上海交通大学：《Design and Manufacturing》课程教学资源（参考文献）Shigley's Mechanical Engineering Design ch3 Load and stress analysis

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 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 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 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 3–1a 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 con￾centrated 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 simply￾supported 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)

Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition 72 I Mechanical Engineering Design positive y direction.It can be shown that differentiating Eq.(3-3)results in dv dM dxdx=q (3-4 Normally the applied distributed load is directed downward and labeled w(e.g.,see Fig.3-6).In this case,w=-g. Equations(3-3)and (3-4)reveal additional relations if they are integrated.Thus, if we integrate between,say,xA and xg,we obtain XB dv= gdx =VB-VA (3-51 which states that the change in shear force from A to B is equal to the area of the load- ing diagram between xa and xB. In a similar manner, dM= V dx Mg-MA (3-61 which states that the change in moment from A to B is equal to the area of the shear- force diagram between xA and xB. Table 3-1 Function Graph of fn (x) Meaning Singularity (Macaulay) Concentrated (-a) K-a)-2=0x≠a Functions moment x-a)-2=±o0×=a (unit doublet) /(x-a)-2dk=(x-o)-1 Concentrated ix-a) K-a-1=0x≠a force x-a-1=+∞X=a (unit impulse) (x-a)-1dx=(x-ao Unit step (x-ajo x-a°= 10X<a 11x≥a x-o°dk=x-al (-o)= 0 Ramp X<a x-ax≥a /x-o!d=-0)2 2 tW.H.Mocoy,"Noten the deflection of beams,"Messenger ofMathematis,vol.48pp.129-30,1919

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGraw−Hill 77 Companies, 2008 72 Mechanical Engineering Design positive y direction. It can be shown that differentiating Eq. (3–3) results in dV dx = d2M dx2 = q (3–4) Normally the applied distributed load is directed downward and labeled w (e.g., see Fig. 3–6). In this case, w = −q. Equations (3–3) and (3–4) reveal additional relations if they are integrated. Thus, if we integrate between, say, xA and xB, we obtain  VB VA dV =  xB xA q dx = VB − VA (3–5) which states that the change in shear force from A to B is equal to the area of the load￾ing diagram between xA and xB. In a similar manner,  MB MA d M =  xB xA V dx = MB − MA (3–6) which states that the change in moment from A to B is equal to the area of the shear￾force diagram between xA and xB. Function Graph of fn (x) Meaning x − a −2 = 0 x = a x − a −2 = ±∞ x = a  x − a −2 dx = x − a −1 Concentrated x − a −1 = 0 x = a force x − a −1 = +∞ x = a (unit impulse)  x − a −1 dx = x − a 0 Unit step x − a 0 =  0 x < a 1 x ≥ a  x − a 0 dx = x − a 1 Ramp x − a 1 =  0 x < a x − a x ≥ a  x − a 1 dx = x − a2 2 † W. H. Macaulay, “Note on the deflection of beams,” Messenger of Mathematics, vol. 48, pp. 129–130, 1919. Concentrated moment (unit doublet) x x – a –2 a x x – a –1 a x x – a 0 a 1 x x – a 1 a 1 1 Table 3–1 Singularity (Macaulay†) Functions

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 73 3-3 Singularity Functions The four singularity functions defined in Table 3-1 constitute a useful and easy means of integrating across discontinuities.By their use,general expressions for shear force and bending moment in beams can be written when the beam is loaded by concentrated moments or forces.As shown in the table,the concentrated moment and force functions are zero for all values ofx not equal to a.The functions are undefined for values of x=a.Note that the unit step and ramp functions are zero only for values of x that are less than a.The integration properties shown in the table constitute a part of the math- ematical definition too.The first two integrations of g(x)for V(x)and M(x)do not require constants of integration provided all loads on the beam are accounted for in (x).The examples that follow show how these functions are used. EXAMPLE 3-2 Derive expressions for the loading,shear-force,and bending-moment diagrams for the beam of Fig.3-5. I Figure 3-5 Solution Using Table 3-1 and g(x)for the loading function,we find Answer q=R1x)-1-Fc-a)-1-F2x-a2)-1+R2x-0-1 (1) Next,we use Eq.(3-5)to get the shear force. Answer V=qdx=R°-F-am°-Fx-am°+R2x-° 2 Note that V =0 atx =0- A second integration,in accordance with Eq.(3-6),yields Answer M=V dx=Ri(x)-Fi(x-ai)-F2(x -az)+Ra(x-1) 3) The reactions R and R2 can be found by taking a summation of moments and forces as usual,or they can be found by noting that the shear force and bending moment must be zero everywhere except in the region 0sxs1.This means that Eq.(2)should give V=0 at x slightly larger than l.Thus R1-F-F2+R2=0 (4④ Since the bending moment should also be zero in the same region,we have,from Eq.(3), Rl-F1l-a1)-F20-a2)=0 5) Equations(4)and(5)can now be solved for the reactions Ri and R2

