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Andres Faria
Andres Faria

Statics Of Rigid Bodies Pytel And Kiusalaas Solutions Manual Chapter 4 Pdf


Statics of Rigid Bodies Pytel and Kiusalaas Solutions Manual Chapter 4




Statics of rigid bodies is a branch of mechanics that deals with the equilibrium of bodies under the action of forces. It is a fundamental topic for engineering students who need to analyze the stability and strength of structures, machines, and systems. One of the popular textbooks for learning statics of rigid bodies is Engineering Mechanics: Statics by Andrew Pytel and Jaan Kiusalaas. This book provides a clear and concise introduction to statics, with numerous examples and exercises to reinforce the concepts and skills. The book also includes a solutions manual that contains detailed solutions to selected problems from each chapter.


In this article, we will review some of the key topics and concepts covered in chapter 4 of the solutions manual, which focuses on coplanar equilibrium analysis. Coplanar equilibrium analysis is a method of solving statics problems involving forces that lie in the same plane. This method involves applying the conditions of equilibrium, which state that the resultant force and moment acting on a body must be zero. The conditions of equilibrium can be expressed as:


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  • Sum of forces in x-direction is zero: $$\sum F_x = 0$$



  • Sum of forces in y-direction is zero: $$\sum F_y = 0$$



  • Sum of moments about any point is zero: $$\sum M_O = 0$$




To apply the coplanar equilibrium analysis, we need to follow these steps:



  • Draw a free-body diagram: A free-body diagram is a sketch that shows all the external forces and moments acting on a body or a part of a body. It is important to label the forces and moments with their magnitudes and directions, as well as the dimensions and angles involved.



  • Select a coordinate system: A coordinate system is a set of axes that defines the directions and signs of the forces and moments. It is usually convenient to choose a coordinate system that aligns with the given or unknown forces or moments.



  • Write the equilibrium equations: Using the conditions of equilibrium, we can write three equations that relate the forces and moments in the x- and y-directions, as well as about any point. We can choose any point as the moment center, but it is often helpful to select a point that passes through one or more unknown forces or moments, so that they do not appear in the equation.



  • Solve for the unknowns: The equilibrium equations form a system of linear equations that can be solved by various methods, such as substitution, elimination, or matrix operations. The solution should satisfy all three equations simultaneously.




To illustrate the coplanar equilibrium analysis, let us look at some examples from chapter 4 of the solutions manual.


Example 4.1




The problem statement is as follows:


The beam shown in Fig. P4.1 supports a load P = 500 N at its end. Determine (a) the reactions at A and B; (b) the internal force system at section C-C.



The solution is as follows:



  • Draw a free-body diagram: We can draw a free-body diagram for the whole beam, as shown below. We assume that the reactions at A and B are vertical forces A_y and B_y, respectively. We also show the load P = 500 N at the end of the beam, as well as the dimensions L = 3 m and h = 2 m.




  • Select a coordinate system: We can choose a coordinate system that has the positive x-axis pointing to the right and the positive y-axis pointing upward.



Write the equilibrium equations: Applying the conditions of equilibrium, we can write three equations as follows:


  • Sum of forces in x-direction is zero: $$\sum F_x = 0$$ Since there are no forces in the x-direction, this equation is trivially satisfied.



  • Sum of forces in y-direction is zero: $$\sum F_y = 0$$ This equation gives: $$A_y + B_y - P = 0$$



  • Sum of moments about any point is zero: $$\sum M_O = 0$$ We can choose any point as the moment center, but it is convenient to choose point A, since it passes through the unknown reaction A_y. This equation gives: $$-B_y L + P (L + h) = 0$$





  • Solve for the unknowns: We can solve for B_y by substituting P = 500 N, L = 3 m, and h = 2 m into the moment equation: $$-B_y (3) + 500 (3 + 2) = 0$$ This gives: $$B_y = \frac500 (5)3 = 833.33 \text N$$ Then, we can solve for A_y by substituting B_y and P into the force equation: $$A_y + 833.33 - 500 = 0$$ This gives: $$A_y = 500 - 833.33 = -333.33 \text N$$ The negative sign indicates that the direction of A_y is opposite to what we assumed.




