-0094-03 ARKANSAS d ///M/11 76 Red NUCLEAR PAGE REY. DATE BY CHK'D ONE NUMBER 37 or 73
- L 2-. l Design Data for the Preliminary. Selection ~of .
l l 3 . Honephsmb Energym.a.m. sjg ,.. VAbsorption S Mastwngdsggg
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j This bulletin presents a rapid method for designing honeycomb
- energy absorption systems. Use of the design curves will provide preliminary solutions for the bcsic requirements of G limit 5 protection.
HONEYCOMB ABSORBS ENERGY BY CRUSHING UNDER LOAD
.This characteristic of honeycomb has proven to be one of the most reliable and, efficient methods of providing "G !.!MIT protection. The action of crushing develops a uniform level of stress near the optimum response desired for energy absorption materials.
. ' pW *.f "r.Tg
?$ " '$ #. kNy
. .,~._,.
STANDARD CORE
.[ rip - Manufactured by both the corrugated and ex-
. 'n' * panded method, this configuration is widely I , ,l
; 1 .' it used for' general energy absorption problems.
ti ,t .. i # 4 ' 2 Aluminum, reinforced plastic, and paper are k* '. the more common materials used in making A -.-
,. 'p .
7}L. I these core materials.
/ .
4 P 4~ .V - M.U( }IQ{d ~ ~f,N, .
- h. . b .8 h :. - _ : $ ' ;;_-%
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j .',.,, " c,) $- f ..
.~
f-:,, i s
<t m Op .
, R4 s .
s s x' ~.[.
.+.
Designed in aluminum for efficient energy ab-
'iQ t I.
-...,s ; - - -
sorption where the spacing requires a thin-wall annular column or small diameter cylinder, ..'.}
~
4
\, Y[,1 TUBE-CORE eliminates the loss of crush strength j l . ,; .. , ; ' ~ "f
'.<7 /.
at the edges, an inherent characteristic of stand-ard core when used in small diameter cylinders. k'y 4/ [N (. - g r,M- k' ~ -: . ' \ , . * . j; ,
- .r . ~ .3. b ,,. s .
l
.? ., tiy
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?. ; .- .S '. ~.'y- .3 -
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- 73 . y r,3 __..- " CROSS. CORE 8
. t n- 1 :.: ,.2. :M. g. . .. .. .
' A$ -
2 - ' : ' -. ..;i::. 7 5 ,, Manufactured in Various SeleCled Cell oKis
#. orientations, foil gages, aluminum alleys, and
-'O[ , . . !. /((;.9.ijnb3E=M.4
.r.
. . ; . . . . n. , u.d. .
3.~ corrugation heights, CROSS. CORE offers a
.__.... multl<ilrectional energy absorption system to E ... pili .. 5. .
'.'Q..'..- .
3O-%.-g'f, 3 %g% *
".g - suit many design ' situations.
-Jg-i .'.[ . '
Y _J/ I !.S -
~~ -
-n or c:.- , y:
S.w'b. . .adW&fW. .._ 6 Registered Trademark. HEXCZ1. 1
~.
. CONTENTS I
'r'oreword . . . . . . . . . .
2 Symbols . . . . . . . .. . . . . . . . . Summary of Design Procedure .......... .. . .3 Design Charts . . . . . ................. Minimum Honeycomb Thickness . . . . . . . ... . . 4 Maximum Honeycomb Crush Strength . . . . . . ...... 5 Safety Factors . . . . . .. .............. 6 Sample Problem . . . .. ............... 6 Typical Honeycomb Crush Strength Properties . . . . ...... 7 Appendix
................ 8 Imd and DeGection . .
Typical Dynamic Responses ............... '9 Formulas . . . . . ................10 Kinetic Energy Equation Solution . . . . . . . . . . 11 Kinetic Energy Absorption Design Chart . '. . . . . . . . . 12 Bibliography . . . . . .. . . . . . . . . . . 13 Prepared by: T, W. Engelbrecht
'i.NO!E Q.h Mt - 03]
b1A:H. Q) PAGE 3 CF l( } SYMBOLS A In' Impact Area a Ft/sec 8 Acceleration or Deceleration Rate F 1.bs. Impact Force fe, PSI Honeycomb Crush Strength G Dimensionless Ratio c/g g Ft/sec 2 Acceleration due to Gravity KE In.-lb. Kinetic Energy m Slugs Mass S In. Stopping or Falling Distance t In. Honeycomb Core Thickness ) Sec. Time v, Ft./Sec. Initial Velocity
- v. Ft/sec. Final Velocity v Ft/sec. Impoet Velecity-Often equal to v, W l.bs. Impact We'ght >
Honeycomb Core Density 4
- e. Lbs./Ft' -
+
M
.}.
