Chapter 6: Structural Analysis Chapter Objectives To show how to determine the forces in the members of a truss using the method of joints and the method of sections. To analyze the forces acting on the members of frames and machines composed of pin-connected members. In this chapter, we shall consider problems dealing with the equilibrium of structures made of several connected parts. These problems call for the determination of the following forces. 1. External forces acting on the structure (reactions). 2. Internal forces forces that hold together the various parts of the structure. Newton s Third Law states, The forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense. In this chapter, we shall consider three broad categories of engineering structures. 1. Trusses. Trusses consist of straight members connected at joints located at the ends of the members. Members of a truss are two-force members. 2. Frames. Frames always contain at least one multi-force member. 3. Machines. Machines are designed to transmit and modify forces, and are structures containing moving parts. Machines, like frames, always contain at least one multi-force member. 6.1 Simple Truss A truss is one major type of engineering structures and is used in bridges and buildings. A truss consists of straight members connected at joints. Truss members are connected at their ends only. No member is continuous through a joint. 6.1
Truss Frame The members of a truss are slender and can support little lateral load. All loads must be applied at the joints and not along the members themselves. In the case of bridge trusses, the dead loads and traffic loads from the deck are first carried by stringers which in turn transmit the loads to floor beams, and then the loads are finally transmitted to the joints of the supporting side trusses (ref. Fig. 6-2 in textbook). Assumptions for Design 1. All loadings are applied at the joints. 2. The members are joined together by smooth pins. Because of these two assumptions, each truss member acts as a two-force member. The member is either in tension or compression. Although the members are actually joined together by means of riveted, bolted, or welded connections, the members are considered to be pinned together. The forces acting at each end of the member reduce to a single force and no couple. Each member is treated as a two-force member. Actual connection (Gusset Plate) The moments that are created at the ends of the members are small and considered insignificant. 6.2
Simple Truss The triangle ABC is considered the basic truss. The truss is said to be a rigid truss meaning that the truss is stable and will not collapse. The only possible deformation involves small changes in the length of its members. A large truss may be obtained by successively adding two members, attaching them to separate existing joints, and connecting them at a new joint. A truss constructed in this manner is called a simple truss. In a simple truss, the total number of members is related to the total number of joints by the following equation. m = 2 n 3 where n = the total number of joints m = the total number of members 6.2 The Method of Joints A truss may be considered as a group of pins and two-force members. We can dismember a truss and draw a free-body diagram for each pin. Since the entire truss is in equilibrium, each pin must be in equilibrium. Consider the truss shown below. 6.3
Each pin represents a concurrent force system. At each pin we can write only two equations of equilibrium (a.k.a. equations of statics). Fx = 0 and Fy = 0 In the case of a simple truss, the number of members and the number of pins (joints) are related by the following equation. m = 2 n 3 where n = the number of pins (joints) m = number of members and the number of unknowns that can be determined by the pins is m + 3 (i.e. 2 n = m + 3). Thus, the forces in all the members plus the two components of the reaction at A (i.e. Ax and Ay) and the vertical reaction at C (i.e. Cy) may be found using the free-body diagrams of the pins. Typically, the entire truss is treated as a rigid body to determine the reactions at the supports. The Method of Joints solution is good if the entire truss is to be analyzed (i.e. all the forces in each member is required). The solution must begin where there are only two unknown forces, usually at one of the supports. 6.4
