aircraft structures for engineers solutions

8/3/2019 Aircraft Structures for Engineers Solutions

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S O L U T I O N S T O P R O B L E M SL E S S O N 1 S T R E N G T H T E R M I N O L O G Y

l.l Nor~l stress and shear stress. or. tension. compression and shearstresses.1.2 fhe tension stress (ref. paragraph 1.1.1) is:

ft = PIA = 186.000/(1.50 x 2.00) = 62.000 p s i1.3 The shear stress (same ref. as above) is:

fs = VIA = 132,000/(3.00) = 44,000 psi1.4 The limit bearing stress (same ref. as above) is:

f = P = 2000 = 80,000 p s ib r y O t -. 2 ' " S . . , . 0 : - " x . . . . . , .' " " 0 ~ 0 ..1.5 The u l tina te factor is 1.5. (ref. paragraph 1.1)

fbru = 1.5 x fbry = 1.5 x 80,000 = 120,000 p s i1.6 From paragraph 1.3.1: E = fie and E = 10.5 x 106 psi approximately.Therefore:

e = fiEe = 35,000 = .00333 i n . lin.1 0 , 5 0 0 . 0 0 0

1.7 From paragraph 1.2.1. e = o i L . Therefore:~ = ~ xL: .00333 x 84 . 2 8 i n .I'hen the length with the load applied would be:L = L1 - 0 : 8 4 - .28 = 8 3 . 7 2 i n .

1.8 E = 10.5 x 106 psi and u ~ 0.33. From paragraph 1.3.3:G == ' ' ' ' ' ' 2 , ( . . . . - + ~ u " T " )E 10.500.000 = 3.95 x 1062 (1. 3 3 ) psi

A l - l


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1 . 9 a ) R e f e r r i n g t o t h e d e f i n i t i o n s o f E t a n d E s i n p a r a g r a p h 1 . 3 . 4 a n d t h et y p i c a l s t r e s s - s t r a i n d i a g r a m i n p a r a g r a p h 1 . 1 . 2 i t c a n b e s e e n t h a ta b o v e t h e p r o p o r t i o n a l l i m i t t h e s l o p e o f t h e c u r v e ( w h i c h i s t h es a m e a s t h e s l o p e o f a t a n g e n t t o t h e c u r v e ) i s s m a l l e r t h a n b e l o wt h e p r o p o r t i o n a l l i m i t w h e r e t h e s l o p e i s t h e m o d u l u s o f e l a s t i c i t y ,E . T h e r e f o r e :

E t i s s m a l l e r t h a n E .b) By d r a w i n g a l i n e , i m a g i n a r y o r r e a l . f r o m t h e p o i n t o n t h e c u r v e t ot h e o r i g i n i t c a n b e s e e n t h a t E s i s g r e a t e r t h a n E a n d :

E s ; s g r e a t e r t h a n E t .

S t r e s s ..

--i r - - .002 S t r a i n

A l - 2


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S O L U T I O N S T O P R O B L E M SL E S S O N 2 R E V I E W O F S T A T I C S

2.1 E i t h e r r e a c t i o n c a n b e f o u n d b y t a k i n g m o m e n t s a b o u t t h e p o i n t o f a c t i o no f t h e o t h e r .: MS = 0 - 7 5 0 x 2 0 . 0 + ( 5 0 x 2 0 . 0 ) 2 0 . 0 + ( 2 5 0 x 1 0 . 0 ) 5.0 - 2 5 . 0 R 1

R 1 = 19001T h e o t h e r r e a c t i o n c a n n o w b e f o u n d e i t h e r b y t a k i n g m o m e n t s a b o u t p o i n t A o rby s u m m i n g v e r t i c a l f o r c e s :

~MA = 0 = 7 5 0 x 5 . 0 + 5 0 x 2 0 . 0 x 5 . 0 + 2 5 0 x 1 0 . 0 x 2 0 . 0 - 25.0 R 2R Z = 23501 o r :

" : F v = 0 = 7 5 0 + 5 0 x 2 0 . 0 + 2 5 0 x 1 0 . 0 - 1 9 0 0 - R 2R 2 = 23501

A f t e r f i n d i n g R 2 b y o n e e q u a t i o n i t c a n b e c h e c k e d by t h e o t h e r e q u a t i o n .2 . 2 A t j o i n t A :

~ F v = 0 = P A B s i n 4 5 + P A C s i n 6 0 - 1 0 0 0. 7 0 7 P A B + . 8 6 6 P A C = 1 0 0 0

~ F h = 0 = P A B c o s 4 5 - P A C c o s 6 0 . 1 U l P A B - . 5 P A C = 0

S o l v i n g t h e s e t w o e q u a t i o n s s i m u l t a n e o u s l y :. 7 0 7 P A S = . 5 P A C. 5 P A C + . 8 6 6 P A C = 1 0 0 0

P A C a 1 0 0 0 a 7 3 2 '( . 5 + . 8 6 6 ). 7 0 7 P A S - . 5 ( 7 3 2 ) a a

PAS a 5 1 8 '

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A t j o i n t B :;F v = 0 = P a D s i n 3 0 + P a E s i n 3 0 - P A S s i n 45

.5 P a D + .5 P S E = 518 s i n 45 ~ 366~h = 0 a P S D c o s 3 0 - P S E c o s 3 0 - P A S c o s 45

.866 Pso - .866 P S E a 51 8 c o s 45 366S o lv i n g s i m u lt a n e o u s l y :

.866 P S D + .866 P S E al866\366 = 634 ( R a t i o i n g E F v e q u a t i o n )CSO ~

.866 P a D - .866 P S E ~ 36 6 ( F h e q u a ti o n )

.866 P s o + .866 P S D 634 + 366 a 1000 ( A d d i n g t h e e q u a t i o n s )P S O = 5771

. 5 x 5 7 7 + .5 P S E = 366P B E 2 1551A t j o i n t C:

Y . f v = 0 a P C F s i n 45 + P C G s i n 45 - P A C s i n 60 . 7 0 7 P C F + . 7 0 7 P C G 7 3 2 s i n 60 634

EFh = 0 = P C F c o s 45 - P C G c o s 45 + P A C c o s 6 0. 7 0 7 P C F - . 7 0 7 P C G - 7 3 2 c o s 60 -366

S o l v i n g s i m u l t a n e o u s ly :2 x . 7 0 7 P C F 634 - 366 268

P C F 1901.707 x 1 90 + .707 P C G 634

P C G 7 0 7 'N o t e : T h i s c a b l e s y s t e m h a s f o u r r e a c t i o n s , w h i c h i s o n e m o r e t h a n t h e t h r e et h a t c a n n o r m a l l y b e f o u n d b y t h e t h r e e e q u a t i o n s o f s t a t i c s .and w o u l dn o r m a l l y b e s t a t i c a 1 1 y i n d e t e n l l i n a t e . T h e r e a s o n t h i s p r o b l e m c o u l d b e s o l v e di s t h a t t h e i n t e r n a l l o a d p a t h w a s s t a t i c a l l y d e t e l " 1 l 1 n a t e .

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T h i s p r o b l e m c o u l d a l s o h a v e b e e n s o l v e d g r a p h i c a l l y . T h i s i s d o n e b yj r a w i n g a f o r c e d i a g r a m a t e a c h j o i n t w i t ht h e l e n g t h s o f t h e f o r c e v e c t o r s p r o p o r t i o n a lt o t h e m a g n i t u d e o f t h e f o r c e a n d a t t h e a n g l eat w h i c h t h e f o r c e a c t s . T h e s k e t c h s h o w s t h ef o r c e d i a g r a m a t p o i n t A .

2.3 L o o k a t a f r e e b o d y o f t h e u p p e r c y l i n d e r .a v a i l a b l e a n d t h e y a c t p e r p e n d i c u l a r t ot h e s u r f a c e o f t h e c y l i n d e r a t t h e p o i n t so f c o n t a c t .

[ F h : 0 : P B c o s 3 0 - P A c o s 6 0P B = P A t E o s 60 .) . .. 5 77 P A\ c o s 3 0

[ F v = 0 = P a s i n 3 0 + P A s 1 n 60 - 200.5 P B + .866 P A = 200.5 ( . 5 7 7 P A l + .866 P A 3200

P A = 1 7 3 #P B 3 . 5 7 7 P A 3 . 5 7 7 x 1 7 3 - lQQ !

N o w l o o k a t a f r e e b o d y o f t h e l o w e r c y l i n d e r .[ F v = 0 = P c s i n 60 - PB s i n 30 - 180

. 86 6 Pc 100 (.50) + 180P c 266'

[ F h . . 0 P o - PB c o s 3 0 - P c c o s 60P o - 1 00 (.866) + 266 (.50)P o -2201

A2-3

1000 16

O n l y two r e a c t i o n s a r e

w~ Z D O Ib ..

30"


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2 . 4 ~ M B = 0 = 36,000 x S 2 + 18,000 x 6 - 60 R1R1 = 33,0001

) : F h : : a 0::1 18,000 - R3 COS0o J = a r c t a n 6 /6 0 .. S. 71 R3 = 18000 = 18,0901c o s 5.71

~ F y = 0 = R + R 2 - R 3 s i n 5.71 - 36,00033,000 + R2 - 18,090 s i n 5.71 36,000

R2 = 480012 . S F i r s t f i n d t h e r e a c t i o n s a t p o i n t s E a n d H t o c o m p l e t e t h e e x t e r n a lb a l a n c e .

t. M H = 0 4000 x 9 + 3000 x 18 + 2000 x 27 - 27 R ER E = 5333#

Z F y = 0 2000 + 3000 + 4000 + 5000 - R E - R HR H = 14,000 5 3 3 3 86671

'.

D e t e r m i n e the i n t e r n a l l o a d s b y b a l a n c i n g e a c h j O i n t , i n d i v i d u a l l y .J o i n t A: 0= a r c t a n 9-2~ = R E :a .Ell! c a m p .A F = A E - 2 0 0 0 3 3 3 3

cos 0 cos 36.87 41661 t e n s

~ Z C X X l 1 0 '

.-~~ ~ ~ : BA EA S = A F s i n 0 4166 s 1 n 36.87 ~ c o m p oJ o i n t E :

E F 0

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J o i n t F :1 : F,. = 0 :0 SF - 4F cosoBF :0 4166 c o s 36.87- '" 3 333* c o m p o~ F h = 0 : : F G - A F s i n G

F G = 4 1 6 6 s i n 3 6 . 8 7 = 25001 t e n sJ o i n t B:~Fv :: 0 :: 3 3 3 3 - 3000 - BG c o s 3 6 . 8 7 BG :: _ . . . . ; 3 . . ; , 3 . ; . , 3_ : : 416i# t e n sc o s 3 6 . B 7 L F l ] : : 0:: 2500 + B G s i n 3 6 . B 7 - B C

B e : : 2500 + 416 s i n 3 6 . 8 7 s ~ c O m p oJ o i n t G :l:F ,. :: 0 = CG - 4 1 6 c o s 3 6 . 8 7

C G = 416 c o s 3 6 . 8 7 s 3331 compo[F h :: 0 = 2 5 0 0 + 416 s i n 3 6 . 8 7 - GH

G H = 2 5 0 0 + 416 s i n 3 6 . 8 7 27501 t e n sJ o i n t C :L F V : : 0 = 4 0 0 0 - 3 3 3 - ' C H cos 3 6 . 8 7

CH : : 3 6 6 7 s~ compoc o s 3 6 . 8 7 1 : F h = 0 ' " 2 7 5 0 - C H s i n 3 6 . 8 7 - C D

C D ' " 2 7 5 0 - 4 5 8 4 s i n 3 6 . 8 7 0J o i n t H :[ F v = 0 3 D H + 4 5 8 4 cos 3 6 . 8 1 _ 8 6 6 7

D H ' " 8 6 6 7 - 4 5 8 4 cos 3 6 . 8 7 50001-J o i n t 0:E F V = 0 ' " 5 0 0 0 - 5 0 0 0T h i s c h e c k s t h e i n t e r n a l b a l a n c e .

A 2- 5

F e .

'2 50 0 IbI ~ O O I .

B ' \ s : C~3331b ..41 1. lb I c . . e r

Z500 lb \ r;.. --"-G~I

2 .1 5 0 Ib


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2 . 6 Z M x = 0 = R S x 15 + R C x 3 - 600 x 615 RS + 3 R c = 3600

Z M y = 0 = R S x 6 + RC x 24 - 600 x 156 RB + 24 RC = 9 0 0 0

C o m b i n e t h e tw o e q u a t i o n s by m u l t i p l y i n g the f i r s t e q u a t i o n by a m i n u s 8 a n da d d i t t o t h e s e c o n d e q u a t i o n .-120 RB - 24 RC = -28,8006 R S + 24 RC = 9 , 0 0 0-114 R S = -19.800

. RB = 174#15 (174) + 3RC = 3600

RC = 3 3 0 1~Fv = 0 = RA + R S + R C - 600

R A = 600 - 174 - 330 z~2.7 (1) [ M y = a = 55.000 x 6.0 + 12.000 x 52.0 - ( R d b s i n 60) 20.0

Rdb = 55.080#(2) [ F z = a = 55.000 - R d b c o s 60 - l A - l S

Z A + Zs 27.460#( 3 ) [ M x = a = 7000 x 60.0 + 10.0 l A - 10.0 l B

Z A - A S = - 4 2 , 0 0 0 1S o l v i n g the e q u a t i o n s f o r EFy a n d E M x s i m u l t a n e o u s l y :

ZA - (27460-ZA) = -42.0002ZA -14,540ZA - 7 2 7 0 1 ,Zs 27460-ZA = 27460-(-7270) 3 4 , 7 3 0 1

N o t e : T h e p o r t i o n s o f ZA a n d Zs d u e to t h e 55,0001 v e r t i c a l l o a d a n d t h e7000# ~ f d e l o a d c o u l d b e d e t e r m i n e d s e p a r a t e l y a n d s u p e r i m p o s e d .

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..

- _I


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(4) ZFx = 0 = 12.000 - R d b s i n 60 + XA + XBX A + Xe = 35.700

(5) J M z = 0 = 7000 x 6.0 + 10.0 XA - 10.0 XeXA - Xe = -4200S o l v i n g t h e e q u a t i o n s f o r E F x a n d r. M z s i m u l t a n e o u s l y :

X B : . 4200 + X AXA + (4200 + XA) = 35.7002 X A = 31.500XA = 15.7501XB = 4200 + XA = 4200 + 12750 = 1 9 . 9 5 0 1

N o t e : T h e p o r t i o n s o f XA a n d Xe d u e to E F x a n d d u e t o E M z c o u l d b e d e t e r m i n e ds e p a r a t e l y a n d s u p e r i m p o s e d .( 6 ) E F y : . 0 = 7000 - YB

YB :: 70001N o t e t h a t i n t h i s p r o b l e m a l l s i x e q u a t i o n s o f s t a t i c s a r e r e q u i r e d b e c a u s et h e r e a r e s i x r e a c t i o n s .

