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Master Statics of Rigid Bodies Chapter 4 with Pytel and Kiusalaas's Solutions Manual PDF Download


- What are the main topics covered in chapter 4 of Pytel and Kiusalaas's book? - How can a solutions manual help students learn better? H2: Equilibrium of a rigid body in two dimensions - Conditions for equilibrium - Free-body diagrams - Equations of equilibrium - Constraints and reactions - Examples and exercises H2: Equilibrium of a rigid body in three dimensions - Conditions for equilibrium - Free-body diagrams - Equations of equilibrium - Constraints and reactions - Examples and exercises H2: Distributed forces: centroids and centers of gravity - Centroids of lines, areas, and volumes - Composite bodies and figures - Theorems of Pappus-Guldinus - Centers of gravity and mass - Examples and exercises H2: Analysis of structures - Trusses: method of joints and method of sections - Frames and machines: method of members and method of sections - Examples and exercises H2: Forces in beams and cables - Internal forces in beams: normal force, shear force, and bending moment - Shear and moment diagrams - Cables under concentrated and distributed loads - Examples and exercises H2: Friction - Laws of dry friction - Problems involving dry friction: wedges, belts, screws, rolling resistance, etc. - Examples and exercises H1: Conclusion - Summary of the main points and concepts from chapter 4 - Benefits of using Pytel and Kiusalaas's solutions manual for learning statics of rigid bodies - Call to action for students to get the solutions manual Table 2: Article with HTML formatting Introduction




Statics of rigid bodies is a branch of mechanics that deals with the analysis of forces acting on objects that are either at rest or moving with constant velocity. It is a fundamental subject for engineering students who need to understand how structures, machines, and systems can withstand external loads without collapsing or breaking. Statics of rigid bodies also provides the basis for studying dynamics, which involves the motion of objects under the influence of forces.




statics of rigid bodies pytel and kiusalaas solutions manual chapter 4 pdf



One of the most popular textbooks for learning statics of rigid bodies is Engineering Mechanics: Statics by Andrew Pytel and Jaan Kiusalaas. This book covers all the essential topics in a clear and concise manner, with numerous examples, illustrations, and problems. Chapter 4 is one of the most important chapters in the book, as it covers several topics that are essential for solving statics problems, such as equilibrium of a rigid body in two and three dimensions, distributed forces, analysis of structures, forces in beams and cables, and friction.


However, learning statics of rigid bodies is not an easy task. It requires a lot of practice, patience, and perseverance. Many students struggle with understanding the concepts, applying the methods, and solving the problems. That is why a solutions manual can be very helpful for students who want to master statics of rigid bodies. A solutions manual provides detailed explanations, step-by-step procedures, tips, tricks, and shortcuts for solving all the problems in the textbook. It also helps students check their answers, identify their mistakes, and improve their skills.


In this article, we will give you an overview of chapter 4 of Pytel and Kiusalaas's book on statics of rigid bodies. We will also show you how you can benefit from using their solutions manual to learn better and faster. If you are a student who wants to ace your statics course, you should definitely read this article till the end.


Equilibrium of a rigid body in two dimensions




One of the first topics covered in chapter 4 is the equilibrium of a rigid body in two dimensions. A rigid body is an idealized object that does not deform or change its shape under the action of forces. Equilibrium means that the net force and the net moment acting on the rigid body are both zero. This implies that the rigid body is either at rest or moving with constant velocity in a straight line.


To analyze the equilibrium of a rigid body in two dimensions, we need to follow these steps:


  • Draw a free-body diagram of the rigid body, showing all the external forces and moments acting on it. These include applied forces, such as weights, loads, and tensions, and reaction forces, such as normal forces, friction forces, and support forces.



  • Write the equations of equilibrium for the rigid body, using the scalar or vector method. The scalar method involves resolving the forces into horizontal and vertical components and summing them up to zero. The vector method involves using the cross product to calculate the moments about a point or an axis and setting them equal to zero.



  • Solve the equations of equilibrium for the unknown forces and moments, using algebraic or graphical methods. The algebraic method involves solving a system of linear equations using substitution, elimination, or matrix methods. The graphical method involves drawing a force polygon and a moment diagram to find the magnitudes and directions of the unknown forces and moments.



