APSC 180 202 - Chapter 5-1, 2, 3 - Shayan Narani - Filled
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Dec 19, 2024
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Chapter 5: Equilibrium of a Rigid Body•Shayan Narani•shayan.narani@ubc.ca•APSC 180 –Winter I 2024Copyright © 2022 Pearson Education, Inc. All Rights Reserved
Chapter 5: Equilibrium of a Rigid BodySection 5.1: Section 5.1 Conditions for Rigid-Body EquilibriumSection 5.2:Free-Body DiagramsSection 5.3: Equations of EquilibriumCopyright © 2022 Pearson Education, Inc. All Rights Reserved
3Students will be able to:a) Identify support reactions,b) Draw a free-body diagram, and,c) Apply equations of equilibrium to solve for unknownsLearning objectivesCopyright © 2022 Pearson Education, Inc. All Rights Reserved
4Section 5.1: ApplicationCopyright © 2022 Pearson Education, Inc. All Rights ReservedThe truck ramps have a weight of 400 lb each.Each ramp is pinned to the body of the truck and held in the position by a cable. How can we determine the cable tension and support reactions?How are the idealized model and the free body diagram used to do this?Which diagram above is the idealized model?
5Section 5.1: ApplicationCopyright © 2022 Pearson Education, Inc. All Rights Reserved
6Section 5.1: Conditions for Rigid-Body EquilibriumCopyright © 2022 Pearson Education, Inc. All Rights ReservedIn contrast to the forces on a particle, the forces on a rigid-body are not usually concurrent and may cause rotationof the body (due to momentscreated by the forces).For a rigid body to be in equilibrium, the net forceas well as the net moment about any arbitrary point O must be equal to zero.()()= 0 no translationand = 0 no rotationOFMForces on a particleForces on a rigid body
7Section 5.1: The Process of Solving Rigid Body Equilibrium ProblemsCopyright © 2022 Pearson Education, Inc. All Rights ReservedFor analyzing an actual physical system, first we need to create an idealized model.Then we need to draw a free-body diagram (F B D) showing all the external (active and reactive) forces.Finally, we need to apply the equations of equilibriumto solve for any unknowns.
8Section 5.2: Free Body Diagrams (1/2)Copyright © 2022 Pearson Education, Inc. All Rights ReservedIdealized modelFree-body diagram (F B D)1.Draw an outlined shape.Imagine the body to be isolated or cut “free” from its constraints and draw its outlined shape.2.Show all the external forces and couple moments. Thesetypicallyinclude: a) applied loads, b) support reactions, and, c) the weight of the body.
9Section 5.2: Free Body Diagrams (2/2)Copyright © 2022 Pearson Education, Inc. All Rights ReservedIdealized modelFree-body diagram (F B D)3.Label loads and dimensions on the F B D:All known forces and couple moments should be labeled with their magnitudes and directions. For the unknown forces and couple moments, use letters like Ax, Ay, MA. Indicate any necessary dimensions.
10Section 5.2: Center of GravityCopyright © 2022 Pearson Education, Inc. All Rights ReservedIdealized modelFree-body diagram (F B D)Center of gravity is defined as the point of application of the weight. When the body is uniform or made from the same material, the center of gravity will be located at the body’s geometric center or centroid.
11Section 5.2: Support Reactions in 2D (1/2)Copyright © 2022 Pearson Education, Inc. All Rights ReservedA few example sets of supports are shown above. As a general rule, if a support prevents translationof a body in a given direction, then a force is developedon the body in the opposite direction.Similarly, if rotation is prevented, a couple momentis exerted on the body in the opposite direction.
12Section 5.2: Support Reactions in 2D (2/2)Copyright © 2022 Pearson Education, Inc. All Rights Reserved
13Example 5.1Copyright © 2022 Pearson Education, Inc. All Rights ReservedGiven: The operator applies a vertical force to the pedal so that the spring is stretched 1.5 in. and the force in the short link at Bis 20 lb.Draw: An idealized model and free-body diagram of the foot pedal.The idealized modelThe free-body diagram
14Section 5.3: Equations of EquilibriumCopyright © 2022 Pearson Education, Inc. All Rights ReservedA body is subjected to a system of forces that lie in the x-y plane. When in equilibrium, the net force and net moment acting on the body are zero (as discussed earlier in Section 5.1). This 2-D condition can be represented by the three scalar equations:xyOF= 0F= 0 M= 0 where point O is any arbitrary point.
15Section 5.3: Alternative Sets Equations of EquilibriumCopyright © 2022 Pearson Education, Inc. All Rights ReservedAlternative IIt is required that the line passing through A and B is not parallel to the y axisF1F3F2F4yxyxFRMR AABCCBAyxFRACB𝐹𝑥= 0𝑀?= 0𝑀?= 0Alternative IIIt is required that points A, B, and C do not lie on the same line𝑀?= 0𝑀?= 0𝑀?= 0
16Example 5.2Copyright © 2022 Pearson Education, Inc. All Rights ReservedGiven: The beam is supported by the roller at A and a pin at B.Find: The reactions at points A and B on the beam.Plan:a.Establishthe x–y axis system.b.Drawa complete F B D of the beam.c.Applythe E-of-E to solve for the unknowns.
17Example 5.2Copyright © 2022 Pearson Education, Inc. All Rights ReservedNote that the distributed load has been reduced to a single force.First, write a moment equation about point B. Why point B?
18Example 5.2Copyright © 2022 Pearson Education, Inc. All Rights ReservedARecall N= 3.713 = 3.71 kN→XYXxYyNow write theF=F= 0 equations.+F= 3.713 sin 30° – B= 0+ F= 3.713 cos30° – 12+B= 0Solving these two equations, we getxyB= 1.86 kN B= 8.78 kN ←↑