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S O L U T I O N S T O P R O B L E M SL E S S O N 3 R E V I E W O F S T R E N G T H O F M A T E R I A L S

3 . 1 = P LAt_ _ 6 _ 5 - . , _ 0 0 _ 0 _ x _ 6 0_ : : 0 . 2 9 7 ; n1.25 x 1 0 . 5 ) ( 106

3 2 fl :: -35,000 x 60. = - 0 . 2 0 i nE 10.5 x 1063 . 3 T h e r e a c t i o n s a r e s t a t i c a l l y i n d e t e r m i n a t e b e c a u s e t h e r e a r e twor e a c t i o n s a n d o n l y o n e e q u a t i o n o f s t a t i c s t h a t a p p l i e s ( E F h = 0 ). H o w e v e r ,a s p o i n t 6 d e f l e c t s . t h e l e n g t h AB m u s t s h o r t e n t h e s a m e a m o u n t a s l e n g t h BCe 1 o n g a t e s . T h u s :

- S A B = 6 S CP A B L A B P s c L S C=E A EP A l ) = P B C ( ~ : ~ ) = P B C (* ) . . 75 P B C

T h i s , t h e n , is t h e s e c o n d e q u a t i o n t h a t i s r e q u i r e d , t h e f i r s t b e i n g :~Fh : 0 = RA + R B - 3 0 , 0 0 0

R A + R S = 3 0 , 0 0 0 = P A B + P S CC o m b i n i n g t h e e q u a t i o n s :

.75 P S C + P S C = 3 0 , 0 0 0P B C = 1 7 ,1401P A S = 3 0 , 0 0 0 - 17,140 = 12,8601

F o r t h e d e f l e c t i o n a t B:1 2 , 8 6 Q x 24 ~ 0 . 1 0 7 i n

1 8 x 1 6 x 1 0 6o r :

6S = P s c LBC = 1 7 , 1 4 0 x 1 8 = 0 . 1 0 7 i n .AE . 1 8 x 16 x 1 0 6A 3 - l


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3.4 The spar cap and strap are forced to elongate the same amount becauset h e y are attached along the length. Therefore:'~ A = 05 ."here subscripts A and 5 denote aluminum an d steel,respectively.

P A x 8 4 =1. SOdO. 4xl06

P s x 8 4. 7 5 x 2 9 .0dOSP A = .717 P s

and by Z F = 0 :P A ? P s 1 7 2 . 0 0 0. 7 1 7 P s + P s a 1 7 2 . 0 0 0

steel P s = 1 0 0 . 2 0 0 1ft = P = 100,200 .. 154,200 psi- AAet .6 5M.S. a Ftu -1 = 160,000 - 1 .. ~~ 154,200

a 1uminum PA:a .717 Ps a .717 X 100,200 .. 71,8001ft:a P = 71,800 5 3 , 2 0 0 psiAnet 1.35M.S Ftu -1 .. 88,000 - 1 ~- Tt 5 3 . 2 0 0

- !

Note: The same answers would result from using the equations:

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3 . 5 F r o m t h e e q u a t i o np r o p o r t i o n a l t o A E / P L .T h u s , f o r t h e s t r a p s t os a m e P ):

6 = P L / A E i t c a n b e s e e n t h a t 6 i s i n v e r s e l yP i s t h e l o a d a n d A E / L i s a m e a s u r e o f s t i f f n e s s .h a v e t h e s a m e s t i f f n e s s { t h a t i s , t h e same f o r t h e

w h e r e T d e n o t e s t i t a n i u m a n dS d e n o te s s t e e lS i n c e t h e l e n g t h s a r e t h e s a m e :

( A E ) r = ( A E l SA T = E S = 2 9 x l 0 6 = 1 . 8 1 3- - 1 6 x 1 0 5A S E TA S = 3 . 7 5 x . 1 5 6 = . 5 8 5 in2A T = 1 . 8 1 3 A S = 1 . 8 1 3 x . 5 8 5 = 1 . 0 6 1 i n 2

3 . 6 F i n d r e a c t i o n s R B a n d R F~ M 2 = U = 5 0 0 x 4 + 2 0 0 0 ) ( 1 0 + 1 0 0 0 x 1 6 + 1 0 0 0 x 2 6 - 2 0 R 1

) { B = 32001~ F v = 0 = 1 0 0 0 + 1 0 0 0 + 2 0 0 0 + 5 0 0 - R l - R 2

1000Ib

th e s h e a r a t t h e l e f t e n d i se q u a l t o t h e a p p l i e d l o a d o f 1 0 0 0l b . A t e a c h a p p l i e d l o a d a n dr e a c t i o n t h e s h e a r c h a n g e s b y t h e~ g n i t u d e o f t h e l o a d . AT h e b e n d i n g m o m e n t i s z e r o~ t t h e f r e e e n d . T h e c h a n g e I nm o m e n t I n s p a n A S i s e q u a l t o t h ea r e a o f t h e s h e a r d i a g r a m f o r t h a ts p a n . T h u s ,

V (Ib)

M B = M A - 10oox6 = 0 - 6 0 0 0 - 6 0 0 0 ' "s i m i l a r l y :

M e = M a + 2 2 0 0 x 4 = - 6 0 0 0 + 8 8 0 0 2 8 0 0 " 'M O = Me'+ 1 2 0 0 x 6 = 2 8 0 0 + 7 2 0 0 1 0 . 0 0 0 ' "M E = M O - 8 0 0 x 6 1 0 , 0 0 0 - 4 8 0 0 5 2 0 0 ' "M F = M E - 1 3 0 0 x 4 2 5 2 0 0 - 5 2 0 0 0

M( in lb)

A 3 - 3

1000lb 2000 Ib 500 \ b

t t l F f - x\ ~OQ 'b

B f3200 IbnoD I \L e o

D E .

-100010,00)

SlOG

x-eoco


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3.7 F i n d r e a c t i o n s by t h e e q u a t i o n s o f s t a t i c s .~ M C = 0 = ( 2 0 0 ' ; 1 2 ) ( 1 ~ ) + l O O O x 1 2 + ( 1 0 0 x 1 2 ) 1 8 - 2 4 R A

R A = 3 8 , 4 0 0 = 160012 4- F v = a ~ 1 0 0 x 1 2 + 1 0 0 0 + 2 0 0 x 1 2 - R A - R C2R C = 1 2 0 0 + 1 0 0 0 + 1 2 0 0 - 1 6 0 0 1 8 0 0 '

F i n d s h e a r a t A . a a n d C b y s u m m i n g v e r t i c a l f o r c e s s t a r t i n g a t p o i n t A :V A = R A . . 16001V a - = V A - 1 0 0 x 1 2= 1 6 0 0 - 1 2 0 0 = 4 0 0 1V a + = V a - - 1 0 0 04 0 0 - 1 0 0 0 = - 6 0 0 'V c = V B + - ~ x 1 22= - 6 0 0 - 1 2 0 0 = -18001

F i n d b e n d i n g m o m e n t a t A . a a n d C :M A = 0Ma = MA + 1600 + 400 12

2..0 + 1 2, 00 0 . . 12 ,0 00 "'( U s i n g a r e a o f s h e a r d i a g r a m ) .M e , . M e + V B + b - w b 20

a 1 2, 00 0 - 6 0 0 X 1 2 - 2 0 0 ( 1 2 ) 2 0 ( U s i n g EMe)6

M('".Ibl

F i n d t h e s h e a r a n d m o m e n t a t c e n t e r o f s p a n a , u s i n g t h e e q u a t i o n s o f s t a t i c s :V = R A - w ! . 1600 - 100 ( 1 2 \ . 10001- 2 2/~ ,. R A (~ ) - w ( ~ )i600 x 6-100(6 ) 3" 7800'

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Sim ilarly. at center of span b u sing the equ ations of paragraph 3.4.2:V = VS+ - w ( b / 2 ) 2 : : -600 - 200(6)2 :: - 900*- 2b 2x12-----~ = MS +(VB+)(b/2) - w(b/2)3 = 12,000-600(6)6b - 200(6)3 = 7800"*6x12

Find shear and bending moment one inch from A using the equa tt ons of sta tic s:~ = RA - w(x) = 1600 - 100(1) = 1500*! : 1 . = RA( 1) - w(x) ~ = 1600 - 100(1) . ! . = 1550"12 2

F ind shear and bending moment one inch from C using the equ ations of paragraph3.4.2:V = VS+ - w(x)2 = -600 - 200(11)2 = -1608'- 2b 2x12~ = MS -(Vs+)(x) - w(x)3:: 12000-600(11) - 200(11)3 = 1703"16b 6x12

3.8 Summ ing vert ical loads (shear) and moment, starting at the free end:V A = 0 , M A = 0VS- =(-wa)a :: -200 x 6 = -1200*'18+ = YS- - P v = ~1200-2000 = -32001MS = wa (a) ! 200(6) 3 = -3600"12Y c = VS+ -(Wb)b :: -3200-400 x 8 a -64001M c MB+(VB+)b-Wb(bX~)

= -3600-3200 X 8 - 400 X 8 X 4 m -42,000"1A t center of span a:

~ = \IIa (x) (!)2 00 (3 ) (1.5) :II 900"'A t center of span b :

M '" Mb' +( VS+) (x) - \l ib( x) (!)= -3600-3200(4) - 400(4)(2) -19.600"'

A3-5

1 2 . D l X J Ib, LU -: 400 IIl/N. ,---

~Ib

1 ' 1 \I i n l b ) I--.......,:::-----------r----~a x

- 4 ' 2 , C O O


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3.9 Balance the beam using the equations of statics:L Fh = 0 = Rh -6000

R h = 60 0 0 1[Me = 0 = 6000 x 5 - 2 4 R A

R A '"' 6000 x 5 125012 4

(.;{)(jj I I : !

A~ J I l ~ _ _ _ _ _ _ _ _ _ _ _ _ _ C~ I ~ ~ - - - - - - - - - - - - ~ - - ~ - -t B , ~~L.L, IlSO I n \ Z S C ;b ~

RB = R A = 1250*

v = - R A = -12501M = 0

v( \ b " l 1--------------------:-., , < .

II

A t A :

A t B - : - \ Z . 5'0 ;.-------------------------------'v = - R A . . -12501M = - R A x 8 ' " ' ( - 1 2 5 0 )= - 1 0 . 0 0 0 " '

At B+: M(,I\'lbJxv = - R A = -12501M = M S - + 6 0 0 0 x 5

= - 1 0 . 0 0 0 + 3 0, 00 0 . . 2 0, 00 0 ' " -10,000At C:

v = - R A :01 -12501M ' " ' S + + ( V s + ) x 1 6

= 2 0 , 0 0 0 - 1 2 5 0 x 1 6 0

A 3 - 6


16/131

S O L U T I O N S T O P R O B L E M SL E S S O N 4 A I R C R A F T E X T E R N A L L O A D S

4.1 A s e x p l a i n e d i n p a r a g r a p h 4 . 1 . 2 :a) +N z a n d l o a d s i t p r o d u c e s a c t d o w n .b) +N x a n d l o a d s i t p r o d u c e s a c t f o r w a r d .c ) "'~y a n d 1 o a d s i t p r o d u c e s a c t 1 e f t .

4 . 2 T h e l o a d f a c t o r N z o n t h e a i r p l a n e i s e q u a l t o t h e t o t a l o f a l l v e r t i c a lf o r c e s ~ c t i n g o n t h e ~ i r c r a f t d i v i d e d b y t h e a i r c r a f t w e i g h t . T h u s := 2 5 2 , 0 0 0 + 3 1 , 7 5 0 - 4 0 , 0 0 0 = 6 . 5N z 37.500

4.3 L o a d f a c t o r s f o r m a s s i t e m s i n t h e f u s e l a g e a n d w i n g o f t h e F - 4 a r e f o u n do n p a g e s 4.5 a n d 4 . 6 r e s p e c t i v e l y .P r o b + N x - N x + N y - N y + N z - H za ) 6 . 0 - 5 . 2 0 . 9 - 0 . 9 B . 5 - 3 . 0b) 6 . 0 - 5 . 2 2 . 4 - 2 . 4 9 . 3 - 4 . 0c ) 6 . 0 - 5 . 2 5 . 8 - 5 . 8 1 2 . 5 - 5 . 1d ) 6 . 0 - 5 . 2 - 4 . 8 * - 4 . 8 * 9 . 2 - 3 . 4e ) 6 . 0 - 6 . 1 1 4 . 0 * - 1 4 . 0 * 1 5 . 5 - 9 . 5

'.


17/131

S O L U T I O N SL E S S O N 5 I N T E R N A L L O A D S

5 . 1 a ) B e n d i n g m o m e n t a n d s h e a rb) T o r s i o nc ) T e n s i o n a n d c o m p r e s s i o nd) S h e a r

5. 2 a ) S e a m a n d t o r q u e b o xb) B e a m , t o r q u e b o x a n d a x i a l m e m b e rc ) B e a m a n d t o r q u e b o xd) B e a me) A x i a l m e m b e rf) S h e a r w e bg ) B e a mh ) B e a m , a x i a l m e m b e ri) A x i a l m e m b e r , s h e a r w e b

5.3 a ) W i n g s h e a r , b e n d i n g a n d t o r q u e , c a u s e d b y ~ i r a n d i n e r t i a l o a d s ,c a u s e s h e a r a n d b e n d i n g i n t h e s p a r . F u e l p r e s s u r e s c a u s e b e n d i n g i n t h e s p a rw e b s a n d s t i f f e n e r s .b) W i n g s h e a r , b e n d i n g a n d t o r q u e c a u s e s h e a r , b e n d i n g a n d a x i a l l o a d i nt h e ri .c ) W i n g s h e a r , b e n d i n g a n d t o r q u e c a u s e a x i a l l o a d a n d s h e a r . A e r o d y n a -m i c a n d f u e l p r e s s u r e c a u s e s h e a r a n d b e n d i n g n o r m a l to t h e p l a n e o f t h e s k i n .d ) A e r o d y n a m i c p r e s s u r e s c a u s e s h e a r o l n d b e n d i n g .e ) A e r o dy n a m i c p r e s s u r e s c a u s e s h e a r a nd b e n d i n g .f ) C o c k p i t p r e s s u r e c a u s e s s h e a r a n d b e n d i n g .g) E n g i n e s t a l l p r e s s u r e s c a u s e s h e a r , b e n d i n g a n d a x i a l l o a d ( t e n s i o n ) h ) A e r o d y n a m i c a n d i n e r t i a l o a d s o n t h e f u s e l a g e f o r w a r d o f F . S . 7 7 . 0 0a r e r e d i s t r i b u t e d b y t h e b u l k h e a d c a u s i n g s h e a r a n d b e n d i n g . I t d i s t r i b u t e sn o s e l a n d i n g g e a r l o a d s f r o m t h e k e e l w e b s , c a u s i n g s h e a r a n d b e n d i n g . C o c k -p i t p r e s s u r e c a u s e s b e n d i n g n o r m a l t o t h e p l a n e o f t h e b u l k h e a d .i) A e r o d y n a m i c a n d e n g i n e bay p r e s s u r e s c a u s e s h e a r , b e n d i n g a n d a x i a ll o a d .

A S . l


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j) A e r o d y n a m i c l o a d s c a u s e s h e a r a n d b e n d i n g .k ) A e r o d y n a m i c l o a d s c a u s e s h e a r . b e n d i n g a n d t o r s i o n a l s h e a r .1 ) S t a b i l a t o r s h e a r . b e n d i n g a n d t o r s i o n d u e to a e r o d y n a m i c l o a d s c a u s ~s h e a r a n d a x i a l l o a d .

A S . 2


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S O L U T I O N S T O P R O B L E M SL E S S O N 6 S T R E S S A N D M A R G I N O F S A F E T Y

6 . 1 T e n s i o n . c o m p r e s s i o n , s h e a r a n d b e a r i n g .6 . 2 T e n s i o n , c o m p r e s s i o n a n d s h e a r .6.3 N o H l i : n i t l o a d ; y e s a t u l t i m a t e l o a d .6.4 D e s i g n l i m i t l o a d ( D L L ) i s 1 0 0 0 l b . D e s i g n u l t i m a t e l o a d ( D U L l = l . s O ( D L L )= 1 5 0 0 l b . F a i l i n g l o a d ( w h i c h , b y d e f i n i t i o n . i s t h e a l l o w a b l e l o a d ) i s 1 5 0 01b .

