Question

The displacement field in a steel machine component (Young’s modulus, E = 210 GPa and Poisson’s
ratio, ? = 0.33) is given (in metres) by the following expressions:
ux = c(x2 + ?) + 2c
uy = 2cx2 − cy
uz = c? + 5c?
where c = 10−4.
For a point (x, y, z) positioned at (1, 1, 1) calculate
a) The strain components; [4 marks]
b) The stress components and sketch a stress element showing how they act; [4 marks]
c) The body force needed to maintain a state of equilibrium. [4 marks]

Answer 1

Solution Given data- E=2100Pa = 210x103MPa 2- 0.33 c=104 Ux=((x+y)+20 Uy= 2 cx²-cy Uz = cy+5cz. point. (161). (6) The stres strain components- Exx Exy Exz Ez tyx Eyy Eyz - tax tzy tzz We have to find out Exx, Eyy, fzz and Exy = tyx, fyz = tzy, tzx= Exz, also Noco, We know Exx= dux dx acu = 2x10YXI. Exx=2x10-4 duy -1004 Eyy = [Eyy = -104 2Uz fzza az 0+56= 5x10-4 €22=5x104

2 Ку" $ук оу dux dy dx 4 = c(+y) +2 у = 24х2 су Uzy+cz = C + 2. С. 27 (1 + 4 ) (144x) е - Sc 25x104 c Хуу = 1+x) 2. x 169 8xy - еду = €уу = 3 ку - 6 х = 2.5 хто ч dy duz dy (2-су) + + dz d dy (y+ Sta) - О+С %yz HD - (09 0.5x o 9 а 2 Eyzatzy [ Eyzz Ezy - ostat dux Josxion 8x = 8x2= dż + duz da 0 +6 821 = a 0 (3) Ex2=Ezx=0 о

2 . Strain components 2 2х10 9 2.5x10 4 2.5 xo-9 -10-4 0.5000s o 5x10-4 0.5X104 2. ?: O 2. - | 0.5x304 2 of 5 (6) Noc, calculation of stress components E=210gpa = 0. 33 — 210x/02м We know E=2a (+o) E 20 а 4- 2(+2) 2(+o. 23) = 7. 95 х 10° мРА stress component- xx Exy 7172. Zук буу 1y- 05. И+ lay 4. = (A+24) + буу+ Xezz - Lame's constant.

2= CH (1-2) (It) (1-27) 0-33X210X103 (10.33 (: - 2x0.33) 0.33X2108103 1.33 X 0.34 - 1532 50.77 MPa Txx = 4+26) Exxt ) Eyy t Xtzz - (153250.77 + 2x 78.94x103) xqx70 4+ 153250.47 X (208) (-104) + 153250.77 x SX10 4 = 62.22 - 15.33+ 76.62 >123.51 MPa Tyg = À Exx + (1+29) Eyy + xtzz = (15-32,50x104 x 2x1074) + (15:32.5%90% +2X78.94 X103) x61004+ 15.3258704 x 5x1074 = 30.65 - 31.113 + 76.62 = 76.162 MPa

022= 16xx + 16уу + (1+24) erz = 15 - 3454/* x 3х1 5 15 345 х/05x (-1) -- 15 34 SP-+ 3x 7 8 94x703) x 510-9 30:45 15. 3 3 1 15 S. S6 с 170 - 89 МРО Now, 2xy = 154 ул = 4 y = =wхох хто = 78.99 x to'xsxo 4 - 347 Иfa ?у? ~ = 4 y = 78 - 31.1°х хи = 1:89 МР. 21, 2 2 2 2 4.8 х = О. Stress components 0,4 Zху 2х7 29x 2. 6 27 174729 6. (23.51 59) 47 7 8 9 29 41 16. (6 2- о (10-89 2. 8 9

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6 Now, for state of equilibrium- 2 Pxx + تها 2xy 1 / 2 2x2 + Bx = 0 ův We know €2250 Exx=2(x, x x = (1+29) Exx + x tyy+X Ezz differential with a Eyys-c, (+24) x26 +0+ 0 = (153250.77 + 2x78-94x103) x 2x104 dox dx 62.22 MPa 2 2 Zху 2y dy 27 (a. Oky) = {(1+4x) c4.9 (from 0 3 / 2 (6.8x2) [ from (11) do 2 (6.0) -(2x2) az W 62.22 + otot Bx = 0. .: body force in x-direction By =-62. 22 AN

Simulary in 7-direction & tay + y Bay + 22 2yz + By = 0. 1 / 2xy = x 3:47 1+48) < af? from 0] 46.4 = 4x10-4x4 2 "Y2 = 0. 22 2 dy byy x203 By = - 2 zxy =-GX 4X10-9 = -78.94 x 4 X 104, (By 2-3 1.57N Now, in 2-direction TX 2x2 + y 2 + 2 2 0 2 2 + B2 = 0, 0 3 Zxz a 2x = 0 pot Ja a dz 22= = (0+070 = 6toton B2=ON Thank you,

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