Question 4.11: Determine the force in each member of the Fink truss shown i...
Determine the force in each member of the Fink truss shown in Fig. 4.26(a).
The Fink truss shown in Fig. 4.26(a) is a compound truss formed by connecting two simple trusses, ACL and DFL, by a common joint L and a member CD.
Static Determinacy The truss contains 27 members and 15 joints and is supported by 3 reactions. Because m + r = 2j and the reactions and the members of the truss are properly arranged, it is statically determinate.
Reactions The reactions at supports A and F of the truss, as computed by applying the three equations of equilibrium to the free-body diagram of the entire truss (Fig. 4.26(b)), are
A_x = 0 A_y = 175 kN \uparrow F_y = 175 kN \uparrow
Joint A The method of joints can now be started at joint A, which has only two unknown forces, F_{AB} and F_{AI} , acting on it. By inspection of the forces acting at this joint (see Fig. 4.26(b)), we obtain the following:
F_{AI} = 391.31 kN (C) F_{AB} = 350 kN (T)Joint I The free-body diagram of joint I is shown in Fig. 4.26(c). Member BI is perpendicular to members AI and IJ, which are collinear, so the computation of member forces can be simplified by using an x axis in the direction of the collinear members, as shown in Fig. 4.26(c).
+\nwarrow \sum{F_{\overline{y}}} =0 -\frac{2}{\sqrt{5}}(50) – F_{BI} = 0
F_{BI} = -44.72 kN
F_{BI} = 44.72 kN (C)
+\nearrow \sum{F_{\overline{x}}} =0 391.31 -\frac{1}{\sqrt{5}}(50) + F_{IJ} = 0
F_{IJ} = -368.95 kN
F_{IJ} = 368.95 kN (C)
Joint B Considering the equilibrium of joint B, we obtain (see Fig. 4.26(b)) the following:
+ \uparrow \sum{F_y}=0 -\frac{2}{\sqrt{5}}(44.72) + \frac{4}{5}F_{BJ} = 0
F_{BJ} = 50 kN (T)
+\longrightarrow \sum{F_x}=0 -350 +\frac{1}{\sqrt{5}}(44.72) + \frac{3}{5}(50) + F_{BC} = 0
F_{BC} = 300 kN (T)
Section aa Since at each of the next two joints, C and J, there are three unknowns (F_{CD}, F_{CG}, and F_{CJ} at joint C and F_{CJ}, F_{GJ}, and F_{JK} at joint J), we calculate F_{CD} by using section aa, as shown in Fig. 4.26(a). (If we moved to joint F and started computing member forces from that end of the truss, we would encounter similar difficulties at joints D and N.)
The free-body diagram of the portion of the truss on the left side of section aa is shown in Fig. 4.26(d). We determine F_{CD} by summing moments about point L, the point of intersection of the lines of action of F_{GL} and F_{KL}.
+\circlearrowleft \sum{M_L}=0 -175(8) + 50(6) + 50(4) + 50(2) + F_{CD}(4) = 0
F_{CD} = 200 kN (T)
Joint C With F_{CD} now known, there are only two unknowns, F_{CG} and F_{CJ} , at joint C. These forces can be determined by applying the two equations of equilibrium to the free body of joint C, as shown in Fig. 4.26(e).
+\uparrow \sum{F_y}=0 \frac{2}{\sqrt{5}}F_{CJ} + \frac{4}{5}F_{CG} = 0
+\longrightarrow \sum{F_x}=0 -300 + 200 – \frac{1}{\sqrt{5}}F_{CJ} + \frac{3}{5}F_{CG} = 0
Solving these equations simultaneously, we obtain
F_{CJ} = -89.5 kN and F_{CG} = 100 kN
F_{CJ} = 89.5 kN (C) F_{CG} = 100 kN (T)
Joints J, K, and G Similarly, by successively considering the equilibrium of joints J, K, and G, in that order, we determine the following:
F_{JK} = 346.6 kN (C) F_{GJ} = 50 kN (T) F_{KL} = 324.21 kN (C) F_{GK} = 44.72 kN (C) F_{GL} = 150 kN (T)Symmetry Since the geometry of the truss and the applied loading are symmetrical about the center line of the truss (shown in Fig. 4.26(b)), its member forces will also be symmetrical with respect to the line of symmetry. It is, therefore, sufficient to determine member forces in only one-half of the truss. The member forces determined here for the left half of the truss are shown in Fig. 4.26(b). The forces in the right half can be obtained from the consideration of symmetry; for example, the force in member MN is equal to that in member JK, and so forth. The reader is urged to verify this by computing a few member forces in the right half of the truss.