To illustrate these steps, let us consider an example from Pytel and Kiusalaas's book:


Example 4.1: Determine the reactions at A and B for the beam shown in Fig. 4.1.


Solution:


  • The free-body diagram of the beam is shown in Fig. 4.2. We assume that A is a pin support and B is a roller support. Therefore, A has two reaction forces, Ax and Ay, and B has one reaction force, By. We also assume that the weight of the beam is negligible compared to the applied loads.



  • The equations of equilibrium for the beam are: $$\sum F_x = 0 \implies A_x = 0$$ $$\sum F_y = 0 \implies A_y + B_y - 500 - 300 = 0$$ $$\sum M_A = 0 \implies -500(2) - 300(5) + B_y(6) = 0$$



  • Solving these equations, we get: $$A_x = 0 \text kN$$ $$A_y = 133.33 \text kN$$ $$B_y = 666.67 \text kN$$



Equilibrium of a rigid body in three dimensions




The next topic covered in chapter 4 is the equilibrium of a rigid body in three dimensions. This is similar to the two-dimensional case, except that we have to consider three orthogonal axes (x, y, z) instead of two (x, y). The conditions for equilibrium are still the same: the net force and the net moment acting on the rigid body must be zero.


To analyze the equilibrium of a rigid body in three dimensions, we need to follow these steps:


  • Draw a free-body diagram of the rigid body, showing all the external forces and moments acting on it. These include applied forces, such as weights, loads, and tensions, and reaction forces, such as normal forces, friction forces, and support forces.



  • Write the equations of equilibrium for the rigid body, using the scalar or vector method. The scalar method involves resolving the forces into x, y, and z components and summing them up to zero. The vector method involves using the cross product to calculate the moments about a point or an axis and setting them equal to zero.



  • Solve the equations of equilibrium for the unknown forces and moments, using algebraic or graphical methods. The algebraic method involves solving a system of linear equations using substitution, elimination, or matrix methods. The graphical method involves drawing a force polygon and a moment diagram to find the magnitudes and directions of the unknown forces and moments.



To illustrate these steps, let us consider an example from Pytel and Kiusalaas's book:


Distributed forces: centroids and centers of gravity




Another topic covered in chapter 4 is the distributed forces: centroids and centers of gravity. A distributed force is a force that acts over a region of space, rather than at a single point. For example, the weight of a body is a distributed force that acts over its entire volume. A centroid is a point that represents the geometric center of a line, an area, or a volume. A center of gravity is a point that represents the resultant of all the weight forces acting on a body.


To find the centroids and centers of gravity of various shapes and bodies, we need to follow these steps:


  • Divide the shape or body into simpler parts, such as rectangles, triangles, circles, etc.



  • Find the centroids of each part using formulas or tables.



  • Find the coordinates of the centroid of the whole shape or body using the principle of moments. This involves multiplying the area or volume of each part by its centroidal coordinate and dividing by the total area or volume.



  • Find the weight of each part by multiplying its volume by its density.



  • Find the coordinates of the center of gravity of the whole body using the principle of moments. This involves multiplying the weight of each part by its centroidal coordinate and dividing by the total weight.



To illustrate these steps, let us consider an example from Pytel and Kiusalaas's book:


Example 4.15: Locate the centroid and center of gravity of the composite area shown in Fig. 4.15.


Solution:


  • The composite area consists of three parts: a semicircle (1), a rectangle (2), and a triangle (3).



  • The centroids of each part are shown in Fig. 4.16. Using formulas or tables, we can find their coordinates as follows: $$\barx_1 = 0 \text m$$ $$\bary_1 = \frac4r3\pi = \frac4(0.5)3\pi = 0.212 \text m$$ $$\barx_2 = \fracb2 = \frac12 = 0.5 \text m$$ $$\bary_2 = \frach2 = \frac12 = 0.5 \text m$$ $$\barx_3 = \fracb3 = \frac13 = 0.333 \text m$$ $$\bary_3 = \frach3 = \frac13 = 0.333 \text m$$



  • The coordinates of the centroid of the composite area are: $$\barx = \frac\sum A_i\barx_i\sum A_i = \frac\pi r^2(0) + bh(0.5) + bh(0.333)\pi r^2 + bh + bh = 0.417 \text m$$ $$\bary = \frac\sum A_i\bary_i\sum A_i = \frac\pi r^2(0.212) + bh(0.5) + bh(0.333)\pi r^2 + bh + bh = 0.425 \text m$$