1
SUMMARY
OF DESIGN PROCEDURE f l Energy absorption problems can be presented in several forms. However, most of these forms can be reduced to the following basic requirements: l
' l. GIVEN-The following information must be given or computed from the general formulas presented on Page 10 of this bulletin.
W - Lbs. - Impacting Weight - v - Ft/sec.- Impact Velocity r_ffit/6' J T( (_ g G a $i- - G Limit y a rAcH- Of PAGE W. CF gf i j I l A - In* - Impacting Area
- 2. MINIMUM THICKNESS-Enter the minimum honeycomb thickness graph on Page 4 and read a value for t minimum.
)
- 3. MAXIMUM CRUSH STRENGTH-Enter the maximum honeycomb crush strength graph on Page 5 and read a value for fe, maximum.
l I YOU,NOW HAVE A PRELIMINARY SOLUTION INDICATING THE FEASIBILITY OF HONEY.
- g COMB IN THIS APPLICATION. THE SOLUTION CAN BE COMPLETED BY SELECTING A HONEY.
l COMB SYSTEM FROM THE DATA PROVIDED ON PAGE 7 AND MAKING THE FINAL CHECK.
- 4. FINAL CHECK-In selecting a honeycomb material. it may be necessary to reduce the actual value of the fe,being used below the maximum value obtained graphleally i. If this is the eqse, care must be taken to also increase the honeycorno core thickness to prevent bottoming out of the pay-load and thus increasing the peak G value. This balancing of the f , value and t. value will be accomplished by equating t, as follows:
- x _O }- x 12 = t in inches for units given on Page 2 t = fe, A 0.7q All final solutions should be checked through this equation.
(1) See Page 12 tor the solution to problems in which G sa_not a design consideration. (2) Standard Core tratenals are supplied with lined crush strengths. (See HEXCEL TSB #120).
l l 4 1 4 PRELIMINARY SOLUTION l roa , 1 m i MINIMUM HONEYCOMB THICKNESS ..,, ,
.< . . 13
.s...~., ~ . . . .
',...~- .
,. n time ,. .....m . . 4 .a u su. se
*n.-y 4. . - . # . . .
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i m me G" !.tMIT'" USE OF T HIS 'CH A R T ___
--o I.5 nF _ k_ 6 T' "' Y p__ .s Impact velocity . . . v p : ,t. { , _\
G 1.imit . . . . . . G l l l
- 1. With the known values of G and Impact velocity v. enter the chart and read a value for the minimum honeycomb thickness required in inches. Any greater honeycomb thickness can be used.
O (1) This solution takes mio consideration the fact that only 709t> of the honeycornb rectenal as available for crushmg. (2) See Page 12 lor the solution to problems m which G is not a design consideration 4
., . i' PRELIMINARY SOLUTION TOR MAXIMUM HONEYCOMB CRUSH STRENGTH I l CRUSH STB 3GTH--lu-PSI samme a m sess 5555Elig5E!ig5SMSi!kMMEgh .
- x. a 4 . 4 s. _ . cms .
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WEIGHT-W.-POUNDS USE OF THIS CHART GM ..G 'Uf00* Y 6~ i G Limit . . .
){ j Impacting Weight . .W . AM. { }FAGE C CF Impact Area . . . .A
- 1. Enter the above chart with the known values of W and G.
- 2. The intersection of the coordinates W and G define an F curve (slanted -
11nes) representing the force of impact.
- 3. Move along this F curve to the intersection of the horizontal line representing the impact area (A) being considered. ,
- 4. At this point on the F eurve, move vertically upward on the chart c d road g the maximum allowable crush strength (fe ,) for this application.
F
- 5. This value of f ,e may now be used to select a honeycomb energy absorption
.materici.
S
~
i SAFETY FACTORS q.
!! should be understood that the method presented here is an idealized system, which will provide solutions ; ' indicating the feasibility and suggested design parameters of a honeycomb energy absorption system. Due to some observed increase in honeycomb crush strength with increasing velocities, pilot systems should be p.
d: signed and tested before final design is adopted and used. This increase in fer at high impact velocities will, in most cases, be covered by multiplying the G volte by 0.8 prior to entering design curves. Naturally, oth:r safety factors normally applied to material properties, etc., should also be included as required. CAUTION Because of the maximum and minimum requirements of a designed en-ergy absorption system, some safety factors may give opposite results. Inspect the net result of all safety factors. SAMPLE PROBLEM Provide an energy absorption system to protect a 150 pound payload against G loads in excess of 30 G's. The total kinetic energy at impact is 2200 foot pounds and the impacting crea of the payload is 75 square inches. Provide a factor of safety of 20% ogainst peak G's.
- 1. GIVEN-W = 150 Lbs.
y = 30 ft/sec. from the kinetic energy graph on Page 11 G = 30 x 0.8 = 24 with safety factor .if' A = 75 In' ..
