Question

2 PROBLEM Consider a unit square of a 2D solid occupying the region os x s1 and os y s1 After loads are applied, the displacements are given by uax and by (a) Sketch the deformed shape for: i) a 0.01, b = 0.015; ii) a = 0.01/x, b -0.01/y; ii) a = o.01y/x, b -0.015x/y (b) Calculate the six strain components for all three cases

Answer 1

Solution Livendata: displacements u= ax a direction v=by – direction of all (9) Shape of defrom where a =0.01, b =0.015 - 11 (11) a=001, b=0.01 (ill) a = 0.014 , b = -0.015 x/y 9 Strain componats is all casts. X x b)

Ron Table is To Uzax va by I cot) I (1.0) 1 (1) b=0.015 U= .012 -0.01 0 0.015 0.0 0 .01 105 Y = 0.01s U=0.01 q=0.0172 -0.dly 0.01 -0.01 0.01 -0.0110.01 -0.010.01 -0.017 -0.011 la=0.019/ x olo 10. oo U= 0.01% o 0.01st 0.01 0.015 V = -0,0152 16=0.05uly 1 Where the value of es and v (i) at a=ool 6=0.015 (U= 0.01%) IV-0.015 y] w at a=ool b=0.07 Ulex =>\0=0.01) V= (y |--00) Sovely 6= -0.015* (3) at a=o.oly = 7 U= 0.017) - -0.0152 the displane ant at different co-ordinates are sham in table. Case 1: Deforned shape at a 0.0i and bc0.05). Y cou t o contest oryginalslupe (1.011-015) Ik deformed shaped (108) 0,02 ('0) & (0.0)

Case 2 is at a=0.01 I b= -0.ol defound shape C0,01,0.99 (1.01,0199) (0.0-0.01) (1.010.01) Case 3: at a= owly & bc - 0.0 ((0.001), deformed slope. fok (1.0, 0.965) (0,07 1,-0.015) Six strain components © ¢xx com Oły2 = po ]-tay @ Eyy- Ozz- 'al on the date uz & Ezz = 2u Dexy. Ya se t au ] Eyx Ezz-Eyz = Exx-02 E Enzo } els seolud Cloons

b=0.015 Case 1 a=0.0 : U=an=0.01x theory 30.012) -0.01 het nie2 (0.012) sp v= by = 0.015 ! - 560159) > 0 ray - Cosy) = o-vis 3x ax da ax Jy sy ::\Exx = load = 001? 2 Eyy = 2 / = 0.0157 ) 3) Exy= 1 2 + 1 1 1 1 1 1 (0+0]=0 4&zz=0/ Note: see in deformed shape 5) Egz = Ezy=0/ ) Ex- Exz=0/ case a aso.el & 6= 0.01 u= ax = (10.01 30.01 v. be tool/y = -0.01 fig only changes inex andry direction dy lav = 0 4 Exx=Ewy = Ezz = Exy = Ezz- Eyz = E + OX all componentar Zero see in deformed shape fis no changes in body but displaced Note: -

Case3 azody f = 0.015 v=bx=( roolsx J =ax = 10-014) * u=ooly = -0.015* ar = -0.05 d = 0 dizon -0.0025) * 1) Exx=00 2) Eyy = 0 3) Exy - à 0.01 -0.0 y ezz=0|| s) Eyz Ergo// 1) E rx-kx2=0): - Note! sie in deformed stape fig, chayes is only in xy diration.

summary

a)

to draw deform shapes displacement at each poin is require

displacement U & V at (0,0) and at a=0.01 & b=0.015 is calculated as U=0 V=0

displacement U & V at (0,1) and at a=0.01 & b=0.015 is calculated as U=0 V=0.015

displacement U & V at (1,0) and at a=0.01 & b=0.015 is calculated as U=0.01 V=0

displacement U & V at (1,1) and at a=0.01 & b=0.015 is calculated as U=0.01 V=0.015

draw the deformed shape @ X+U in x axis and Y+V in y axis at each above points

similarly calculate all U&V values and draw shapes

b)

using shear and displacement relation formulas calculate all values

for more details refer above images

Thank you