Example Method of Joints Given: The truss shown. Find: Force in each member. First, find the reactions at the supports. Fx = 0 = Cx Cx = 0 MC = 0 = 1,000 (12) + 2,000 (24) 6 Ey 6Ey = 1,000 (12) + 2,000 (24) = 12,000 + 48,000 = 60,000 Ey = + 10,000 lb The direction that was assumed is correct. Ey = 10,000 lb Fy = 0 = - 2000 1000 + Cy + Ey Cy = 3000 Ey = 3,000 10,000 Cy = - 7,000 lb The direction that was assumed is not correct. Cy = 7,000 lb Next, draw a free-body diagram for each of the joints and solve for the forces. Assume tension (pulling on the joint) tension is good. For Joint A: Fy = 0 = - 2000 (4/5) AD AD = - (5/4) 2000 = - 2500 lb AD = 2500 lb (C) Fx = 0 = AB + (3/5) AD AB = - (3/5) AD = - (3/5) (- 2500) AB = + 1500 lb AB = 1500 lb (T) 6.5
For Joint D: Fy = 0 = (4/5) BD (4/5) 2500 BD = + 2500 lb BD = 2500 lb (T) Fx = 0 = (3/5) 2500 + (3/5) BD + DE DE = - (3/5) 2500 - (3/5) BD DE = - (3/5) 2500 - (3/5) (+2500) = - 3000 lb DE = 3000 lb (C) For Joint B: Fy = 0 = - 1000 - (4/5) 2500 (4/5) BE (4/5) BE = - 1000-2000 BE = - 3750 lb BE = 3750 lb (C) Fx = 0 = - 1500 + BC - (3/5) 2500 + (3/5) BE BC = + 1500 + (3/5) 2500 (3/5) (- 3750) BC = + 5250 lb BC = 5250 lb (T) For Joint E: Fy = 0 = - (4/5) 3750 + (4/5) CE + 10,000 (4/5) CE = (4/5) 3750 10,000 = - 7,000 CE = - 8,750 lb CE = 8,750 lb (C) Check (using Joint E): Fx = 0 = (3/5) 3750 + 3000 + (3/5) CE 0 = (3/5) 3750 + 3000 + (3/5) (- 8750) 0 = 2250 + 3000 5250 = 0 OK Also, check Joint C: Fx = 0 = - 5250 + ( 3/5 ) 8750 = 0 Fy = 0 = - 7000 + ( 4/5 ) 8750 = 0 OK OK 6.6
6.3 Zero-Force Members Using Joint C: Fy = 0 = (4/5) BC and BC = 0 In order to determine zero force members in a plane truss: 1. Examine the unloaded joints. 2. If a joint is unloaded, determine if there are more than three members framing into the joint. 3. For three members framing into one joint, if two of the three members are collinear, then the force in the third member is zero. Zero force members are not useless. 1. These members may carry loads when the loading conditions change. 2. These members are needed to support the weight of the truss. 3. These members help to maintain the truss in the desired shape. 6.7
6.4 The Method of Sections If the force in only one member or if the forces in only a few members are desired, the Method of Sections is a more efficient method of solution. In practice, the portion of the truss to be analyzed is obtained by passing a section through three members of the truss, at least one of which is the desired member. The section is a line that is drawn which divides the truss into two completely separate parts, but does not intersect more than three members if possible. Then, if the entire truss is in equilibrium, then any portion of the truss is in equilibrium also. Consider the truss shown at the right. To find the internal forces in members BD, BE, and CE, make a cut through these members. Assuming tension in these members: Use ME = 0 to find BD. Use Fy = 0 to find BE. Use MB = 0 to find CE. 6.8
Example Method of Sections Given: The truss shown. Find: Force in members BC, BK, LK. Find the reaction at the left support. MG = 0 = - Ay (180) + 2 (150) + 4 (120) + 4 (90) + 4 (60) + 2 (30) 180 Ay = 1440.0 Ay = 8.0 k For the portion of the truss left of the cut, find the forces in members BC, BK, and LK. MB = 0 = - 8 (30) + 40 LK LK = + 6.0 k LK = 6.0 k (T) MK = 0 = - (3/ 10 ) BC (40) (1/ 10 ) BC (30) + 2 (30) 8 (60) = - 37.95 BC 9.49 BC 420 47.44 BC = - 420 BC = - 8.85 BC = 8.85 k (C) MO = 0 = - (3/5) BK (40) (4/5) BK (120) - 2 (120) + 8 (90) = - 24 BK 96 BK + 480 120 BK = 480 BK = + 4.0 k BK = 4.0 k (T) Check: Fx = 0 = (3/ 10 ) BC + (3/5) BK + LK = (3/ 10 ) (- 8.85) + (3/5) (4.0) + 6.0 = - 8.40 + 2.40 + 6.0 = 0 OK 6.9
Example Method of Sections (K truss) Given: The truss shown. Find: Force in upper chord EJ, lower chord GH, and diagonals FJ & FH. Find the reactions at the supports. MV = 0 = - 80 Ay + 8 (60) + 8 (50) Ay = + 11.0 k Ay = 11.0 k (as assumed) Certain trusses will require more than a single cut. For the K-truss, a vertical cut crosses four members; thus, there are four unknowns. But using a different cut, forces in a couple of the required members may be found. For Cut 1: MG = 0 = - 11 (20) EJ (20) EJ = - 11.0 k EJ = 11.0 k (C) ME = 0 = GH (20) - 11 (20) GH = + 11.0 k GH = 11.0 k (T) 6.10
