Required information a 32 kn tensile load p is applied to a...

Required information A 3.2-kN tensile load (P) is applied to a test coupon made from 1.6-mm flat steel plate (E = 200 GPa, v= 0.30). 12 mm 50 mm P B Determine the resulting change in the thickness of portion AB. The resulting change in the thickness of portion AB is x 10-9 m.

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Transcript

00:01 Given data l is equal to 6000 mm and b is equal to 150, p is equal to 20 mm and p is equal to 300 into 10 to power 3 newton.

00:27 B is equal to 200 in gp is equal to 2 into 10 to power 5 newton per mm square and m is equal to 0 .25.

00:50 So according to the diagram and direction the diagram that's for x axis, y axis and z axis and that's p for both sides and t here.

01:02 After that the stress in in x direction stress in x direction so half sigma x is equal to p upon a c upon s is equal to p upon b after putting the values 300 into 10 to power 3 upon 150 into 20 so half sigma sigma x is equal to 100 newton per mm.

01:45 Similarly, half sigma y is equal to 0.

01:49 And half sigma z is equal to 0.

01:52 Changing length.

01:55 Changing length.

01:57 Level is equal to half sigma x upon b minus m.

02:06 Sigma y upon b minus m.

02:11 Half sigma z upon b.

02:14 Into m is equal to 100 upon 2 into 10 to the power 5 into 6000 is equal to 3 mm.

02:30 Because of half sigma y and half sigma z is equal to 0.

02:37 So change in width tell sp is equal to half sigma z upon e minus m.

02:49 1 .5s y upon e minus m .5s z upon e into b.

02:57 So, putting the values minus 0 .25 into 100 upon 2 into 10 to the power 5 into 150 is equal to minus 0 .01875 mm.