~ = (P a l l / P u l t ) - 1 ' : :( 1 5 0 0 / 1 5 0 0 ) - 1 = Q6 . 5 F t u F t y . F c y , F s u , F b r y , F b r u . E . G . a n d \J.6.6 M I L - H D B K - 5 " M i l i t a r y S t a n d a r d i z a t i o n H a n d b o o k - M e t a l l i c ~ a t e r i a l s a n dE l e m e n t s f o r A e r o s p a c e V e h i c l e S t r u c t u r e s " .ti.7 C o m p r e s s i o n s t r e s s e s - d u e t o c r i p p l i n g a n d / o r c o l u m n b u c k l i n g .

S h e a r s t r e s s e s d u e t o b u c k l i n g o f t h e s h e a r w e b .6 . 8 N o n e , b e c a u s e t h e m e c h a n i c a l p r o p e r t i e s a r e t h e m a x i m u m a l l o w a b l es t r e s s e s f o r t h e m a t e r i a l .6 . 9 M c D o n n e l l r e p o r t s M AC 338 a n d M AC 339, o t h e r c o m p a n y s t r e n g t h m a n u a l s , .a n d v a r i o u s c o l l e g e t e x t b o o k s a n d h a n d b o o k s .6.10 a) f P 9500 .t = A = ~ = 50,500 PSl

F t u ' : :61,000 p s iM~ ':: Ft u -1 - 61,000 -1 ,. +.21.~. ~ - 50,500

b) ft = P = 2 5 0 0 0 ' : :0 1 0 0 p s iA ~ F tu = 8 6 . 0 0 0 p s iM . S. = F t u - 1- f t = 8 6 , 0 0 0 1 0 78 0 . 1 0 0 - ' "+.

A 6 - 1


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6.11

C) f s = " 6500 = 2 6 , 000 p s iA ~F s u = 3 1 . 0 0 0 p s i ( u s i n g e x t r u s i o n a l l o w a b l e )

F 3 1 , 0 0 0M . S . = ~ - 1 = 2 0 . 0 0 0 - 1 = + . 1 9sd) f b r = P = P = 3 5 0 0 = 2 2 4 . 4 0 0 p s iA T I t . 3 1 2 x . 0 5 0

F b r u = 2 6 1 , 0 0 0 p s i ( u s i n g T i - 6 A l - 4 V a l l o w a b l e )

M . S . F b 2 6 1 , 0 0 0= r u -1 = 224,400 -1 2 +.16rt;;:-

I M a t e r i a l F t u F b r uV i " A I S I 3 0 1 half hard 1 5 1 3 0 4 ~k : 2 ' 1 2 5 k s i 4 1 3 0 s t e e l 1 2 5 2 5 1~ . . 1 2 5 =~ 1 5 1 .828

2 5 13 0 4 = .826

R e q u i r e d t ... 1 2 5 = . 1 5 1 3 i n c h m i n i m u m~

A 6 - 2


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SO L UT IO N S T O P RO B LE M SLE SSO N 7 SE CT IO N P RO PE RT IE S

.

7.l F irst , find the effec tive area. T hen determ ine the stress from th eequation f P IA . Find the allowab le stress from the tab le in Lesson 6and determ ine the margin of safety from the equation M .S. a (F /f) -1 .a ) At ,. . 1 0 [3 . 0 0 - . 5 0 - 2x. 2S) ] . 2 0 1 " 2

ft ~ 1 0 , 5 0 0 = 5 2 , 5 0 0 p si F t u 5 9 ks1- 'J .2 0M S. ,. ( Ftu / ft ) -1 ,. (5V 5 2 5 ) -1 a !.:..!l

b} A c ,. . 1 0 ( 3 . 0 0 - . 5 0 ) - .25 in 2fc ~ P - 1 0 , 5 0 0 = 4 2 , 0 0 0 p si F e - F tu 5 9 ksi- ~ .25~ - ( Fe/ fc) -1 - (59/ 4 2 ) -1 - +.40

c} A s = A t a . 2 0 in 2fs = V a 7 2 0 0 , .3 6 , 0 0 0 p si F su - 37 ks i- 'J --:1'0'M.S. - (FSu/fS) -1 a(37/36) -1 "!.:.Ql

7.2 Selec t a differential area dA having allpoints equ ally distant from the base.dA = b dy

Then .using t h e equation Ay - [Y. dA:

y * ; : \ y dy i n (b)[f2 J : k [ t ]y = h!Using the equation Ix - J ' y 2dA:r x / y2 b dy b ,/ y2 dy b [ f1:o 0 JIx = bh3j

A 7 - 1

yI


22/131

C e n t r o i d a l I a I x - A d 2 s I x - A y 2 ' " ~ 3 - ( b h ) ( ~ 2 )[ s b h 3IT

7.3 D i v i d e e a c h s e c t i o n i n t o a m i n i m u m n u m b e r o f r e c t a n g u l a r e l e m e n t s ,n e g l e c t i n g t h e s m a l l f i l l e t r a d i u s a r e a s , a n d s o l v e , u s i n g t h e e q u a t i o n sg i v e n i n L e s s o n 7 . T h i s i s m o s t c o n v e n i e n t l y d o n e u s i n g a t a b l e t h a t c o m -b i n e s t h e c a l c u l a t i o n s o f E x a m p l e s 2 a n d 4 ( p a r a g r a p h s 7 . 3 . 2 and 7 . 4 . 3 ) .a) A n g l eC D I h = b h 3 a . 1 2 ( 1 . 5 ) 3 a . 0 3 3 7 5 i n 4IT 12 . C D I n a 1 . 3 8 ( . 1 2 ) 3 = . 0 0 0 2 0 ; n 412

E l e m e n t b t A y A y A y 2 I h1 1 . 5 . 1 2 . 1 8 . 7 5 . 1 3 5 . 1 0 1 2 5 . 0 3 3 7 52 1 . 3 8 . 1 2 . 1 6 5 6 . 0 6 . 0 0 9 9 . 0 0 0 6 0 . 0 0 0 2 0l : :mt .I~~ . I O I a S . O ~ ~ ~ S

A ' " . 3 4 5 6 in2y ( '" x b e c a u s e o f s y m m e t r y ) a ! : ( A y ) a . 1 4 4 9 4 1 9 i nl : A 3~S6I n ( , . Iy b e c a u s e o f s y m m e t r y )

= l : I h + l : ( A y 2 ) - A ( ; ) 2 ' " . 0 3 3 9 5 + . 1 0 1 8 5 - . 3 4 5 6 ( . 4 1 9 ) 2 a 0 7 5 1 i n 4'" I h b y s y r n m et r y .

P h ' " Pv .,. VT ,r.om . 4 6 6 i nV~3a56 'b ) T e eC D I n ' " b h 3 a . 1 2 ( 1 . 5 ) 3 a . 0 3 3 7 5 1 n 4IT 12 t h 1 . 8 8 ( . 1 2 ) 3 , . . 0 0 0 2 7 i 412E l e m e n t b t A Y A y A y 2 I h

1 1 . 5 . 1 2 . 1 8 . 7 5 . 1 3 5 . 1 0 1 2 5 . 0 3 3 7 52 1 . 8 8 . 1 2 . 2 2 5 6 . 0 6 . 0 1 3 5 4 . 0 0 0 8 1 . 0 0 0 2 7l : .~O!5 . I Q S a . I O ~ O o . C 3 a O ~A ' " . 4 0 5 6 1 n 2

y . '" ~.(A ) . 1 4 8 5 4 . 3 6 6 i n=tr: . a o s s

TI1.5

X a 0 b e c a u s e if t h e r e i s a n a x i s o f s y m m e t r y . t h e c e n t r o i d l i e s o n t h a tiil'S':"A 7 - 2


23/131

[ ( A y 2 ) - A ( y ) 2 = . 0 3 4 0 2 + . 1 0 2 0 6 - . 4 0 5 6 ( . 3 6 6 ) 2 = . 0 8 1 7 i n . 4I = = .'12(2.0)3 + 1.38(.12)3 = .08 + .0002 = .0802 in.41 2 1 22=~ : h = y ~ = .4488 in.~ = V : v y : ~ ~ ~ g. 4 4 4 7 i n .

(1) I h = bh3 = . 2 0 ( 1 . 5 ) 3 = . 0 5 6 2 5 i n 4IT 12( 2 ) I h = . 8 0 ( . 2 0 ) 3 = . 0 0 0 5 3 i n 41 2(3) I h = 1.2 ( . 2 0 ) 3 = . 0 0 0 8 i41 2

y I r . - . , _ . ,T.~1

c ) C h a n n e 1Because this section is not symmetrical, it.is necessary to tabulate x terms as well asy terms.

~---...J-;cf.-/.~ -1Ele, b t A Y A y A y 2 I h x A x A x 2 I v .1 1 . 5 . 2 0 . 3 0 . 7 5 . 2 2 5 . 1 6 8 7 5 . 0 5 6 2 5 . 1 0 . 0 3 . 0 0 3 . 0 0 12 . 8 0 . 2 0 .16 1.4 . 2 2 4 . 3 1 3 6 . 0 0 0 5 3 . 6 0 . 0 9 6 . 0 5 7 6 . 0 0 8 5 33 1.2 . 2 0 . 2 4 . 1 0 . 0 2 4 . 0 0 2 4 . o o o a o . 8 0 . 1 9 2 . 1 5 3 6 . 0 2 8 8~ :47! ~ 5 . L l S ' S B :Jm :TIl! . L l : JS j jA = . 7 0 i n 2

i= Z ( A y ) = . 4 7 3 = . 6 7 6 i n .[A -:iO-x = [ ( A x ) = . 3 1 8 = . 4 5 4 i n .E A --:iOIh = dh + d A y 2 ) - A ( y ' ) 2 = . 0 5 7 5 8 + . 4 8 4 7 5 - . 7 0 ( . 6 7 6 ) 2 = ~ ;n4I v = E I v + E ( A x 2) - A ( x ) 2 = . 0 3 8 3 3 + . 2 1 4 2 - . 7 0 ( . 4 5 4 ) 2 z . 1 0 8 2 ; n 4

A 7 - 3


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'h = \ !I;= \1.2225 = . 5 6 3 & i n .I -i V-'iOj =_J_ i~ = ~ . 1 0 8 2 = . 3 9 3 2 i n .V i! - .70

7. 4 Th is p r o b l e m i s s o l v e d t h e same w a y a s p r o b l e m 7 . 3 . ~ A.,Y l = . . y . s i n 4 5 +fs i n 4 5 " , . . 5 7 3 i n . ~/S( A~/,'1!~(4~, -):.Y2 = 1.38 s i n 4 5 + .1 2 s i n 4 5 . . . 5 3 0 i n . ~ ) " Yz ~" ",.-z 2 - + * - . ~ , ~ - - ) : .F i n d Ih o f t h e t w o e l e m e n t sf r o m t h e e q u a t i o n o f p a r a g r a p h 7 . 4 . 4 .E l e m e n t C D l m a x = t b 3 = . 1 2 ( l . 5 ) 3 = . 0 3 3 75 i n 41 2 1 2

1 m i n = b t 3 = 1 . 5 { . 1 2 ) 3 = . 00 0 2 1 6 i n 41 2 1 21 4 50 = I m a x c o s 2 4 5 + 1 m i n s i n 2 4 5

= . 0 1 6 8 7 5 + . 0 0 0 1 0 S s . 0 1 6 9 8 3 i n 4E l e m e n t l m a x = t b 3 = . 1 2 ( 1 . 3 S } 3 = . 0 2 6 2 S 0 i n 4I T 1 2

I = b t 3 = 1 . 3 8 ( . 1 2 ) 3 = . 0 0 0 1 9 9 i n 4m 1 n 1 2 1 21 4 5 = l m a x c o s 2 4 5 + I m i n s i n 2 4 5

= . 0 1 3 1 4 0 + . 0 0 0 0 9 9 . . . 0 1 3 2 3 9 i n 4E l e . b t A y A y A y 2 I h1 l . 5 0 . 1 2 . 1 S . 5 7 3 . 1 0 3 1 4 . 0 5 9 1 0 . 0 1 6 9 8 32 1 . 3 8 . 1 2 . 1 6 5 6 . 5 3 0 . 0 8 7 7 7 . 0 4 6 5 2 . 0 1 3 2 3 9~ . 1 D g I . I D 5 6 2 . 1 : I 3 D 2 2 2-'J = E ( A y ) , . . 1 9 0 9 1 = . 5 5 2 i nto( 3456

i . 1 h + E ( A y 2 ) - A ( y ) 2 , . . 0 3 0 2 2 2 + . 1 0 5 6 2 - . 3 4 5 6 { . 5 5 2 ) 2x' =

o= . 0 3 05 3 6 i n 4= ,fT = ~ . 0 3 0 5 3 6 = . 2 9 7 i n .V 1 : .3456

A 7 - 4


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7 . 5 T h i s p r o b l e m i s s o l v e d t h e s a m e w a y a s p r o b l e m 7 . 3 T h e a r e a m a y b ed i v i d e d i n t o e l e m e n t s i n a n y o f s e v e r a l w a y s b u t t h e a n s w e r s w i l l be t h esam e ,d ) A n g l e(1) I h = . 1 2 ( 1 . 3 8 ) 3 = . 0 2 6 2 8 1 i n 412Ele. b t A Y A y A y 2 Ih1 1. 3 8 . 1 2 . 1 6 5 6 . 8 6 . 1 4 2 4 1 6 . 1 2 2 4 7 8 . 0 2 6 2 8 12 1.5 0 . 1 7 . 2 5 5 . 0 8 5 . 0 2 1 6 7 5 . 0 0 1 8 4 2 . 0 0 0 6 1 4-:4m . 1 5 ~ 0 9I . I 2 ~ ~ 2 0 . 0 2 5 S 9 5

( 2 ) I h = 1 . 5 0 ( . 1 7 ) 3 = . 0 0 0 6 1 4 i n 412 _, 1--.12C D T1 .3 3A = . 4 2 0 6 in2

