Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition 82 I Mechanical Engineering Design Answer Substitutingp=64.3 into Eq.(3-9)again yields t =0,indicating that-24.03 MPa is also a principal stress.Once the principal stresses are calculated they can be ordered such that o1 >02.Thus,o1 =104.03 MPa and o2 =-24.03 MPa. Since for o=104.03 MPa,p=-25.7,and since is defined positive ccw in the transformation equations,we rotate clockwise 25.7 for the surface containing o1.We see in Fig.3-11c that this totally agrees with the semigraphical method. To determine t1andt,we first use Eq.(3-ll)to calculate中，： .=am-（2,） Ox-dy 80 =tan- =19.3°，109.3° 2(-50) Forp=19.3°，Eqs.(3-8)and(3-9)yield Answer g=80+0+80-0 2 2 os[2(19.3]+(-50)sim[2(19.3)]=40.0MPa T=- 2sim2(19.31+(-50)cos219.31=-64.0MPa 80-0 Remember that Eqs.(3-8)and (3-9)are coordinate transformation equations.Imagine that we are rotating the x,y axes 19.3 counterclockwise and y will now point up and to the left.So a negative shear stress on the rotatedx face will point down and to the right as shown in Fig.3-11d.Thus again,results agree with the semigraphical method. For中=109.3°，Eqs.(3-8)and(3-9)give a=40.0 MPa and t=+64.0MPa. Using the same logic for the coordinate transformation we find that results again agree with Fig.3-11d. 3-7 General Three-Dimensional Stress As in the case of plane stress,a particular orientation of a stress element occurs in space for which all shear-stress components are zero.When an element has this particular ori- entation,the normals to the faces are mutually orthogonal and correspond to the prin- cipal directions,and the normal stresses associated with these faces are the principal stresses.Since there are three faces,there are three principal directions and three prin- cipal stresses o1.o2.and o3.For plane stress,the stress-free surface contains the third principal stress which is zero. In our studies of plane stress we were able to specify any stress state or,oy,and txy and find the principal stresses and principal directions.But six components of stress are required to specify a general state of stress in three dimensions,and the problem of determining the principal stresses and directions is more difficult.In design,three-dimensional transformations are rarely performed since most maxi- mum stress states occur under plane stress conditions.One notable exception is con- tact stress,which is not a case of plane stress,where the three principal stresses are given in Sec.3-19.In fact,all states of stress are truly three-dimensional,where they might be described one-or two-dimensionally with respect to specific coordi- nate axes.Here it is most important to understand the relationship amongst the three principal stresses.The process in finding the three principal stresses from the six
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGraw−Hill 87 Companies, 2008 82 Mechanical Engineering Design Substituting φp = 64.3◦ into Eq. (3–9) again yields τ = 0, indicating that −24.03 MPa is also a principal stress. Once the principal stresses are calculated they can be ordered such that σ1 ≥ σ2. Thus, σ1 = 104.03 MPa and σ2 = −24.03 MPa. Answer Since for σ1 = 104.03 MPa, φp = −25.7◦, and since φ is defined positive ccw in the transformation equations, we rotate clockwise 25.7° for the surface containing σ1. We see in Fig. 3–11c that this totally agrees with the semigraphical method. To determine τ1 and τ2, we first use Eq. (3–11) to calculate φs: φs = 1 2 tan−1 −σx − σy 2τxy = 1 2 tan−1 − 80 2(−50) = 19.3◦ , 109.3◦ For φs = 19.3◦, Eqs. (3–8) and (3–9) yield Answer σ = 80 + 0 2 + 80 − 0 2 cos[2(19.3)] + (−50)sin[2(19.3)] = 40.0 MPa τ = −80 − 0 2 sin[2(19.3)] + (−50) cos[2(19.3)] = −64.0 MPa Remember that Eqs. (3–8) and (3–9) are coordinate transformation equations. Imagine that we are rotating the x, y axes 19.3° counterclockwise and y will now point up and to the left. So a negative shear stress on the rotated x face will point down and to the right as shown in Fig. 3–11d. Thus again, results agree with the semigraphical method. For φs = 109.3◦, Eqs. (3–8) and (3–9) give σ = 40.0 MPa and τ = +64.0 MPa. Using the same logic for the coordinate transformation we find that results again agree with Fig. 3–11d. 3–7 General Three-Dimensional