**Problem 910**

A timber beam AD, 6 in. thick by 10 in. high and loaded as shown in Figure P-910, is pinned at its lower end and supported by a horizontal cable CE. Compute the maximum compressive stress developed in the beam.

**Solution 910**

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$6T = 4(6000) + 12(3000)$

$T = 10,000 ~ \text{lb}$

$\Sigma F_H = 0$

$A_h = T = 10,000 ~ \text{lb}$

$\Sigma F_V = 0$

$A_v = 6000 + 3000 = 9000 ~ \text{lb}$

$P_1 = A_h \cos \theta = 10,000(4/5) = 8000 ~ \text{lb}$

$P_2 = A_v \sin \theta = 9000(3/5) = 5400 ~ \text{lb}$

$P_3 = 6000 \sin \theta = 6000(3/5) = 3600 ~ \text{lb}$

$P_4 = T \cos \theta = 10,000(4/5) = 8000 ~ \text{lb}$

$P_5 = 3000 \sin \theta = 3000(3/5) = 1800 ~ \text{lb}$

$V_1 = A_h \sin \theta = 10,000(3/5) = 6000 ~ \text{lb}$

$V_2 = A_v \sin \theta = 9000(4/5) = 7200 ~ \text{lb}$

$V_3 = 6000 \cos \theta = 6000(4/5) = 4800 ~ \text{lb}$

$V_4 = T \sin \theta = 10,000(3/5) = 6000 ~ \text{lb}$

$V_5 = 3000 \cos \theta = 3000(4/5) = 2400 ~ \text{lb}$

$P_1 + P_2 = 8000 + 5400 = 13,400 ~ \text{lb}$

$V_2 - V_1 = 7200 - 6000 = 1200 ~ \text{lb}$

At Point *B*

$\sigma_a = 223.33 ~ \text{psi compression}$

$\sigma_f = \dfrac{6M}{bd^2} = \dfrac{6(6000)(12)}{6(10^2)}$

$\sigma_f = 720 ~ \text{psi at extreme tension and compression fibers}$

$\sigma_B = \sigma_a + \sigma_f = 223.33 + 720$

$\sigma_B = 943.33 ~ \text{psi}$

At Point *C*

$\sigma_a = 163.33 ~ \text{psi compression}$

$\sigma_f = \dfrac{6M}{bd^2} = \dfrac{6(12,000)(12)}{6(10^2)}$

$\sigma_f = 1440 ~ \text{psi at extreme tension and compression fibers}$

$\sigma_C = \sigma_a + \sigma_f = 163.33 + 1440$

$\sigma_C = 1603.33 ~ \text{psi}$

Thus, maximum compressive stress is that at *C*

$\sigma_{max} = 1603.33 ~ \text{psi}$ *answer*