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Quiz 1 Solutions

Quiz 1 Solutions. Low 39% High 112% Average 82% Mode 67%. 5.1 & 5.2 Objectives : a ) Identify support reactions b) Draw a free-body diagram.

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Quiz 1 Solutions

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  1. Quiz 1 Solutions Low 39% High 112% Average 82% Mode 67%

  2. 5.1 & 5.2 Objectives: a) Identify support reactions b) Draw a free-body diagram. Chapter 5: EQUILIBRIUM OF A RIGID BODY & FREE-BODY DIAGRAMS

  3. In contrast to the forces on a particle, the forces on a rigid-body are not usually concurrent and may cause rotation of the body (due to the moments created by the forces). Forces on a particle For a rigid body to be in equilibrium, the net force as well as the net moment about any arbitrary point O must be equal to zero.  F = 0 (no translation) and  MO = 0 (no rotation) Forces on a rigid body CONDITIONS FOR RIGID-BODY EQUILIBRIUM (Section 5.1)

  4. The truck ramp has a weight of 400 lb. The ramp is pinned to the body of the truck and held in the position by the cable. APPLICATIONS

  5. Two smooth pipes, each having a mass of 300 kg, are supported by the tines of the tractor fork attachment. APPLICATIONS (continued)

  6. 2) Draw a free-body diagram (FBD) showing all the external (active and reactive) forces. THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMS 1) Create an idealized model 3) Apply the equations of equilibrium to solve for any unknowns.

  7. Free-body diagram (FBD) • 1. Draw an outlined shape.Here the body is isolated or cut “free” from constraints. FREE-BODY DIAGRAMS (Section 5.2) Idealized model • Show all the external forces and couple moments. • a) applied loads, b) support reactions, c) the weight of the body.

  8. FREE-BODY DIAGRAMS Idealized model Free-body diagram • Label loads and dimensions on the FBD: • Known forces and couple moments should be labeled with their magnitudes and directions. • For unknown forces and couple moments, use letters like Ax, Ay, MA, etc.. • Indicate any necessary dimensions.

  9. SUPPORT REACTIONS IN 2-D Support reactions are given in your textbook (Table 5-1). Rules: If a support prevents translationof a body in a given direction, thena force is developedon the body in the opposite direction. If rotation is prevented, a couple momentis exerted on the body in the opposite direction.

  10. Given: 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 B is 20 lb. Draw: An idealized model and free- body diagram of the foot pedal. EXAMPLE

  11. If the directions of the force and the couple moments are both reversed, what will happen to the beam? A) The beam will lift from A.B) The beam will lift at B.C) The beam will be restrained. D) The beam will break. Think about it… A

  12. Draw a FBD of member ABC, which is supported by a smooth collar at A, roller at B, and link CD. Look at Table 5-1 Example

  13. Example (continued) FBD

  14. Think About It… A rigid body is subjected to forces as shown. This body can be considered as a ______ member. A) Single-force B) Two-force C) Three-force D) Six-force B

  15. Think About It… • The three scalar equations  FX =  FY =  MO = 0, are ____ equations of equilibrium in two dimensions. • A) Incorrect B) The only correct • C) The most commonly used D) Not sufficient C

  16. Objectives: a) Apply equations of equilibrium tosolve for unknowns, and, b) Recognize two-force members. 5.3 EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMEBERS

  17. APPLICATIONS A 850 lb of engine is supported by three chains, which are attached to the spreader bar of a hoist. You need to check to see if the breaking strength of any of the chains is going to be exceeded. We can determine the force acting in each of the chains

  18. A 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). Here: EQUATIONS OF EQUILIBRIUM (Section 5.3)  Fx = 0  Fy = 0  MO = 0 where point O is any arbitrary point. Notethat these equations are the ones most commonly used for solving 2-D equilibrium problems.

  19. The solution to some equilibrium problems can be simplifiedif we recognize members that are subjected to forces at only two points (e.g., at points A and B). TWO-FORCE MEMBERS & THREE FORCE-MEMBERS (Section 5.4) If we apply the equations of equilibrium to such a member, we can quickly determine that the resultant forces at A and B must be equal in magnitude and act in the opposite directions along the line joining points A and B.

