Solution to Problem 238 Statically Indeterminate

Problem 238
The lower ends of the three bars in Fig. P-238 are at the same level before the uniform rigid block weighing 40 kips is attached. Each steel bar has a length of 3 ft, and area of 1.0 in.², and E = 29 × 10⁶ psi. For the bronze bar, the area is 1.5 in.² and E = 12 × 10⁶ psi. Determine (a) the length of the bronze bar so that the load on each steel bar is twice the load on the bronze bar, and (b) the length of the bronze that will make the steel stress twice the bronze stress.

Solution 238

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(a) Condition: P_st = 2P_br
$\Sigma F_V = 0$

$2P_{st} + P_{br} = 40$

$2(2P_{br}) + P_{br} = 40$

$P_{br} = 8 \, \text{kips}$

$P_{st} = 2(8) = 16 \, \text{kips}$

$\delta_{br} = \delta_{st}$

$\left( \dfrac{PL}{AE} \right)_{br} = \left( \dfrac{PL}{AE} \right)_{st}$

$\dfrac{8000 \, L_{br}}{1.5 (12 \times 10^6)} = \dfrac{16000 \, (3 \times 12}{1.0 (29 \times 10^6)}$

$L_{br} = 44.69 \, \text{in}$

$L_{br} = 3.72 \, \text{ft }$ answer

(b) Condition: σ_st = 2σ_br
$\Sigma F_V = 0$

$2P_{st} + P_{br} = 40$

$2(\sigma_{st} \, A_{st}) + \sigma_{br} \, A_{br} = 40$

$2 \, [ \, (2 \sigma_{br}) \, A_{st} \, ] + \sigma_{br} \, A_{br} = 40$

$4 \sigma_{br} (1.0) + \sigma_{br} (1.5) = 40$

$\sigma_{br} = 7.27 \, \text{ksi}$

$\sigma_{st} = 2(7.27) = 14.54 \, \text{ksi}$

$\delta_{br} = \delta_{st}$

$\left( \dfrac{\sigma L}{E} \right)_{br} = \left( \dfrac{\sigma L}{E} \right)_{st}$

$\dfrac{7.27(1000) \, L_{br}}{12 \times 10^6} = \dfrac{14.54(1000)(3 \times 12)}{29 \times 10^6}$

$L_{br} = 29.79 \, \text{in}$

$L_{br} = 2.48 \, \text{ft}$ answer

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