# Problem 005-ms | Method of Sections

**Problem 005-ms**

The structure shown in Figure T-07 is pinned to the floor at A and H. Determine the magnitude of all the support forces acting on the structure and find the force in member BF.

**Solution 005-ms**

$\Sigma M_A = 0$

$1.2R_H = 1.2(100) + 4.5(80)$

$R_H = 400 \, \text{ kN upward}$ *answer*

$\Sigma M_H = 0$

$1.2A_V = 4.5(80)$

$A_V = 300 \, \text{ kN downward}$ *answer*

$\Sigma F_H = 0$

$A_H = 80 \, \text{ kN to the left}$ *answer*

If you wish to check the reaction forces at A and H, pass a cutting plane through members AB, AG, and AH. As shown in the figure above, the cutting plane b-b cut these members. Take the section above b-b and replace the cut members by internal force to maintain equilibrium. Solve for the internal force in members AB, AG, and AH. With these forces known, you can then check the reactions through joints A and H. The free body diagram of the section above b-b is similar to the free body diagram of section above a-a. See the solution of section a-a below for easy reference of doing the section b-b.

**From section above a-a**

$\Sigma F_H = 0$

$\frac{4}{\sqrt{41}}F_{BF} = 80$

$F_{BF} = 128.06 \, \text{ kN tension}$ *answer*