The force required to fracture the material is 8437.5 lb and the deflection of the sample at fracture is 0.030758 in.
Let's determine the force required to fracture the material and the deflection of the sample at fracture, assuming that no plastic deformation occurs.
We are given the following information:
Flexural strength (F) = 45,000 psi
Flexural modulus (E) = 18 × 10^6 psi
Width (b) = 0.5 in
Height (h) = 0.375 in
Length (L) = 8 in
Distance between supports (d) = 5 in
We can use the following formula to calculate the force required to fracture the material:
F = (48Ebh^2) / (6d^2)
Where:
F is the force required to fracture the material (lb)
E is the flexural modulus (psi)
b is the width of the sample (in)
h is the height of the sample (in)
L is the length of the sample (in)
d is the distance between supports (in)
Plugging in the given values, we get:
F = (48 * 18 × 10^6 * 0.5 * 0.375^2) / (6 * 5^2)
F = 8437.5 lb
We can use the following formula to calculate the deflection of the sample at fracture:
δ = Fd^3 / (48Ebh^3)
Where:
δ is the deflection of the sample at fracture (in)
F is the force required to fracture the material (lb)
E is the flexural modulus (psi)
b is the width of the sample (in)
h is the height of the sample (in)
L is the length of the sample (in)
d is the distance between supports (in)
Plugging in the given values, we get:
δ = (8437.5 * 5^3) / (48 * 18 × 10^6 * 0.5 * 0.375^3)
δ = 0.030758 in
Therefore, the force required to fracture the material is 8437.5 lb and the deflection of the sample at fracture is 0.030758 in.