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.
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) ΣFx = 0: There is no external horizontal force acting on the truss. Therefore, the horizontal reaction at D will be zero, i.e. Dx = 0kN
(ii) ΣFy = 0 -> Ay + Dy - 25 -10 - 20 - 15 = 0 -> Ay + Dy = 70;
(iii) ΣMz = 0: Considering z-axis perpendicular to the plane and passing through joint A. Take moment of all the forces about z-axis.
ΣMz = 0 -> Ay x 0 + Dy x 6 - 20 x 2 - 15 x 4 - 10 x 2 + 25 x 0 =0 -> Dy = 120/6 -> Dy = 20kN
Therefore, from equation (i) we get Ay = 70 - 20 -> Ay =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) ΣFx = 0 -> FBC + FGF + FGC cos45 = 0
(v) ΣFy = 0 -> 50 - 25 - 20 -10 - FGC sin45 = 0 -> FGC sin45 = - 5 -> FGC = - 7.072kN
(vi) ΣMG = 0 -> FBC x 2 - 50 x 2 + 25 x 2 = 0 -> FBC = 25kN
Thus, substituting the value of FBC and FGC in equation (iv) we get FGF = - 20kN
Given: A pin-jointed truss is given in Figu...