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Method of sections for Analysis of pin-jointed plane Truss  

Problem 3-2

Using method of sections determine the forces in the members BC, GC and GF of the pin-jointed plane truss shown in figure 3-2(a)

Figure 3-2(a)


This truss is determinate as it satisfies the condition of determinacy of plane truss:      m = 2j - 3

Reactions at the support:

Support A is on the roller, therefore it will have only vertical reaction and no  horizontal reaction. Support D being hinged will experience both horizontal and vertical reactions. (refer to figure 3-2(b)).

Figure 3-2 (b)

(i) There is no horizontal force acting on the truss therefore the horizontal reaction at D will be zero to satisfy Σ Fx = 0;

(ii) Applying the condition of equilibrium Σ Fy = 0; we get

         Ay+ Dy - 25 -10 - 20 - 15 = 0;

         Ay+ Dy  = 70;

(iii) Σ Mz = 0; Considering z-axis perpendicular to the plane and passing through joint A. Take moment of all the forces about z-axis (taking clock-wise negative and anticlock-wise positive);

        Ay x 0 + Dy x 6  - 20 x 2 - 15 x 4  - 10 x 2 + 25 x 0 =0;

        we get Dy = 120/6 = 20 kN;

        Therefore Ay = 70 - 20 = 50 kN

Figure 3-2(c)


Calculation of member forces by method of sections

Method of sections is useful when we have to calculate the forces in some of the members, not all. It is expalined in this example. 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 3-2(c). 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 the figure.

Σ Fx = 0; => FBC + FGF + FGC cos45 = 0;            (i)

Σ Fy = 0; => 50 - 25 - 20 -10 - FGC sin45 = 0;       (ii)

                      FGC sin45 = - 5;     FGC = - 7.072 kN;

Σ M = 0 about point G

           FBC x 2 - 50 x 2 + 25 x 2 = 0;                       (iii)

           FBC = 25 kN;

Substituting the value of FBC and FGC in eq. (i)  we get

            FGF = - 20 kN

negative value indicates that the force is opposite to the assumed direction.

Result of Member forces calculations
Member Force (k2N) Nature of force
GC 7.07 Compressive
BC 25 Tensile
GF 20 Compressive

Problem 3-1 Solving Truss by method of joints

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Last updated on Wednesday January 22, 2014