Question

A rectangular steel eyebar 3/4 inches thick and 24 ft long between centers of end pins is subjected to an axial tensile load of 50,000 lb. If this bar must not elongate more than 0.20 inches, calculate
(a) the width of the bar and
(b) tensile stress developed at maxium elongation, using required width calculated in (a)

Answer 1

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

Answer 2

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) $E = 29\times10^6 psi$. If a 288 inches long bar must no elongate more than 0.2 inches, then its strain must be:

$\epsilon = \frac{\Delta L}{L} = \frac{0.2}{288} = 0.000694$

Then the tensile stress must not exceed

$\sigma = \epsilon E = 0.000694*29*10^6 = 20139 psi$

If the tensile load is 50000 lb then the cross-sectionarea must not be larger than

$A = F/\sigma = 50000 / 20139 = 2.48 in^2$

If it's 3/4 inches thick then the width must be

2.48 / (3/4) = 3.31 inches