Stress As in the case of plane stress, a particular orientation of a stress element occurs in space for which all shear-stress components are zero. When an element has this particular orientation, the normals to the faces are mutually orthogonal and correspond to the principal directions, and the normal stresses associated with these faces are the principal stresses. Since there are three faces, there are three principal directions and three principal stresses σ1, σ2, and σ3. For plane stress, the stress-free surface contains the third principal stress which is zero. In our studies of plane stress we were able to specify any stress state σx , σy , and τxy and find the principal stresses and principal directions. But six components of stress are required to specify a general state of stress in three dimensions, and the problem of determining the principal stresses and directions is more difficult. In design, three-dimensional transformations are rarely performed since most maximum stress states occur under plane stress conditions. One notable exception is contact stress, which is not a case of plane stress, where the three principal stresses are given in Sec. 3–19. In fact, all states of stress are truly three-dimensional, where they might be described one- or two-dimensionally with respect to specific coordinate axes. Here it is most important to understand the relationship amongst the three principal stresses. The process in finding the three principal stresses from the six
Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition Load and Stress Analysis 83 Figure 3-12 Mohr's circles for three- dimensional stress. 12 (a) (b) stress components ox.oy.o.ty.ty.and x,involves finding the roots of the cubic equation' o3-(ax+0y+0)a2+(ax0,+0xa:+00:-t-t-t)o -(OxOyo:2txy ty:tax -0x ty -dyti -d:tiy)=0 (3-15) In plotting Mohr's circles for three-dimensional stress,the principal normal stresses are ordered so that2>03.Then the result appears as in Fig.3-12a.The stress coordinates o.r for any arbitrarily located plane will always lie on the bound- aries or within the shaded area. Figure 3-12a also shows the three principal shear stresses and T Each of these occurs on the two planes,one of which is shown in Fig.3-12b.The fig- ure shows that the principal shear stresses are given by the equations 2=2 (3-16) 2 2g=2a3 2 t =01-03 2 Of course,Tmax=Ti/3 when the normal principal stresses are ordered (o1>02>03). so always order your principal stresses.Do this in any computer code you generate and you'll always generate Tmax. 3-8 Elastic Strain Normal strain e is defined and discussed in Sec.2-1 for the tensile specimen and is given by Eq.(2-2)as e=8/1,where 8 is the total elongation of the bar within the length I.Hooke's law for the tensile specimen is given by Eq.(2-3)as o=Ee 3-17 where the constant E is called Young's modulus or the modulus of elasticity. For development of this equation and further elaboration of three-dimensional stress transformations see: Richard G.Budynas,Advanced Strength and Applied Stress Analysis,2nd ed.,McGraw-Hill,New York, 1999,pp.46-78. Note the difference between this notation and that for a shear stress,say..The use of the shilling mark is not accepted practice,but it is used here to emphasize the distinction
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis 88 © The McGraw−Hill Companies, 2008 Load and Stress Analysis 83 stress components σx , σy , σz, τxy , τyz, and τzx , involves finding the roots of the cubic equation1 σ3 − (σx + σy + σz)σ2 + σxσy + σxσz + σyσz − τ 2 xy − τ 2 yz − τ 2 zx σ − σxσyσz + 2τxy τyzτzx − σx τ 2 yz − σy τ 2 zx − σzτ 2 xy = 0 (3–15) In plotting Mohr’s circles for three-dimensional stress, the principal normal stresses are ordered so that σ1 ≥ σ2 ≥ σ3. Then the result appears as in Fig. 3–12a. The stress coordinates σ , τ for any arbitrarily located plane will always lie on the boundaries or within the shaded area. Figure 3–12a also shows the three principal shear stresses τ1/2, τ2/3, and τ1/3. 