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. The pin support isnot capable of exerting a couple. Thus a pin support can exert a force on a body inany direction. The force of the pin support 0 on the beam 1 at point A, Figure 5.3, isexpressed in terms of its components in planexyxyxy--0001xF01yF01yF01xF11=10yF=10xFlink 1 link 0AAAFig. 5.3 Pin joint forcesF01= F01xı + F01yj.4 5 EquilibriumThe 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= −F01yThe pin supports are used in mechanical de-vices that allow connected links to rotate relative to each other.Roller SupportFigure 5.4(a) represents a roller support which is a pin support mounted on rollers.xy01yF110link 1(a) (b)(c)smoothFig. 5.4 Roller supportThe roller support 0 can only exert a force normal (perpendicular) to the surface1 on which the roller support moves freely, Fig. 5.4(b)F01= F01yj.The roller support cannot exert a couple about the axis of the pin and it cannot ex-ert a force parallel to the surface on which it translates. Figure 5.4(c) shows otherschematic representations used for the roller support. A plane link on a smooth sur-face can also modeled by a roller support. Bridges and beams can be supported inthis way and they will be capable of expansion and contraction.Fixed SupportFigure 5.5 shows a fixed support or built-in support. The body is literally built into awall. A fixed support 0 can exert two components of force and a couple on the link 1F01= F01xı + F01yj, and M01= M01zk.5.2 Supports 5xy10101xF01yF01zMlink 1Fig. 5.5 Fixed support5.2.2 Three-Dimensional SupportsBall and Socket SupportFigure 5.6 shows a ball and socket support, where the supported body is attached toa ball enclosed within a spherical socket. The socket permits the body only to rotatefreely. The ball and socket support cannot exert a couple to prevent rotation. Thexyz1221zF21xF21yFlink 11Fig. 5.6 Ball and socket supportball and socket support can exert three components of forceF21= F21xı + F21yj + F21yk.Bearing SupportThe type of bearing shown in Fig. 5.7(a) supports a circular shaft while permitting itto rotate about its axis, z-axis. In the most general case, as shown in Fig. 5.7(b),the bearing can exert a force on the supported shaft in each coordinate direc-tion, F21x, F21y, F21z, and can exert couples about axes perpendicular to the shaft,M21x, M21y, but cannot exert a couple about the axis of the shaft. Situations can oc-cur in which the bearing exerts no couples, or exerts no couples and no force parallelto the shaft axis as shown in Fig. 5.7(c). Some radial bearings are designed in thisway for specific applications.6 5 Equilibriumbearingzyxzyx1221zF21xF21yFlink 1121yM21xMzyx21xF21yF(c)1(a)(b)Fig. 5.7 Bearing support5.3 Free-Body DiagramsFree-body diagrams are used to determine forces and moments acting on simplebodies in equilibrium. The beam in Figure 5.8(a) has a pin support at the left endA and a roller support at the right end B. The beam is loaded by a force F and aABFABFCMM(a)(b)CFig. 5.8 Free-body diagram of a beammoment M at C. To obtain the free-body diagram first the beam is isolated from itssupports. Next, the reactions exerted on the beam by the supports are shown on thethe free-body diagram, Figure 5.8. Once the free-body diagram is obtained one canapply the equilibrium equations.The steps required to determine the reactions on bodies are1. 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)5.3 Free-Body Diagrams 7One 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.Free-Body Diagrams for Kinematic