2 Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hfll Mechanical Engineering Companies,2008 Design,Eighth Edition Load and Stress Analysis Chapter Outline 3-1 Equilibrium and Free-Body Diagrams 68 3-2 Shear Force and Bending Moments in Beams 71 3-3 Singularity Functions 73 3-4 Stress 75 3-5 Cartesian Stress Components 75 3-6 Mohr's Circle for Plane Stress 76 3-7 General Three-Dimensional Stress 82 3-8 Elastic Strain 83 3-9 Uniformly Distributed Stresses 8q 3-10 Normal Stresses for Beams in Bending 85 3-11 Shear Stresses for Beams in Bending 90 3-12 Torsion 95 3-13 Stress Concentration 105 3-14 Stresses in Pressurized Cylinders 107 3-15 Stresses in Rotating Rings 110 3-16 Press and Shrink Fits 110 3-17 Temperature Effects 111 3-18 Curved Beams in Bending 112 3-19 Contact Stresses 117 3-20 Summary 121 67
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis 72 © The McGraw−Hill Companies, 2008 Load and Stress Analysis Chapter Outline 3–1 Equilibrium and Free-Body Diagrams 68 3–2 Shear Force and Bending Moments in Beams 71 3–3 Singularity Functions 73 3–4 Stress 75 3–5 Cartesian Stress Components 75 3–6 Mohr’s Circle for Plane Stress 76 3–7 General Three-Dimensional Stress 82 3–8 Elastic Strain 83 3–9 Uniformly Distributed Stresses 84 3–10 Normal Stresses for Beams in Bending 85 3–11 Shear Stresses for Beams in Bending 90 3–12 Torsion 95 3–13 Stress Concentration 105 3–14 Stresses in Pressurized Cylinders 107 3–15 Stresses in Rotating Rings 110 3–16 Press and Shrink Fits 110 3–17 Temperature Effects 111 3–18 Curved Beams in Bending 112 3–19 Contact Stresses 117 3–20 Summary 121 3 67
Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition 68 Mechanical Engineering Design One of the main objectives of this book is to describe how specific machine components function and how to design or specify them so that they function safely without failing structurally.Although earlier discussion has described structural strength in terms of load or stress versus strength,failure of function for structural reasons may arise from other factors such as excessive deformations or deflections. Here it is assumed that the reader has completed basic courses in statics of rigid bodies and mechanics of materials and is quite familiar with the analysis of loads,and the stresses and deformations associated with the basic load states of simple prismatic elements.In this chapter and Chap.4 we will review and extend these topics briefly. Complete derivations will not be presented here,and the reader is urged to return to basic textbooks and notes on these subjects. This chapter begins with a review of equilibrium and free-body diagrams associated with load-carrying components.One must understand the nature of forces before attempting to perform an extensive stress or deflection analysis of a mechanical com- ponent.An extremely useful tool in handling discontinuous loading of structures employs Macaulay or singularity functions.Singularity functions are described in Sec.3-3 as applied to the shear forces and bending moments in beams.In Chap.4,the use of singularity functions will be expanded to show their real power in handling deflections of complex geometry and statically indeterminate problems. Machine components transmit forces and motion from one point to another.The transmission of force can be envisioned as a flow or force distribution that can be fur- ther visualized by isolating internal surfaces within the component.Force distributed over a surface leads to the concept of stress,stress components,and stress transforma- tions(Mohr's circle)for all possible surfaces at a point. The remainder of the chapter is devoted to the stresses associated with the basic loading of prismatic elements,such as uniform loading,bending,and torsion,and topics with major design ramifications such as stress concentrations,thin-and thick-walled pressurized cylinders,rotating rings,press and shrink fits,thermal stresses,curved beams, and contact stresses. 3-1 Equilibrium and Free-Body Diagrams Equilibrium The word system will be used to denote any isolated part or portion of a machine or structure-including all of it if desired-that we wish to study.A system,under this definition,may consist of a particle,several particles,a part of a rigid body,an entire rigid body,or even several rigid bodies. If we assume that the system to be studied is motionless or,at most,has constant velocity,then the system has zero acceleration.Under this condition the system is said to be in eguilibrium.The phrase static equilibrium is also used to imply that the system is at rest.For equilibrium,the forces and moments acting on the system balance such that ∑F=0 3-1) ∑M=0 (3-21 which states that the sum of all force and the sum of all moment vectors acting upon a system in equilibrium is zero
