Answer:
a) 76,800 Psi
b) 22.14 × 10⁶ Psi
Explanation:
Data provided in the question:
Length of the block, w = 8 in
Width of the block, h = 0.25
Distance between the supports, L = 4 in
Applied force, F = 400 lb
Deflection at the time of failure, δ = 0.037 in
Now,
a) Flexural strength = [tex]\frac{3FL}{2wh^2}[/tex]
= [tex]\frac{3\times400\times4}{2\times0.5\times0.25^2}[/tex]
= 76,800 Psi
b) Flexural modulus = [tex]\frac{FL^3}{4wh^3\delta}[/tex]
= [tex]\frac{400\times4^3}{4\times0.5\times0.25^3\times0.037}[/tex]
= 22.14 × 10⁶ Psi
