E ≔ 25 GPa Modulus of Elasticity ν ≔ 0.2 Poisson's Ratio σx ≔ 120 MPa Stress in the X-Direction εy ≔ 0 Strain in the y-direction (no change in height) x ≔ 100 mm Length y ≔ 40 mm Height z ≔ 25 mm Width Using the equation for Stress in the X-Direction, we solve for Strain the X-Direction εx + ⎛⎝ν ⋅ εy⎞⎠ solve , εx 4.608 ⋅ MPa εx ≔ σx ＝ E ⋅ ―――― ――― → ――――= 0.005 GPa 1 − ν2 We use this to solve for stress in the Y-Direction εy + ν ⋅ εx σy ≔ E ⋅ ――― = 24 MPa 1 − ν2 Answer for (a) We now know the strains in the x and y direction, but we need to know the strain the zdirection to get the full deformation of the cube −ν εz ≔ ――⋅ ⎛⎝εx + εy⎞⎠ = −0.001 1−ν Now we can use the equations for strain to find the change in dimensions xnew ≔ x − ⎛⎝εx ⋅ x⎞⎠ = 99.539 mm ynew ≔ y = 40 mm znew ≔ z − ⎛⎝εz ⋅ z⎞⎠ = 25.029 mm Using this, we can calculate the change in area of the top face, as well as the change in the volume of the block. ∆AABCD ≔ ⎛⎝xnew ⋅ znew⎞⎠ − (x ⋅ z) = −9 mm 2 Answer for (b) ∆V ≔ ⎛⎝xnew ⋅ ynew ⋅ znew⎞⎠ − (x ⋅ y ⋅ z) = −346.131 mm 3 Answer for (c)