Solution to Problem 241 Statically Indeterminate

For L₁:
$(75 - 74.98)(12) = \dfrac{P_1 (74.98 \times 12)}{0.05(29 \times 10^6)}$

$P_1 = 386.77 \, \text{lb}$
 

For L₂:
$(75 - 74.99)(12) = \dfrac{P_2 (74.99 \times 12)}{0.05(29 \times 10^6)}$

$P_2 = 193.36 \, \text{lb}$
 

Let
P = P₃ (Load carried by L₃)
P + P₂ (Total load carried by L₂)
P + P₁ (Total load carried by L₁)
 

$\Sigma F_V = 0$

$(P + P_1) + (P + P_2) + P = W$

$3P + 386.77 + 193.36 = 1500$

$P = 306.62 lb = P_3$
 

$\sigma_3 = \dfrac{P_3}{A} = \dfrac{306.62}{0.05}$

$\sigma_3 = 6132.47 \, \text{ psi}$           answer

Bring L₁ into L₂ = 74.99 ft
$\delta = \dfrac{PL}{AE}$

$(74.99 - 74.98)(12) = \dfrac{P_1 (74.98 \times 12)}{0.05(29 \times 10^6)}$

$P_1 = 193.38 \, \text{lb}$
 

Let P = P₂ (Load carried by L₂)
 

P + P₁ (Total load carried by L₁)
 

$\Sigma F_V = 0$

$(P + P_1) + P = 500$

$2P + 193.38 = 500$

$P = 153.31 \, \text{lb}$
 

$P + P_1 = 153.31 + 193.38$

$P + P_1 = 346.69 \, \text{lb}$
 

$\sigma = \dfrac{P + P_1}{A} = \dfrac{346.69}{0.05}$

$\sigma = 6933.8 \, \text{ psi}$           answer