2 Each of these occurs on the two planes, one of which is shown in Fig. 3–12b. The figure shows that the principal shear stresses are given by the equations τ1/2 = σ1 − σ2 2 τ2/3 = σ2 − σ3 2 τ1/3 = σ1 − σ3 2 (3–16) Of course, τmax = τ1/3 when the normal principal stresses are ordered (σ1 > σ2 > σ3), so always order your principal stresses. Do this in any computer code you generate and you’ll always generate τmax. 3–8 Elastic Strain Normal strain is defined and discussed in Sec. 2-1 for the tensile specimen and is given by Eq. (2–2) as = δ/l, where δ is the total elongation of the bar within the length l. Hooke’s law for the tensile specimen is given by Eq. (2–3) as σ = E (3–17) where the constant E is called Young’s modulus or the modulus of elasticity. 1/2 1/3 2/3 3 2 1 (a) (b) 1/2 1 2 Figure 3–12 Mohr’s circles for threedimensional stress. 1 For development of this equation and further elaboration of three-dimensional stress transformations see: Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., McGraw-Hill, New York, 1999, pp. 46–78. 2 Note the difference between this notation and that for a shear stress, say, τxy . The use of the shilling mark is not accepted practice, but it is used here to emphasize the distinction
Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis I©The McGraw-Hil 89 Mechanical Engineering Companies,2008 Design,Eighth Edition 84 Mechanical Engineering Design When a material is placed in tension,there exists not only an axial strain,but also negative strain (contraction)perpendicular to the axial strain.Assuming a linear, homogeneous,isotropic material,this lateral strain is proportional to the axial strain.If the axial direction is x,then the lateral strains are ey =e=-vex.The constant of pro- portionality v is called Poisson's ratio,which is about 0.3 for most structural metals. See Table A-5 for values of v for common materials. If the axial stress is in the x direction,then from Eq.(3-17) Ox Ox Ex Ey=E2=-V- (3-18) E E For a stress element undergoing ox,oy,and o:simultaneously,the normal strains are given by 6r=E[ox-(oy+o)】 1 6,=E[a,-Wax+)】 (3-19列 1 :=E:-(a+,】 Shear strain y is the change in a right angle of a stress element when subjected to pure shear stress,and Hooke's law for shear is given by t=Gy (3-20) where the constant G is the shear modulus of elasticiry or modulus of rigidity. It can be shown for a linear,isotropic,homogeneous material,the three elastic con- stants are related to each other by E=2G(1+v) (3-21) 3-9 Uniformly Distributed Stresses The assumption of a uniform distribution of stress is frequently made in design.The result is then often called pure tension,pure compression,or pure shear,depending upon how the external load is applied to the body under study.The word simple is some- times used instead of pure to indicate that there are no other complicating effects. The tension rod is typical.Here a tension load F is applied through pins at the ends of the bar.The assumption of uniform stress means that if we cut the bar at a section remote from the ends and remove one piece,we can replace its effect by applying a uni- formly distributed force of magnitude oA to the cut end.So the stress o is said to be uniformly distributed.It is calculated from the equation F 0=A (3-22) This assumption of uniform stress distribution requires that: The bar be straight and of a homogeneous material The line of action of the force contains the centroid of the section The section be taken remote from the ends and from any discontinuity or abrupt change in cross section
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGraw−Hill 89 Companies, 2008 84 Mechanical Engineering Design When a material is placed in tension, there exists not only an axial strain, but also negative strain (contraction) perpendicular to the axial strain. Assuming a linear, homogeneous, isotropic material, this lateral strain is proportional to the axial strain. If the axial direction is x, then the lateral strains are y = z = −νx . The constant of proportionality v is called Poisson’s ratio, which is about 0.3 for most structural metals. See Table A–5 for values of v for common materials. If the axial stress is in the x direction, then from Eq. (3–17) x = σx E y = z = −ν σx E (3–18) For a stress element undergoing σx , σy , and σz simultaneously, the normal strains are given by x = 1 E σx − ν(σy + σz) y = 1 E σy − ν(σx + σz) (3–19) z = 1 E σz − ν(σx + σy ) Shear strain γ is the change in a right angle of a stress element when