A truss is a coplanar system of structural members joined together at their ends to form a stable framework. If small changes in the lengths of the members due to loads are neglected, the relative positions of the joints cannot change.
Characteristics of Trusses
Three bars pinned together to form a triangle represents the simplest type of truss. Some of the more common types of roof trusses are shown in Fig. 5.46.
The top members are called the upper chord; the bottom members, the lower chord; and the verticals and diagonals web members.
The purpose of roof trusses is to act like big beams, to support the roof covering over long spans. They not only have to carry their own weight and the weight of the roofing and roof beams, or purlins, but cranes, wind loads, snow loads, suspended ceilings, and equipment, and a live load to take care of construction, maintenance, and repair loading. These loads are applied at the intersection of the members, or panel points, so that the members will be subjected principally to axial stresses—tension or compression.
Bow’s Notation
For simple designation of loads and stresses, capital letters are placed in the spaces between truss members and between forces. Each member and load is then designated by the letters on opposite sides of it. For example, in Fig. 5.47a, the upper chord members are AF, BH, CJ, and DL. The loads are AB, BC, and CD, and the reactions are EA and DE. Stresses in the members generally are designated by the same letters but in lowercase.
Method of Joints
A useful method for determining the stresses in truss members is to select sections that isolate the joints one at a time and then apply the laws of equilibrium to each.
FIGURE 5.46 Common types of roof trusses.
Considering the stresses in the cut members as external forces, the sum of the horizontal components of the forces acting at a joint must be zero, and so must be the sum of the vertical components. Since the lines of action of all the forces are known, we can therefore compute two unknown magnitudes at each joint by this method. The procedure is to start at a joint that has only two unknowns (generally at the support) and then, as stresses in members are determined, analyse successive joints.
Let us, for illustration, apply the method to joint 1 of the truss in Fig. 5.47a. Equating the sum of the vertical components to zero, we find that the vertical component of the top-chord must be equal and opposite to the reaction, 12 kips (12,000 lb). The stress in the top chord at this joint, then, must be a compression equal to 12 x 30⁄18 20 kips. From the fact that the sum of the horizontal components must be zero, we find that the stress in the bottom chord at the joint must be equal and opposite to the horizontal component of the top chord. Hence the stress in the bottom chord must be a tension equal to 20 x 24⁄30 16 kips.
Moving to joint 2, we note that, with no vertical loads at the joint, the stress in the vertical is zero. Also, the stress is the same in both bottom chord members at the joint, since the sum of the horizontal components must be zero.
Joint 3 now contains only two unknown stresses. Denoting the truss members and the loads by the letters placed on opposite sides of them, as indicated in Fig. 5.47a, the unknown stresses are SBH and SHG. The laws of equilibrium enable us to
FIGURE 5.47 Method of joints applied to the roof truss shown in (a). Stresses in members at each joint are determined graphically in succession (b) to (e)
write the following two equations, one for the vertical components and the second for the horizontal components:
Both unknown stresses are assumed to be compressive; i.e., acting toward the joint. The stress in the vertical does not appear in these equations, because it was already determined to be zero. The stress in FA, SFA, was found from analysis of joint 1 to be 20 kips. Simultaneous solution of the two equations yields SHG = 6.7 kips and SBH = 13.3 kips. (If these stresses had come out with a negative sign, it would have indicated that the original assumption of their directions was incorrect; they would, in that case, be tensile forces instead of compressive forces.)
All the force polygons in Fig. 5.47 can be conveniently combined into a single stress diagram. The combination (Fig. 5.47ƒ) is called a Maxwell diagram.
Method of Sections
An alternative method to that described in Art. 5.9.3 for determining the stresses in truss members is to isolate a portion of the truss by a section so chosen as to cut only as many members with unknown stresses as can be evaluated by the laws of equilibrium applied to that portion of the truss. The stresses in the cut members are treated as external forces. Compressive forces act toward the panel point and tensile forces away from the joint. Suppose, for example, we wish to find the stress in chord AB of the truss in Fig. 5.48a. We can take a vertical section XX close to panel point A. This cuts not only AB but AD and ED as well. The external 10-kip (10,000-lb) loading and 25- kip reaction at the left are held in equilibrium by the compressive force C in AB, tensile force T in ED, and tensile force S in AD (Fig. 5.48b). The simplest way to find C is to take moments about D, the point of intersection of S and T, eliminating these unknowns from the calculation.
from which C is found to be 60 kips. Similarly, to find the stress in ED, the simplest way is to take moments about A, the point of intersection of S and C:
from which T is found to be 53.3 kips.
FIGURE 5.48 Stresses in truss members cut by section XX, shown in (a), are determined by method of sections (b).
On the other hand, the stress in AD can be easily determined by two methods. One takes advantage of the fact that AB and ED are horizontal members, requiring AD to carry the full vertical shear at section XX. Hence we know that the vertical component V of S 25 = 10 = 10 5 kips. Multiplying V by sec (Fig. 5.48b), which is equal to the ratio of the length of AD to the rise of the truss (15⁄9), S is found to be 8.3 kips. The second method—presented because it is useful when the chords are not horizontal—is to resolve S into horizontal and vertical components at D and take moments about E. Since both T and the horizontal component of S pass through E, they do not appear in the computations, and C already has been computed. Equating the sum of the moments to zero gives V = 5, as before.
Some trusses are complex and require special methods of analysis. (Norris et al., ‘‘Elementary Structural Analysis,’’ 4th ed., McGraw-Hill Book Company, New York).