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGraw−Hill 73 Companies, 2008 68 Mechanical Engineering Design One of the main objectives of this book is to describe how specific machine components function and how to design or specify them so that they function safely without failing structurally. Although earlier discussion has described structural strength in terms of load or stress versus strength, failure of function for structural reasons may arise from other factors such as excessive deformations or deflections. Here it is assumed that the reader has completed basic courses in statics of rigid bodies and mechanics of materials and is quite familiar with the analysis of loads, and the stresses and deformations associated with the basic load states of simple prismatic elements. In this chapter and Chap. 4 we will review and extend these topics briefly. Complete derivations will not be presented here, and the reader is urged to return to basic textbooks and notes on these subjects. This chapter begins with a review of equilibrium and free-body diagrams associated with load-carrying components. One must understand the nature of forces before attempting to perform an extensive stress or deflection analysis of a mechanical component. An extremely useful tool in handling discontinuous loading of structures employs Macaulay or singularity functions. Singularity functions are described in Sec. 3–3 as applied to the shear forces and bending moments in beams. In Chap. 4, the use of singularity functions will be expanded to show their real power in handling deflections of complex geometry and statically indeterminate problems. Machine components transmit forces and motion from one point to another. The transmission of force can be envisioned as a flow or force distribution that can be further visualized by isolating internal surfaces within the component. Force distributed over a surface leads to the concept of stress, stress components, and stress transformations (Mohr’s circle) for all possible surfaces at a point. The remainder of the chapter is devoted to the stresses associated with the basic loading of prismatic elements, such as uniform loading, bending, and torsion, and topics with major design ramifications such as stress concentrations, thin- and thick-walled pressurized cylinders, rotating rings, press and shrink fits, thermal stresses, curved beams, and contact stresses. 3–1 Equilibrium and Free-Body Diagrams Equilibrium The word system will be used to denote any isolated part or portion of a machine or structure—including all of it if desired—that we wish to study. A system, under this definition, may consist of a particle, several particles, a part of a rigid body, an entire rigid body, or even several rigid bodies. If we assume that the system to be studied is motionless or, at most, has constant velocity, then the system has zero acceleration. Under this condition the system is said to be in equilibrium. The phrase static equilibrium is also used to imply that the system is at rest. For equilibrium, the forces and moments acting on the system balance such that F = 0 (3–1) M = 0 (3–2) which states that the sum of all force and the sum of all moment vectors acting upon a system in equilibrium is zero.
Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition Load and Stress Analysis 69 Free-Body Diagrams We can greatly simplify the analysis of a very complex structure or machine by successively isolating each element and studying and analyzing it by the use of free-body diagrams. When all the members have been treated in this manner,the knowledge can be assembled to yield information concerning the behavior of the total system.Thus,free-body diagram- ming is essentially a means of breaking a complicated problem into manageable segments, analyzing these simple problems,and then,usually,putting the information together again. Using free-body diagrams for force analysis serves the following important purposes: The diagram establishes the directions of reference axes,provides a place to record the dimensions of the subsystem and the magnitudes and directions of the known forces,and helps in assuming the directions of unknown forces. The diagram simplifies your thinking because it provides a place to store one thought while proceeding to the next. The diagram provides a means of communicating your thoughts clearly and unam- biguously to other people. Careful and complete construction of the diagram clarifies fuzzy thinking by bringing out various points that are not always apparent in the statement or in the geometry of the total problem.Thus,the diagram aids in understanding all facets of the problem. The diagram helps in the planning of a logical attack on the problem and in setting up the mathematical relations. The diagram helps in recording progress in the solution and in illustrating the methods used. The diagram allows others to follow your reasoning,showing all forces. EXAMPLE 3-1 Figure 3-la shows a simplified rendition of a gear reducer where the input and output shafts AB and CD are rotating at constant speeds and respectively.The input and output torques(torsional moments)are T=240 Ibf.in and To respectively.The shafts are supported in the housing by bearings at A,B.C,and D.The pitch radii of gears G and G2 are ri=0.75 in and r2=1.5 in,respectively.Draw the free-body diagrams of each member and determine the net reaction forces and moments at all points. Solution First,we will list all simplifying assumptions. 1 Gears G and G2 are simple spur gears with a standard pressure angle=20 (see Sec.13-5). 2 The bearings are self-aligning and the shafts can be considered to be simply supported. 3 The weight of each member is negligible 4 Friction is negligible. 5 The mounting bolts at E,F,H,and are the same size. The separate free-body diagrams of the members are shown in Figs.3-1b-d.Note that Newton's third law,called the law of action and reaction,is used extensively where each member mates.The force transmitted between the spur gears is not tangential but at the pressure angle中.Thus,N=Ftan中
