i · 2021. 2. 23. · equilibrium: + vars: fab, fac, frsc, fco, fbo →t att:...
TRANSCRIPT
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