Method of sections for Truss member force calculation

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)

Solution:

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 F_x = 0;

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

         A_y+ D_y - 25 -10 - 20 - 15 = 0;

         A_y+ D_y  = 70;

(iii) Σ M_z = 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);

        A_y x 0 + D_y x 6  - 20 x 2 - 15 x 4  - 10 x 2 + 25 x 0 =0;

        we get D_y = 120/6 = 20 kN;

        Therefore A_y = 70 - 20 = 50 kN

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).

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 F_x = 0; => F_BC + F_GF + F_GC cos45 = 0;            (i)

Σ F_y = 0; => 50 - 25 - 20 -10 - F_GC sin45 = 0;       (ii)

                      F_GC sin45 = - 5;     F_GC = - 7.072 kN;

Σ M = 0 about point G

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

           F_BC = 25 kN;

Substituting the value of F_BC and F_GC in eq. (i)  we get

            F_GF = - 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

You can also visit the following links of solved examples

Problem 3-1 Solving for Truss member forces by method of joints

Problem 7-5 Deflection of pin-jointed plane truss by unit load method

Example 5-2 SFD and BMD of Simple beam

Example 5-1 SFD and BMD for Cantilever

Exmple 5-4 SFD & BMD for Overhanging beam

More Solved Examples
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