Answer:
a) 3.31 inches
b) 20139 psi
Explanation:
L = 24 ft = 24*12 = 288 inches
Let the elastic modulus of the steel bar (structured ASTM-A36) [tex]E = 29\times10^6 psi[/tex]. If a 288 inches long bar must no elongate more than 0.2 inches, then its strain must be:
[tex]\epsilon = \frac{\Delta L}{L} = \frac{0.2}{288} = 0.000694[/tex]
Then the tensile stress must not exceed
[tex]\sigma = \epsilon E = 0.000694*29*10^6 = 20139 psi[/tex]
If the tensile load is 50000 lb then the cross-sectionarea must not be larger than
[tex]A = F/\sigma = 50000 / 20139 = 2.48 in^2[/tex]
If it's 3/4 inches thick then the width must be
2.48 / (3/4) = 3.31 inches
Answer:
a) w = 3.19 in
b)  σ = 20849.17 psi
Explanation:
Given
t = 3/4 in
L = 24 ft = 288 in
P = 50,000 lb
ΔL = 0.20 in
E = 30022811.71 lb/in²
a) We can apply the formula
ΔL = P*L/(A*E)  ⇒  A = P*L/(ΔL*E)
⇒  A = 50,000 lb*288 in/(0.20 in*30022811.71 lb/in²)
⇒  A = 2.398 in²
Then we have
A = w*t  ⇒   w = A/t
⇒   w = 2.398 in² /0.75 in
⇒   w = 3.198 in
b)  σ = P/A  ⇒    σ = 50,000 lb/2.398 in²
⇒    σ = 20849.17 psi
