Problem I

:# .

ETTEN--

→x

÷m#¥ #

b -- D= 4 , A- 5 i E - 20 MPs: , 0=0.3

f- 5- Kips = 5000lb

Fxirz F-* And Fg:{ Ft.

25005 =- 2500

a)FBO

%¥¥¥:*

Egad -buns : Extreme Reactions

→Efx -- Axt Fx - O → Axe - Fx CD

T' I Fyi Ay -16g -Fy :O L2)

#Ms .- Cyb - 2bFy :O→ Cy-- 2kg (3)

*EIMps -Cyb -Ay .2b=o→ Ay - - { b (4)

hence

A-a- Fx , Cy -- Zfy And Aye - Fy

Intense resultants

FBA OB

FBA %} FBOrap

FBCf. FBO* FBCi¥¥

A-•→ ← ⇒o*•0→ ← → a-•→

Ax Fac Fac Fac T FEO Feo Feo the ff*F×Cy

Equilibrium :+ Vars : FAB, Fac , Frsc, Fco , FBO→t

Att : IF×= -Feo - {Frsotfx -- OG)

→ I Fyi -Fytrzfo - O'6)

FxFeo From (5) and CG)↳FB FB D= Zpgfy For - tzotfx -- - ⇐FathAE C

FF EFI, = - Fac + Fco = O (7) .

*¥•→ -Fac T Eeo IL FI = Cgt FBC -- O (8)

CyFrom (7)And (8)FAE Foos Fx - dry FYFrsc : - Cy -- - 2. Fg

At A→

Ay Eff -- Fac +Ait FBI = 0 (9)

µ, FBA*Fyi Ay -135 Frsa =D ( so)

A-•a)60'

from Los)'and Cio)

Ax Fac FBA'¥Ay=t¥%

Stresses . f- F/a A- 5

-

A-B = IrsFI = ¥325517 Jooorg psiA Tension

TAC = - IsEft Efc. . If 0+25%1 .. 2000oz psitension

(BC = -2F# e -Zg .

2500 i -1000 psicompression

CBO = If, Fay

- I 000Ffg psi tension

T.co = - ↳Eft Efc. - If 0+25%1 .- 2000oz psitension

B) member BCUsing generalized Hook 's IsarTy --Gx -- Lxy - Eyz = 2×2=0

2-

hence , the Axial strainis [

Ex -- E. Cox - ol.gg/errzfD E s

O -5= EE -- - 'foe:O. -kEf .- ex -s al : Exxl :÷I:E÷z+iothe change in Arts is is 5E - Ey -- Eze - DEE - 0.3×-5×10=-9*1,5×15Eft

¥ .tn#:::::::::ii::::- CA-Ao) -- tolled) - Ao

-

Idle da ⇐ bitted -Ao1- - I = 511*15×15532-5=¥f5xio4E

: in:*::*::*.

IA = Aol ft ET - Aa

= 5C It I. sxios)-

-

5

= I . 5×10-4 in'

① A e- L2- old - OED

'

= Lt CZ (I ← 20Ex - 0¥)= K2 (f- X- 20 Ex)

Iff e - 204

Lists -1L Eye Ez]

DA -

- Holly t Ey)

HW2P2Lin KCC)

i

block CB )

1) Enix - tins .ca Eaxismsxoe

'

- 19900×0.31= 8500×0.93

= 6169lb'

- 7505lb

2) f- Eq=E¥ F- EE -

- Edd

*÷÷÷÷i. ÷÷÷÷:÷ .3) Mom's done:n UF ALF't 0.002in

OWE sin-40%98-1)⇐ 3.133×10

>in

= 9.95×15"rsd0775in'(33¥18)

F- 0,0570 = 1,305×1 -04rad= 0,00750

C B L• O

Uu } Ubl is' ?L

4) compatibility : check displacements

Y} %→ Ub - §. Ue

for : Uce aint'→ has .- 2g him

U B. =2.39×502

zubm-xitfsilnsxtisp-dfor.UA-- Upw" -a no = Szuba't

He -- S . 222×10-3 ⇐ aim-1Allowable d-sptaanat

Hence the maximum allowable forcewill be defineded by UnixFBD✓ TFL 16 12 12

T to TRAFB p

I⇐ Ma - O - 40 E -24Fs t 12 P-- O

Pe 40F4FB12

For : Us-- UBM" → Frs Fmf =7905lb

↳ue -

- 5.22×1 -03in y her 6 the fork on link

OC is :

Fat = E . Uf → Fei 10×106*0.31*5510320= 809.1 bb

Hence , the maximum allowable force is i

p= 40×8809.lt 24.7905

IT⇐ta

Problem 3 a

"simple compression : b-

92

Ex! - oee-sei.ve: ¥E

cc c TagEye - OIC → Eye - UE .z

.

c

E

£I= IcE fats

b)Equi biaxial extension : ←Fse :-. # fr: - or's].

- crops ↳ej-ziai-oriy.E.gr &

EE - E. C. ofgieajjfiezr'

E

C) EE -

- Exb defied

÷÷÷÷÷÷÷÷÷t:¥¥÷¥¥Pt. aft'

PF - ¥baDc⇒o.es/Pb...-Aazp'⇒ o .- as

-

. 500lbd) PE -afar. *{IF 4in'

-9- EE - 50.0=50 lb Ab 0.4in

'

4normal strains : is

Ext. Ej .- ¥007. tz.oIay.s.es 5-0=8.68"7.12103 0,4

E! I -¥2 Tbs- 50,0-4=4.736×10-2

Change in Volume✓To- It Ex)GeEy ) (ft Ez)Tt Ext Eye Ez e ExEye ExEze Ey Eze E×EyEzIt puts -20 e A-of-404 -Dyers-032K¥Jef-14 -120+99-2 ✓

3 IbE

✓= ✓° ⇒ 0--0,5

OV-- V- Veein

Problem 4 Ty-

8 of

KIKI ×

,

ExEyEzVxy/8xzI> O LO SO LO O O

t - -

Ex -- E. Lox -0cg. f- f. Lox-08370°

- et

Eg -

- tetra -ucrxe.no/z5f.--zL8-urxJL0Ez--tEEE-0Gxery5f--&f-010×81) > OTu

Kisi Eff co- -

Vyxzi 2×2/6=0yz.IE//z/I-- 0no