For Cut 2: Fy = 0 = 11.0 8.0 + ( 1/ 2 ) FJ ( 1/ 2 ) FH 0 = 3.0 + ( 1/ 2 ) FJ - ( 1/ 2 ) FH Fx = 0 = ( 1/ 2 ) FJ + ( 1/ 2 ) FH 11.0 + 11.0 0 = ( 1/ 2 ) FJ + ( 1/ 2 ) FH FJ = - FH Using the first equation, substitute - FH for FJ and solve for FH. Fy = 0 = 11.0 8.0 + ( 1/ 2 ) FJ ( 1/ 2 ) FH 0 = 11.0 8.0 + ( 1/ 2 ) (- FH) ( 1/ 2 ) FH 0 = 3.0 (2/ 2 ) FH FH = 3.0 ( 2 /2) = + 2.12 FH = 2.12 k (T) FJ = - FH = - 2.12 k FJ = 2.12 k (C) 6.11
6.5 Space Trusses A space truss consists of straight members joined together at their extremities to form a three-dimensional (3D) configuration. The most elementary (or basic ) rigid space truss consists of six members to form a tetrahedron. Starting from the basic space truss, a simple space truss is obtained by adding three new members, all joined at one end at a new joint, and joined to three existing joints. In a simple space truss, the number of members and the number of joints are related by the following equation. m = 3 n 6 where m = the number of members n = the number of joints Assumptions for Design The members of a space truss may be treated as two-force members provided the external loading is applied at the joints and the joints consist of ball-and-socket connections. The conditions of equilibrium for each joint are expressed by the following three equations. Fx = 0 Fy = 0 Fz = 0 6.12
6.6 Frames and Machines A frame is a structure that contains at least one multi-force member; that is, a member acted upon by three or more forces. Note the contrast to truss members that consist of all two-force members. Analyzing a Frame Consider the crane shown. Using the three equations of equilibrium, we can determine the tension T in the cable, and the components of the reaction at A. MA = 0 yields T Fx = 0 yields Ax Fy = 0 yields Ay Free-Body Diagrams In order to determine the internal forces holding the various parts of the frame together, we must dismember the frame and draw a free-body diagram for each part. Points B, C, and E are pin connections and, therefore, are replaced by horizontal and vertical restraining forces. As we assign directions to the components of the reactions, Newton s third law (equal and opposite forces) must be satisfied. 6.13
Check for determinacy. There are 3 equations for each free-body diagram. There are 4 free-body diagrams; thus, we can solve for as many as 12 unknowns. In this problem there are only 9 unknowns. The two components for each of the forces at pins B, C, and E. The two components of the reaction at A. The tension in the cable. Equations of Equilibrium Using the equations of equilibrium, the solution follows as outlined below. Using the entire structure as a free-body diagram, solve for the components of the reaction at A (that is, Ax and Ay) and the tension T in the cable. Then using a free-body diagram of member ABCD: MB = 0 yields Cx MC = 0 yields Bx Using a free-body diagram of member CEF: MC = 0 yields Ey ME = 0 yields Cy Using a free-body diagram of member BE: Fx = 0 yields Ex Fy = 0 yields By Frames Which Cease to be Rigid When Detached from Their Supports The reactions cannot be completely determined from the free-body diagram of the entire frame. Thus, we must dismember the frame even to find the external forces. Equilibrium equations are said to be necessary, but not sufficient for a nonrigid structure. 6.14
Example Frame Analysis Given: Frame shown. Find: Components of the forces acting on each member of the frame. Find the reactions at the supports. MA = 0 = - 360 (15) 240 (33) + 12 Ex 12 Ex = 5400 + 7920 = 13,320 Ex = 1110 lb Fx = 0 = Ax + Ex Ax = - Ex = - 1110 lb Ax = 1110 lb Fy = 0 = Ay 360-240 Ay = + 600 Ay = 600 lb Find the components of the forces acting on each member. FBD 1 MB = 0 = - 600 (6) + 18 Dy 18 Dy = + 3600 Dy = + 200 Dy = 200 lb on ABD MD = 0 = - 600 (24) - 18 By 18 By = - 14,400 By = - 800 By = 800 lb on ABD 6.15
FBD 2 MC = 0 = - 12 Bx + 18 By + 360 (9) 240 (9) 12 Bx = 18 (- 800) + 360 (9) 240 (9) = - 14,400 + 3240 2160 = - 13,320 Bx = - 1110 Bx = 1110 lb on ABD Fx = 0 = - Bx + Cx Cx = Bx = - 1110 Cx = 1110 lb on BC Fy = 0 = - By 360 + Cy - 240 Cy = By + 360 + 240 = - 800 + 360 + 240 = - 200 Cy = 200 lb on BC FBD 3 MC = 0 = 1110 (24) 12 Dx Dx = 2220 lb on CDE 6.16
Machines Machines are structures designed to transmit or modify forces. The solution involving forces on a machine is similar to that for a frame. The solution will generally involve the use of one or more free-body diagrams. The free-body diagrams should be chosen to include the input forces and the reactions to the input forces. 6.17