
MECHANICS  CASE STUDY SOLUTION

1.0 inch Specimen Cube Inserted
into Test Frame


A simple cube of material is inserted into a snug fitting test frame and then
a 3,000 lb load is applied on the top surface. The sides of the test frame are
stiff and will not deflect at the test load level. Also, the sides are smooth
so there is no friction forces as the specimen is loaded.
The objective in testing the specimen is to determine the Poisson's ratio of
the material. The Young's modulus, E, is known, 2,000 ksi. Also, the total deflection
after loading is 0.001 inches.





Determining Poisson's ratio

Cube Under Loading


A good place to start solving this problem is the full 3D Hooke's law equations.
While these look complex, many terms are easily determined. The vertical stress
can be calculated as
σ_{z} = F/A =
3000 lb/(1 in)^{2} = 3,000 psi = 3.0 ksi
The other two horizontal stresses are unknown, but must be the same due to
symmetric geometry and loading. Thus,
σ_{x} = σ_{y} =
p




Problem Diagram 

Next, the three strains can be determined. The vertical strain can be found
since the deflection is known, giving
ε_{z} = δ/L
= (0.001 in) / (1.0 in) = 0.001
The other two strains are also known since the outside walls do not deflect
(due to their high stiffness). Thus,
ε_{x} = ε_{y} =
0.0
Also, Young's modulus is known, E = 2,000 ksi.




Graph of Function
f(ν) = (1ν) / [(1+ν)(12ν)] 

Simplifying the 3D equations to two equations give,
Next, substituting specific values reduce these equations to
The negative signs represent compression for both strain and stress. There are
now two unknowns, p and ν. The second equation can be
rearranged to give,
This can be manipulated to give
3ν^{2} + 0.5ν 
0.5 = 0
Using the quadratic equation, the positive root gives
ν = 0.3333
This equation can also be solved numerically by
graphical methods or by using numerical programs like MathCad, Mathematica, MathLab,
or even Excel. The equation is graphed at the left which also gives
ν = 0.3333






Stresses



The stresses in the x and y directions can now be determined,
p = σ_{x} = σ_{y} =
1.5 ksi = 1,500 psi