,17

~~~a j_f--/.SO -II

y = .: (Al) = . 1 6 4 0 9 1 = . 3 9 in- ? A . 4 2 0 5::h + L ( A y 2 ) - A G ' ) 2 = . 0 2 6 8 9 5 + . 1 2 4 3 2 0 - . 4 2 0 6 ( . 3 9 ) 2

= . 0 8 7 2 4 in4

( l ) I h = l . S 9 ( . 0 6 3 ) 3 = . 0 0 0 0 3 9 i n 41 2( 2 ) I h = . 8 0 ( . 2 0 ) 3 = . 0 0 0 5 3 3 i n 41 2(3 ) I h = .2 0 ( 1 . 5 0 ) 3 = . 0 5 6 2 5 i n 41 2(4) I h = l . 2 0 ( . 2 0 ) 3 = . O O O S i n 41 2

b) C h a n n e l

E le b t A y A y A y 2 Ih1 l . 8 9 . 0 6 3 . 1 1 9 - . 0 3 1 5 - . 0 0 3 7 4 9 . 0 0 0 1 1 8 . 0 0 0 0 3 92 . S O . 2 0 . 1 6 - . 1 6 3 - . 0 2 6 0 8 . 0 0 4 2 5 1 . 0 0 0 5 3 33 1.5 0 . 2 0 . 3 0 - . 8 1 3 - . 2 4 3 9 . 1 9 8 2 9 1 . 0 5 6 2 5 - I-4 1 . 2 0 . 2 0 . 2 4 - 1 . 4 6 3 - . 3 5 1 1 2 . 5 1 3 3 3 7 . 0 0 0 8:m- - . b 2 ~ 9 ~ ~ . 7 I 5 9 ~ 7 . 0 5 7 5 2 2A = . 8 1 9 i n 2-y = [ ( A y ) = - . 6 2 4 8 4 9 = - . 7 6 3 i n[A .819I = ~ I h + E ( A y 2 ) - A ( i ) 2 = . 0 5 7 6 2 2 + . 7 1 5 9 9 7 - . 8 1 9 ( . 7 6 3 ) 2

= . 7 9 7 2 1 n 4

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(1) Ih = .0018 in4( 2) I h = .000032 i n4(3) Ih = .045733 in4(4 ) Ih = .000933 in4

j.1:f7.6 T h i s p r o b l e m i s s o l v e d the s a m e w a y a s p r o b l e m 7.5A g a i n , t h e r e a r e m a n y w a y s t h e a r e a c a n b e d i v i d e di n t o e l e m e nt s .

I E 1 e I b t A Y Ay Ay2 Ih1 : .80 .30 .24 -.15 -.036 .0054 .00182 i 1.80 .060 .108 -.03 -.00324 .000097 .0000323 I 1.40 .20 .28 -.70 - .196 .1372 .0457334 I 1.40 .20 .28 -1.50 -.42 .63 .000933I ~ - .o~~2 i t . 7725~? .1jila4~SA = .908 i n 2-y = ; : ( f { ) = -.65524 = -.722 in. . . .908I = ;: Ih + E (Ay2) - A(y)2 = .048498 + .772697 - .90S(. 722)2

= .3479 1n47 . 7 T h i s p r o b l e m i s m o s t c o n v e n i e n t l y s o l v e d u s i n ga t a b l e s i m i l a r t o t h o s e o f p r e v i o u s p r o b l e m s .

(l) Ih = 2.0(.125)3 = .0003 in41 2(2) Ih = .125(l.0)3 = .0104 in412(3) Ih = .070(3.65)3 = .2837 in41 2(4) Ih = 2.0(.100)3 = .0002 in41 2(5) Ih = 2.0( .100)3 :: .0002 in412

L f . 1

.100

(6) Ih = 1.8(.060)3 = .0000 1n412

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\E1 e I b t Area y Ay Ay2 I h1 i 2.0 .125 .25 4.0375 1.0094 4.0754 .00032 I 1.0 .125 .125 3.475 .4344 1. 5095 .01043 i 3 .65 .070 .2555 2.025 .5174 1.0477 .28374 I 2.0 .100 .20 .15 .0300 .0045 .0002'I5 I 2.0 .100 .20 .05 .0100 .0005 .00020 l . B .060 .108 .03 .0032 .0001 ---I r. IJSs 2.00~~ 1 5 . 5 j 7 7 . 29itSa ) Area A = 1.1385 in2b) Centroid i= [(Ay) = 2.0044 = 1.76 in.::A 1.13Mc) Moment of inertia I = LIh + !;(Ay 2) - ACy }2

= .2948 + 6.6377 - 1.1385(1.76)2= 3.406 in4

d ) First moment of area of the skin, Q = Ay.Q 5 , 6 = {Ay}S ~ (AY)6 where y~iS measured from the centroid.

= .2 0 (1.76 - .05) + .108(1 .76 - .03)= .5288 in3

e) First moment of area of inner tee cap Q = Ay.Ql,2 = (AY)1 + (AY}2

= .25(4.0375 - 1.76) + .125(3.475-1.76)= ,7838 in3

Note&: For calculating first moment of area, y is either ( 7 - Ytable) or(y table - 7),

A7-7


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P r o b l e m 8 . 1

S O L U T I O N S T O P R O B L E M SL E S S O N # 8 F A S T E N E R S A N D J O I N T S

f

12

9

h

jF a s t e n e r

M e m b e r P a 1 1

5 2 05 7 5

1 5 5 02 0 5 3~ U 89 7 4 0

S o u r c e ( p a g e n o . a r e i n M A C 3 3 9 )P g . 1 . 1 3 b o t t o m t a b l e f o r e / D = 2P g . 1 . 1 3 b o t t o m t a b 1 e f o r e / D = 1. 5P g . 1 . 1 4 t a b l eP g . 1 . 1 3 b o t t o m t a b l e f o r e / D ~ 2P g . 1 . 1 5 b o t t o m t a b l eN o t r e q u i r e d p e r N o t e 2 o n p g . 1 . 1 5P g . 1 . 1 5 . 0 1 A i r F o r c e v a l u eP g . 1 . 1 3 b o t t o m t a b l e , e / D = 2P g . 1.2 3 s h e a r s t r e n g t h t a b l e & f : : : . . .P b r a1 1 = D t F b r u = . 1 8 7 5 x . 0 9 0 x 1 6 3 . 0 0 ~P g . 1 . 3 8 f o r a 110 b o l t 1\P b r a l l = D t F b r u = . 1 8 7 5 x . 0 5 0 . x 2 6 1 . 0 0 0 ~P g . 1 . 3 9 . 0 1 f o r a 5 / 3 2 i n . h i - 1 0 k l\P b r a l l = D t F b r u = . 1 5 6 x . 0 6 3 x 1 2 6 . 0 0 0 ~P g . 1.45 f o r a 3 / 1 6 " s t e e l j o - b o l t ~P b r a l l = D t F b r u = . 2 0 0 x . 0 6 3 x 1 6 3 , 0 0 0 ~( N o t e t h a t t h e df a , o f a 3/16 j o - b o l ti s . 2 0 0 " )P g . 1 . 4 5 f o r a 3/16" a l u m i n u m j o - b o l t( S a m e a s p r o b l e m 8 . 1 h )P b r z D t F b r u = . 3 7 5 X . 1 9 0 X 1 2 1 ' 0 0 0 ~P b r : l l D t F b r u . 3 7 5 x . 1 9 0 X l 1 4 , O O ~P g . 1 . 4 5 . 0 1 s h e a r s t r e n g t h

J o i n t P a 1192711

4 7 0 #778#

5 2 0 #26901

238011

123811

1 7 6 5 *

15501

81201

a

12

1 0 9 99 2 74 7 05 5 1778

T h e v a l u e 29201 i n t h e u p p e r t a b l e i s b e l o w t h e s m a l l l e t t e r" a " w h i c h i n d i c a t e s t h a t i t i s h i g h e r t h a n t h e s t r e n g t h f o r a1 6 0 k s i f a s t e n e r a s e x p l a i n e d i n N o t e 2 o f M A C 3 3 9 . p g . 1 . 2 3 .

121212

2 6 9 02 7 5 0

F b r u f r o m t h e t a b l e f n L e s s o n N o . 6 .

12

2 3 8 02 4 4 71 3 8 41 2 3 81 7 6 52 0 5 3

b

c

d

e

1212

1212

A 8 - 1


29/131

P r o b l e m 8 . 2M e m b e r Pa 1 1 S o u r c e ( p a g e n o . a r e m M A C 339) J o in t P ~ l l

a 1 676 P g . 1.12 J 6762 109~ P g . 1.13 121.000 +10993 181 1 .080 x 1875 x 1 "1 n f lF a s t e n e r 1180 P g . 1.12 o r 1.13 ( p e r s h e a r f a c e )b 1 2035 ,050 x ,156 x 261.00~2 1591 .050 x .156 x 204.00 2 22l2#3 2212 .100 x .156 x 141,80F a s t e n e r 1820 P g . 1.22 ( p e r s h e a r f a c e )c 1 3780 .125 x .250 x 121.00~ 3780_2 5748 .190 x .250 x 121,00 +46603 9813 .250 x .250 x 157,00 ~#F a s t e n e r 4660 P g . 1.39

M A C 339 g i v e s a v a l u e o f 11801 f o r a 006 r i v e t i n .080. b u tt h a t i s f o r a s i n g l e s h e a r j o i n t . B e c a u s e t h i s j o i n t i sl o a d e d i n d o u b l e s h e a r , t h e l o a d i n m e m b e r 3 i s n o t l i m i t e dt o t h e s i n g l e s h e a r v a l u e .F b r u f r o m t h e t a b l e i n L e s s o n N o . 6 .B e c a u s e e/ D i s l e s s t h a n 1.5, t h e F b r u m u s t b e c a l c u l a t e d .O n p g . 12.12 o f M A C 339. e n t e r t h e c h a r t w i t h e/ D = 1.2.M o v e h o r i z o n t a l l y t o t h e R ! D c u r v e a n d f r o m t h a t p o i n tm o v e v e r t i c a l l y t o r e a d F b r ! F t u = 1.02. F r o m t h e t a b l e i nL e s s o n N o . 6 , F t u = 139,000 p s i . T h e r e f o r e :

F b r = 1.02 x 139.000 = 141.800 p s iffi N o t e t h a t m e m b e r 2 c a n p i c k u p 5748. b u t t h e f a s t e n e r i so n l y c a p a b l e o f t r a n s f e r r i n g 46601.

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P r o b l e m 8 . 3

_ ~ t _ R _ o o m T e m p e r a t u r e A t E 1 e v a teo T e m p e r a t u r eM e m b e r P C B S o u r c e J O l n t ! F a c t o r : : I o u r c e t O ~ : & J O l n tP a 11 x 2 P i t11a 1 4 7 0 P g 1 . 1 4 4701 . 9 1 2 P g 1 . 8 3 , 2 0 2 4 4 2 9 4 2 9 *2 5 7 5 P g 1.1 3 . 7 7 6 P g 1 . 8 3 , 7 1 7 8 4 4 6R i v e t 5 9 6 P g 1.13 . 8 8 5 P g 1 . 8 1 , 2 0 1 7 5 2 7b 1 4 7 0 P g 1 . 1 4 4701 . 8 4 8 P g 1 . 8 3 , 2 0 2 4 3 9 92 5 7 5 P g 1 . 1 3 . 4 5 0 P g 1 . 8 3 , 7 1 7 8 2 5 9 2 5 9 #K i v e t 5 9 6 P g l.1 3 . 6 5 0 P g 1 . 8 1 , 2 0 1 7 3 8 7c 1 4 2 9 0 P g 1 . 3 9 . 0 6 4 2 9 0 . 7 5 3 Pg 1.84 ~ 3 2 3 0 3 2 3 0~ 4 7 8 8 . 0 6 3 x . 2 5 x 3 0 4 , 0 0 0 + 4 6 6 0 . 6 5 L e s s o n ' 8 1 3 1 1 2 + 3 1 1 23 1 0 0 4 0 . 1 6 0 x . 2 5 x 2 5 1 , 0 0 0 -mt!I . 8 7 5 L e s s o n # 8 2 8 7 8 5 b " J l ' 2 ' ,W i - 1 0 k 4 6 6 0 . 8 4 3 9 1 4

P a g e n u m b e r s a r e f r o m M A C 3 3 9 .F o r 3 0 1 s t . s t . 1 / 2 h a r d t h e t e m p . f a c t o r s a r e . 6 7 f o r6 0 0 0 a n d . 5 9 f o r 8 0 0 0 F o r 6 5 0 , t h e f a c t o r i s. 6 7 - ( 6 5 0- 6 0 0) ( . 67 - . 5 9 ) = . 6 5 .8 0 0 - 6 0 0F o r a l l o y s t e e l t h e t e m p f a c t o r s a r e . 9 2 f o r 6 0 0 a n d . 7 4 f o r 8 0 0 .F o r 6 5 0 0 t h e f a c t o r ; s . 9 2 - ( 6 5 0 - 6 0 0 ) ( . 9 2 - . 7 4 ) = . 8 7 5 .B O O - 6 0 0

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8.4 a) The loads in the fasteners are proportional to the fastenercross-sectional areas which are proportional to the squares of theu iameters.

F a s t en e r no. 0 O Z1 .1875 .03522 .25 .0625

~Fastener III P1 = (~p = .0352 x 5000 = 180111- \ ~ (02 ) .0977 -Fastener #2 P2 = ( 02 ) p = ~ x 5000 = 3199#

- L : (02) .0977 -b) Following the same procedure as for problem a) above:

Fastener no. 0 0 21 .156 .02432 .156 .02433 .1875 .03524 .1875 .0352-:TTn

R iv ets II I & 1 1 2 PI = P2 = ( L)p = .0243 x 2750 = 562#E (02 ) .1190

R; vets #3 & #4 P3 = P4 = ( L)p = .0352 x 2750 = 813#E(02) .1190c) Again, the load in the fasteners is proportional to the diameters

squared. Find the proportion of the load P in each fastener and thenfind the location of P by the laws of statics.

Fastener no. D 0 2 O Z I r . ( D 2 )1 .260 .0676 .50262 .164 .0269 .20003 .200 .0400 .2974~

~M3 = Pa = .5026Px2.0+.2000Px1.00 = 1.205 Pa = 1.205 in.d) The allowable load for the joint is the lowest load that produces azero margin of safety on the critical fastener. For each fastenerfind the total load P that would cause failure of that fastener. Thelowest of these three loads is Pall.

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C D 0/0astener no. 02/r.(02) PSal1*1 .5026 4500 89532 .2000 1680 84003 .2974 2620 8910

* From MAC 339 pages 1.45.01Pall = 8 400* based on fastener *2Pl = .5026 P = .5026 x 8400 = 4222*P z = .2000 P = .2000 x 8400 = 1680*P3 = .2974 P = .2974 x 8400 = 2498*

8.5 Find the centroid of the fastener pattern and the moment about thatcentroid.a) Because of symmetry, the centroid of this pattern is at pointshown. The distance from thecentroid to each fastener is:

"c " a s

e (2.0 2 + 1 .5 2 )1/2 = 2.5 in.P = M e A

1 : (e2A )= M because all fasteners~ have the same A and E.

b) 8ecause the structure can onlycarry axial loads, the momentmust be reacted by a couplea 5 shown. Then:

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c) Because the pattern is not symmetrical, the centroid must becalculated. The value 1.0 can beused for rivet areas because allrivets are the same size.

- _ ~ (A x ) _ 2(1.Ox2.0)+1.0x4.0 = 1.60"x - ~ - 5 . 0e = G _ x - x ) 2 + (y_y)~1/2

MeA Mep = -L (e 2 A ) - ~

I~o _ l _-~ C~

0, ~'&c - - _ o fy I. 6 " " i0 0

- = ~:(_!tl :' 2 (1 .0 x2 .0 ) : :1 08 "Y ~ 5 . 0 .

For each fastener, the values of e and p are calculated in thefollowing table.

