Based on our discussion in Section 3.4, we can define a plane truss as internally stable if the number and geometric arrangement of its mem- bers is such that the truss does not change its shape and remains a rigid body when detached from the supports. The terminternalis used here to refer to the number and arrangement of members contained within the truss. The instability due to insu‰cient external supports or due to im- proper arrangement of external supports is referred to asexternal.
Basic Truss Element
The simplest internally stable (or rigid) plane truss can be formed by connecting three members at their ends by hinges to form a triangle, as shown in Fig. 4.9(a). This triangular truss is called the basic truss ele- ment. Note that this triangular truss is internally stable in the sense that it is a rigid body that will not change its shape under loads. In contrast, a rectangular truss formed by connecting four members at their ends by hinges, as shown in Fig. 4.9(b), is internally unstable because it will
FIG.4.8 Truss Bridge with Bolted Connections Rio Vista Bridge
Courtesy of the State of California, Caltrans, Dist 4, Photographer John Huseby.
SECTION 4.2 Arrangement of Members of Plane Trusses—Internal Stability 95
change its shape and collapse when subjected to a general system of co- planar forces.
Simple Trusses
The basic truss elementABC of Fig. 4.10(a) can be enlarged by attach- ing two new members,BDandCD, to two of the existing jointsBandC and by connecting them to form a new jointD, as shown in Fig. 4.10(b).
As long as the new joint D does not lie on the straight line passing through the existing joints B andC, the new enlarged truss will be in- ternally stable. The truss can be further enlarged by repeating the same procedure (as shown in Fig. 4.10(c)) as many times as desired. Trusses constructed by this procedure are called simple trusses. The reader should examine the trusses depicted in Figs. 4.4 and 4.5 to verify that each of them, with the exception of the Baltimore truss (Fig. 4.4) and the Fink truss (Fig. 4.5), is a simple truss. The basic truss element of the simple trusses is identified asABC in these figures.
A simple truss is formed by enlarging the basic truss element, which contains three members and three joints, by adding two additional members for each additional joint, so the total number of membersmin a simple truss is given by
mẳ3ỵ2ðj3ị ẳ2j3 (4.1)
in which jẳtotal number of joints (including those attached to the supports).
FIG.4.9
FIG.4.10 Simple Truss
Compound Trusses
Compound trusses are constructed by connecting two or more simple trusses to form a single rigid body. To prevent any relative movement between the simple trusses, each truss must be connected to the other(s) by means of connections capable of transmitting at least three force components, all of which are neither parallel nor concurrent. Two ex- amples of connection arrangements used to form compound trusses are shown in Fig. 4.11. In Fig. 4.11(a), two simple trussesABCandDEF are connected by three members, BD;CD, andBF, which are nonparallel and nonconcurrent. Another type of connection arrangement is shown in Fig. 4.11(b). This involves connecting the two simple trusses ABC and DEF by a common joint C and a member BD. In order for the compound truss to be internally stable, the common jointCand jointsB andDmust not lie on a straight line. The relationship between the total number of membersmand the total number of joints jfor an internally stable compound truss remains the same as for the simple trusses. This relationship, which is given by Eq. (4.1), can be easily verified for the compound trusses shown in Fig. 4.11.
Internal Stability
Equation (4.1) expresses the requirement of the minimum number of members that a plane truss of j joints must contain if it is to be in- ternally stable. If a plane truss containsmmembers and jjoints, then if
m<2j3 the truss is internally unstable mb2j3 the truss is internally stable
(4.2)
FIG.4.11 Compound Trusses
SECTION 4.2 Arrangement of Members of Plane Trusses—Internal Stability 97
It is very important to realize that although the foregoing criterion for internal stability is necessary, it is not su‰cient to ensure internal stability. A truss must not only contain enough members to satisfy the mb2j3 condition, but the members must also be properly arranged to ensure rigidity of the entire truss. Recall from our discussion of sim- ple and compound trusses that in a stable truss, each joint is connected to the rest of the structure by at least two nonparallel members, and each portion of the truss must be connected to the remainder of the truss by connections capable of transmitting at least three nonparallel and nonconcurrent force components.
Example 4.1
Classify each of the plane trusses shown in Fig. 4.12 as internally stable or unstable.
Solution
(a) The truss shown in Fig. 4.12(a) contains 20 members and 12 joints. Therefore, mẳ20 and 2j3ẳ2ð12ị 3ẳ21. Sincemis less than 2j3, this truss does not have a su‰cient 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 andEFGH, connected by two parallel members,BEandDG. These two horizontal members cannot prevent the rela- tive 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 memberDEhas now been added to prevent the relative displacement between the two portionsABCDandEFGH. The entire truss now acts as a single rigid body. Addition of memberDE increases the number of members to 21 (while the number of joints re- mains the same at 12), thereby satisfying the equationmẳ2j3. 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 increasingm to 25, while j remains constant at 12. Becausem>2j3, the truss is internally stable. Also, since the diÔerencem ð2j3ị ẳ4, 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,BGandDE, from panelBE, thereby decreasingmto 23; jremains constant at 12. Althoughm ð2j3ị ẳ2—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 mẳ26 and jẳ15, thereby yielding m<2j3. This is also clear from the diagram of the truss which shows that the portionsABEandCDEof the truss can rotate with respect to each other. The diÔerencem ð2j3ị ẳ 1 indicates that this truss has one less member
than required for internal stability. Ans.
(f ) In Fig. 4.12(f ), a memberBChas been added to the truss of Fig. 4.12(e), which prevents the relative movement of the two portionsABEandCDE, thereby making the truss internally stable. Asmhas now been increased to 27, it
satisfies the equationmẳ2j3 for jẳ15. Ans.
(g) The tower truss shown in Fig. 4.12(g) has 16 members and 10 joints. Becausem<2j3, the truss is internally unstable. This is also obvious from Fig. 4.12(g), which shows that memberBCcan rotate with respect to the rest of the
continued
FIG.4.12
structure. This rotation can occur because jointCis connected by only one member instead of the two required to com-
pletely constrain a joint of a plane truss. Ans.
(h) In Fig. 4.12(h), a memberAChas been added to the truss of Fig. 4.12(g), which makes it internally stable. Here
mẳ17 and jẳ10, so the equationmẳ2j3 is satisfied. Ans.
SECTION 4.2 Arrangement of Members of Plane Trusses—Internal Stability 99