Lecture Notes on Internal Forces - Statics | ME 2560, Assignments of Statics

Download Lecture Notes on Internal Forces - Statics | ME 2560 and more Assignments Statics in PDF only on Docsity! CH. VIII ME2560 STATICS Internal Forces 59 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 8.1a, subjected to two forces F1 and F2, and the support reactions Ax, Ay, and By (figure 8.1b). 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 8.1c. 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 NC and Vc and the resultant couple moment MC 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 NC is obtained by applying ΣFx = 0; VC is obtained directly from ΣFy = 0; and MC is determined by summing moments about point C, ΣMC = 0, in order to eliminate the moments of the unknowns NC and VC. 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 8.2a. In three Figure 8.1. Determination of the internal forces in a beam. CH. VIII ME2560 STATICS Internal Forces 60 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 8.2b. Here Ny is the normal force, and Vx and Vz are shear force components. My is a torsional or twisting moment, and Mx and Mz 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 8.3a 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 8.3b. 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.