  20. In the cases above, members AB can be considered as two-force members, provided that their weight is neglected. This fact simplifies the equilibrium analysis of some rigid bodies since the directions of the resultant forces at A and B are thus known (along the line joining points A and B). EXAMPLE OF TWO-FORCE MEMBERS

  21. STEPS FOR SOLVING 2-D EQUILIBRIUM PROBLEMS 1. If not given, establish a suitable x - y coordinate system. 2. Draw a free body diagram (FBD) of the object under analysis. 3. Apply the three equations of equilibrium to solve for the unknowns.

  22. IMPORTANT NOTES 1. If there are more unknowns than the number of independent equations, then we have a statically indeterminate situation.We cannot solve these problems using just statics. 2. The order in which we apply equations may affect the simplicity of the solution. For example, if we have two unknown vertical forces and one unknown horizontal force,then solving  FX = 0 first allows us to find the horizontal unknown quickly. 3. If the answer for an unknown comes outas negative number, then the sense (direction) of the unknown force is opposite to that assumed when starting the problem.

  23. Given: The 4kN load at B of the beam is supported by pins at A and C . Find: The support reactions at A and C. Plan: EXAMPLE A 1. Put the x and y axes in the horizontal and vertical directions, respectively. 2. Determine if there are any two-force members. 3. Draw a complete FBD of the boom. 4. Apply the E-of-E to solve for the unknowns.

  24. EXAMPLE A (Continued) FBD of the beam: AY 4 kN 1.5 m 1.5 m AX C A B 45° FC Note: Upon recognizing CD as a two-force member, the number of unknowns at C are reduced from two to one. Now, using E-of E, we get, + MA = FC sin 45 1.5 – 4  3 = 0 Fc = 11.3 kN  + FX = AX + 11.31 cos 45 = 0; AX = – 8.00 kN  + FY = AY + 11.31 sin 45 – 4 = 0; AY = – 4.00 kN Note that the negative signs means that the reactions have the opposite direction to that shown on FBD.

  25. Given: The jib crane is supported by a pin at C and rod AB. The load has a mass of 2000 kg with its center of mass located at G. Assume x = 5 m. Find: Support reactions at B and C. GROUP PROBLEM SOLVING Plan: a) Establish the x – y axes. b) Draw a complete FBD of the jib crane beam. c) Apply the E-of-E to solve for the unknowns.

  26. FBD of the beam GROUP PROBLEM SOLVING (Continued) FAB 4 3 4 m 5 0.2 m Cx 5 m 2000(9.81) N Cy First write a moment equation about Point C. Volunteer Needed! +  MC = (3 / 5) FAB 4 + (4 / 5) FAB 0.2 – 2000(9.81)  5 = 0 FAB = 38320 N = 38.3 kN

  27. FBD of the beam GROUP PROBLEM SOLVING (Continued) FAB 4 3 4 m 5 0.2 m Cx 5 m 2000(9.81) N Cy FAB = 38320 N = 38.3 kN Now solve the FX and FY equations.  + FX = Cx– (4 / 5) 38320 = 0  + FY = – Cy+ (3 / 5) 38320 – 2000(9.81) = 0 Volunteer Needed! Solving these two equations, we getCx = 30656 N or 30.7 kNand Cy = 3372 N or 3.37 kN

  28. 5.5 Objectives: a) Identify support reactions in 3-D and draw a free body diagram, and, b) apply the equations of equilibrium. 3-D FREE-BODY DIAGRAMS, EQUILIBRIUM EQUATIONS, CONSTRAINTS AND STATICAL DETERMINACY

  29. Ball-and-socket joints and journal bearings are often used in mechanical systems. To design the joints or bearings, the support reactions at these joints and the loads must be determined. APPLICATIONS

  30. The tie rod from point A is used to support the overhang at the entrance of a building. It is pin connected to the wall at A and to the center of the overhang B. APPLICATIONS(continued) If A is moved to a lower position D, will the force in the rod change or remain the same? By making such a change without understanding if there is a change in forces, failure might occur.

