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INTERNAL FORCES
8.1 INTERNAL FORCES DEVELOPED IN STRUCTURAL MEMBERS
Designing any mechanical member requires an investigation of the loads acting within the element to ensure that the material will be able to support such loads. The determination of these internal loads can be achieved by using the method of sections.
Considering a simple supported beam, as shown in figure 8.1 a , subjected to two forces F 1 and F 2 , and the support reactions Ax, Ay, and By (figure 8.1 b ). The determination of the internal loadings acting on the cross section at C can be achieved by considering that an imaginary section passes through the beam cutting it into two segments. By doing this, the internal loadings at the section C become external on the free–body diagram of each segment, as shown in figure 8.1 c.
Since segments AC and CB were in equilibrium before the beam was sectioned, the equilibrium of each segment is maintained provided the rectangular force components N C and V c and the resultant couple moment M C are developed at the section. These loadings must be equal in magnitude and opposite in direction on each of the segments (Newton’s third law). The magnitude of each of these loadings can be calculated by applying the three equations of equilibrium to either segment AC or CB. A direct solution for N C is obtained by applying Σ Fx = 0; V C is obtained directly from Σ Fy = 0; and M C is determined by summing moments about point C , Σ MC = 0, in order to eliminate the moments of the unknowns N C and V C.
The force components N , acting normal to the beam at the cut section, and V , acting tangent to the section, are the normal or axial force and the shear force , respectively. The couple moment M is the bending moment , as presented in figure 8.2 a. In three
Figure 8.1. Determination of the internal forces in a beam.
dimensions, a general internal force and couple moment resultant will act at the section. The x, y, z components of these loadings are shown in figure 8.2 b. Here N y is the normal force , and V x and V z are shear force components. M y is a torsional or twisting moment , and M x and M z are bending moment components. For most applications, these resultant loadings will act at the geometric center or centroid ( C ) of the section’s cross-sectional area. Although the magnitude for each loading generally will be different at various points along the axis of the member, the method of sections can always be used to determine their values.
FREE–BODY DIAGRAMS
Trusses are composed of two-force members that only support normal loads. On the other hand, frames and machines are composed of multiforce members, and so each of these members will generally be subjected to internal normal, shear, and bending loadings. For example, in order to determine the internal loadings in the frame shown in figure 8.3 a at the sections cut by the line H , G , and F , the initial step would require to draw a free–body diagram of the top portion of this section as shown in figure 8.3 b. At each point where a member is sectioned there is an unknown normal force, shear force, and bending moment, therefore, it is not possible to apply the three equations of equilibrium to this section in order to obtain these nine unknowns. Instead, to solve this problem it is necessary to disassemble the frame and determine the reactions at the connections of the members. Once this is done, each member may then be sectioned at its appropriate point, and the three equations of equilibrium can be applied to determine N , V , and M. For example, the free-body diagram of segment DG , figure 8.3c, can be used to determine the internal loadings at G provided the reactions of the pin, Dx and Dy are known.
Figure 8.2. Normal and shear forces and torsional and bending moments.
Sign Convention
Although the choice of a sign convention is arbitrary, in the majority of engineering applications the convention explained next is followed. As schematically shown in figure 8.5, the positive directions are denoted by an internal shear force that causes clockwise rotation of the member on which it acts, and by an internal moment that causes compression or pushing on the upper part of the member. Also, positive moment would tend to bend the member if it were elastic, concave upward. Loadings that are opposite to these are considered negative
Homework No. 8.2:
7–43, 7–49, 7–57, 7–59.
Figure 8.4. Shear force and bending moment diagrams.
Figure 8.5. Beam sign convention.
8.3 ANALYSIS OF STRUCTURAL MEMBERS USING ATLAS
ATLAS is free program that determines the shear force and bending moment diagrams on beams. The program allows the application of point and linearly distributed loads. The software allows the use of simple supported beams as well as cantilevered beams.
The first step in using the software consists in defining the length of the beam. Once this has been done, select the supports you need and place them at the desired location. The location of a support can be modified by double clicking on it. Placing a load is done by simply selecting the icon corresponding to load and clicking on the location where the load is to be applied. Modification of the magnitude of the load is done in a similar way as that for moving a support. Finally, in order to apply a linearly distributed load, select the load icon and click on the initial location of the load, drag the mouse without releasing the left button towards the direction required. Release the button once the end of the distributed load has been reached. The values of the distributed load can be modified similarly as to a point load. Application of couple moments is not allowed by the software.
Homework No. 8.3: Use ATLAS to solve: 7–43, 7–49, 7–57, 7–59 (do not apply the 180 ft-lb moment)–Print your results.
