Vector Mechanics For Engineers Statics Twelfth Edition Chapter

Vector Mechanics For Engineers: Statics Twelfth Edition Chapter 4 Equilibrium of Rigid Bodies ©View Stock/Getty Images RF © 2019 Mc. Graw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of Mc. Graw-Hill Education.

Contents Introduction Free-Body Diagram Reactions for a Two-Dimensional Structure Rigid Body Equilibrium in Two Dimensions Practice Sample Problem 4. 1 Sample Problem 4. 4 Practice Statically Indeterminate Reactions and Partial Constraints Rigid Body Equilibrium in Three Dimensions Reactions for a Three-Dimensional Structure Sample Problem 4. 8 © 2019 Mc. Graw-Hill Education.

Introduction • For a rigid body, the condition of static equilibrium means that the body under study does not translate or rotate under the given loads that act on the body. • The necessary and sufficient conditions for the static equilibrium of a body are that the forces sum to zero, and the moment about any point sum to zero: • Equilibrium analysis can be applied to two-dimensional or threedimensional bodies, but the first step in any analysis is the creation of the free body diagram. © 2019 Mc. Graw-Hill Education.

Free-Body Diagram The first step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free body diagram. • Select the body to be analyzed and detach it from the ground all other bodies and/or supports. • Indicate point of application, magnitude, and direction of external forces, including the weight of the body. • Indicate point of application and assumed direction of unknown forces from reactions of the ground and/or other bodies, such as the supports. • Include the dimensions that will be needed, such as those necessary to compute the moments of the forces. Access the text alternative for these images. © 2019 Mc. Graw-Hill Education.

4. 1 EQUILIBRIUM IN TWO DIMENSIONS © 2019 Mc. Graw-Hill Education.

Reactions for a Two-Dimensional Structure • Reactions consisting of a single force with a known line of action. © 2019 Mc. Graw-Hill Education.

Reactions at Supports and Connections for a Two-Dimensional Structure • Reactions consisting of a force of unknown direction and magnitude and a couple of unknown magnitude. © 2019 Mc. Graw-Hill Education.

Practice 1 The frame shown supports part of the roof of a small building, and the tension in cable BDF is known to be 150 k. N. Your goal is to draw the free body diagram (FBD) for the frame. On the following page, you will choose the most correct FBD for this problem. First, you should draw your own FBD. Access the text alternative for this image. © 2019 Mc. Graw-Hill Education.

Practice Choose the most correct FBD for the original problem. Discuss with a neighbor why each choice is correct or incorrect. © 2019 Mc. Graw-Hill Education. 2 A B C D Access the text alternative for these images.

Practice A Choose the most correct FBD for the original problem. Discuss with a neighbor why each choice is correct or incorrect. © 2019 Mc. Graw-Hill Education. 3 B Answer is: B is the most correct, though C is also correct. A & D are incorrect; why? C D Access the text alternative for these images.

Rigid Body Equilibrium in Two Dimensions • For known forces and moments that act on a two-dimensional structure, the following are true: • Equations of equilibrium become: where A can be any point in the plane of the body. • The 3 equations can be solved for no more than 3 unknowns. • The 3 equations cannot be augmented with additional equations, but they can be replaced. Access the text alternative for these images. © 2019 Mc. Graw-Hill Education.

Sample Problem 4. 1 1 Strategy: Draw a free-body diagram to show all of the forces acting on the crane, then use the equilibrium equations to calculate the values of the unknown forces. Modeling: A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B. Access the text alternative for these images. © 2019 Mc. Graw-Hill Education.

Sample Problem 4. 1 Analysis: 2 • Determine B by solving the equation for the sum of the moments of all forces about A. • Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces. • Check the values obtained. © 2019 Mc. Graw-Hill Education.

Sample Problem 4. 1 Reflect and Think: You can check the values obtained for the reactions by recalling that the sum of the moments of all the external forces about any point must be zero. For example, considering point B, you can show © 2019 Mc. Graw-Hill Education. 3

Sample Problem 4. 4 1 Strategy: • Discuss with a neighbor the steps for solving this problem. • Create a free-body diagram for the frame and cable. The frame supports part of the roof of a small building. The tension in the cable is 150 k. N. • Apply the equilibrium equations for the reaction force components and couple at E. Determine the reaction at the fixed end E. Access the text alternative for this image. © 2019 Mc. Graw-Hill Education.

Sample Problem 4. 4 • Which equation is correct? Modeling: A. B. C. Analysis: • Apply one of the three equilibrium equations. Try using the condition that the sum of forces in the x-direction must sum to zero. © 2019 Mc. Graw-Hill Education. 2 D. E.

Sample Problem 4. 4 3 • Which equation is correct? Modeling: A. B. C. Analysis: • Apply one of the three equilibrium equations. Try using the condition that the sum of forces in the x-direction must sum to zero. © 2019 Mc. Graw-Hill Education. D. E. • What does the negative signify? • Discuss why the others are incorrect.

Sample Problem 4. 4 4 • Which equation is correct? A. B. C. D. • Now apply the condition that the sum of forces in the y-direction must sum to zero. © 2019 Mc. Graw-Hill Education. E.

Sample Problem 4. 4 5 • Which equation is correct? A. B. C. D. • Now apply the condition that the sum of forces in the y-direction must sum to zero. E. • What does the positive signify? • Discuss why the others are incorrect. © 2019 Mc. Graw-Hill Education.

Sample Problem 4. 4 6 • Three good points are D, E, and F. Discuss what advantage each point has over the others, or perhaps why each is equally good. • Assume that you choose point E to apply the sum-of-moments condition. Write the equation and compare your answer with a neighbor. • Finally, apply the condition that the sum of moments about any point must equal zero. • Discuss with a neighbor which point is the best for applying this equilibrium condition, and why. © 2019 Mc. Graw-Hill Education. • Discuss with a neighbor the origin of each term in the above equation and what the positive value of ME means.

