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)