, 2. MINIMUM THICKNESS-Entering the minimum honeycomb thickness chart on Page 4 with V = 30 ft/sec. and G = 24, read a value:
t, minimum = 10 inches
- 3. MAXIMUM CRUSH STRENGTH-Entering the maximum crush strength graph on Page 5 with G = 24 and W = 150 lbs., intersect F curve 3700.
Moving along this curve to the intersection of A = 75 in' and then reading vertically to the top of the chart: f e, maximum = 49 ps! *
- 4. SELECT A HONEYCOMB-Deciding to use a standard expanded aluminum honeycomb, we select a 3/16 3003 .0007 honeycomb material from Hex.
cel's TSB #120 Mechanical Properties of Hexcel Honeycomb Materials". This material has a listed crush strength of 43 psi.
- 5. FINAL CHECK-Running a final check for t. from the check equation given on Page 3.
Wv' O.5 I I'
- f,,A
- 6 Tg x 12 l
t = 150 x 30' x 0.5 x 12 = 11.2 inches . ? G.V! 4,,,_E- M W ' 43 x 75 0.7 x 32.2 g( l
- 6. USE-AL 3/16 3003 .0007 with fer = 43 psi at t. = 11.2 inches 6
l
'TY PIC AL HONEYCOMB CRUSH STdENGTH PROPERTIES !
l
- )
i ! i Due to the number of vanables available in the manufacture of honeycomb energy absorption systems. it is I suggested that the actual selection of a specific material. cell size, foil gage, etc. be left to the honeycomb ! manufacturer. Hercel's Applications Engineers are available to interpret your specifications and select the mest economical combinations suitable to your design requirements. Configurations such as TUBE. CORE and CROSSCORE are always supplied as custom designed materials to meet the crush strength and thickness values required. To provide a preliminary selection of the weight of honeycomb required to satisfy your design conditions, the graph below presents a plot of crush strength vs. density for various honeycomb configurations and ca. terials. Actual fe, values for expanded core materials in the one-toelght pound density range may be found in Hexcel's Technical Service Bulletin No.120, Mechanical Properties of Hexcel Honeycomb Materials'" DYNAMIC CRUSH STRENGTH It should be emphasized that the dynamic crush strengih appears to be-alunction of the impact velocity and is usucily higher than the static crush strength shown below. For preliminary design. an increase of 307, in the static value shown below may be used as the dynamic crush strength as a first approximation. HONEYCOMB CRUSH STRENGTH VS. DENSITY 4000 m- w ; ~4; w. w w+
++
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l APPENDDC LOAD AND DEFLECTION
, load and deflection characteristir s encountered in statically crushing honeycomb are graphically shown
( the figure below. This curve indicates the value of the resistance to loading offered by the honeycomb es a compressive load is overcome and crushing starts and continues. Three items on this curve are of special crest: A. COMPRESSIVE PEAK-Honeycomb systems which are not precrushed or in which the initial contact area has not been reduced exhibit this peak force level at impact. This undesirable peak can be easily eliminated by proper design. B. MAXIMUM CRUSH IIVEL- A horizorital line has been drawn across this curve representing a maximum crush level. If this hypothetical line had been developed as a design maximum, then the actual honeycomb fer - value must remain equal to or below this level for crushing to take place. C. MINIMUM STOPPING DISTANCE-The vertical une drawn down this graph represents a hypothetical stopping distance minimum established in conjunction with the maximum crush level value. This value represents a minimum value for the maximum crush level line and any crush level selected below the maximum limit will require an increase in this mini-mum thickness.
- PEAK -
l f,, MAL l u O e-1
'hildl8 i %-6 -N f.*A 9 Ij
.1 l
i g DEFLECTION W The octual crush strength of honeycomb is nonnolly given as the crverage value of the lood diagram inflections. 8
APPENDIX J TY PIC AL DYN AMIC RESPONSES The accelerometer traces shown below are actual plots of deceleration vs. time (ordinate and abscissa re-spectfully) for three conditions of impact.
. Condition #1 This square trace is the pattern normally ob.
- a .-- - tained from the crushing of a constant crea of I
[f)vhWj honeycomb material. g
%. . h. . . ._
' l i i j . . - . .
J
~ '
i ;
) ,
j' Condition #2 I ,4 I I '
- e-The sloped trace is a pattern developed by
. . . .1l ~ ,, .
crushing a honeycomb section in which the crea of material is increasing with the deBec- ! i, j pf. , tion.
}e 3 dl i i n r . . ,,
Yjjid Q b.,D M d pTTA0H. ( PAGE to CF \( 4 1 ... .- - - . .