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis 78 © The McGraw−Hill Companies, 2008 Load and Stress Analysis 73 3–3 Singularity Functions The four singularity functions defined in Table 3–1 constitute a useful and easy means of integrating across discontinuities. By their use, general expressions for shear force and bending moment in beams can be written when the beam is loaded by concentrated moments or forces. As shown in the table, the concentrated moment and force functions are zero for all values of x not equal to a. The functions are undefined for values of x = a. Note that the unit step and ramp functions are zero only for values of x that are less than a. The integration properties shown in the table constitute a part of the math￾ematical definition too. The first two integrations of q(x) for V(x) and M(x) do not require constants of integration provided all loads on the beam are accounted for in q(x). The examples that follow show how these functions are used. EXAMPLE 3–2 Derive expressions for the loading, shear-force, and bending-moment diagrams for the beam of Fig. 3–5. a1 a2 l F1 F2 R1 R2 x y q O Solution Using Table 3–1 and q(x) for the loading function, we find Answer q = R1x −1 − F1x − a1 −1 − F2x − a2 −1 + R2x − l −1 (1) Next, we use Eq. (3–5) to get the shear force. Answer V =  q dx = R1x 0 − F1x − a1 0 − F2x − a2 0 + R2x − l 0 (2) Note that V = 0 at x = 0−. A second integration, in accordance with Eq. (3–6), yields Answer M =  V dx = R1x 1 − F1x − a1 1 − F2x − a2 1 + R2x − l 1 (3) The reactions R1 and R2 can be found by taking a summation of moments and forces as usual, or they can be found by noting that the shear force and bending moment must be zero everywhere except in the region 0 ≤ x ≤ l. This means that Eq. (2) should give V = 0 at x slightly larger than l. Thus R1 − F1 − F2 + R2 = 0 (4) Since the bending moment should also be zero in the same region, we have, from Eq. (3), R1l − F1(l − a1) − F2(l − a2) = 0 (5) Equations (4) and (5) can now be solved for the reactions R1 and R2. Figure 3–5

Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition 74 I Mechanical Engineering Design EXAMPLE 3-3 Figure 3-6a shows the loading diagram for a beam cantilevered at A with a uniform load of 20 Ibf/in acting on the portion 3 in sxs7 in,and a concentrated counter- clockwise moment of 240 lbf.in at x=10 in.Derive the shear-force and bending- moment relations,and the support reactions Mi and R1. Solution Following the procedure of Example 3-2,we find the load intensity function to be q=-M1x-2+R1x)-1-20x-3)°+20x-7)°-240(x-10)-2(1 Note that the 20(x-7)0 term was necessary to"turn off"the uniform load at C. Integrating successively gives Answers V=-M1x)-1+R1(x)°-20x-3)1+20x-7)1-240-10-21 M=-M1(x)°+R1x)1-10-3)2+10x-7)2-240G-10)0 (3) The reactions are found by makingx slightly larger than 10 in,where both Vand M are zero in this region.Equation(2)will then give -M1(0)+R1(1)-20(10-3)+20(10-7)-240(0)=0 Answer which yields R=80 Ibf. From Eq.(3)we get -M1(1)+80(10)-10(10-3)2+10(10-7)2-240(1)=0 Answer which yields M=160 Ibf.in. Figures 3-6b and c show the shear-force and bending-moment diagrams.Note that the impulse terms in Eq.(2).-M(x)and-240(x-10)1,are physically not forces Figure 3-6 (a)Loading diagram for a beam cantilevered at A. 7 in (b)Shear-force diagram. 3in- 240 lbf-in (c)Bending-moment diagram. (a) V(Ibf) Step Ramp (b) M(Ibf-in) Step 240 Parabolic 80- Ramp 160 Slope=80 Ibf-in/in

