Question 4.8: Determine the forces in members CJ and IJ of the truss shown...

Section aa As shown in Fig. 4.23(a), a section aa is passed through members IJ, CJ, and CD, cutting the truss into two portions, ACI and DGJ. The left-hand portion, ACI, will be used to analyze the member forces.

Reactions Before proceeding with the calculation of member forces, we need to determine reactions at support A. By considering the equilibrium of the entire truss (Fig. 4.23(b)), we determine the reactions to be A_x = 0, A_y = 100 kN \uparrow , and G_y = 100 kN \uparrow .

Member Forces The free-body diagram of the portion ACI of the truss is shown in Fig. 4.23(c). The slopes of the inclined forces, F_{IJ} and F_{CJ} , are obtained from the dimensions of the truss given in Fig. 4.23(a) and are shown on the free-body diagram. The unknown member forces are determined by applying the equations of equilibrium, as follows.

Because F_{CJ} and F_{CD} pass through point C, by summing moments about C, we obtain an equation containing only F_{IJ}:

+\circlearrowleft \sum{M_C}=0                -100(8) + 40(4) – \frac{4}{\sqrt{17}}F_{IJ} (5) = 0

The negative answer for F_{IJ} indicates that our initial assumption about this force being tensile was incorrect. Force F_{IJ} is actually a compressive force.

Next, we calculate F_{CJ} by summing moments about point O, which is the point of intersection of the lines of action of F_{IJ} and F_{CD}. Because the slope of member IJ is 1:4, the distance OC = 4(IC) = 4(5) = 20 m (see Fig. 4.23(c)). Equilibrium of moments about O yields

+\circlearrowleft \sum{M_O}=0                100(12) – 40(16) – 40(20) + \frac{3}{\sqrt{13}}F_{CJ} (20) = 0

Checking Computations To check our computations, we apply an alternative equation of equilibrium, which involves the two member forces just determined.

+ \uparrow \sum{F_y} = 100 – 40 – 40 – \frac{1}{\sqrt{17}}(132) + \frac{3}{\sqrt{13}}(14.42) = 0                           Checks