The answers are:



  • (a) The reactions at A and B are A_y = -333.33 N and B_y = 833.33 N.



  • (b) The internal force system at section C-C consists of a normal force N, a shear force V, and a bending moment M. To find these forces, we can draw a free-body diagram for the left part of the beam, as shown below. We also show the reactions A_y and B_y that we found in part (a).




Applying the conditions of equilibrium, we can write three equations as follows:


  • Sum of forces in x-direction is zero: $$\sum F_x = 0$$ This equation gives: $$N = 0$$



  • Sum of forces in y-direction is zero: $$\sum F_y = 0$$ This equation gives: $$-A_y + V - B_y = 0$$ Substituting A_y = -333.33 N and B_y = 833.33 N, we get: $$V = -333.33 + 833.33 = 500 \text N$$



  • Sum of moments about any point is zero: $$\sum M_O = 0$$ We can choose any point as the moment center, but it is convenient to choose point C, since it passes through the unknown shear force V. This equation gives: $$-A_y h - M + B_y L = 0$$ Substituting A_y = -333.33 N, B_y = 833.33 N, h = 2 m, and L = 3 m, we get: $$M = -333.33 (2) + 833.33 (3) = 1666.67 \text Nm$$





The internal force system at section C-C is N = 0, V = 500 N, and M = 1666.67 Nm.



Example 4.2




The problem statement is as follows:


The truss shown in Fig. P4.2 supports a load P at joint E. Determine (a) the reactions at A and D; (b) the internal forces in members AB and BC.



The solution is as follows:



  • Draw a free-body diagram: We can draw a free-body diagram for the whole truss, as shown below. We assume that the reactions at A and D are vertical forces A_y and D_y, respectively, and horizontal forces A_x and D_x, respectively. We also show the load P at joint E, as well as the dimensions L = 4 m, h = 2 m, and a = 1 m.




  • Select a coordinate system: We can choose a coordinate system that has the positive x-axis pointing to the right and the positive y-axis pointing upward.



Write the equilibrium equations: Applying the conditions of equilibrium, we can write three equations as follows:


  • Sum of forces in x-direction is zero: $$\sum F_x = 0$$ This equation gives: $$A_x + D_x = 0$$



  • Sum of forces in y-direction is zero: $$\sum F_y = 0$$ This equation gives: $$A_y + D_y - P = 0$$



  • Sum of moments about any point is zero: $$\sum M_O = 0$$ We can choose any point as the moment center, but it is convenient to choose point A, since it passes through three unknown reactions A_x, A_y, and D_x. This equation gives: $$-D_y L + P (L - a) = 0$$





  • Solve for the unknowns: We can solve for D_y by substituting P = 500 N, L = 4 m, and a = 1 m into the moment equation: $$-D_y (4) + 500 (4 - 1) = 0$$ This gives: $$D_y = \frac500 (3)4 = 375 \text N$$ Then, we can solve for A_y by substituting D_y and P into the force equation: $$A_y + 375 - 500 = 0$$ This gives: $$A_y = 500 - 375 = 125 \text N$$ Finally, we can solve for A_x and D_x by substituting either one of them into the force equation: $$A_x + D_x = 0$$ This gives: $$A_x = -D_x$$ Since there are no horizontal forces acting on the truss, we can assume that A_x and D_x are zero. Alternatively, we can use another equilibrium equation, such as the sum of moments about point D, to verify that A_x and D_x are zero.




The answers are:



  • (a) The reactions at A and D are A_x = D_x = 0 N, A_y = 125 N, and D_y = 375 N.



  • (b) The internal forces in members AB and BC are axial forces F_AB and F_BC, respectively. To find these forces, we can use the method of joints, which involves isolating each joint and applying the conditions of equilibrium. We can start with joint B, as shown below. We assume that the forces in the members are tensile (positive) if they point away from the joint, and compressive (negative) if they point toward the joint.




Applying the conditions of equilibrium, we can write two equations as follows:


Sum of forces in x-direction is zero: $$\sum F_x = 0$$ This equation gives: $$F_AB \cos \theta - F_BC \cos \theta = 0$$ where


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