# Solution to Problem 241 Statically Indeterminate

**Problem 241**

As shown in Fig. P-241, three steel wires, each 0.05 in.2 in area, are used to lift a load W = 1500 lb. Their unstressed lengths are 74.98 ft, 74.99 ft, and 75.00 ft.

(a) What stress exists in the longest wire?

(b) Determine the stress in the shortest wire if W = 500 lb.

**Solution 241**

_{1}= 74.98 ft; L

_{2}= 74.99 ft; and L

_{3}= 75.00 ft

Part (a)

_{1}and L

_{2}into L

_{3}= 75 ft length:

(For steel: E = 29 × 10

^{6}psi)

$\delta = \dfrac{PL}{AE}$

For L_{1}:

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

$P_1 = 386.77 \, \text{lb}$

For L_{2}:

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

$P_2 = 193.36 \, \text{lb}$

Let

P = P_{3} (Load carried by L_{3})

P + P_{2} (Total load carried by L_{2})

P + P_{1} (Total load carried by L_{1})

$\Sigma F_V = 0$

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

$3P + 386.77 + 193.36 = 1500$

$P = 306.62 lb = P_3$

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

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

Part (b)

P

_{1}+ P

_{2}= 580.13 lb > 500 lb (L

_{3}carries no load)

Bring L_{1} into L_{2} = 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 = P_{2} (Load carried by L_{2})

P + P_{1} (Total load carried by L_{1})

$\Sigma F_V = 0$

$(P + P_1) + P = 500$

$2P + 193.38 = 500$

$P = 153.31 \, \text{lb}$

$P + P_1 = 153.31 + 193.38$

$P + P_1 = 346.69 \, \text{lb}$

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

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

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