continum mechanics sheet

7/27/2019 Continum Mechanics Sheet

1/8

| "* Lyx 'Lzxoii = lr*y oy TryLT xz Trrz Crz

s2=str+sj+sZs2=si+s"2

S, =lo*lmryx*nr",Sr=lrrr*morlnt,S" =ltr"+mryz+noz

Sn = I Sr*mSr*nS,Sn = 12 o, + m2 o, + n2 o" t Z(lm trx * mntr" # nl rz*)Sn=l2or*m2orl n2o3 In terms of principle stressesS, = (ft.st:$S, = U2m2(oroz)2 +mzn2(or- os)2 +n212(q- ot)z ], Int"rrns of principle stresses

-Ito'o-Izoo-Ir:0It=ox*o"*o,Iz = (ttry + tj" * r3r) - (oro, * oro" + oror)Is = 6aoy62 + 2(t rrtrzt "r) - orty"' - 6yTr*2 - orT ryz

To check your answer :It=otl02*o312= - (o1.o2+ 62.o3+ o3.o1llz= oroz,6tThe same procedure isapplied foroz and o3 just replace oqby o2or o3

I(o, - or) + m(rr,) + n(t",) = ol(t-)* m(on - or)+n(r"n)=oI (rr") + m(r"") * n(o, - or) = o12+mz*n2=1

....(1)

....(2)

;:..(3)o; = 121o, * m21o" * n?1 o, + Z(t rm rr o * m fl 1r y 2 + I pp, *)oy = lf o* * mjo, + nlo, + 2(l.rmg r" * rn2n2r yz + l2nrt"r)o2 = l!o, t m!o" * n!o, + 2(Irmg*" I m3r\r yz + Isn3r rr)Tyryr=ll2o**ntfli2oy*n1n2o"*(I1mr+ lrmr)rrr*(mp2+m2nr)ryz*(lp2*12n1)t",xytTr = ll3or*rrt1.rrtsly *n1n36, t (11m. * Irml)tr, * (mflz*msn1)ryz * (11n3 * lsnl)t"*Ty,z,=Irl"o**rtL2rrlscy*n2nso"1(l2ms*Itm")trr*(mrn"*m3n2)Tyz*(I2ns*Isn2)t",

r,yt = os =)(o, + or) +i@, - or)cos(Zl) * t,rsin(Zl)r*trt = t, =I(o, - or)sin(zl) - trrcos(2l)ot=:(r, * d! zort)=eh|pz=?pt+900

tan(Zlr) =-W0r=|pt+450


2/8

s@&@E[@N @F @@M@ E@UA8[@N

Is = o2soyo, * 2(t rrir"t u) - (o*ryz ') - (o"t*n')') - (ort,*

79o =;sln

1f= t [L * 2,13J2 sin

-,( sr n\\-'zF)O1 (r'*?)l

It=ox*or*o"

I, = (ttry + tj" + tZ*) - (o*o, * oro, + o"or)

L2Js: Is * 5Il2 * Uri

o, =I[], + zJ3]rsin(cps)

o, =I[r, * z,{ursin (eo - ?)1


3/8

@HAPTMEW\M@


4/8

o* = #;)ftt - v)e* * u(e, + t,)lov = #,,l{t - u)t, * v(e*+ r,)l o,

:hlT - v)e" * v(e*+ r,)]o" = # lO - v)e" * v(e,+ rn)J

orfrftr*ve2l

G -TtY - t'YTxy 2 exy

o^ =* (o1* o2 t os)W =}(ore, + o"e, * o"E, * rxyTxy I ryzyy" * tr"Tr")wu =l(o1q * o2e2 | qes)

DTor=Tor='i

oz=Txz=Tyz=O

@EAPEERE@RM

e,=i Io* - u(ov + o)]tr=I lo, - v(o, + o)le,=l lor - v(or + o)l

eptstress = ep!strainepzstress = qpzstrain

The principle angles of stress are equal to principlestrain angles

o=g? fo1* o2* osf or A= (er * e2 * e3)

) or=v(or+ or)


5/8

@ffiAFEWWIITE

e- = induced strain

,, =Zr, The process is done by steps eachnew strain

Yield strength: stress in which the relation between stress and strain islineafultimdte strengtft: stress associated with max load without fracture.