subjected to pure shear stress, and Hooke’s law for shear is given by τ = Gγ (3–20) where the constant G is the shear modulus of elasticity or modulus of rigidity. It can be shown for a linear, isotropic, homogeneous material, the three elastic constants are related to each other by E = 2G(1 + ν) (3–21) 3–9 Uniformly Distributed Stresses The assumption of a uniform distribution of stress is frequently made in design. The result is then often called pure tension, pure compression, or pure shear, depending upon how the external load is applied to the body under study. The word simple is sometimes used instead of pure to indicate that there are no other complicating effects. The tension rod is typical. Here a tension load F is applied through pins at the ends of the bar. The assumption of uniform stress means that if we cut the bar at a section remote from the ends and remove one piece, we can replace its effect by applying a uniformly distributed force of magnitude σA to the cut end. So the stress σ is said to be uniformly distributed. It is calculated from the equation σ = F A (3–22) This assumption of uniform stress distribution requires that: • The bar be straight and of a homogeneous material • The line of action of the force contains the centroid of the section • The section be taken remote from the ends and from any discontinuity or abrupt change in cross section
Budynas-Nisbett:Shigley's1.Basics 3.Load and Stress Analysis The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition Load and Stress Analysis 85 For simple compression,Eq.(3-22)is applicable with F normally being con- sidered a negative quantity.Also,a slender bar in compression may fail by buckling. and this possibility must be eliminated from consideration before Eq.(3-22)is used.3 Use of the equation F 三A (3-23) for a body,say,a bolt,in shear assumes a uniform stress distribution too.It is very difficult in practice to obtain a uniform distribution of shear stress.The equation is included because occasions do arise in which this assumption is utilized. 3-10 Normal Stresses for Beams in Bending The equations for the normal bending stresses in straight beams are based on the fol- lowing assumptions: 1 The beam is subjected to pure bending.This means that the shear force is zero, and that no torsion or axial loads are present. 2 The material is isotropic and homogeneous. 3 The material obeys Hooke's law. 4 The beam is initially straight with a cross section that is constant throughout the beam length. 5 The beam has an axis of symmetry in the plane of bending. 6 The proportions of the beam are such that it would fail by bending rather than by crushing,wrinkling,or sidewise buckling. 7 Plane cross sections of the beam remain plane during bending. In Fig.3-13 we visualize a portion of a straight beam acted upon by a positive bending moment M shown by the curved arrow showing the physical action of the moment together with a straight arrow indicating the moment vector.The x axis is coincident with the neutral axis of the section,and the xz plane,which contains the neutral axes of all cross sections,is called the neutral plane.Elements of the beam coincident with this plane have zero stress.The location of the neutral axis with respect to the cross section is coincident with the centroidal axis of the cross section Figure 3-13 Straight beam in positive bending. See Sec.4-11
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis 90 © The McGraw−Hill Companies, 2008 Load and Stress Analysis 85 For simple compression, Eq. (3–22) is applicable with F normally being considered a negative quantity. Also, a slender bar in compression may fail by buckling, and this possibility must be eliminated from consideration before Eq. (3–22) is used.3 Use of the equation τ = F A (3–23) for a body, say, a bolt, in shear assumes a uniform stress distribution too. It is very difficult in practice to obtain a uniform distribution of shear stress. The equation is included because occasions do arise in which this assumption is utilized. 3–10 Normal Stresses for Beams in Bending The equations for the normal bending stresses in straight beams are based on the following assumptions: 1 The beam is subjected to pure bending. This means that the shear force is zero, and that no torsion or axial loads are present. 