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# 6.9: Method of Sections: Problem Solving II

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Method of Sections: Problem Solving II

### 6.9: Method of Sections: Problem Solving II

Consider an arbitrary truss structure composed of diagonal, vertical, and horizontal members fixed to the wall. To calculate the force acting on members CB, GB, and GH, method of sections can be used. The loads and lengths of the horizontal and vertical members are known parameters, as shown in the figure.

To begin, a cut is made along a plane intersecting CB, GB, and GH members, and a free-body diagram of the right side section is drawn.

The moment equilibrium equation about point G is applied.

The result gives the force along member CB as 4 kN, with a positive sign indicating tension in the member.

Now, FCB can be expressed using a slope triangle in BCG, while FGH can be expressed using a slope triangle in CEG. Considering the summation of vertical and horizontal forces, the force equilibrium equations can be written as the following:

The value of FCB is substituted, and the force equilibrium equations are solved simultaneously. The result yields the force FGH as -7.454 kN and FGB as 3.772 kN. The negative sign of FGH indicates that the force is compressive, while the positive sign for FGB indicates the tensile force.