Sample Problem 4. 4 7 Reflect and Think: The cable provides a fourth constraint, making this situation statically indeterminate. This problem therefore gave us the value of the cable tension, which would have been determined by means other than statics. We could then use three available independent static equilibrium equations to solve for the remaining three reactions. © 2019 Mc. Graw-Hill Education.

Practice 4 A 2100 -lb tractor is used to lift 900 lb of gravel. Determine the reactions at each of the two rear wheels and two front wheels. • First, create a free body diagram. Discuss with a neighbor what steps to take to solve this problem. • Second, apply the equilibrium conditions to generate three equations, and use these to solve for the desired quantities. Access the text alternative for these images. © 2019 Mc. Graw-Hill Education.

Practice 5 • Draw the free body diagram of the tractor (on your own first). • From among the choices, choose the best FBD, and discuss the problem(s) with the other FBDs. A. C. © 2019 Mc. Graw-Hill Education. B. D.

Practice 6 • Draw the free body diagram of the tractor (on your own first). • From among the choices, choose the best FBD, and discuss the problem(s) with the other FBDs. Answer is: A. B. C. D. © 2019 Mc. Graw-Hill Education.

Practice 7 Now let’s apply the equilibrium conditions to this FBD. • Start with the moment equation: Discuss with a neighbor: Points A or B are equally good because each results in an equation with only one unknown. © 2019 Mc. Graw-Hill Education. • What’s the advantage to starting with this instead of the other conditions? • About what point should we sum moments, and why?

Practice 8 Assume we chose to use point B. Choose the correct equation for A. B. C. D. © 2019 Mc. Graw-Hill Education.

Practice 9 Assume we chose to use point B. Choose the correct equation for A. B. C. Answer is D. FA= 650 lb, so the reaction at each rear wheel is 325 lb © 2019 Mc. Graw-Hill Education.

Practice 10 Naapply the final equilibrium condition, Why was the third equilibrium condition, not used? © 2019 Mc. Graw-Hill Education.

What if…? 1 • Now suppose we have a different problem: How much gravel can this tractor carry before it tips over? • Discuss with a neighbor how you would solve this problem. • Hint: Think about what the free body diagram would be for this situation… © 2019 Mc. Graw-Hill Education.

Statically Indeterminate Reactions and Partial Constraints • More unknowns than equations: statically indeterminate. • Fewer unknowns than equations: partially constrained. Access the text alternative for these images. © 2019 Mc. Graw-Hill Education. • Equal number unknowns and equations but improperly constrained.

4. 2 TWO SPECIAL CASES © 2019 Mc. Graw-Hill Education.

4. 2 A Equilibrium of a Two-Force Body If a two-force body is in equilibrium, the two forces must have the same magnitude, the same line of action, and opposite sense. The same can be said about the resultants of several forces applied at only two points. © 2019 Mc. Graw-Hill Education.

4. 2 B Equilibrium of a Three-Force Body If a three-force body is in equilibrium, the lines of action of the three forces must be either concurrent or parallel (the exception and special cases). © 2019 Mc. Graw-Hill Education.

4. 3 EQUILIBRIUM IN THREE DIMENSIONS © 2019 Mc. Graw-Hill Education.

Rigid Body Equilibrium in Three Dimensions • Six scalar equations are required to express the conditions for the equilibrium of a rigid body in the general three dimensional case. • These equations can be solved for no more than 6 unknowns, which generally represent reactions at supports or connections or unknown applied forces. • The scalar equations can be conveniently obtained by first applying the vector forms of the conditions for equilibrium: © 2019 Mc. Graw-Hill Education.

Reactions at Supports and Connections for a Three-Dimensional Structure 1 © 2019 Mc. Graw-Hill Education.

Reactions at Supports and Connections for a Three-Dimensional Structure 2 © 2019 Mc. Graw-Hill Education.

Sample Problem 4. 8 1 Strategy: Draw a free-body diagram of the sign, and express the unknown cable tensions as Cartesian vectors. Then determine the cable tensions and the reaction at A by writing and solving the equilibrium equations. A sign of uniform density weighs 270 lb and is supported by a ball-and-socket joint at A and by two cables. Determine the tension in each cable and the reaction at A. Access the text alternative for this image. © 2019 Mc. Graw-Hill Education.

Sample Problem 4. 8 Analysis: Modeling: Since there are only 5 unknowns, the sign is partially constrained. All forces intersect with the x axis, so the equation is not useful to the solution. © 2019 Mc. Graw-Hill Education. 2

Sample Problem 4. 8 • Apply the conditions for static equilibrium to develop equations for the unknown reactions. © 2019 Mc. Graw-Hill Education. 3 Solve the 5 equations for the 5 unknowns,

Sample Problem 4. 8 4 Reflect and Think: Cables can only act in tension, and the free-body diagram and Cartesian vector expressions for the cables were consistent with this. The solution yielded positive results for the cable forces, which confirms that they are in tension and validates the analysis. © 2019 Mc. Graw-Hill Education.

What if…? 2 Could this sign be in static equilibrium if cable BD were removed? Discuss with your neighbor, and be sure to provide the reason(s) for your answer. The sign could not be in static equilibrium because TEC causes a moment about the y axis (due to the existence of TECz) that must be countered by an equal and opposite moment. This can only be provided by a cable tension that has a z component in the negative z direction, such as what TBD provides. © 2019 Mc. Graw-Hill Education.

End of Chapter 4 © 2019 Mc. Graw-Hill Education.