  31. Support reactions are given in your text book (Table 5-2). SUPPORT REACTIONS IN 3-D (Table 5-2) As a general rule, if a support prevents translation of a body in a given direction, then a reaction force acting in the opposite direction is developed on the body. Similarly, if rotation is prevented, a couple moment is exerted on the body by the support.

  32. A single bearing or hinge can prevent rotation by providing a resistive couple moment. However, it is usually preferred to use two or more properly aligned bearings or hinges. Thus, in these cases, only force reactions are generated and there are no moment reactions created. IMPORTANT NOTE

  33. EQUATIONS OF EQUILIBRIUM (Section 5.6) As stated earlier, when a body is in equilibrium, the net force and the net moment equal zero, i.e.,  F = 0 and  MO = 0 . • These two vector equations can be written as six scalar equations of equilibrium (E of E). These are • FX =  FY =  FZ = 0 • MX =  MY =  MZ = 0 The moment equations can be determined about any point. Usually, choosing the point where the maximum number of unknown forces are present simplifies the solution.

  34. CONSTRAINTS AND STATICAL DETERMINACY (Section 5.7) Redundant Constraints: When a body has more supports than necessary to hold it in equilibrium, it becomes statically indeterminate. A problem that is statically indeterminate has more unknowns than equations of equilibrium.

  35. Here, we have 6 unknowns but there is nothing restricting rotation about the AB axis. In some cases, there may be as many unknown reactions as there are equations of equilibrium. However, if the supports are not properly constrained, the body may become unstable for some loading cases. IMPROPER CONSTRAINTS

  36. Given: The rod, supported by thrust bearing at A and cable BC, is subjected to an 80 lb force. Find: Reactions at the thrust bearing A and cable BC. Plan: EXAMPLE B a) Establish the x, y and z axes. b) Draw a FBD of the rod. c) Write the forces using scalar equations. d) Apply scalar equations of equilibrium to solve for the unknown forces.

  37. EXAMPLE B (continued) FBD of the rod: • Applying scalar equations of equilibrium in appropriate order, we get • F X = AX= 0 ; AX = 0 • F Z = AZ+ FBC – 80 = 0 ; • M Y = – 80 ( 1.5 ) + FBC ( 3.0 ) = 0 ; • Solving these last two equations:FBC = 40 lb, AZ = 40 lb

  38. EXAMPLE (continued) = 40 lb Now write scalar moment equations about point A • ∑M X = ( MA) X + 40 (6) – 80 (6) = 0 ; (MA ) X= 240 lbft • M Z = ( MA) Z = 0 ; (MA ) Z= 0

  39. Given: A rod is supported by smooth journal bearings at A, B, and C. Assume the rod is properly aligned. Find: The reactions at all the supports for the loading shown. Plan: GROUP PROBLEM SOLVING a) Draw a FBD of the rod. b) Apply scalar equations of equilibrium to solve for the unknowns.

  40. GROUP PROBLEM SOLVING (continued) A FBD of the rod: • Applying scalar equations of equilibrium • F Y = 450 cos 45 + CY= 0 ; CY = – 318 N • M Y = CZ(0.6) – 300 = 0 ; CZ = 500 N • M Z = – BX ( 0.8 ) – ( – 318 ) ( 0.6 ) = 0 ; BX = 239 N

  41. GROUP PROBLEM SOLVING (continued) • ∑ M X = BZ ( 0.8 ) – 450 cos 45 (0.4) – 450 sin 45 ( 0.8 + 0.4 ) • + 318 ( 0.4 ) + 500 ( 0.8 + 0.4 ) = 0 ; BZ = – 273 N • F X = AX + 239 = 0 ; AX = – 239 N • F Z = AZ – ( – 273 ) + 500 – 450 sin 45 = 0 ; AZ = 90.9 N

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