  • Assuming that the density of the material is $\rho$, we can find the weight of each part as follows: $$W_1 = \rho g A_1 = \rho g (\pi r^2)$$ $$W_2 = \rho g A_2 = \rho g (bh)$$ $$W_3 = \rho g A_3 = \rho g (\frac12bh)$$



  • The coordinates of the center of gravity of the composite body are: $$\barx_G = \frac\sum W_i\barx_i\sum W_i = \frac\rho g (\pi r^2)(0) + \rho g (bh)(0.5) + \rho g (\frac12bh)(0.333)\rho g (\pi r^2) + \rho g (bh) + \rho g (\frac12bh) = 0.417 \text m$$ $$\bary_G = \frac\sum W_i\bary_i\sum W_i = \frac\rho g (\pi r^2)(0.212) + \rho g (bh)(0.5) + \rho g (\frac12bh)(0.333)\rho g (\pi r^2) + \rho g (bh) + \rho g (\frac12bh) = 0.425 \text m$$



Analysis of structures




Another topic covered in chapter 4 is the analysis of structures. A structure is a system of connected parts that is designed to support loads without excessive deformation or failure. Some common types of structures are trusses, frames, and machines. A truss is a structure composed of slender members joined together at their ends by pins or welds. A frame is a structure composed of rigid members joined together by pins, welds, or hinges. A machine is a structure composed of rigid or flexible members that are designed to transmit forces and produce motion.


To analyze the forces in the members of structures, we need to follow these steps:


  • Draw a free-body diagram of the whole structure, showing all the external forces and moments acting on it. These include applied forces, such as weights, loads, and tensions, and reaction forces, such as normal forces, friction forces, and support forces.



  • Write the equations of equilibrium for the whole structure, using the scalar or vector method. The scalar method involves resolving the forces into horizontal and vertical components and summing them up to zero. The vector method involves using the cross product to calculate the moments about a point or an axis and setting them equal to zero.



  • Solve the equations of equilibrium for the unknown reaction forces and moments, using algebraic or graphical methods. The algebraic method involves solving a system of linear equations using substitution, elimination, or matrix methods. The graphical method involves drawing a force polygon and a moment diagram to find the magnitudes and directions of the unknown reaction forces and moments.



  • Draw free-body diagrams of each member or section of the structure, showing all the internal and external forces and moments acting on it. The internal forces are the forces that each member exerts on its adjacent members. The external forces are the same as those shown in the free-body diagram of the whole structure.



  • Write the equations of equilibrium for each member or section of the structure, using the scalar or vector method. The scalar method involves resolving the forces into horizontal and vertical components and summing them up to zero. The vector method involves using the cross product to calculate the moments about a point or an axis and setting them equal to zero.



  • Solve the equations of equilibrium for the unknown internal forces and moments, using algebraic or graphical methods. The algebraic method involves solving a system of linear equations using substitution, elimination, or matrix methods. The graphical method involves drawing a force polygon and a moment diagram to find the magnitudes and directions of the unknown internal forces and moments.



To illustrate these steps, let us consider an example from Pytel and Kiusalaas's book:


Example 4.23: Determine the force in each member of the truss shown in Fig. 4.23.


Solution:


  • The free-body diagram of the whole truss is shown in Fig. 4.24. We assume that A is a pin support and B is a roller support. Therefore, A has two reaction forces, Ax and Ay, and B has one reaction force, By.



  • The equations of equilibrium for the whole truss are: $$\sum F_y = 0 \implies A_y + B_y - 30 - 10 = 0$$ $$\sum M_A = 0 \implies -20(4) - 30(8) - 10(12) + B_y(16) = 0$$



  • Solving these equations, we get: $$A_x = 20 \text kN$$ $$A_y = 5 \text kN$$ $$B_y = 35 \text kN$$



  • The free-body diagrams of each member of the truss are shown in Fig. 4.25. We assume that all the members are in tension, unless proven otherwise. We also label the internal forces as FAB, FAC, FBC, etc.