Fas tener e2 e Me P1 1.602 + .802 = 3.20 1.789 17,890 1 1 1 8 H2 .402 + .802 = .80 .894 8,940 -mJ3 2.402 + .802 = 6.40 2.530 25,300 TmJ4 1.602 + 1.202 = 4.00 2.000 20,000 ~5 .402 + 1.202 = 1.60 1.265 12,650 -mIl" Y b ' : m -

d) Separate the unsymmetrical appliedanti symmetric components. Thesymmetric load is the 60001 loadand the antisymrnetric load is themoment caused by the load not bei nga t the cen tro i d

load into its symmetric and

M = 6000 x .50 = 30001'0 _L

~ 1,00

~ooo ~ ~300o:'~ 0 __L_

ooFind the fastener loads due tothe synvne tri c load.

P = 6000 = 15001P - , -Fi nd the fastener loads due to the antisymmetri c load.

Pm = MeA = ME (e 2 A) Ee _( l . 5 2 + l.02)1/2 1.803 in.=

Pm. 3000 &41614xl.A03Combine symmetrical and antisymmetrical loads.

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F o r f a s t e n e r s n o . 1 a n d 2 ::) = 9 0 - a r c t a n (l.0/l.5) = 5 6 . 3 1

B y t h e l a w o f c o s i n e s :p 2 = p p 2 + P m 2 + 2 P p P m c o s 8= 1 5 0 0 2 + 4 1 6 2 + 2 ( 1 5 0 0 ) ( 4 1 6) c o s 5 6 . 3 1 = 3 , 1 1 5 , ' 3 0 0P I = P 2 = 1 7 6 5 #

F o r f a s t e n e r s n o . 3 a n d 4 :I) :: 9 0 + a r c t a n (1.011.5) = 1 2 3 . 6 9

B y t h e l a w o f c o s i n e s :p 2 = p p 2 + P m 2 + 2 P p P m c o s 8: : 1 5 0 0 2 + 4 1 6 2 + 2 ( 1 5 0 0 ) ( 4 1 6) c o s 1 2 3 . 69 - = 1 , 7 3 0 , 8 0 0P 3 = P 4 = 1 3 1 6 #

8 . 6 al F o r i n f i n i t e l y s t i f f f a s t e n e r s , t h e e l o n g a t i o n o f b o t h s t r a p s b e t w e e nf a s t e n e r s n o . 1 a n d 2 m u s t b e t h e s a m e . T h e e q u a t i o n f o r e l o n g a t i o ni s :, P L" = A r

B e c a u s e t h e L, A . E a n d 6 o f b o t h s t r a p s a r ef a s t e n e r s n o . l a n d 2 , t h e ? m u s t a l s o b et h e s a m e . T h i s i s o n l y p o s s i b l e i f 50S o ft h e l o a d i s t r a n s f e r r e d b y t h e e n df a s t e n e r . T h e s a m e i s t r u e f o r t h e o t h e re n d o f t h e s p l i c e . T h e r e f o r e , t h e l o a dd i s t r i b u t i o n i s a s s h o w n i n t h e t a b l e .

t h e s a m e b e t w e e n

2 03 04 . 5 ?

b) T h e r e f e r e n c e d t a b l e s h o w s t h a t f o r f o u r f a s t e n e r s , e a c h e n d f a s t e n e rw i l l t r a n s f e r 3 7 . 5 % o f t h e l o a d . T h e r e f o r e ,t h e l o a d d i s t r i b u t i o n i s a s s h o w n i n t h et a b l e .

A S - 7

2 . l2SP3 . l 2 S P4 .37SP


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c ) F o r i n f i n i t e l y s t i f f f a s t e n e r s :6a = 6b

a n d 6 = P L a s s h o w n i n L e s s o n #3.A E( ~ ) a = (~)b. 2 0 P L = . S O P Lwt;t wt;;r. 2 0 = . S Ot; tb

t b = . S O = 41; -:1nd ) T h e e n d f a s t e n e r s c a r r y a l l t h e l o a d . a s t h e l o a d i n c r e a s e s . u n t i lt h e b e a r i n g y i e l d s t r e s s ( F b r y ) i s r e a c h e d . A s t h e l o a d c o n t i n u e s t oi n c r e a s e , t h e h o l e s i n t h e b e a r i n g - c r i t i c a l p a r t w i l l e l o n g a t es l i g h t l y . T h i s y i e l d i n g a d d s t o t h e d e f l e c t i o n = ~ o f t h e s t r a p ,

a l l o w i n g t h e t o t a l d e f l e c t i o n o f t h e t w o s t r a p s to r e m a i n e q u a l e v e nt h o u g h t h e l o a d s a r e n o t e q u a l . T h i s p e r m i t s t h e m i d d l e f a s t e n e r s t ol o a d u p u n t i l t h e y . a l s o , r e a c h t h e y i e l d p o i n t . A s t h e l o a di n c r e a s e s f u r t h e r , t h e f a s t e n e r l o a d s Increase e q u a l l y u nt i l t h ef a i l u r e p o i n t i s r e a c h e d . T h e r e f o r e , j u s t b e f o r e f a i l u r e :Pl = P 2 = P3 = P4

B . 7 a ) T h e l o a d o n t h e c l i p ; s t h e r e a c t i o n a t o n e e n d o f t h e b e a m .P = w L = S O x 2 0 . 0 = B a a l'2 2

T h e r i v e t l o a d s a r e f o u n d b y t h e l a w s o f s t a t i c s w i t h t h e l o a da p p l i e d a t t h e a p e x o f t h e c l i p a n g l e .E M2 = 0 = .40 x B O O - l.6PlP I = 2 0 0 *

E F v = 0 = P 2 - s a o - 2 0 0P 2 = 1 0 0 0 #

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~ 3 = 0 = . 6 0 x 8 0 0 - . 9 0 P 4P 4 = 533#F v = a = P 3 - 8 0 0 - 5 3 3P 3 = 1 3 3 3 #

F o r e a c h r i v e t l o c a t i o n . f i n d t h e s m a l l e s t r i v e t t h a t i s s t r o n ge n o u g h i n s h e a r a n d c h e c k t h e a l l o w a b l e l o a d i n t h e c l i p a n d i n t h em a t i n g s t r u c t u r e .A

R i v e t P T r i a l R i v e t P a l l m M a t i n g P a l l &n F a s t e n e r M . S .N O . i n C l i p S t r u c t u r e S t r u c t u r e P a 11 &1 2 0 0 A D 4 (BJ4) 3 8 8 . 1 2 5 7 1 7 8 - 7 6 3 8 8 3 8 8 +.942 1 0 0 0 0 0 6 ( C X 6 ) 1 1 1 6 . 1 2 5 7 1 7 8 - 7 6 1 1 8 0 1 1 1 6 + . 1 23 1 3 3 3 D D S (CXB) 1 5 0 1 . 0 4 0 7 1 7 8 - 7 6 1 6 3 0 2 1 5 0 1 +.134 5 3 3 A D 5 ( B J 5 ) 5 9 6 . 0 4 0 7 1 7 8 - 7 6 5 7 5 5 7 5 + . 0 8

~ F r o m M A C 3 3 9 P g . 1 . 1 2 a n d 1 . 1 3 e x c e p t a s s h o w n .& . P a l l = t D F b r u = . 0 4 0 x . 2 5 0 x 1 6 3 . 0 0 0 = 1 6 3 0 ' w h e r e F b r u i s f r o m t h et a b l e i n L e s s o n N o . 6 .it.:~.S. = (Pall/P)-lb) T h e l o a d i s 8 0 0 # a s i n P r o b l e m a l . T h e r i v e t l o a d s a r e f o u n d a sf o l l o w s . T h e v e r t i c a l a p p l i e d l o a d i s d i v i d e d e q u a l l y b e t w e e n t h et w o r i v e t s i n a f l a n g e b e c a u s e t h e y a r e o f e q ~ a l s t i f f n e s s . T h em o m e n t i s r e a c t e d a s a c o u p l e o n t h e t w o r i v e t s . T h e t o t a l l o a d o ne a c h r i v e t i s t h e r e s u l t a n t o f t h e s e t w o l o a d s .

P \ I

~ - f L t W O O ~t P vf I , -

F o r r i v e t s ' 1 a n d ' 2 :P v = P = 8 0 0 = 4 0 0 1" 2 " T:-:M2 = 0 = 8 0 0 x . 4 0 - !h

1.6R e s u l t a n t l o a d :

P I = P 2 = ( 4 0 0 2 + 2 0 0 2 ) 1 / 2 = 4 4 7 '

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F o r r i v e t s # 3 ~ n d # 4 :

M4 = 0 = 800 x .6 0 - fh- : - g O f'h

P v = P = 800 = 400#'1 Z

P h = 3 0 0 1 1P3 = P4 = (4002 + 3002)1/2 = 500# P . , ;

R i v e t P T r i a l R i v e t P a l l &a t i n 9 P a l l &n F a s t e n e r ~.N o . i n C l i p S t r u c t u r e S t r u c t u r e P;1l11 & 2 447 A D S (BJ5) 594 .125 7178-76 596 594 + . 3 33 & 4 500 A D S (BJ5) 594 .040 7178-76 575 575 + .15

& , F r o m MAC 339 Pg . 1.12 a n d 1.13.&M.S . = (Pall/+1

: . M 2 = 0 = .30 X 1500 - 2.0 P hP h = 2251

c) T h e v e r t i c a l l o a d i s d i v i d e d e q u a l l yb e t w e e n t h e t h r e e r i v e t s b e c a u s e t h e ya r e o f e q u a l s t i f f n e s s .P v = 1500 = 500*--r

R e s u l t a n t r i v e t l o a d s a r e :P I = P 3 = (5002 + 2 2 5 2 )1/2 = ~P2 = P v = 500*

.~

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8.3 ~) Because the clip has only a singl~ bolt rather than a row of bolts,the effective bolt spacing is the clip width which is 1 inch. Bolthe~d clearance is O.lB in. and thickness is .090.~nter the chart on MAC339 page 12.01 with t = .090. Go upvertically to the curve for bolt spacing = 1.0" and from there to theright horizontally to the curve for bolt head clearance = .18(interpolating between the .15 and .20 curves). Move verticallydownward from this point and read yield load per bolt of 600 lbs.

Pallult: 1.5 Pallyield = 1.5 x 600 = 9001

b) This problem is solved using MAC339 page 12.02. The parameters ~re:Tl=T2=.090, W=1.0 inch and e=.435 inch.Enter the chart with T1=.090. go up to the T2=.090 curve. right tothe 0=.25 curve and down to read ultimate load of 1.5 kips. This isthe a l l owab Ie load only if the clip has the following geometric pro-portions shown on MAC339 page 12.02: W=4D and e=1.75D. For thisclip, W=1.0=4 X .25=40 and e=.438=1.75 x .25=1.75D. Therefore. it isnot necessary to use the factors K1 and K2 of MAC339 page 12.00.

Pallult = 1.5 kips = 12QQ!c) This problem is solved using MAC339 page 12.01.01. The parameters


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C o m p u t e t h e v a l u e s o f :T I . 0 9 0- ,......- - - - = --- ..:...;.:~ -- --H - R / 2 - D / 2 . 4 3 5 - . 1 8 0 / 2 - . 2 5 / 2

~ = . 0 9 0 = . 3 6D . 2 5

= .409

T l . D 9 0- = - = . 5 0 , 'R . 1 8 0E n t e r t h e c u r v e i n M A C 3 3 9 o n p a g e 1 2 . 0 2 . 0 2 w i t h T lH - R / 2 - 0 / 2.409. G o u p v e r t i c a l l y t o t h e T1/D = . 3 6 l i n e . t h e n l e f th o r i z o n t a l l y t o t h e ll/R = .50 l i n e . M o v e d o w n f r o m t h i s poin t annr e a d t h e v a l u e o f P / D 2 = 5 3 k s i, S o l v e f o r P a l l = 5 3 ( 0 ) 2 = 5 3 ( . 2 5 ) L= 3 . 3 1 k i p s = 3 3 1 0 l b s . B o l t s p a c i n g = 4 D = 4 x . 2 5 = 1 . 0 .

=

P a l l o w u l t = 3 3 1 0 l b s .e ) E n t e r t h e c h a r t o n M A C 3 3 9 p a g e 1 2 . 0 2 w i t h T 1 = . 2 5 , go up t o t h ei n t e r p o l a t e d c u r v e f o r T2 = . 1 2 5 , r i g h t to t h e 0 = . 3 8 c u r v e ; j n d d o w n

t o r e a d . u l t i m a t e l o a d o f 4 . 6 k i p s . A s i n p r o b l e m b ) . t h i s i s t h ea l l o w a b l e l o a d o n l y i f W = 4 D = 4 x . 3 7 5 = 1 . 5 i n c h a n d i fe = 1 . 7 5 D = 1 . 7 5 x . 3 7 5 = . 6 5 6 i n c h .T h e a c t u a l v a l u e s a r e W = 1 . 2 5 a n d e = . 5 5 . T h e r e f o r e . u s i n g t h ee q u a t i o n s o n M A C 3 3 9 p a g e 1 2 . 0 0 :

n = W I D = ( 1 . 2 5 / . 37 5 )= 3 . 33 3K l = ( n - 1 ) / 3 = ( 3 . 3 3 - 1 ) / 3 = . 7 7 8m = e 1 0 = (. 551 . 3 7 5 ) = 1.4 6 7K 2 . = 1 . 7 5 / m = ( 1 . 7 5 / 1 . 4 6 7 ) = 1 . 1 9 3P a l l u 1 t = P c h a r t x K 1 x K 2 = 4 6 0 0 x . 7 7 8 x 1 . 1 9 3 = 4 2 7 0 ~

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3. 9 a) The allowable load is the allowable bearing load:

Fbr is the allowable bearing stress based on R I D or H I D , whichever iscritical.R/D = 1.00/1.00 = 1.0Fbr/Ftu = .78 (from MAC339 pg. 12.12)Ftu = 61,000 psi (from table in Lesson No.6)Fbr = .78x 61,000 = 47,600 psi (based on R I D )W I D = 2 . 0 0 / 1 . 0 0 = 2.0 (W = 2R .'. not critical)Pbrall = DtFbr = 1.00 x .750 x 47,600 = 35,700#

b) Tr,e allowable load is the allowable bearing load in member C D or C DDr the fastener shear strength, whichever is the lowest.~~ember C D allowable bearing load:e / D = . 3 5 / . 2 5 = 1.40 = R I Drbr/Ftu = 1.23 (from MAC339 pg. 12.12)Ftu = 86,000 psi (from table in Lesson No.6)Fbr = 1.23 Ftu = 1.23 x 86,000 ~ 105,800 psii~ember C D Pbrall = DtFbr = .250 x .10 x 105,800 = 2645#i-lember C D all owabl e beari ng load:e l D = .30/.25 = 1.20 = R I DFbr/Ftu = 1.02 (from MAC339 pg. 12.12)Ftu = 59,000 psi (from table in Lesson No.6)Fbr = 1.02 Ftu = 1.02 x 59,000 = 60,200 psir~ember Pbra 1 1 = DtFbr = .250 x .160 x 60.200 = 2408~Check the bol t:PSa 1 1 = 4650# per M A C 339 pg. 1.33

