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 loading 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 shearforce 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 mathematical 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 counterclockwise moment of 240 lbf · in at x = 10 in. Derive the shear-force and bendingmoment 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 convention 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 tensile 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 orthogonal 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