**Problem 864**

An 18-ft beam, simply supported at 4 ft from each end, carries a uniformly distributed load of 200 lb/ft over its entire length. Compute the value of EIδ at the middle and at the ends.

**Solution 864**

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$R = \frac{1}{2}(200)(18) = 1800 ~ \text{ lb}$

$M_A = M_E = 0$

$M_B = M_D = -200(4)(2) = -1600 ~ \text{ lb}\cdot\text{ft}$

$M_C = 5(1800) - 200(9)(4.5) = 900 ~ \text{ lb}\cdot\text{ft}$

**At the middle (labeled as point C), use the span B-C-D**

$L_2 = 5 ~ \text{ft}$ ← span from C to D

$M_B L_1 + 2M_C(L_1 + L_2) + M_DL_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_B}{L_1} + \dfrac{h_D}{L_2} \right)$

$-1600(5) + 2(900)(5 + 5) - 1600(5) + \dfrac{200(5^3)}{4} + \dfrac{200(5^3)}{4} = 6EI \left( \dfrac{h_C}{5} + \dfrac{h_C}{5} \right)$

$-8000 + 18,000 - 8000 + 6250 + 6250 = 2.4EI \, h_C$

$14,500 = 2.4EI \, h_C$

$EI \, h_C = \dfrac{18,125}{3}$

The positive sign indicates that points B and D are above point C, hence,

$\delta_C = \dfrac{18,125}{3EI} ~ \text{ downward}$ *answer*

**At the ends (labeled as points A and E), use the span A-B-D**

$L_2 = 10 ~ \text{ft}$ ← span from B to D

$M_A L_1 + 2M_B(L_1 + L_2) + M_DL_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_A}{L_1} + \dfrac{h_D}{L_2} \right)$

$0 + 2(-1600)(4 + 10) - 1600(10) + \dfrac{200(4^3)}{4} + \dfrac{200(10^3)}{4} = 6EI \left( \dfrac{h_A}{4} + 0 \right)$

$-44,800 - 16,000 + 3200 + 50,000 = 1.5EI \, h_A$

$-7600 = 1.5EI \, h_A$

$EI \, h_A = -\dfrac{15,200}{3}$

The negative sign indicates that points A is below point B, hence,

$\delta_A = \dfrac{15,200}{3EI} ~ \text{ downward}$ *answer*