  • The equations of equilibrium for each member of the truss are: Member AB: $$\sum F_x = 0 \implies FAB - 20 = 0 \implies FAB = 20 \text kN (T)$$ Member AC: $$\sum F_x = 0 \implies FAC \cos 45 - A_x = 0 \implies FAC = \fracA_x\cos 45 = \frac20\cos 45 = 28.28 \text kN (T)$$ Member BC: $$\sum F_y = 0 \implies FBC + FAC \sin 45 - A_y = 0 \implies FBC = A_y - FAC \sin 45 = 5 - 28.28 \sin 45 = -15 \text kN (C)$$ Member CD: $$\sum F_x = 0 \implies FCD + FAC \cos 45 - FBC \cos 45 - FDE \cos 45 = 0$$ Member CE: $$\sum F_y = 0 \implies FCE + FDE \sin 45 - B_y = 0$$ Member DE: $$\sum M_C = 0 \implies -FDE(4) + 30(4) +10(8) = 0 \implies FDE = \frac30(4) +10(8)4 = 35 \text kN (T)$$ Member EF: $$\sum M_E = 0 \implies -FEF(4) + FDE(4) +10(4) = 0 \implies FEF = FDE +10 =35 +10=45\text kN (T)$$



  • Solving these equations, we get: $$FCD = -15.86 \text kN (C)$$ $$FCE = -17.68 \text kN (C)$$



Forces in beams and cables




Another topic covered in chapter 4 is the forces in beams and cables. A beam is a long and slender member that is subjected to transverse loads, such as weights, pressures, and distributed forces. A cable is a flexible and elastic member that is subjected to tensile loads, such as weights and tensions. Both beams and cables are important structural elements that are used in bridges, buildings, cranes, etc.


To analyze the forces in beams and cables, we need to follow these steps:


  • Draw a free-body diagram of the beam or cable, showing all the external forces and moments acting on it. These include applied forces, such as weights, loads, and tensions, and reaction forces, such as normal forces, friction forces, and support forces.



  • Write the equations of equilibrium for the beam or cable, using the scalar or vector method. The scalar method involves resolving the forces into horizontal and vertical components and summing them up to zero. The vector method involves using the cross product to calculate the moments about a point or an axis and setting them equal to zero.



  • Solve the equations of equilibrium for the unknown reaction forces and moments, using algebraic or graphical methods. The algebraic method involves solving a system of linear equations using substitution, elimination, or matrix methods. The graphical method involves drawing a force polygon and a moment diagram to find the magnitudes and directions of the unknown reaction forces and moments.



  • Draw a free-body diagram of a segment or section of the beam or cable, showing all the internal and external forces and moments acting on it. The internal forces are the forces that each segment exerts on its adjacent segments. The external forces are the same as those shown in the free-body diagram of the whole beam or cable.



  • Write the equations of equilibrium for the segment or section of the beam or cable, using the scalar or vector method. The scalar method involves resolving the forces into horizontal and vertical components and summing them up to zero. The vector method involves using the cross product to calculate the moments about a point or an axis and setting them equal to zero.



  • Solve the equations of equilibrium for the unknown internal forces and moments, using algebraic or graphical methods. The algebraic method involves solving a system of linear equations using substitution, elimination, or matrix methods. The graphical method involves drawing a shear force diagram and a bending moment diagram to find the magnitudes and directions of the unknown internal forces and moments.



To illustrate these steps, let us consider an example from Pytel and Kiusalaas's book:


Example 4.31: Draw the shear force and bending moment diagrams for the beam shown in Fig. 4.31.


Solution:


  • The free-body diagram of the beam is shown in Fig. 4.32. We assume that A is a pin support and B is a roller support. Therefore, A has two reaction forces, Ax and Ay, and B has one reaction force, By.



  • The equations of equilibrium for the beam are: $$\sum F_x = 0 \implies A_x = 0$$ $$\sum F_y = 0 \implies A_y + B_y - 20 - 10 - 10 = 0$$ $$\sum M_A = 0 \implies -20(2) - 10(4) - 10(6) + B_y(8) = 0$$



  • Solving these equations, we get: $$A_x = 0 \text kN$$ $$A_y = 10 \text kN$$ $$B_y = 30 \text kN$$



  • The free-body diagram of a segment of the beam is shown in Fig. 4.33. We assume that V is the shear force and M is the bending moment at any cross section of the beam.



The equations of equilibrium for the segment of the beam are: $$\sum F_x = 0 \implies V + dV - dF_x


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