$\Sigma M_{D} = 0$

$8R_1 = 200(5) + 400(1)$

$R_1 = 175 \, \text{ lb}$

$\Sigma M_{A} = 0$

$8R_2 = 200(3) + 400(7)$

$R_2 = 425 \, \text{ lb}$

By ratio and proportion

$\dfrac{y_C}{1} = \dfrac{2125}{5}$

$y_C = 425 \, \text{ lb}\cdot\text{ft}$

From the conjugate beam

$\Sigma M_D = 0$

$8F_1 + \frac{1}{2}(4)(1600)[1 + \frac{2}{3}(4)] = \frac{1}{2}(3)(525)[5 + \frac{1}{3}(3)] + \frac{1}{2}(5)(2125)[\frac{2}{3}(5)]$

$F_1 = 1337.5 \, \text{ lb}\cdot\text{ft}^2$

$\Sigma M_A = 0$

$8F_2 + \frac{1}{2}(4)(1600)[3 + \frac{1}{3}(4)] = \frac{1}{2}(3)(525)[\frac{2}{3}(3)] + \frac{1}{2}(5)(2125)[3 + \frac{1}{3}(5)]$

$F_2 = 1562.5 \, \text{ lb}\cdot\text{ft}^2$

Consider the section to the left of B in conjugate beam

$M_B = \frac{1}{2}(3)(525)[\frac{1}{3}(3)] - 3F_1$

$MB = 787.5 - 3(1337.5)$

$M_B = -3225 \, \text{ lb}\cdot\text{ft}^3$

Thus, the deflection at B is

$EI ~ \delta_B = M_B$

$EI ~ \delta_B = 3225 \, \text{ lb}\cdot\text{ft}^3$ *answer*

Consider the section to the right of C in conjugate beam

$M_C = \frac{1}{2}(1)(y_C)[\frac{1}{3}(1)] - 1F_2$

$M_C = \frac{1}{2}(1)(425)[\frac{1}{3}(1)] - 1(1562.5)$

$M_C = -1491.67 \, \text{ lb}\cdot\text{ft}^3$

Thus, the deflection at C is

$EI ~ \delta_C = M_C$

$EI ~ \delta_C = -1491.67 \, \text{ lb}\cdot\text{ft}^3$

$EI ~ \delta_C = 1491.67 \, \text{ lb}\cdot\text{ft}^3$ downward *answer*

## dito po sa part ng solution

dito po sa part ng solution

ΣMA=0

8F2+(1/2)(4)(1600)[(1/3)(4)]=(1/2)(3)(525)[(2/3)(3)]+(1/2)(5)(2125)[3+(1/3)(5)]

F2=1562.5 lb⋅ft2

hindi po ba

ΣMA=0

8F2+(1/2)(4)(1600)[3+(1/3)(4)]=(1/2)(3)(525)[(2/3)(3)]+(1/2)(5)(2125)[3+(1/3)(5)]

F2=1566.25 lb⋅ft2

## You are correct. Update has

In reply to dito po sa part ng solution by Mark123 (not verified)

You are correct, fortunately, the answer for F2 is correct. Update has been made for the moment arm. Thank you.