# Problem 003-mm | Method of Members

**Problem 003-mm**

For the structure shown in Fig. (MM)-P003 below, determine the reactions at A and D and the internal force in member CF.

**Solution 003-mm**

$\Sigma M_D = 0$

$2A_x = 3(60) + 6(30)$

$A_x = 180 \, \text{ kN}$

$\Sigma M_A = 0$

$2D_x = 3(60) + 6(30)$

$D_x = 180 \, \text{ kN}$

From the FBD of member ABC

$\Sigma F_x = 0$

$\frac{3}{5}F_{CF} = A_x$

$\frac{3}{5}F_{CF} = 180$

$F_{CF} = 300 \, \text{ kN}$

$\Sigma M_B = 0$

$3A_y = 1.5(\frac{4}{5}F_{CF})$

$3A_y = 1.2(300)$

$A_y = 120 \, \text{ kN}$

From FBD of the whole system

$\Sigma F_V = 0$

$D_y = A_y + 60 + 30$

$D_y = 120 + 60 + 30$

$D_y = 210 \, \text{ kN}$

**Summary**

$A_x = 180 \, \text{ kN leftward}$

$A_y = 120 \, \text{ kN downward}$

$D_x = 180 \, \text{ kN rightward}$

$D_y = 210 \, \text{ kN upward}$

$F_{CF} = 300 \, \text{ kN tension}$