Method of sections

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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:

Figure 1: The pin-jointed plane truss considered.

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.

Figure 2: Section (1) - (1) passing through GF, GC and BC.

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_GCcos45 = 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