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Please determine the forces in the members BC, GC and GF of the pin-jointed plane truss shown in Figure 1 using the method of sections.

**Given:**

**Solution:**

__Reactions at the supports__

Support A will have only vertical reaction and no horizontal reaction since it is on the roller. Support D is hinged and therefore will experience both horizontal and vertical reactions. Furthermore, the truss considered is determinate as it satisfies the condition of determinacy of plane truss: m = 2j – 3.

(i) ΣF_{x} = 0: There is no external horizontal force acting on the truss. Therefore, the horizontal reaction at D will be zero, i.e. **D _{x} = 0kN**

(ii) ΣF_{y} = 0 -> A_{y} + D_{y} - 25 -10 - 20 - 15 = 0 -> A_{y} + D_{y} = 70;

(iii) ΣM_{z} = 0: Considering z-axis perpendicular to the plane and passing through joint A. Take moment of all the forces about z-axis.

ΣM_{z} = 0 -> A_{y} x 0 + D_{y} x 6 - 20 x 2 - 15 x 4 - 10 x 2 + 25 x 0 =0 -> D_{y} = 120/6 -> **D _{y} = 20kN**

Therefore, from equation (i) we get A_{y} = 70 - 20 -> **A _{y} =50kN**

__Calculation of member forces by method of sections__

As it is explained in this example, method of sections is useful when we have to calculate the forces in some of the members, not all. We cut the truss into two parts through section (1) - (1) passing through GF, GC and BC. We consider the equilibrium of the left side part of the truss as shown in Figure 2. While considering equilibrium of this part we look only for the support reactions, the external forces acting on the truss, and the member forces which are cut by the section.

Let's assume the unknown forces in members GF, GC and BC as tensile as shown in Figure 2. It should be outlined that possible negative values indicate that the force is opposite to the assumed direction.

(iv) ΣF_{x} = 0 -> F_{BC} + F_{GF} + F_{GC }cos45 = 0

(v) ΣF_{y} = 0 -> 50 - 25 - 20 -10 - F_{GC} sin45 = 0 -> F_{GC} sin45 = - 5 -> **F _{GC} = - 7.072kN**

(vi) ΣM_{G} = 0 -> F_{BC} x 2 - 50 x 2 + 25 x 2 = 0 -> **F _{BC} = 25kN**

Thus, substituting the value of F_{BC} and F_{GC} in equation (iv) we get **F _{GF} = - 20kN**

Structural AnalysisTruss deflection using the unit load method

Buckling

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Confinement (structural)

Elasticity and Inelasticity

Fatigue

Moment distribution

Nonlinear analysis

Structural stability

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