Thus, the joint strength is:Pall = 2408*

AS-13


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c ) f h e allowable lo~d i s the allowable bearing load on the lug.K/D = .65/.50 = 1.30Ftu = 88,000 psi per table in Lesson No.6F o r = 1.13 Ftu = 1.13 x 88.000 ='99,400 psi (from M A C 3 3 9 pq. 12.12)For W / D the grain direction is long transverse, as shown on M A C 3 3 gpage 12.12. The table on M A C 339 page 12.13 specifies the use Q f luacurve *2 for a 7178-T6 extrusion with LT grain direction. Therefore.on M A C 339 page 12.12:

W I D = 1.20/.50 = 2.40Fbr = 1.34 Ftu = 1.34 x 88,000 = 117,900 psi

If;1 was equal to zero the allowable load would be:Po = DtFbrmin = .50 x .375 x 99,400 = 13,640#

However:8 = 90 - 35 = 55Ps/P o = .828 per M A C 339 pg. 12.16.Ps = .828 Po = .828 x 18,640 = 15,430#

8.10 a) The allowable load is found by using the chart on page 12.15 of r ~ A C339. The parameters are Fbrl' Fbr2' e/O and the degree of clamp-up.For lugs ( ! ) . the female or outside lugs:

R / D = .90/.50 = 1 . SFbrl = 1.58 Ftu from M A C 339 pg. 12.12Ftu = 61,000 psi from table in Lesson No.6Fbrl = 1.58 x 61.000 = 96.400 psiw = 1 . S 0 = 2 R .. w Id , i s not cri t ica 1

" \ \\.\\AS-14 \


42/131

2-'

For lug , .~~he center or male lug:R I D = .60i~50 = 1.20Fbr2 = 1.02 Ftu = 1.02 x 61.00 = 62,200 psiW I D is not critical as for lugs C DThe maximum gap a t worst tolerance is:x-d = (1. 56+.03) - (1.50-.03) = .12

Because the male lug can be on one side of the slot the entire qa pcan be on one side. This would create a more critical bending momenton the bolt than for the male lug being centered, so use:e = .12elD = .12/.50 = .24

Enter the chart on page 12.15 of M A C 339 with Fbr = 62.2 ksi andmove upward to an interpolated curve for Fbrt = 96.4 ksi. From thispoint, move horizontally to the curve labelell "160 ksi unclamped"(the second curve from the ri ght) and then downward from that poi ntto an interpolated curve for elD = .24. From this point movehorizontally to the curve labeled "unclamped" and from there downwardto read K = .88. Then:Bolt Pall = KVallVall = 18,65U# from M A C 339 pg. 1.33Bol t Pall = .88 x 18,650 = 16,400#

Check 1 ug C D :Lug (!) Pall = 2DtFbr1 = 2 x .500 x .42 x 96,400 = 40,500#

Check 1 ug :LugPall = DtFbr2 = .50 x 1.50 x 62,200 = 46,650#

The allowable load is the lowest of these three allowable loads.Pall = 16,400#

AS-15


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b ) Tne torque is correct, as shown on r ~ A C 339 page 1.34, for c l amoinq-upwith a tension nut. Follow the same procedure as above except whengoing to the left from the Fbrl" curve, go to the "160 ksi cl am pe dtension nut" curve (the third from last curve). From that point movedownward to .the same interpolated e/D = .24 curve and from therehori zontally to the "clamped tension nut" curve and then downward toread K = 1.34. Then:Bolt Pall = 1.34 Vall = 1.34 x 18,650 = 24,990#

Bec~use the allowible lug bearing loads are larger than thisallowable bolt load :Pa 11 = 2 4 , 9 9 0 1 1

c) Permissible clamp-up (gap) is determined from M A C 339 page 12.14.The parameters are LIt, material and grain direction. The table onM A C 339 page 12.13 tells which curve to use on page 12.14 for variousalloys and grain directions. For 2024-T4 plate. curve number 5 isused regard1ess of grain direction.

LI t = 2.5/.42 = 5.95Enter the chart on M A C 339 page 12.14 with LIt = S.9S. Move upwardsto curve 5 and, from there, move to the left and read:

x- d = .10tAllowable qa p > (x-d) = .10t = .10 x .38 = .038 inch. The nominalgap is .06 and the maximum gap is .12

.'. C lamp-up is n ot pe rmissible .d) The gap in the sketch is .06 + .06, so the maximum gap is .12 inch.

The required lug length for tliisgap is found by reversing theprocedure of the preceding problem.x- d = .12~ = .12/.42 = .286t

Enter the chart on MAC 339 page 12.14 with (x-d)/t = .286 and movehorizontally to curve 5. From that point move downward to read:LI t = 10.1 Then:L = 10.1 t = 10.1 x .42 ~ 4.24 inches

A8-16


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- ,- cfSOLUT ION TO PROBLEMS

L E S S O N 9 C R I P P L I N G

P RO BLE M 9 .1

o Fir st find th e crit ical cr ipplina load an d str ass for t ha an gle ifit is alclad 7075-T 6 she e t me t al.

t& EDGE FC~l81118nt b bit C ON Do bt btFcc

~1.45 .1 14.5 1 e.f. 31.000 .145 4495.95 .1 9.5 1 a.f. 43.000 .095 !Qll7i4o 8580

it. Ref: M A C 339P I . 16.16 ~ C lad din g 2.5% pe r sid .P cc E bt Fcc 8580 lbs.

E bt F 'F cc ~ 35.750 p.icc E bt . 2 4o N ext . fin d t he cr itical crippUna load an d .t r ess for th e ang le ifit is bare 7075-T 6 sh ee t me tal

F 80.000 psi } fr om table in 19n 6cy ,E O 10.3 ]I; 106 psiF will be found for each flan ae usinl t h e curve on pag e 16.14 ofccM A C 339:


45/131

F cy( B . ) E D G E F c/ llE lem ent b t E t C O N D o F cy F cc bt btl' cc

f f i 1. 4 5 . 1 1.2 7 1 e. f . . 47 3 7 , 6 0 0 . 1 4 5 5 4 5 2. .5 . 1 . 8 4 1 e. f . .6 5 52 . 0 00 .0 9 5 4 9 40T40 1 0 m~ R e f: MA C 3 3 9. P g. 1 6 .1 4

Fcc= E bt F 1 03 9 2 lbs.ccE bt F cc - l Q d 2 l - 4 3. 3 0 0 p siE bt . 2 4

P cc

o Fin ally .find t h e crit ica l cr ip plin g loa d a n d st r e ss for th e a ng leif it is a 7 0 75-T 6 e xt ~ion .E D G E

FC ~lem ent b t bit C O N D o bt bt F cc

f f i 1.45 .1 1 4 : - 5 1 f 34 . 8 00 . 1 4 5 5 0 46. 9 5 . 1 9 . 5 1 f. 4 9. 0 0 0 . 0 95 4 6 55-:140 9 7 O i '~ Re f: MA C 3 39 . P g. 1 6 . 15Pcc - E bt F cc - 97 0 1 lbs.

E bt F .F ~ cc - 2lQ! - 4 0 , 4 2 0 PS1cc E bt . 24


46/131

P RO BLE M 9 .2o F ir st , fin d t h e cr it ica l cr ip p lin g lo a d a n d st r e ss for t h e 7 0 7 5-T 6e xt rud .d t

E D G BFC~l e m e n t b t bI t C O N D o b t btFcc

~1 . 4 6 9 . 1 5 9 . 8 1 . f . 4 7 , 0 0 0 . 2 2 0 1 0 . 3 4 01 . 0 063 1 5 . 9 1 f . 3 2 , 0 0 0 . 0 6 3 2 . 0 1 61 . 0 . 0 6 3 1 5 . 9 1 f . 3 2 , 0 0 0 . 0 6 3 2 . 0 1 6:346 1 4 . 3 7 2

&e f: MA C 3 3 9 , P g . 1 6 . 1 5Pcc 1 : bt F 1 4 , 3 7 2 lbs.ccF 1 : bt P cc 1 4 , 3 7 2 4 1 , 5 3 7 p s icc 1 : bt . 3 4 6

o N e xt , fin d t h e cr it ica l cr ip p lin g loa d a n d st r e ss for t he 7 07 5-T 6e xt rud ed c ha nn .l .

E D G B pc~l e m e n t b t bit C O N D o bt bt Pcc

~. 8 2 3 0 8 0 1 0 . 3 1 f 4 6 , 0 0 0 . 0 6 6 3 , 0 3 61.43 1 5 9 . 5 No e . f . 7 8 , 0 0 0 . 2 1 4 1 6 , 6 9 2. 8 2 3 0 8 0 1 0 . 3 1 f . 4 6 , 0 0 0 . 0 6 6 3 , 0 3 6. 3 4 6 2 2 , 7 6 4

~ R e f: HAC 3 3 9 , PI' 1 6 . 1 3P 1 : b t r 2 2 , 7 6 4 lb cc ccF t b t ' cc 2 2 . 7 6 4 6 3 , 7 9 2 p . icc I bt . 3 4 6


47/131

P RO BLE M 9 .3o F i n d t h e F e y an d E for e xt r ud e d 70 7 5-T 6 m a t e r i al a t 3 0 0F ( 1 0 h r .exposur e )

1 ) R oom t e m p e r a t ur e p rop e r t i es fr om t a b le in l e sson 6Fcy 6 1, 0 0 0 p si R ef: MI L-HD B K -5 , P g . 3-2 8 7

6E 1 0 . 4 x 1 0 p si2) T em pe ra tu re r ed uct io n f act or s

K F . 7 7 (MI L-H D BK -5, P g . 3-2 9 1)cyKE . 9 0 ( MI L-H D BK -5, P g . 3-2 93 )

F e y ~ Fe Y RT K e y 6 1 , 00 0 ( .7 7 ) - 4 6 , 97 0 p si6 6

E ~T K E 1 0 .4 x 1 0 ( . 9 0) 9. 3 6 x 10 p siF cc w il l b e foun d for e a ch fl a n g e usin g t h e cur ve on pa g e 1 6. 1 4 ofM A C 3 3 9 .

Fc y (~) E D G E FCfoE l e m e n t b t E t C O N D o F e y Fcc b t bt l ce

1. 4 69 . 1 5 . 6 9 1 e . f. . 7 6 3 5, 6 9 7 . 2 2 0 7 , 85 31 . 0 . 0 6 3 1 . 1 3 1 e . f. . 5 1 2 3, 9 5 5 . 0 63 1 , 5 0 91 . 0 . 06 3 1 . 13 1 e . f. . 51 2 3, 9 5 5 . 0 6 3 1 . 5 0 9:346 1 0 , 8 7 1

~ R e f: MAC 3 3 9, P g . 16 . 1 4pcc E bt P 10 87 1 lb s.ccF E b t F cc. 1Q!Il 31 ,41 9 p sie c E b t . 3 4 6

-,


48/131

9 . 4 U sa p. 1 6 . 2 8 .0 9 of MA C 33 9 :

EDGEElement b t bI t C O N D o s e : F e e F e cb t

~1 . 4 6 9 . 15 9 . 8 1 a . f. .2 2 0 8 3 , 00 0 1 8 , 2 6 01 . 0 . 0 6 3 1 5 . 9 1 a.f. . 0 6 3 5 6 , 0 0 0 3 , 5 2 81 . 0 . 0 6 3 1 5 . 9 1 a.f. ~ 56 , 0 0 0 3152 8. 3 46 2 5 ,3 1 6

p ~ E bt F 2 5 ,3 1 6 lb s.cc ccE b t FF __ . . .; : c~e2 5 , 31 6 7 3 ,1 7 0 psice E b t . 34 6

9 . 5 Usa p . 16 . 2 8 . 09 of MA C 3 3 9 :

EDGEElement b t b I t C O N D o b t Fe e F e e bti. 4 6 9 . 1 5 0 9. 8 1 e . f . . 2 2 0 72 , 0 0 0 1 5, 8 4 01 . 0 . 06 3 1 5 .9 1 a . f. . 06 3 4 9 , 00 0 3 ,0 8 71 . 0 .0 6 3 1 5. 9 1 a . f. . 0 6 3 4 9 , 0 0 0 31 8 7T46 2 2 , 0 1 4P E bt F 2 2 ,0 1 4 1 bs.ec ccFe e E b t Fcc. 2 2 , 0 14 63 . 6 2 0 p siE b t . 3 46


49/131

- -,\

9.6C D ,

EDGEEl e me n t b t bit C O N D o bt Fcc Fccbt

~

1.00 .080 12.5 1 e .f. .0 80 39 .000 3.1201.00 .08 0 12.5 1 e .f. .080 39.000 3.1201.425 .0 70 2 0.4 N o e . f .100 65 .000 6.50 0. 965 .070 13 .8 1 e .f. .068 36.000 2.448:m 15.188E bt FF ~ cc ,.cc ---=.E bt 1 5,188~ ~ 46. 300 psi.328 in

De t e r m in e t h e e ffect ive wid t h for t h e uppe r skins:( 1 E ) skin ,. 10.0 x 103KSr ,. 96.7 =:. 1 00J(~E) stiff s: 7 x lOJKSr

Usin g p. 1 6.13 of MA C 339:2w~ .. 35. 5t

For t h e uppe r skins wh ich ar e on e e d g e fr e e :w 2w_! .35 (~) .35 (3 5.5) ,. 12. 4t t

w = (12.4 )(.050 in ) . 62 in .e

11'-9


50/131

- - /.. .. .. c ;. -

T h e t otal wid t h of skin to use for t h e upper skins is:w 2(.62 in ) + 1.00 in. 2.24 in.AS wt (2.24 in)(.050 in .) .112 in 2

De t e rmin e t he e ffective wid t h for t h e lower skin :103KSI 101.5 .:= 100

1 = = i i : i i = a . . . . iii1"3"KS~IUsing p , 16.13

2w-!35.5tw 1 (35.5)(.050 in ) .89 in .e 2

A6 2(we )t 2(.89 in )(.OsO in ) .089 in 2So t h e tot al allowable cripplin g load is:Pcc. (46300 psi) (.328 in2 + .112 in2 + .089 in2) 24490 1b


51/131

9 . 7 .

EDGEEl e me n 1 b t bIt C O N D o bt Fcc Fccbt

1 1 . 0 0 . 08 0 1 2 . 5 1 a.f. . 0 8 0 3 9 . 0 0 0 3 . 1 2 02 1 . 0 0 .0 8 0 1 2 . 5 1 e .f. . 0 8 0 3 9 . 0 0 0 3 . 1 2 03 1 . 4 25 .0 7 0 2 0 . 4 N o e .f. . 1 0 0 6 5 . 0 0 0 6 . 5 0 04 . 9 65 . 0 7 0 1 3. 8 1 a.f. . 0 6 8 3 6 . 0 0 0 2 . 4 4 8':Jii 1 5 . 1 8 8F cc E F bt # cc 1 5 . 1 8 8 2 -46.300 p aiE b t . 3 2 8 inDe t e r min e t h e e ffe ct iv e w id t h for t h e uppe r ski ns:

Eski n -10.0 x 1 0 3 ICSI - 9 6 . 7 ~ 1 00~E) st iff b o . 7 x 1 0 3XSI

Usin g p. 1 6 . 1 3 of MA C 3 3 9 :2w_ _ " , ! - 3 5 . 5t

For t h e upp ar skin s wh ich ar e on e e d .e fr e e :w (2v_! .3 5 --1). 1 2 . 4t t

Si nce t h e se ar a ch a m-m ill e d skin s. d e t e r mi n e fr om wh a t poi n t t h e e ffect i vew id t h sh ould be m easur e d .( 1 ) ~(.O sOin ) - ( 1 2 . 4 ) ( . 0 5 0 i n ) - . 6 2 i n .t( 2) ~(. 0 2 0 in ) + .5 0 i n . - ( 1 2 . 4 ) ( . 0 7 0 in ) + . 50 i n - . 7 4 8 in .t

