Question 4.9: Determine the forces in members FJ, HJ, and HK of the K trus...
Determine the forces in members FJ, HJ, and HK of the K truss shown in Fig. 4.24(a) by the method of sections.
From Fig. 4.24(a), we can observe that the horizontal section aa passing through the three members of interest, FJ, HJ, and HK, also cuts an additional member FI, thereby releasing four unknowns, which cannot be determined by three equations of equilibrium. Trusses such as the one being considered here with the members arranged in the form of the letter K can be analyzed by a section curved around the middle joint, like section bb shown in Fig. 4.24(a). To avoid the calculation of support reactions, we will use the upper portion IKNL of the truss above section bb for analysis. The free-body diagram of this portion is shown in Fig. 4.24(b). It can be seen that although section bb has cut four members, FI, IJ, JK, and HK, forces in members FI and HK can be determined by summing moments about points K and I , respectively, because the lines of action of three of the four unknowns pass through these points. We will, therefore, first compute F_{HK} by considering section bb and then use section aa to determine F_{FJ} and F_{HJ}.
Section bb Using Fig. 4.24(b), we write
+ \circlearrowleft \sum{M_I} = 0 -25(8) – F_{HK}(12) = 0
F_{HK} = -16.67 kN
F_{HK} = 16.67 kN (C)
Section aa The free-body diagram of the portion IKNL of the truss above section aa is shown in Fig. 4.24(c). To determine F_{HJ}, we sum moments about F, which is the point of intersection of the lines of action of F_{FI} and F_{FJ} . Thus,
+ \circlearrowleft \sum{M_F} = 0 -25(16) – 50(8) + 16.67(12) – \frac{3}{5}F_{HJ}(8) – \frac{4}{5}F_{HJ}(6) = 0
F_{HJ} = -62.5 kN
F_{HJ} = 62.5 kN (C)
By summing forces in the horizontal direction, we obtain
+ \longrightarrow \sum{F_x} = 0 25 + 50 – \frac{3}{5}F_{FJ} – \frac{3}{5}(62.5) = 0
F_{FJ} = 62.5 kN (T)
Checking Computations Finally, to check our calculations, we apply an alternative equilibrium equation, which involves the three member forces determined by the analysis. Using Fig. 4.24(c), we write
+ \circlearrowleft \sum{M_I} = -25(8) – \frac{4}{5}(62.5)(6) + \frac{4}{5}(62.5)(6) + 16.67(12) = 0 Checks