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGraw−Hill 79 Companies, 2008 74 Mechanical Engineering Design EXAMPLE 3–3 Figure 3–6a shows the loading diagram for a beam cantilevered at A with a uniform load of 20 lbf/in acting on the portion 3 in ≤ x ≤ 7 in, and a concentrated counter￾clockwise moment of 240 lbf · in at x = 10 in. Derive the shear-force and bending￾moment relations, and the support reactions M1 and R1. Solution Following the procedure of Example 3–2, we find the load intensity function to be q = −M1x −2 + R1x −1 − 20x − 3 0 + 20x − 7 0 − 240x − 10 −2 (1) Note that the 20x − 70 term was necessary to “turn off” the uniform load at C. Integrating successively gives Answers V = −M1x −1 + R1x 0 − 20x − 3 1 + 20x − 7 1 − 240x − 10 −1 (2) M = −M1x 0 + R1x 1 − 10x − 3 2 + 10x − 7 2 − 240x − 10 0 (3) The reactions are found by making x slightly larger than 10 in, where both V and M are zero in this region. Equation (2) will then give −M1(0) + R1(1) − 20(10 − 3) + 20(10 − 7) − 240(0) = 0 Answer which yields R1 = 80 lbf. From Eq. (3) we get −M1(1) + 80(10) − 10(10 − 3) 2 + 10(10 − 7) 2 − 240(1) = 0 Answer which yields M1 = 160 lbf · in. Figures 3–6b and c show the shear-force and bending-moment diagrams. Note that the impulse terms in Eq. (2), −M1x−1 and −240x − 10−1, are physically not forces (a) (b) D A B C y q x 10 in 7 in 3 in R1 M1 20 lbf/in 240 lbfin x V (lbf) O Step Ramp (c) x M (lbfin) O –160 80 Parabolic Step 80 240 Ramp Slope = 80 lbfin/in Figure 3–6 (a) Loading diagram for a beam cantilevered at A. (b) Shear-force diagram. (c) Bending-moment diagram.

80 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 75 and are not shown in the V diagram.Also note that both the M and 240 Ibf.in moments are counterclockwise and negative singularity functions;however,by the con- vention shown in Fig.3-2 the M and 240 lbf.in are negative and positive bending moments,respectively,which is reflected in Fig.3-6c. 3-4 Stress When an internal surface is isolated as in Fig.3-2b,the net force and moment acting on the surface manifest themselves as force distributions across the entire area.The force distribution acting at a point on the surface is unique and will have components in the normal and tangential directions called normal stress and tangential shear stress, respectively.Normal and shear stresses are labeled by the Greek symbols o and t, respectively.If the direction of o is outward from the surface it is considered to be a ten- sile stress and is a positive normal stress.If o is into the surface it is a compressive stress and commonly considered to be a negative quantity.The units of stress in U.S. Customary units are pounds per square inch(psi).For SI units,stress is in newtons per square meter (N/m);I N/m=1 pascal (Pa). 3-5 Cartesian Stress Components The Cartesian stress components are established by defining three mutually orthogo- nal surfaces at a point within the body.The normals to each surface will establish the x,y,z Cartesian axes.In general,each surface will have a normal and shear stress. The shear stress may have components along two Cartesian axes.For example,Fig. 3-7 shows an infinitesimal surface area isolation at a point O within a body where the surface normal is the x direction.The normal stress is labeled or.The symbol o indicates a normal stress and the subscript x indicates the direction of the surface normal.The net shear stress acting on the surface is(x)oet which can be resolved into components in the y and z directions,labeled as ty and Tx,respectively (see Fig.3-7).Note that double subscripts are necessary for the shear.The first subscript indicates the direction of the surface normal whereas the second subscript is the direction of the shear stress. The state of stress at a point described by three mutually perpendicular surfaces is shown in Fig.3-8a.It can be shown through coordinate transformation that this is suf- ficient to determine the state of stress on any surface intersecting the point.As the Figure 3-7 Stress components on surface normal to x direction

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis 80 © The McGraw−Hill Companies, 2008 Figure 3–7 Stress components on surface normal to x direction. Load and Stress Analysis 75 and are not shown in the V diagram. Also note that both the M1 and 240 lbf · in moments are counterclockwise and negative singularity functions; however, by the con￾vention shown in Fig. 3–2 the M1 and 240 lbf · in are negative and positive bending moments, respectively, which is reflected in Fig. 3–6c. 3–4 Stress When an internal surface is isolated as in Fig. 3–2b, the net force and moment acting on the surface manifest themselves as force distributions across the entire area. The force distribution acting at a point on the surface is unique and will have components in the normal and tangential directions called normal stress and tangential shear stress, respectively. Normal and shear stresses are labeled by the Greek symbols σ and τ , respectively. If the direction of σ is outward from the surface it is considered to be a ten￾sile stress and is a positive normal stress. If σ is into the surface it is a compressive stress and commonly considered to be a negative quantity. The units of stress in U.S. Customary units are pounds per square inch (psi). For SI units, stress is in newtons per square meter (N/m2 ); 1 N/m2 = 1 pascal (Pa). 3–5 Cartesian Stress Components The Cartesian stress components are established by defining three mutually orthogo￾nal surfaces at a point within the body. The normals to each surface will establish the x, y, z Cartesian axes. In general, each surface will have a normal and shear stress. The shear stress may have components along two Cartesian axes. For example, Fig. 3–7 shows an infinitesimal surface area isolation at a point Q within a body where the surface normal is the x direction. The normal stress is labeled σx . The symbol σ indicates a normal stress and the subscript x indicates the direction of the surface normal. The net shear stress acting on the surface is (τx )net which can be resolved into components in the y and z directions, labeled as τxy and τxz, respectively (see Fig. 3–7). Note that double subscripts are necessary for the shear. The first subscript indicates the direction of the surface normal whereas the second subscript is the direction of the shear stress. The state of stress at a point described by three mutually perpendicular surfaces is shown in Fig. 3–8a. It can be shown through coordinate transformation that this is suf- ficient to determine the state of stress on any surface intersecting the point. As the Q y x z x xy xz (x)net

Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hil 81 Mechanical Engineering Companies,2008 Design,Eighth Edition 76 Mechanical Engineering Desigr Figure 3-8 (a)General three-dimensional stress.(b)Plane stress with "cross-shears"equal. (a) (b) dimensions of the cube in Fig.3-8a approach zero,the stresses on the hidden faces become equal and opposite to those on the opposing visible faces.Thus,in general,a complete state of stress is defined by nine stress components,ox,oy.o.Txy. Txz,Tyx,Tyz,Tax,and Tay. For equilibrium,in most cases,"cross-shears"are equal,hence tyx Txy tiy=Tyz Txz Ta (3-7 This reduces the number of stress components for most three-dimensional states of stress from nine to six quantities,ox,oy.o.txy,ty,and ta. A very common state of stress occurs when the stresses on one surface are zero. When this occurs the state of stress is called plane stress.Figure 3-8b shows a state of plane stress,arbitrarily assuming that the normal for the stress-free surface is the z direction such that o=a=y=0.It is important to note that the element in Fig.3-8b is still a three-dimensional cube.Also,here it is assumed that the cross-shears are equal such that tyx =Txy,and tyz=tay =tx:=ta=0. 3-6 Mohr's Circle for Plane Stress Suppose the dx dy dz element of Fig.3-8b is cut by an oblique plane with a normal n at an arbitrary angle counterclockwise from the x axis as shown in Fig.3-9.This section is concerned with the stresses o and t that act upon this oblique plane.By summing the forces caused by all the stress components to zero,the stresses o and t are found to be a=sycos 20+txy sin 2 2 2 (3-8) t=xax sin20+tsy cos2 2 (3-91 Equations (3-8)and (3-9)are called the plane-stress transformation equations. Differentiating Eq.(3-8)with respect toand setting the result equal to zero gives tan2φp= 2Uxy (3-10) Ox-Oy

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGraw−Hill 81 Companies, 2008 76 Mechanical Engineering Design dimensions of the cube in Fig. 3–8a approach zero, the stresses on the hidden faces become equal and opposite to those on the opposing visible faces. Thus, in general, a complete state of stress is defined by nine stress components, σx , σy , σz, τxy , τxz, τyx , τyz, τzx , and τzy . For equilibrium, in most cases, “cross-shears” are equal, hence τyx = τxy τzy = τyz τxz = τzx (3–7) This reduces the number of stress components for most three-dimensional states of stress from nine to six quantities, σx , σy , σz, τxy , τyz, and τzx . A very common state of stress occurs when the stresses on one surface are zero. When this occurs the state of stress is called plane stress. Figure 3–8b shows a state of plane stress, arbitrarily assuming that the normal for the stress-free surface is the z direction such that σz = τzx = τzy = 0. It is important to note that the element in Fig. 3–8b is still a three-dimensional cube. Also, here it is assumed that the cross-shears are equal such that τyx = τxy , and τyz = τzy = τxz = τzx = 0. 3–6 Mohr’s Circle for Plane Stress Suppose the dx dy dz element of Fig. 3–8b is cut by an oblique plane with a normal n at an arbitrary angle φ counterclockwise from the x axis as shown in Fig. 3–9. This section is concerned with the stresses σ and τ that act upon this oblique plane. By summing the forces caused by all the stress components to zero, the stresses σ and τ are found to be σ = σx + σy 2 + σx − σy 2 cos 2φ + τxy sin 2φ (3–8) τ = −σx − σy 2 sin 2φ + τxy cos 2φ (3–9) Equations (3–8) and (3–9) are called the plane-stress transformation equations. Differentiating Eq. (3–8) with respect to φ and setting the result equal to zero gives tan 2φp = 2τxy σx − σy (3–10) y y x y yx xy xy xy x y xy xz x x y x y z x z yz zy z x (a) (b) Figure 3–8 (a) General three-dimensional stress. (b) Plane stress with “cross-shears” equal