1 1 9 - / /


52/131

/."Since t h e e ffect ive wid t h of crit e ria (1 ) is le ss t h a n (2 ) . m e asur e t heeffe ct iv e wid th off of th e fa ste n e r.

So: A 5 2[( .5 0 in ) (. 0 5 0 in ) + ( . 12 in) ( . 02 0 in)] + ( .0 50 " )( 1. 0" ) . 10 5 in 2

D e t er m ine t h e e ffect ive wid t h for t h e lower skin:fr om pr oble m 9. 6 :

w2 _! 3 5 . 5tw_! 1 7 . 7 5t.

C he ck c ri te ri a. :(1 ) ~(. 0 5 0 in) (1 7 . 75 ) ( .0 5 0 in) . 89 in .t( 2) ( w. )( . 0 20 in ) + . 5 0 in. ( 1 7. 7 5 )( . 0 20 in ) + . 50 in . 8 6 in .t

Since cr it er ia ( 2) is le ss th a n 0 ) . m ea sure th e e ffe ctiv e wid th off of t hee nd of th e ch a m-m ill lan d .A 6 2[( . 5 0 in ) (. 0 5 0 in ) + 07 . 7 5) ( . 02 0 in) ( . 02 0 in) } 06 4 in 2

Th e tot a l allowa ble cr ipp ling loa d is:Pcc ( 4 63 0 0 psi) (. 3 2 8 in 2 + . 10 5 in 2 + . 0 6 4 in2 ) 2 3 01 0 lb

fJ9-IZ


53/131

9. 8 D ete rm in e t he cent roid of the section;

E lem ent b t A rea y A y A'l2 I hr r 2 .0 .1 25 .2 5 4 . 0 3 7 5 1.0094 4.0754 .0003~) 1.0 .1 25 .1 25 3 . 4 7 5 .4344 1.5095 .0104i s < 3 .65 .0 70 .2 555 2 .0 25 .5 174 1 .04 77 .2837I.0 .1 00 . 20 .1 5 . 03 00 . 0 0 4 5 . 0 0 0 22 .0 . 1 0 0 . 20 .0 5 . 010 0 .0 00 5 .0 00 2) 1 .8 . 06 0 . 108 .0 3 . 0 0 3 2 . 0 0 0 1 .- - -1. 138 5 2 .0 04 4 6.6377 :mi

a ) A re a A 1 .13 85 in2

b ) Centroid y . E ( A y ) 2 .0 044 1 . 7 6 in .tA 1.1385

& .90" - 15 ( thick ness)


54/131

F or t h e upp e r sid . of t h e se ct ion:

E D G EE lem ent b t bit C O N D o bt Fcc Fccbt

a 1 . 0 0 . 1 2 5 8 1 e. f. . 1 2 5 36 . 5 00 4 . 5 63b 1 .0 0 . 1 2 5 8 1 e. f. . 1 2 S 3 6 . 5 0 0 4 . 5 6 3c 1 .0 6 3 . 1 2 S 8. 5 1 f . 1 3 3 3 5 . 00 0 4 . 6 55d 1 . 6 9 .070 2 4 . 1 No e . f .1lS S7.500 6 , 7 8 5e . 4 0 . 07 0 5 . 7 1 a . f. . 0 2 8 7 4 . 0 0 0 2 . 0 7 2:ill 2 2 , 6 3 8

E F btFcc cct bt - 2 2 .6 3 8 4 2 . 79 0 p si S 'z .qF or t ha lowe r sid e of t h e se ction :

E D G E[Ele m en 1 b t bit C O N D o bt Fcc Fccbth 1 . 5 3 .0 7 0 21 . 9 N o f. . 10 7 6 2 . 00 0 6. 6 3 4f 1 . 00 . 1 0 10 1 a .f. .1 0 4 7 . 00 0 4 . 7 00g 1 .0 0 . 1 0 1 0 1 a. f. . 1 0 4 7 . 0 0 0 4 . 7 0 0:307 1 6 , 0 3 4 t Fccbt 16 . 0 34~ 52 . 2 30 pa it bt . 3 0 7 in

E .kin ~ 1 0 0(~E) .tiff

u p . 1 6. 1 3 of HA C 33 9 t o d. t . rm in e . ffe ctiv e wid th of skin :2w__,! 3 3 .Stwe 3 5 ( 33 .S ) 1 1 . 7t

D e t er m ina wh e th e r t o m e a sur a off of fast e ne r or ch am-m ill lan d :( 1 ) (1 1 . 7) ( . 10 0 in) 1 . 17 in .( 2) ( 1 1 .7 ) ( .0 6 0 in ) + . 5 0 in . 1. 2 0 in .Since (1) < ( 2 ) , m e a sur e e ffe ctiv e widt h off of faa t e ne r .

11 9 -/1/-


55/131

Askin ( 2 . 00 in ) (. 1 00 in ) + 2 (. 6 7 in ) (. 0 6 0 in) . 2 80 in 2T his ar e a of t he skin is a ssum ed to a ct a t th e a v er a g e st iffe n er st r ess:Fee. 5 2 2 30 psiPee. ( 5 2 23 0 p si) ( .3 0 7 in 2 + . 28 0 in2 ) 3 0 6 59 lb

119-15


56/131

L E S S O N 10 - P R O B L ! K S O L U T I O N S

1. F ir st d et er mi ne f

A 2~RA VG t 2 w (. 234 i n )( .03 2 i n ) . 047 in 23 3 4I = wRA VG t w(. 234 in ) (. 032 in ) .0 013 inf .00 13 in 4 .166 in .047 in 2

T h is st r ut wil l be t r e a t e d as a pin -e n d e d column .

L' L s 5 in~,. 5 i n 3 0.1! .166 in

U si n g p. 16. 31 of MAC 33 9. n ot e t h a t t h is val ue of L'/p fal ls i n t h e sh or tc ol um n r eg io n.

T o d e t e r min e t h e al low abl e cr ipp lin g st r e ss for t h is se ct ion , r e fe r t o t h ein se t of t h e uppe r fi gur e or p . 3-404 of M I L - B D B ~ - 5 D :

Q. .50 in 15.6t ,. . 032 i n

so F , . 49 KS1cc

RIO -I


57/131

c

Usin l p . 16 .3 1 of HA C 33 9, be lin on th e ve rt ical axis at F 48 KSI .ccFollow a pa ra bolic cur ve , int er pola tina be twee n th e p lott ed curv es, to t hev alue of L ' 30 .1 on th e horizont al axis. F ollowinl a h or iz on ta l lin e"7 .back t o t he v er tica l axis fr om th is poin t, give s F .: .4Z.S ' KSI .c

Alte rn at iv ely, th e eq ua tion for th e J oh nson cur ve could be used t o g et t hes a m e resu lt:

Z .Fe Fcc - ( Fcc) t I : . : ) 2 . 4 8 KSI - (4 8 K SI) 2 (3 0. 1) 24 1 1 ' 1E \ f 2 64( 3. 14 ) (1 0. 2x10 p si)

F ""42.8 K,;)lc .

D et er mi ne Al lo wa bl e Loa d:

Pcr (4 2. 5 KSI) (. 04 7 in2 ) 2 00 0 I b

2. f will be the same a s in pr oble m 1.L' L 15 in

L' 90.3fTh e allowa ble cripp ling str e. s, F 48 KSI (fr om p roblem 1 )cc

R IO - !


58/131

Usin g p. 16.31 of HAC 3 39. be g in a t Fcc 48 KSI. r e a d ove r on t h e Joh n son,cur ve . follow it alon g t h e Eule r cur ve un t il t h e value of b_ 90.3 is" 1r e ach e d . R e a d in g ove r h or iz on t ally fr om t h is poin t g ive s an Fc 12 KSI.Alt e r n a t ive ly t h e Eule r cur ve e q uat ion can be us.d t o d e t e r min e Fc:

Fc 1 T 2 E (3.14)2( ID.hIOt. eSI) 12.3 KSI(~r' (90.3)2T h e allowable load is

P (12 KSI)(.047 in 2) 564 lbcr

3. a) De t e r min e se ct ion pr ope r t ie s

A (.399"){ . 399") - (.335")( .335 11) .047 in 2I(.399 11)4 - (.335")4 00106 in 4

12 12..; :a (I.j.00106 .150 inJ 4 A .0 47

b) De t e r min e t h e allowable cr ipplin g st r e ss:

Each of t h e four si ~e s of t h e se ct ion can be con sid e r e d t o be an o-e d g e -fr e . e le m e n t wit h b .378 11 an d t .032".

1110-3


59/131

I

T h i s g iv e . ~ 11 .8. & n t e r in l t h e cr ip p1i n l cur ve or pa g e 1 6.19 oftMAC 3 39 g iv e .

U sin g t h e cur v e on p. 16.3 1 at MA C 33 9:

L'. 5.0 " 33 .3f . 1 5 0 "F 48.0 XSIcP (48. 0 KSI )(.0 47 in 2 ) 22S6 l b ul ti ma tecr

4. Solution:

.1 66 in ( Se e p ro bl em 1)

. IL _b_ ~ 3 .49 in.[C /2.05

L' 3 .49 in 21 .0f .1 66 inr e fe r r in l t o t h e i n .e t of t h e up pe r fi gur e on p. 3-404 of M I L - 5 D :

~ .50 in 15 .6t .032 i n

F 48 KSIcc

1110 - " "


60/131

Usinl p. 1 6 . 3 1 of MAC 33 9.

F 45.8 KSIcP (45.8 ksi)(.047 in 2) 2153 lbcr

5. Solut ion : . 1 6 6 in ( Se e p ro bl em 1)

L I .. 1.. .. 5 in 2.5"r: rrL' 2.50 in 15.1f .166 in

r e fer r in g t o t h e in set of t h e uppe r figur e on p. 3-4 0 4 of M I L - S O :

D .50 in .. 1 5 . 6t .032 in

F .. 48 KSIcc

using p . 1 6. 31 of MA C 3 39 :

F 4 7 KSIc

P (47 KSI)(.047 in2) 2209 lbcr

F I'O - S


61/131

6. Solution: f .166 in (s p1:'oblUl1)L' l . .J :!! 10 inf C [ : 2 s.~ 10 in 60.2f . 1 6 6 in

re fe r rin g t o t he inse t of t he upper figure on p . 3-4 0 4 of M I L - 5 0 :

Q . 5 0 in 15.6t . 0 3 2 in

F 4 8 KSIcc

Usin g p. 1 6 . 3 1 of MA C 3 3 9 :

F 27 KSIc

T he ultimat e allowable load is:

Pcr (27 KSI)(.047 in 2) 1269 lb

"/0 - b


62/131

;;/

7. S o l u t i o n : ( 1118 -T~L,- r { , ! a uT ~ . ' O O t/ A t l A a.o"\0 1 I 1I1S-T~..L. A - A :

Z.O

D e t e r m i n e t h e a l l ow a b l e c r i p p l i n g s t r e s s f o r t h e s e ct i o n :

b . . .1. 9 0 6 " = 1 0 . 2t . 1 8 7 5 "

Si n ce t h i s i s a n e q ua l l ~g a n g l ~ s t i ff e n e r , t h e a v e r a g e s t r e ss f o r t h et o t a l s e c t i o n i s e q ua l t o F co m p ut e d f o r a n y l e g . F r o m p . 1 6 . 2 3 o fc cMA C 3 3 9 :

F = 48.0 KSIc c

R I O - 7


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D e t e r m i n e f :F i r at d e t er m i n e t h e m ome n t of in e r t ia a n d a r e a of t h e st iffe n e r-w e bc om b in a t i on :

' 1 . . 1 6E - 1 0. 7 x 1 0 psiUsi n g p . 16 . 1 3 of MA C 3 3 9

2 w = 3 5 . 5et

W = 1/ 2 (3 5 . 5 ) ( . 1 0 i n ) 1 . 7 8 i nei ,E 1 e b l t A Y I Ay A l I

1 2 . 00 I . 1 88 . 3 7 6 1 .1 0 , . 4 1 4 .4 5 5 4 .1 2 5 3i2 1.8 1 . 18 8 . 3 40 . 1 9 4 I . 06 6 . 0 1 2 8 I . 0 0 1 03 3 . 5 6 .1 0 0 . 35 6 .0 5 0 . 0 1 8 . 0 0 0 9 . 0 0 0 3 y=!:t.. ... 4 98 = .46')"A 1 . 0 7 2 .1 . 0 7 2 . 4 9 8 . 4 6 9 1 .1 2 6 6

IfIO - g


64/131

So ~. 10 in 17.2f .582 infr om figur e on p. 16.31 at MAC 339:,an d usin g F 48.0 KS1 an d I :.. 17.2ccg i ve s F 46.5 KSIc

P (46.5 KSI)(1.072 in 2) 49,850 lb P (Joh n soncr J l oad )f

c ri ti ca l c ol um n

Usin g fig . 16.38 wit h P1 0 g ive s P2 1.92P z PE

Usin g t h e Joh n son cr it ical column load t o d e t e r min e t h e allowable columnload for a unifor mly un load e d column . P2:

P 2 ( ~ ) ( P J),Th e st r ess pr oduce d by t h is load is we ll be y on d t h e e last ic limit as we ll,

P 2 (1.92)(49850 lb) 95710 lb

an d much h ig h er t h an t h e cr ipplin g load. So t h e allowable load for t hissect ion is t h e cr ipplin g load :

P (48.0 KSI)(1.073 in 2) 51.5 Kipsc8 . Solut ion : For 2024-T 3

F 34.0 KSIcyA2 2'1fr t - (!1l)( .451 in )( .049 in ) .139 in2avg12 'If AVG3 t 'If(.451in)3(.049 in ) .014 in 4E2 10.2 x 106 psi


65/131

F or 43 40 St e e l En d Fit t in g :F 2 40 KS1cyAl 1f(.37~ in)l- .110 in2

'3'75")41 1 .1r ( '=r= ' - 9.71 x 10 -4 in 446El 29.0 x 10 pa i

E It


66/131

F r om p r o bl e m 8 :P - 4 0 0 0 1 bcr

F or a s im p l y s up p o r t e d b e a m

t 100 \ k tI::= = i ) : : : : : J J ) ~ ( ~ r : :5 0 = 1 1 0=~S D \b

y ..o( 1 0 0 1 b ) ( 1 8 i n ) . 0 8 5 1 i n (Ref:

4 8 ( 1 0 . 2 x 1 06 p si )


67/131

S H E A R F L O W H O M E W O R K P R O B L E M SS o l u t 1 o n s

011.1 a)A

~ ~b"-

~~A 1 "- "- "- "- . . . . 8.300lb C f e .

S o l u t i o n : q a = 3 0 0 l b = l O a l b / i n3 . 0 i n

T a k e M o m e n t s a b o u t P t . B :t8 = 2 A A O B q d - 2 A A C B Q a . . 0( 6 ; n ) ( 5 i n ) q d = ( 6 i n ) ( ' 3 ; n ) ( 1 0 0 l b / i n )

q d = 6 0 l b / i nq c = 6 0 l b / i n

S u m m o m e n t s a b o u t P t . A :~ M a = 2 A A O B Q b - 2 A A B C Q c = a

( 5 i n ) { 6 ; n ) o b = ( 3 i n } { 6 ; n ) ( 6 0 l b / i n )q b . . 3 6 l b / i n

1 # 1 1 - I


68/131

1 1 . 1 b ) c._~=t='==---".

O - - - - - = = = = : . - - - _ . J . A

S o l u t i o n :S i n c e s h e a r f l o w s o n n o n - p a r a l l e l s i d e s o f a t r a p e z o i d a r e e q u a l .

q d ' " 7 5 0 l b / i nS u m M o m e n t s a b o u t D :tM o = 2 A U C S q c - 2 A U B A Q b = 0

(4 i n ) ( 6 ; n ) q c = (4 ; n ) ( 1 0 ; n ) ( 7 5 0 l b / i n )q c = 1 2 5 0 1b / i n

S u m M o m e n t s a b o u t S: .t M S = 2 A D C S q d - 2 A O B A Q a ' " 0(4 i n ) ( 6 i n ) ( 7 5 0 ~ ) = ( 1 0 i nH4 i n ) ~

i nQ a = 4 5 0 l b / i n

RII-l


69/131

011.1 c )

A l t b% ~1 '" '" . . . . . . . . . . - . . . . . . . . . .C . B' r c .

S o l u t i o n :N o t e t h a t t h i s s h e a r p a n e l h a s i d e n t i c a l h e i g h t s t o t h o s e i n 1 1 . 1 a ) b u t

w i t h a d i f f e r e n t b a s e .S u m M o m e n t s a b o u t 8 :

E M S = 2 A A D S q d - 2 A A C S Q a = 0( 5 ; n ) ( 8 i n) ( q d ) = ( 8 i n ) ( 3 ; n ) ( 10 0 1 b / i n )