L6=- AoOr=Y=l

Luo"=ELt=z= instantenous load

Yield strength =Y = k.G)" n: strain hardening expojlent, L^o,K=- A'o^= xQ)

Tensile strength o*=#=

Total normal strain ,,=Ir, The process is done by steps eachstep produces new strain, =he)e=Ine=zh(+)e=-e=t(*)

used when no necking occursused when necking occursused when there is cold-working

e=In(l*e)

PE = (+) *Loovo*n=w)*roozo

lAo - Ar\ RAf =1 'l;-\ro/t00re2 -W= | SdeJz,


6/8

@ffiAPBBNFOUM

k: max alowable shaer stress (T."*)Y : tenstle yield strength " str e s s"

Omax- Omin=Yot-Os=Y =2k(oroz)z * (oz - os)2 + (oz - ot)z = 2Yzv ='lgk


7/8

@EAFEBREE@ffiW

Detbrmine ttre ptastic stieil aiffi inalt-'P ,Assune tn9 Cylinder to be fabricated of a subjected only to iniernal pressureplastib material of yield strength oy anO ulfimate:strenLgth... ,,, !," r:: :i


8/8

@MAPUBREU@HS

rr;i:

'::;titit|ltl:rii;aszi,:;tti){tl:|&Iir:;tE'::lGi.lilu-.. . {s:'P.orl::h:!,: !!'.',:r::.a:::Ai'l:iiilu-

l:1:ll1:.:::&0)!:::Pa. :::1rII r;;t,i

t:*u$

Critical speed 1fov 5 \t@"=\p'sa'+"")Radial stress z b2(3+u)oe=or=pa'-- This is for solid andannular discs

Angular stress os=pazT\,*",#) This is for annular disconlY

P*r+ial,yielding

Critical speed lt's above critical speed of initial yield and lower than critical speed ofcomolete vield.Radial stress o, = *(-t * 5)o, * p,'+(n' + r' -'#-*)

Angular stresso, = -fi\*5)o,+ paz!@'* c2 -#r'.ry)

oaz a3-c3 c-aor=T-i-* , o, ,T=CCritical speed A1

I

(q,o" --!-" '\;\- p b3+a3)Radial stress *=('-:-'#-,!.# o^,

F : { ;lrii.:

i;l;;:iri:l;i:t !:

Determine the maximum deflection due to an applied force P acting on simplv supportedrectangular beam. as shown below.

Max deflection umax =Icn",, ...';W, ,:i,"* , ."l+#ri:ii' .1'r:lllTllr:r l\ r : :r r,il tipft pf,,rectangula,r,,cross,rsectio,nr:ris:su,bjdrtedito.load N;,as 5ho.wn,.,b.elow,.derive,gen,eralrelationship involving N and tvt which govern first the case of initialyielding , and then fullyplastic deformation , for the stiaight part of the line of length Lr , I :::: I ::::::: ::t::: ::::. "';li;: ;ij" i;,' H".:Ii.. ,".1r.1., . ...ri:. l;i. .,#; . "in. .a', , "'ffiiution'''' :;.+,1 rliiii, l:t, r:. rt,' r::i:::. : :r tr -:i#.r, - iii;i'i::' :';:i NtrMr_,-T-_ INv Mv

i,,,:r," lltJitli:, "'ry

iDeteimine the miximum,allowable plastic stress and'stain in the frame sustaining a verticbl:load P, as shown below .Aiiume ttrai(q=a5 ) Cnd that each element is constructed of analuminum alloy with following properties:oi = 350 tntPa,k = 840 MPa,n = 0:2:,,1o =.,.3',,fr

:43T{-'':. I An..ttbeam;..,ihowq.,bet0w;.i.isi$ubi.e,cted,.tg.pt{-a bending resuftlng fiom ehd coq,ples. .,Determine the moment causing initialyielding and that resulting ih,complete plasticdeformation

ti_),F,ffiffi$:::ttttt:;:tffi1::U:::t:t;s"sgr r 1.,4i': :t :':f#i4IH #trfj.1i',' ;{a#d:,: l1F;':.:: : ""l:$flTii{Lf '. "," ffii' .; ., . r." ";-, - :"$T,

., ffil'.,''": . ':r:r::i

continum mechanics sheet

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