Chapter 5Equilibrium5.1 Equilibrium EquationsA body is in equilibrium when it is stationary or in steady translation relative to aninertial reference frame. The following conditions are satisfied when a body, actedupon by a system of forces and moments, is in equilibrium:1. the sum of the forces is zero∑F = 0. (5.1)2. the sum of the moments about any point is zero∑MP= 0, ∀P. (5.2)If the sum of the forces acting on a body is zero and the sum of the moments aboutone point is zero, then the sum of the moments about every point is zero.Proof. The body shown in Figure 5.1, is subjected to forces FAi, i = 1,...,n, andcouples Mj, j = 1,...,m. The sum of the forces is zero∑F =n∑i=1FAi= 0,and the sum of the moments about a point P is zero∑MP=n∑i=1rPAi×FAi+m∑j=1Mj= 0,where rPAi=−→PAi, i = 1,...,n. The sum of the moments about any other point Q is∑MQ=n∑i=1rQAi×FAi+m∑j=1Mj=12 5 EquilibriumMQPFAiMjrPAirQAirQP...Mm......Ai1AAn1FA1...FAnrQA1rPA1rQAi=rQP+ rPAiFig. 5.1 Body subjected to forces FAiand couples Mjn∑i=1(rQP+ rPAi) ×FAi+m∑j=1Mj=rQP×n∑i=1FAi+n∑i=1rPAi×FAi+m∑j=1Mj=rQP×0 +n∑i=1rPAi×FAi+m∑j=1Mj=n∑i=1rPAi×FAi+m∑j=1Mj=∑MP= 0.A body subjected to concurrent forces F1, F2, . . ., Fnand no couples. If the sum ofthe concurrent forces is zero,F1+ F2+ ... + Fn= 0,the sum of the moments of the forces about the concurrent point is zero, so the sumof the moments about every point is zero. The only condition imposed by equilib-rium on a set of concurrent forces is that their sum is zero.5.2 Supports 35.2 Supports5.2.1 Planar supportsThe reactions are forces and couples exerted on a body by its supports. The follow-ing force convention is defined: Fi jrepresents the force exerted by link i on link j.Pin supportFigure 5.2 shows a pin support. A beam 1 is attached by a smooth pin to a groundbracketbracketpinpinbeambeampin supportbeamschematic representationside viewFig. 5.2 Pin jointbracket 0. The pin passes through the bracket and the beam. The beam can rotateabout the axis of the pin. The beam cannot translate relative to the bracket becausethe support exerts a reactive force that prevents this movement. Thus a pin supportcan exert a force on a body in any direction. The force of the pin support 0 on thebeam 1 at point A, Figure 5.3, is expressed in terms of its components in planeF01= F01xı + F01yj.The directions of the reactions F01xand F01yare positive. If one determine F01xorF01yto be negative, the reaction is in the direction opposite to that of the arrow. Theforce of the beam 1 on the pin support 0 at point A, Figure 5.3, is expressedF10= F10xı + F10yj = −F01xı −F01yj,where F10x= −F01xand F10y= −F01y. The pin support is not capable of exerting acouple.4 5 Equilibriumxyxyxy--0001xF01yF01yF01xF11=10yF=10xFFBD of link 1 FBD of link 0AAAFig. 5.3 Pin joint forcesRoller supportFigure 5.4 represents a roller support which is a pin support mounted on rollers.The roller support 0 can only exert a force normal (perpendicular) to the surface 1xy01yF110FBD of link 1Fig. 5.4 Roller support5.3 Free-body diagrams 5on which the roller support moves freelyF01= F01yj.The roller support cannot exert a couple about the axis of the pin and it cannot exerta force parallel to the surface on which it translates.Fixed supportFigure 5.5 shows a fixed support or built-in support. The body is literally built into axy10101xF01yF01zMFBD of link 1Fig. 5.5 Fixed supportwall. A fixed support 0 can exert two components of force and a couple on the link1F01= F01xı + F01yj, and M01= M01zk.5.2.2 Three-dimensional supportsBall and socket supportFigure 5.6 shows a ball and socket support, where the supported body is attachedto a ball enclosed within a spherical socket. The socket permits the body only torotate freely. The ball and socket support cannot exert a couple to prevent rotation.The ball and socket support can exert three components of forceF21= F21xı + F21yj + F21yk.5.3 Free-body diagramsFree-body diagrams are used to determine forces and moments acting on simplebodies in equilibrium.The beam in Figure 5.7(a) has a pin support at the left end A and a roller support at6 5 Equilibriumxyz1221zF21xF21yFFBD of link 1Fig. 5.6 Ball and socket supportthe right end B. The beam is loaded by a force F and a moment M at C. To obtainABFABFCMM(a)(b)CFig. 5.7 Free-body diagram of a beamthe free-body diagram first the beam is isolated from its supports. Next, the reac-tions exerted on the beam by the supports are shown on the the free-body diagram,Figure 5.7. Once the free-body diagram is obtained one can apply the equilibriumequations.The steps required to determine the reactions on bodies are5.3 Free-body diagrams 71. draw the free-body diagram, isolating the body from its supports and showingthe forces and the reactions;2. apply the equilibrium equations to determine the reactions.For two-dimensional systems, the forces and moments are related by three scalarequilibrium equations∑Fx= 0, (5.3)∑Fy= 0, (5.4)∑MP= 0, ∀P. (5.5)One can obtain more than one equation from Eq. (5.5) by evaluating the sum of themoments about more than one point. The additional equations will not be indepen-dent of Eqs. (5.3)-(5.5). One cannot obtain more than three independent equilibriumequations from a two-dimensional free-body diagram, which means one can solvefor at most three unknown forces or couples.5.3.1 Free-body diagrams for kinematic chainsA free-body diagram is a drawing of a part of a complete system, isolated in orderto determine the forces acting on that rigid body. The vector Fi jrepresents the forceexerted by link i on link j and Fi j= −Fji.Figure 5.8 shows the joint reaction forces for a pin joint, Fig. 5.8 (a), and a sliderjoint, Fig. 5.8 (b).Figure 5.9 shows various free-body diagrams that are considered in the analysisof a slider-crank mechanism Fig. 5.9 (a). In Fig. 5.9 (b), the free body consists ofthe three moving links isolated from the frame 0. The forces acting on the systeminclude an external driven force F, and the forces transmitted from the frame at jointA, F01, and at joint C, F03. Figure 5.9 (c) is a free-body diagram of the two links 1and 2 and Fig. 5.9 (d) is a free-body diagram of the two links 0 and 1. Figure 5.9(e)is a free-body diagram of crank 1 and Fig. 5.9 (f) is a free-body diagram of slider 3.The force analysis can be accomplished by examining individual links or a subsys-tem of links. In this way the joint forces between links as well as the