F34

'90 kN

FIGURE 2.18 Example — method of joints, planar truss.

FIGURE 2.18 Example — method of joints, planar truss.

After solving for joints 1 and 2 one proceeds to take a section around joint 3 at which there are now two unknown forces, namely, F34 and F35. Summation of the vertical components at joint 3 gives

Substituting for F23, one obtains F35 = 63.6 kN (compressive). Summing the horizontal components and substituting for F13 one gets

Therefore,

The next joint involving two unknowns is joint 4. When we consider a section around it, the summation of the vertical components at joint 4 gives

Now, the forces in all the members on the left half of the truss are known and by symmetry the forces in the remaining members can be determined. The forces in all the members of a truss can also be determined by making use of the method of sections.

2.3.2 Method of Sections

In this method, an imaginary cutting line called a section is drawn through a stable and determinate truss. Thus, a section subdivides the truss into two separate parts. Since the entire truss is in equilibrium, any part of it must also be in equilibrium. Either of the two parts of the truss can be considered and the three equations of equilibrium, Y1 = 0, Y1 = 0, and Y1M = 0, can be applied to solve for member forces.

The example considered in Section 2.3.1 is once again considered (Figure 2.19). To calculate the force in the member 3-5, F35, a section AA should be run to cut the member 3-5 as shown in the figure. It is only required to consider the equilibrium of one of the two parts of the truss. In this case, the portion of the truss on the left of the section is considered. The left portion of the truss as shown in Figure 2.19 is in equilibrium under the action of the forces namely, the external and internal forces. Considering the equilibrium of forces in the vertical direction one can obtain

Therefore, F35 is obtained as

The negative sign indicates that the member force is compressive. The other member forces cut by the section can be obtained by considering the other equilibrium equations, namely, Y1M = 0. More sections can be taken in the same way to solve for other member forces in the truss. The most important advantage of this method is that one can obtain the required member force without solving for the other member forces.

135 kN

FIGURE 2.19 Example — method of sections, planar truss.

FIGURE 2.20 Compound trusses: (a) compound roof truss, (b) compound bridge truss, and (c) cantilevered construction.

FIGURE 2.20 Compound trusses: (a) compound roof truss, (b) compound bridge truss, and (c) cantilevered construction.

2.3.3 Compound Trusses

A compound truss is formed by interconnecting two or more simple trusses. Examples of compound trusses are shown in Figure 2.20. A typical compound roof truss is shown in Figure 2.20a in which two simple trusses are interconnected by means of a single member and a common joint. The compound truss shown in Figure 2.20b is commonly used in bridge construction and, in this case, three members are used to interconnect two simple trusses at a common joint. There are three simple trusses interconnected at their common joints as shown in Figure 2.20c.

The method of sections may be used to determine the member forces in the interconnecting members of compound trusses similar to those shown in Figure 2.20a and Figure 2.20b. However, in the case of a cantilevered truss the middle simple truss is isolated as a free-body diagram to find its reactions. These reactions are reversed and applied to the interconnecting joints of the other two simple trusses. After the interconnecting forces between the simple trusses are found, the simple trusses are analyzed by the method of joints or the method of sections.

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