. .. 1 l
g Condition #3
- ! 1 This trace shows the sharp "G" level increase produced when inadequate honeycomb thick-
. . ....~. . ...[ ....-
- ! ( ness allows the payload to bottom out. Main-
; ! ! y .
taining the designed t, mhtimum should pre-vent this type of failure. t' [l
- t 1 t o 9
-,-a,,.- - - , . - - . , . . . ~ , , - - . . , , . , . , - - . - , , - - - - - - , - - . - . _ , , - - - , , - - - - -
'~ APPENDIX , FORMULAS ( ! The following formulas are those commonly encountered in simple energy absorption calculations. Gen:ral: < * *
- 1. Kinetic Energy, KE = % mv' 4. G = c/g
- 2. Mass, m = W/g 5. Velocity, v8: = ve' + 2a5
- 6. Distance travelled, S = vet + % at'
- 3. Dynamic Force, F = mo ,
Honeycomb Energy Absorption:
- 7. Stopping Distance, S. From 5. above for e v equal to impact velocity and v.
equal to zero, vs = gg3 ' and from 4. above, o = Gq. Therefore: S = 29G 1
- 8. Minimum core thickness, t,. Assuming 70?e of the total honeycomb thickness is available for crushing, then S = 0.7t, and therefore, from 7.
t, l. I*.- 0.7 29G This formula has been presented graphically on Page 3 of this bulletin. e
- 9. Crush strength, f,,, Since f,, = F/A and from 3. above:
.__ i F = ma =WE = WG @-6 M - 48'
' WfinlE._
- n-.-A0H. ( PAGE (\ CF (f then fe, = WG .
I This formula has been presented graphically ! on Page 4 of this bulletin. It can be shown that: KE = fe, AS from 1; above
*V' WGS - WGS = FS = le, AS KE = % mv' 2g 2g Setting S equal to 0.7t. and solving for t, will produce the final check equation presented on Page 3.
- 1. Formulas based on acceleration be.ng a constant t
- 2. ,g == 32.2 Wsee for earth ennionmer.l only
1
\ ,
i i
. APPENDIX 4
KINETIC ENERGY EQU ATION S O L UTION 1 --- ,
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- 1000 10.000 100.000 1.000A00 o/
IINETIC ENERGY EE--TOOT POUNDS
- Motted for g = 32.2 ft/sec. (For earth environment only) ,
11 l _ -- _ __ _ . _ _ _ _ _ _ _ _ _ _ _o
1 l e 1 I l s ! KINETIC ENERGY ABSORPTION j sis chart effers a solution to energy absorption problems in which no limiting G value is required (i.e. stop ! payload in a spectfic distance with no rebound - - G not critical). The chart graphically solves the equa- l in: ~ ~
-1 KE = le, AS = le, A 0.7t. _ -! A s e. @ - P. % s 9 e - o $ l "W M ( PAGE 17 CF If j USE OF THIS CHART 1 Ent:r the chart with the known value of KE in foot-pounds and total core thickness t. in inches. The inter-section of the coordinates KE and t, defines a diagonal line.
Move along this diagonal to the intersection of the horizontal line representing the impact area A being considerod. At this point on the diagonal, move vertically upward on the chart and read the crush strength fe, required. HONEYCOMB CRUSH STRENGTH-fe,-PSI 1000 100 10 14.000 10L000 ,J l l t i 1 1 If e f f'! t I I I f i i [U 'if i f I! I f i I f I # I 83 i I I 3 a 11 lla Ii Il!IIl I I l' 'i d 'I t ill 11 I I I I I I II I I I I I l'i'!Iill : II I I I l'ii II!!IIII I I I I I I f
- 3 5 8 II I I e.
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cc: ORIGINAL DOCunENT FILE k in , _~ _ _ _ _ . . . _ . , _. . _ ,
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; , REV, 09-01-6 e Calculation Submittal
\
, T0* Desig n ngineeg g Clerk (DEC) Date: "
FROM: __ C,a sM i The purpose of this submittal is to identify thethis of status date.of the attached calculation /TCR as
~ '
CALCUIATION No. *
# i REY.
STATUS REMARES: 1 l i Status Codes: P (Pending) i A Not currently installed (As-Installed) OPP Qualifies Existing Installed Condition (Operability) Operability Evaluation for Past i OPC Condition only Operability Evaluation in S C njunction with Current Condition (Superseded) No Longer Valid. Historical Record needs that identiffurther clarification. Provide remarks if claseffleation o Also, for status of P or OPC provide remarkscategories planned - -yDCP when calculation status is expected to change (Le., what actions ar
- f. Condition Report and Action item f, etc.
understood that any schedule information provided will It be should be pre). liminary. cc: ANO DCC (Ref. - Cale. Listed Above) CsJc. File ! # APet. A3 KANSAS NUCLEA3 ONN}}