2 The material is isotropic and homogeneous. 3 The material obeys Hooke’s law. 4 The beam is initially straight with a cross section that is constant throughout the beam length. 5 The beam has an axis of symmetry in the plane of bending. 6 The proportions of the beam are such that it would fail by bending rather than by crushing, wrinkling, or sidewise buckling. 7 Plane cross sections of the beam remain plane during bending. In Fig. 3–13 we visualize a portion of a straight beam acted upon by a positive bending moment M shown by the curved arrow showing the physical action of the moment together with a straight arrow indicating the moment vector. The x axis is coincident with the neutral axis of the section, and the xz plane, which contains the neutral axes of all cross sections, is called the neutral plane. Elements of the beam coincident with this plane have zero stress. The location of the neutral axis with respect to the cross section is coincident with the centroidal axis of the cross section. 3 See Sec. 4–11. Figure 3–13 Straight beam in positive bending. M M x y z
Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition 86 Mechanical Engineering Desigr Figure 3-14 Compression Bending stresses occording to Eq.3-241. Neutral axis.Centroidal axis The bending stress varies linearly with the distance from the neutral axis,y,and is given by My 0x=一1 (3-241 where I is the second moment of area about the z axis.That is I=ydA (3-25) The stress distribution given by Eq.(3-24)is shown in Fig.3-14.The maximum magni- tude of the bending stress will occur wherey has the greatest magnitude.Designating as the maximum magnitude of the bending stress,and c as the maximum magnitude of y Mc Omax= 1 (3-26a Equation(3-24)can still be used to ascertain as to whetherox is tensile or compressive. Equation(3-26a)is often written as M 0max=乙 (3-266) where Z I/c is called the section modulus. EXAMPLE 3-5 A beam having a T section with the dimensions shown in Fig.3-15 is subjected to a bending moment of 1600 N.m that causes tension at the top surface.Locate the neu- tral axis and find the maximum tensile and compressive bending stresses. Solution The area of the composite section is A =1956 mm2.Now divide the T section into two rectangles,numbered I and 2,and sum the moments of these areas about the top edge. We then have 1956c1=12(75)(6)+12(88)(56) and hence ci =32.99 mm.Therefore c2 =100-32.99=67.01 mm. Next we calculate the second moment of area of each rectangle about its own cen- troidal axis.Using Table A-18,we find for the top rectangle h= 2bh3=27512=1.080×10mm 1
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGraw−Hill 91 Companies, 2008 86 Mechanical Engineering Design The bending stress varies linearly with the distance from the neutral axis, y, and is given by σx = − My I (3–24) where I is the second moment of area about the z axis. That is I = y2 d A (3–25) The stress distribution given by Eq. (3–24) is shown in Fig. 3–14. The maximum magnitude of the bending stress will occur where y has the greatest magnitude. Designating σmax as the maximum magnitude of the bending stress, and c as the maximum magnitude of y σmax = Mc I (3–26a) Equation (3–24) can still be used to ascertain as to whether σmax is tensile or compressive. Equation (3–26a) is often written as σmax = M Z (3–26b) where Z = I/c is called the section modulus. EXAMPLE 3–5 A beam having a T section with the dimensions shown in Fig. 3–15 is subjected to a bending moment of 1600 N · m that causes tension at the top surface. Locate the neutral axis and find the maximum tensile and compressive bending stresses. Solution The area of the composite section is A = 1956 mm2. Now divide the T section into two rectangles, numbered 1 and 2, and sum the moments of these areas about the top edge. We then have 1956c1 = 12(75)(6) + 12(88)(56) and hence c1 = 32.99 mm. Therefore c2 = 100 − 32.99 = 67.01 mm. Next we calculate the second moment of area of each rectangle about its own centroidal axis. Using Table A-18, we find for the top rectangle I1 = 1 12 bh3 = 1 12(75)123 = 1.080 × 104 mm4 Compression Neutral axis, Centroidal axis Tension x c y y Figure 3–14 Bending stresses according to Eq. (3–24)