Question 4.1: Classify each of the plane trusses shown in Fig. 4.12 as int...
Classify each of the plane trusses shown in Fig. 4.12 as internally stable or unstable.
(a) The truss shown in Fig. 4.12(a) contains 20 members and 12 joints. Therefore, m = 20 and 2j – 3 = 2(12) – 3 = 21. Since m is less than 2j – 3, this truss does not have a sufficient number of members to form a rigid body; therefore, it is internally unstable. A careful look at the truss shows that it contains two rigid bodies, ABCD and EFGH, connected by two parallel members, BE and DG. These two horizontal members cannot prevent the relative displacement in the vertical direction of one rigid part of the truss with respect to the other. Ans.
(b) The truss shown in Fig. 4.12(b) is the same as that of Fig. 4.12(a), except that a diagonal member DE has now been added to prevent the relative displacement between the two portions ABCD and EFGH. The entire truss now acts as a single rigid body. Addition of member DE increases the number of members to 21 (while the number of joints remains the same at 12), thereby satisfying the equation m = 2j – 3. The truss is now internally stable. Ans.
(c) Four more diagonals are added to the truss of Fig. 4.12(b) to obtain the truss shown in Fig. 4.12(c), thereby increasing m to 25, while j remains constant at 12. Because m > 2j – 3, the truss is internally stable. Also, since the difference m – (2j – 3) = 4, the truss contains four more members than required for internal stability. Ans.
(d) The truss shown in Fig. 4.12(d) is obtained from that of Fig. 4.12(c) by removing two diagonals, BG and DE, from panel BE, thereby decreasing m to 23; j remains constant at 12. Although m – (2j – 3) = 2—that is, the truss contains two more members than the minimum required for internal stability—its two rigid portions, ABCD and EFGH, are not connected properly to form a single rigid body. Therefore, the truss is internally unstable. Ans.
(e) The roof truss shown in Fig. 4.12(e) is internally unstable because m = 26 and j = 15, thereby yielding m < 2j – 3. This is also clear from the diagram of the truss which shows that the portions ABE and CDE of the truss can rotate with respect to each other. The difference m – (2j – 3) = -1 indicates that this truss has one less member than required for internal stability. Ans.
(f ) In Fig. 4.12(f ), a member BC has been added to the truss of Fig. 4.12(e), which prevents the relative movement of the two portions ABE and CDE, thereby making the truss internally stable. As m has now been increased to 27, it satisfies the equation m = 2j – 3 for j = 15. Ans.
(g) The tower truss shown in Fig. 4.12(g) has 16 members and 10 joints. Because m < 2j – 3, the truss is internally unstable. This is also obvious from Fig. 4.12(g), which shows that member BC can rotate with respect to the rest of the structure. This rotation can occur because joint C is connected by only one member instead of the two required to completely constrain a joint of a plane truss. Ans.
(h) In Fig. 4.12(h), a member AC has been added to the truss of Fig. 4.12(g), which makes it internally stable. Here m = 17 and j = 10, so the equation m = 2j – 3 is satisfied. Ans.