# C_5 - Chapter 5 Equilibrium 5.1 Equilibrium Equations A...

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Chapter 5 Equilibrium 5.1 Equilibrium Equations A body is in equilibrium when it is stationary or in steady translation relative to an inertial reference frame. The following conditions are satisﬁed when a body, acted upon 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 M P = 0 , P . (5.2) If the sum of the forces acting on a body is zero and the sum of the moments about one 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 F Ai , i = 1 ,..., n , and couples M j , j = 1 ,..., m . The sum of the forces is zero F = n i = 1 F Ai = 0 , and the sum of the moments about a point P is zero M P = n i = 1 r PAi × F Ai + m j = 1 M j = 0 , where r PAi = -→ PA i , i = 1 ,..., n . The sum of the moments about any other point Q is M Q = n i = 1 r QAi × F Ai + m j = 1 M j = 1

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2 5 Equilibrium M Q P F Ai M j r PAi r QAi r QP ... M m A i 1 A A n 1 F A 1 F An r QA 1 r PA 1 r QAi = r QP + r PAi Fig. 5.1 Body subjected to forces F Ai and couples M j n i = 1 ( r QP + r PAi ) × F Ai + m j = 1 M j = r QP × n i = 1 F Ai + n i = 1 r PAi × F Ai + m j = 1 M j = r QP × 0 + n i = 1 r PAi × F Ai + m j = 1 M j = n i = 1 r PAi × F Ai + m j = 1 M j = M P = 0 . A body subjected to concurrent forces F 1 , F 2 , , F n and no couples. If the sum of the concurrent forces is zero, F 1 + F 2 + ... + F n = 0 , the sum of the moments of the forces about the concurrent point is zero, so the sum of 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 3 5.2 Supports 5.2.1 Planar Supports The reactions are forces and couples exerted on a body by its supports. The follow- ing force convention is deﬁned: F ij represents the force exerted by link i on link j . Pin Support Figure 5.2 shows a pin support. A beam 1 is attached by a smooth pin to a ground bracket bracket pin pin beam beam pin support beam schematic representation side view Fig. 5.2 Pin joint bracket 0. The pin passes through the bracket and the beam. The beam can rotate about the axis of the pin. The beam cannot translate relative to the bracket because the support exerts a reactive force that prevents this movement. The pin support is not capable of exerting a couple. Thus a pin support can exert a force on a body in any direction. The force of the pin support 0 on the beam 1 at point A , Figure 5.3, is expressed in terms of its components in plane x y x y x y - - 0 0 01 x F 01 y F 01 y F 01 x F 1 1 = 10 y F = 10 x F link 1 link 0 A A A Fig. 5.3 Pin joint forces F 01 = F 01 x ı + F 01 y j .

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4 5 Equilibrium The directions of the reactions F 01 x and F 01 y are positive. If one determine F 01 x or F 01 y to be negative, the reaction is in the direction opposite to that of the arrow. The force of the beam 1 on the pin support 0 at point A , Figure 5.3, is expressed F 10 = F 10 x ı + F 10 y j = - F 01 x ı - F 01 y j , where F 10 x = - F 01 x and F 10 y = - F 01 y The pin supports are used in mechanical de- vices that allow connected links to rotate relative to each other.
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## This note was uploaded on 08/29/2011 for the course MECH 2110 taught by Professor Clark,b during the Spring '08 term at Auburn University.

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C_5 - Chapter 5 Equilibrium 5.1 Equilibrium Equations A...

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