a d = 6 0 l b / i n f o r h o r i z o nt a l e q u i l i b r i u mQ c = 6 0 l b / 1 n

S u m M o m e n t s a b o u t A :E M A = 2 A A D 8 Q b - 2 A A C S Q c = 0

( 5 i n ) ( 8 i n ) q b = ( 3 1 n ) ( 8 ; n ) ( 6 0 1 b / i n )q b = 3 6 l b / i n

R e c a l l t h a t t h e v a l u e s o f t h e s h e a r f l o w s f o u n d i n 1 1 . 1 a ) a r e e x a c t l yt h e s a m e a s f o u n d h e r e . T h i s d e m o n s t r a t e s t h a t t h e v a l u e o f t h e s h e a r f l o w sa r e d e p e n d e n t o n l y o n t h e r a t i o o f t h e h e i g h t s o n o p p o s i t e e n d s o f t h et r a p e z o I d .

111/-3


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11.1 d) ~ = = = ' = = = = ~ B

S o l u t i o n :Q c = 2 5 0 0 1b = ? O O 1 b /1n5 . 0 0 i n

S u m M o m e n t s a b o u t S :I: e = 2 A A O S Q d - 2 A s o c o c = 0( 1 1 i n ) ( l i n ) q d = ( 5 i n) ( l l 1 n ) ( 5 0 0 l b / i n)

q d = 2 5 0 0 l b / i no b = 2 5 0 0 l b / i n

S u m ~ m e n t s a b o u t 0 :E M O = 2 A A S O Q a - 2 A S O C Q b . .0

( 1 1 i n ) ( l i n ) Q a = ( 1 1 i n) ( 5 1 n ) ( 2 5 0 0 l b / i n )O a = 1 2, 50 0 1 b /1 n

c . 2500 Ib

N o t 1 c e t h a t q a = 2 5 Q c l T h i s e x a m p l e s h o w s t h a t s h e a r p a n e l s w i t h o n e s i d ev e r y s h o r t s h o u l d b e a v o i d e d .

IUJ-~


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11.1 e)

S o l u t 1 o n :1 ) F i n d t h e t o t a l a r e a o f t h e p a n e l

A ~ A A F B + A B C E F + A C D EA = ( 3 . 0 ; n ) ( 2 . 0 i n ) + ( 2 . 0 i n ) ( 1 / 2 X 3 . 0

2i n + 4 . 0 i n ) + ( 4 . 0 1 n ) ( 4 . 0 i n)

2

. 2 ) F i nd A A B D :A A B O = ( 3 . 0 i n ) ( 8 . 0 i n ) = 1 2 . 0 i n 2

2

3 ) F i n d A g C O :A B C O = A - A A B O = 1 8 . 0 i n 2 - 1 2 . 0 ; n 2 z 6 . 0 ; n 2

4 ) S u m M o m e n t s a b o u t D :r r 1 0 = q a ( 2 A A B O ) - q b ( 2 A B C D ) = 0

q a = ( 2 0 0 1 b / i n ) 12 .0 ; n 2 = 1 0 0 l b / i n2 4 . 0 ; n 2

5) F i n d A A C O :A A C O = ( 1 / 2 X 4. 0 i n) ( 8 . 0 i n ) = 1 6 . 0 i n 2

6 ) F i n d A A B C :A A B C ~ A - A A C O : lS.J in2 - 1 6 . 0 1 n 2 = 2 . 0 i n 2

7) S u m M o m e n t s a b o u t A :E M A ~ Q b ( 2 A A B C ) - o c ( ~ A A C ~ ) = 0

Q c = ( 2 0 0 1 b / i n ) 4.0 ; n 2 = 2 5 l b / i n3 2 . 0 ; n 2

8 ) S u m M o m e n t s a b o u t B :E M e = Q d 2 A A B O - q c 2 . A B C D = 0

Q d = ( 2 5 l b / i n ) 1 2 . 0 1 n 2 z 1 2 . 5 1 b / i n2 4 . 0 i n 2

111/-5

= 1 8 . 0 i n 2


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11.2 a) 8000 lb

~ +) \~b 000 ~tvlb.....=======:i-B O O O I b t

~ t l l~=----,\t l~t : : : : I i = = = = = = = = = = = = = = : : z , ~ i 8 D O O \b = R 3RlSince this beam is rectangular, the shear flow is the same on all edges.

q = 8000 lb = 1600 lb/in5 in

R2 = (1600 lb/in)(17 in) = 27,200 lbf


73/131

11.2 b)

T h i s p r o b l e m c a n b e s o l v e d b y t h e m e t h o d o f s u p e r p o s 1 t 1 o n .

I r O O O Ibb ~ l b ~ ~ t = 1==i ~ ..~- R~' - - - - : : ; : : = = - - - ' I ~ ~Q30 Ib= R]R SS a l a n c i n g t h e l e f t c a p :

q ( 5 i n ) = 6 9 3 0 1 bQ = 1 3 8 6 l b / 1 n

B a l a n c i n g t h e u p p e r c a p :R4 = ( 1 3 8 6 l b / 1 n ) ( 1 7 i~) - 4 0 0 0 l bR4 = 19,562 l b

B a l a n c i n g t h e l o w e r c a p :R 5 = ( 1 3 8 6 l b / 1 n ) ( 1 7 . C in)

_ R 5 = 23,562 1b

nil - 7


74/131

~ '300 Ib ~em Ib (_1 1 . 2 c ) ,Z3850 i r\ l b ,(+ ,,,,

1 1100 'bF .R - _ I -I [ l i l t q O O \b1 2 0 0 J ~ 1 1 < r z % 1 L l

R Z F-1 .B a l a n c i n g t h e r i g h t m o s t c a p. 9 0 0 l b = ( q l ) ( 6 i n )q l = 1 5 0 1 b / 1 n

B a l a n c i n g t h e u p p e r a n d l o w e r c a p s :F l = F2 1& ( 1 5 0 l b / 1 n ) ( 1 0 . 5 i n )

F 1 . . 1 5 7 5 l bF 2 . . 1 5 7 5 l b

8 a l a n c i n g t h e m i d d l e c a p :3 0 0 l b + ( 1 5 0 1 b / i n ) ( 6 1 n ) . . ( q 2 ) ( 6 i n )

Q 2 = 2 0 0 l b / 1 nB a 1 a n c 1 ~ g t h e u p p e r c a p ~

R l = Q2(lZQ i n ) + F 1 = ( 2 0 0 1 b / i n ) ( 1 2 . 0 i n ) + 1 5 7 5 1 bR 1 = 3 9 7 5 1 b

B a l a n c i n g t h e L o w e r C a p :R2 .. ( q 2 ) ( 1 2 . 0 in) + F Z = ( Z O O l b / 1 n } ( l Z . 0 in) + 1 5 7 5 lbR2 = 3 9 7 5 l b

AIJ- 8


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1 1 . 2 d) Ibb5D Ib .. ... .~ +

.. .~.. ... .

lO O IbR I - -~ Q O O \b~ = = = = = = = T = = = = = = ~

~oo IbqDO lb

B a l a n c i n g t h e r i g h t m o s t c a p :9 0 0 l b = q l ( 6 i n )

Q l = 1 5 0 l b / 1 nB a l a n c i n g t h e u p p e r a n d l o w e r c a p s :

F l = ( 1 5 0 l b / 1 n ) ( 1 0 . 5 i n )F 2 = F 1 = 1 5 7 5 1 bB a l a n c i n g t h e m i d d l e c a p :

( 1 5 0 1 b / i n ) ( 6 i n ) = 3 0 0 l b + q 2 (6 1 n )q 2 = 1 0 0 l b / 1 n

B a l a n c i n g t h e l o w e r c a p :R 2 = F 2 + ( q 2 ) ( 1 2 . 0 i n ) = 1 5 7 5 1 b + ( 1 0 0 l b / 1 n ) ( 1 2 . 0 i n )R 2 = 2 7 7 5 l b

B a l a n c i n g t h e u p p e r c a p :R l ' " F l + ( 0 2 ) ( 1 2 . 0 in) = 1 5 7 5 1 b + ( 1 0 0 l b / i n ) ( 1 2 . 0 i n )R l = 2 7 7 5 l b

1111-'


76/131

1 1 . 3 a ) t ZO O IbI I I I I It l t X l lb t 100 Ib

B a l a n c i ng t h e e n d c a p s :q 1 (5 i n ) = 10 0 1b

q l ' " 2 0 l b / i nq 2 (5 i n ) = 1 0 0 1 b

Q 2 = 2 0 l b / i nB a l a n c i n g t h e u p p e r a n d l o w e r c a p s

F l = ( 1 0 1 n ) ( 2 0 l b / i n ) = 2 0 0 1 bF2 = ( 1 0 1 n ) ( 2 0 l b / i n ) = 2 0 0 l bF1 = 2 0 0 1bF2 = 200 l b

C h e c k i n g t h e c e n t e r v e r t i c a l m e ~ b e ~ :?q 1 (5.0 ; n) + 02 (5.0 ; n ) = 2'JO 1 b

( 2 0 l b / i n ) ( S . O i n ) + p.) l~/in)(:.O i n ) :a 2 0 0 l b

/11/-/0


77/131

1 1 . 3 b ) ;3000 \b

I S O O i J I I!2 .IQ 5 ' ' " t 405 Ib

B a l a n c i n g t h e C a p s :l e f t C a p :(5 ; n ) q l = 2 1 9 5 l b

0 1 = 4 3 9 l b / i nR 1 g h t C a p :q 2 ( 5 i n ) = 4 0 5 l b

q 2 = 8 1 l b / i nU p p e r C a p A b o v e l e f t H a n d P a n e l :

( 4 3 9 l b / i n ) ( 6 i n ) = F lF , = Z b 3 4 l bU p p e r C a p A b o v e R i g h t H a n d P a n e l :

q 2 ( 1 4 i n ) = F 2(81 l b / i n ) ( 1 4 i n ) ' " F 2 = 1 1 3 4 l b

B a l a n c e J o i n t a b o v e C e n t e r S t i f f e n e r

l o w e r C a p B e l o w R 1 g h t H a n d P a n e l :( q 2 ) (14 i n) ' " F 3( 8 1 l b / i n) ( 1 4 i n) = F 3 = 1 1 3 4 l b

C h e c k i n g t h ' e l o w e r C a p B e 1 o w t h e l e f t H a n d P a n e 1 :( 0 1 ) ( 6 . 0 i n ) - 1 5 0 0 l b ~ F 3( 4 3 9 l b / i n ) ( 6 . 0 i n ) - 1 5 0 0 l b = 1 1 3 4 l b = F 3

/1 / / -11


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".3 c) , tODD \b * 4rOC() Ib~ I I I ~2.4++ Ib+++ Ib

=_Fz.- __F+- __- - + Z O O O \ b I~~~ t~J ttJ [ 3 J t * 1 1 ~ z ~ i j+44 \bi - - 12.44+-F- -- - F3- IbIB a l a n c e t h e L e f t C a p :

4 4 4 l b = ( q l ) ( 4 i n )B a l a n c e t h e R i g h t C a p :

2 4 4 4 - l b = q 3 ( 4 i n )

q l = 1 1 1 l b / ; n q 3 = 6 11 l b / i n

8 a l a n c e t h e l e f t c e n t e r c a D :( q l + q 2 ) ( 4 i n ) = l O O O l b

q 2 = 38 9 l b l 1 n

B a l a n c e t h e l o w e r c a p s :(111 l b / i n ) ( 4 i n ) = F l = 4 4 4 l bB y S ymm et ry :

F 2 = F l = 4 4 4 l bD e t e rm i n e F 3 :

(389 l b / i n ) { 9 i n ) - 4~4 lb = F 3 = 3057 l bA g a i n , b y S ymm etry:

F 3 = F4 = 3057 1b

111 / - /2 .


79/131

11.4 a ) bOO \b"I Z . O O O ; ,, I b( + ~ C a p A r e a s = . 0 8 0 i n 2I0 !bO D lb

P I -;- b O D Ib1~ D e t e r m l n e t h e U p p e r a n d~ 1 1 'tl l~f F L o w e r C a p L o a d s a t W a l l :I1~~ 1 1 ~l I~ EM o = ( 6 0 0 l b ) ( 2 0 i n ) - P 1 (6 i n ) = a

! . , o o IbP, :z 2 0 0 0 l b.-+- B y S y m m e t r y :Z .O O C l \ b ~ P z P 2 :z 2 0 0 0 l b

S i n c e t h e m i d d l e c a p i s a t t h e n e u t r a l a x i s , i t c a n n o t p i c k u p l o a d .S o b a l a n c i n g t h e l o w e r c a p : B a l a n c i n g t h e u p p e r c a p :

2 0 0 0 l b = ( Q 2 ) ( 2 0 i n ) 2 0 0 0 l b = Q l ( 2 0 i n )q 2 = 1 0 0 l b / i n a 1 = 100 1b/ i n

8 a l a n c i n g t h e r i g h t c a p s :600 1b - q 1 (3 ; n) = F .

F l = 3 0 0 1 0N o t e t h a t t h e s h e a r f l o w w o u l d h a v e b e e n t h e s a m e i f t h e c e n t e r s t i f f e n e r h a dn o t b e e n t h e r e . T h i s e x a m p l e i l l u s t r a t e s t h a t h a v i n g a s t i f f e n i n g m e m b e r a tt h e n e u t r a l a x i s i s u s e l e s s , u n l e s s a n e x t e r n a l l o a d ; s t o b e " d u m p e d i n " a ti t s e n d . o r if i t i s t o b e u s e d a s a D a n e l b r e a k e r f o r d i a g o n a l t e n s i o n w e b s( S e e L e s s o n 1 3 ) .

su-is


80/131

- III

bO O \b

l l o O O Ib

M o me n t o f I n e r t i a := 2 ( A 1 Y l 2 + A 2 Y 2 2 )= 2 [ ( . 0 8 0 1 n 2 ) ( 4 . 5 0 i n ) 2 + ( . 0 5 0 1 n 2 ) ( 1 . 5 0 i n ) 2 ]= 3 . 4 7 i n 4

A n a l t e r n a t e m e t h o d f r o m t h e o n e p r e s e n t e d i n t h e t e x t c a n b e u s e d t od e t e r m i n e t h e p a n e l s h e a r f l o w s w h i c h h a s t h e a d v a n t a g e o f k n o w i n g t h e t o t a lc a p l o a d s i n i t i a l l y :

~ \b P l - - ~ = = = = = = = = = = - = = = = = = = ~P 2 ~ ~ = = = = = = = = = = = = = = = = = = ~v tP 2 --F===================~P 1 _!::===================~8 y G e o m e tr i c a l C o n s i d e r a t i o n s :

V = 6 0 0 l b

L o a d i n U p p e r a n d L o w e r C a p s :a1 = ! : 1 = ( 1 2 . 0 0 0 1n .1b ) (4 . 50 ;n )I 3 . 4 7 i n 2

P l = ( 1 5 5 6 0 os i)( . 0 8 0 i n 2 ) = 1 2 4 5 l bL o a d i n t h e M i d d l e C a p s :< r 2 = ~ = ( 1 2 1 0 0 0 in 1 b ) ( 1 5 0 i n )I 3 . 4 7 i n 2P 2 = ( 5 1 9 0 p S i ) ( 0 . 5 0 i n 2 ) = 2 6 0 l b .

p... b O OD e t e r m i n e S h e a r F l o w s : a ~8 a 1a n c i n g t h e U p p e r C a p : n ~ p 1 . 1 . . _ _-==='U== I == ::::;::- -.11 ~ 1( q 1 ) ( 20 . 0 i n ) = P l = 1 2 4 5 l b r .

q l = 6 2 . 3 1 b / i n F l o . . . . F ISa lanc1ng a ~;ddle Cap: ~I 1 , , -_===


81/131

1 1 . 5 a ) z . o o \b/ In . ~C o & c,

A C U T P IOUTc . B c . .

z o o l o /II\.

' n ' I J .- - - -

[ ] 1

I. F 1 r s t d e t e r m i n e t h e i n t e r n a l s h e a rf l o w o n p an e l s a d j a c e n t t o c u t o u t :F r o m t h e s k e t c h a b o v e :( % ) ( 1 0 in) = ( 2 0 0 1 b / i n ) ( 1 5 tn)

% = 3 0 0 1 b / 1 nF r o m t h e s k e t c h a t l e f t :( q a ) ( 1 " " i n ) = ( 2 0 0 l b / ; n ) ( 21 tn)

0 a = 3 0 0 l b / i n... p~, - II. S e t u o f r e e b o d y s t a t 1 c s e q u at i o n s :** 1 : P 1 + ( q c)( 7 1n) = ( 2 00 1b / in ) ( 7 1 n )* 2 : P 2 + ( q e ) ( 5 i n ) = ( 2 0 0 l b/ i n ) ( 5 i n )

3 : 2 P 2 + ( 2 0 0 l b / i n ) ( 5 i n ) = ( 3 0 0 l b / 1 n ) ( S i n )* 4: P 2 + (qc) ( 5 i n ) =

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