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis 74 © The McGraw−Hill Companies, 2008 Load and Stress Analysis 69 Free-Body Diagrams We can greatly simplify the analysis of a very complex structure or machine by successively isolating each element and studying and analyzing it by the use of free-body diagrams. When all the members have been treated in this manner, the knowledge can be assembled to yield information concerning the behavior of the total system. Thus, free-body diagramming is essentially a means of breaking a complicated problem into manageable segments, analyzing these simple problems, and then, usually, putting the information together again. Using free-body diagrams for force analysis serves the following important purposes: • The diagram establishes the directions of reference axes, provides a place to record the dimensions of the subsystem and the magnitudes and directions of the known forces, and helps in assuming the directions of unknown forces. • The diagram simplifies your thinking because it provides a place to store one thought while proceeding to the next. • The diagram provides a means of communicating your thoughts clearly and unambiguously to other people. • Careful and complete construction of the diagram clarifies fuzzy thinking by bringing out various points that are not always apparent in the statement or in the geometry of the total problem. Thus, the diagram aids in understanding all facets of the problem. • The diagram helps in the planning of a logical attack on the problem and in setting up the mathematical relations. • The diagram helps in recording progress in the solution and in illustrating the methods used. • The diagram allows others to follow your reasoning, showing all forces. EXAMPLE 3–1 Figure 3–1a shows a simplified rendition of a gear reducer where the input and output shafts AB and C D are rotating at constant speeds ωi and ωo, respectively. The input and output torques (torsional moments) are Ti = 240 lbf · in and To, respectively. The shafts are supported in the housing by bearings at A, B, C, and D. The pitch radii of gears G1 and G2 are r1 = 0.75 in and r2 = 1.5 in, respectively. Draw the free-body diagrams of each member and determine the net reaction forces and moments at all points. Solution First, we will list all simplifying assumptions. 1 Gears G1 and G2 are simple spur gears with a standard pressure angle φ = 20° (see Sec. 13–5). 2 The bearings are self-aligning and the shafts can be considered to be simply supported. 3 The weight of each member is negligible. 4 Friction is negligible. 5 The mounting bolts at E, F, H, and I are the same size. The separate free-body diagrams of the members are shown in Figs. 3–1b–d. Note that Newton’s third law, called the law of action and reaction, is used extensively where each member mates. The force transmitted between the spur gears is not tangential but at the pressure angle φ. Thus, N = F tan φ
Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hil Mechanical Engineering Companies,2008 Design,Eighth Edition 70 Mechanical Engineering Design urT,=240bfin R in (a)Gear reducer (b)Gear box 1.5in =240bfin (c)Input shaft (d)Output shaft Figure 3-1 (a)Gear reducer;(b-d)free-body diagrams.Diagrams are not drawn to scale. Summing moments about the x axis of shaft AB in Fig.3-ld gives ∑Mx=F0.75)-240=0 F =320 lbf The normal force is N=320 tan 20=116.5 lbf. Using the equilibrium equations for Figs.3-lc and d,the reader should verify that: RAy =192 Ibf,RA:=69.9 Ibf,RBy 128 Ibf,RB:=46.6 Ibf,RCy 192 Ibf,Rc:= 69.9 Ibf,Rpy =128 Ibf,Rp:=46.6 Ibf,and To =480 Ibf.in.The direction of the output torque T is oppositebecause it is the resistive load on the system opposing the motion. Note in Fig.3-16 the net force from the bearing reactions is zero whereas the net moment about the x axis is 2.25 (192)+2.25(128)=720 lbf.in.This value is the same as Ti+To=240+480 =720 Ibf.in,as shown in Fig.3-la.The reaction forces RE,RF,RH,and RI.from the mounting bolts cannot be determined from the equilibrium equations as there are too many unknowns.Only three equations are available,∑Fy=∑Fz=∑Mx=0.In case you were wondering about assumption 5,here is where we will use it (see Sec.8-12).The gear box tends to rotate about the x axis because of a pure torsional moment of 720 lbf.in.The bolt forces must provide
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGraw−Hill 75 Companies, 2008 70 Mechanical Engineering Design Summing moments about the x axis of shaft AB in Fig. 3–1d gives Mx = F(0.75) − 240 = 0 F = 320 lbf The normal force is N = 320 tan 20° = 116.5 lbf. Using the equilibrium equations for Figs. 3–1c and d, the reader should verify that: RAy = 192 lbf, RAz = 69.9 lbf, RBy = 128 lbf, RBz = 46.6 lbf, RCy = 192 lbf, RCz = 69.9 lbf, RDy = 128 lbf, RDz = 46.6 lbf, and To = 480 lbf · in. The direction of the output torque To is opposite ωo because it is the resistive load on the system opposing the motionωo. Note in Fig. 3–1b the net force from the bearing reactions is zero whereas the net moment about the x axis is 2.25 (192) + 2.25 (128) = 720 lbf · in. This value is the same as Ti + To = 240 + 480 = 720 lbf · in, as shown in Fig. 3–1a. The reaction forces RE , RF , RH , and RI , from the mounting bolts cannot be determined from the equilibrium equations as there are too many unknowns. Only three equations are available, Fy = Fz = Mx = 0. In case you were wondering about assumption 5, here is where we will use it (see Sec. 8–12). The gear box tends to rotate about the x axis because of a pure torsional moment of 720 lbf · in. The bolt forces must provide (a) Gear reducer 5 in 4 in C A I E B D F H G2 G1 0 T0 i , Ti 240 lbfin (c) Input shaft B Ti 240 lbfin G1 r1 RBz RBy 1.5 in 1 in N A F RAz RAy (d) Output shaft D G2 r2 T0 RDz RDy N F C RCz RCy (b) Gear box z y x 5 in 4 in C A I E D F H B RDy RBy RBz RDz RCy RAy RAz RF RH RCz RI RE Figure 3–1 (a) Gear reducer; (b–d) free-body diagrams. Diagrams are not drawn to scale.
Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis ©The McGraw-Hil Mechanical Engineering Companies,2008 Design,Eighth Edition Load and Stress Analysis 71 an equal but opposite torsional moment.The center of rotation relative to the bolts lies at the centroid of the bolt cross-sectional areas.Thus if the bolt areas are equal:the center of rotation is at the center of the four bolts,a distance of (4/2)2+(5/2)2=3.202 in from each bolt:the bolt forces are equal (Rg RF=R=RI=R),and each bolt force is perpendicular to the line from the bolt to the center of rotation.This gives a net torque from the four bolts of 4R(3.202)=720.Thus,RE =RF =RH=RI=56.22 Ibf. 3-2 Shear Force and Bending Moments in Beams Figure 3-2a shows a beam supported by reactions R and R2 and loaded by the con- centrated forces F1,F2,and F3.If the beam is cut at some section located at x =x and the left-hand portion is removed as a free body,an internal shear force V and bending moment M must act on the cut surface to ensure equilibrium (see Fig.3-2b).The shear force is obtained by summing the forces on the isolated section.The bending moment is the sum of the moments of the forces to the left of the section taken about an axis through the isolated section.The sign conventions used for bending moment and shear force in this book are shown in Fig.3-3.Shear force and bending moment are related by the equation V=d dx (3-3) Sometimes the bending is caused by a distributed load g(x).as shown in Fig.3-4; g(x)is called the load intensity with units of force per unit length and is positive in the Figure 3-2 Free-body diagram of simply supported beam with Vand M shown in positive directions. Figure 3-3 Sign conventions for bending and shear. Negative s Figure 3-4 q(x) Distributed load on beam
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis 76 © The McGraw−Hill Companies, 2008 Load and Stress Analysis 71 an equal but opposite torsional moment. The center of rotation relative to the bolts lies at the centroid of the bolt cross-sectional areas. Thus if the bolt areas are equal: the center of rotation is at the center of the four bolts, a distance of (4/2)2 + (5/2)2 = 3.202 in from each bolt; the bolt forces are equal(RE = RF = RH = RI = R), and each bolt force is perpendicular to the line from the bolt to the center of rotation. This gives a net torque from the four bolts of 4R(3.202) = 720. Thus, RE = RF = RH = RI = 56.22 lbf. 3–2 Shear Force and Bending Moments in Beams Figure 3–2a shows a beam supported by reactions R1 and R2 and loaded by the concentrated forces F1, F2, and F3. If the beam is cut at some section located at x = x1 and the left-hand portion is removed as a free body, an internal shear force V and bending moment M must act on the cut surface to ensure equilibrium (see Fig. 3–2b). The shear force is obtained by summing the forces on the isolated section. The bending moment is the sum of the moments of the forces to the left of the section taken about an axis through the isolated section. The sign conventions used for bending moment and shear force in this book are shown in Fig. 3–3. Shear force and bending moment are related by the equation V = d M dx (3–3) Sometimes the bending is caused by a distributed load q(x), as shown in Fig. 3–4; q(x) is called the load intensity with units of force per unit length and is positive in the Figure 3–2 Free-body diagram of simplysupported beam with V and M shown in positive directions. Figure 3–3 Sign conventions for bending and shear. Figure 3–4 Distributed load on beam. Positive bending Positive shear Negative shear Negative bending x y q(x) x1 x1 y y F1 F2 F3 F1 x x R1 R2 R1 V M (a) (b)