2 STRESS AND STRAIN Z Gzz 0以 Gyz Oxz Oyy Figure 1.1.Nine components of stress acting on an Gyx infinitesimal element. where i and are iterated over x,y,and z.Except where tensor notation is required,it is simpler to use a single subscript for a normal stress and denote a shear stress by r. For example,.ox≡Oxx and txy≡axy- 1.2 STRESS TRANSFORMATION Stress components expressed along one set of orthogonal axes may be expressed along any other set of axes.Consider resolving the stress component,oy=Fy/Ay,onto the x'and y axes as shown in Figure 1.2. The force,Fy,acts in the y direction is Fy=F cos and the area normal to y is Ay=Ay/cose,so oy=Fy/Ay Fy cose/(Ay/cose)=ay cos20. (1.4a) Similarly ty'x=Fx/Ay Fy sine/(Ay/cos0)=ay cos0 sin0. (1.4b) Note that transformation of stresses requires two sine and/or cosine terms. Pairs of shear stresses with the same subscripts that are in reverse order are always equal (e.g.,tij=tji).This is illustrated in Figure 1.3 by a simple moment balance Figure 1.2.The stresses acting on a plane,A',under a normal stress,cy
2 STRESS AND STRAIN σzz x y z σzx σzy σxz σxx σxy σyx σyy σyz Figure 1.1. Nine components of stress acting on an infinitesimal element. where i and j are iterated over x, y, and z. Except where tensor notation is required, it is simpler to use a single subscript for a normal stress and denote a shear stress by τ . For example, σx ≡ σx x and τx y ≡ σx y . 1.2 STRESS TRANSFORMATION Stress components expressed along one set of orthogonal axes may be expressed along any other set of axes. Consider resolving the stress component, σy = Fy/Ay , onto the x and y axes as shown in Figure 1.2. The force, Fy , acts in the y direction is Fy = Fy cos θ and the area normal to y is Ay = Ay/ cos θ, so σy = Fy/Ay = Fy cos θ/(Ay/ cos θ) = σy cos2 θ. (1.4a) Similarly τyx = Fx/Ay = Fy sin θ/(Ay/ cos θ) = σy cos θ sin θ. (1.4b) Note that transformation of stresses requires two sine and/or cosine terms. Pairs of shear stresses with the same subscripts that are in reverse order are always equal (e.g., τ i j = τ ji). This is illustrated in Figure 1.3 by a simple moment balance y x Fy Fy Fx Ay x y θ Figure 1.2. The stresses acting on a plane, A , under a normal stress, σy
1.2 STRESS TRANSFORMATION 3 Txy y Figure 1.3.Unless ty=tyx,there would not be a mo- Tyx ment balance. T on an infinitesimal element.Unless ti;=tii,there would be an infinite rotational acceleration.Therefore tij Tji. (1.5) The general equation for transforming the stresses from one set of orthogonal axes (e.g.,n,m,p)to another set of axes (e.g.,i,j,k),is 33 (1.6) n=lm=l Here,the term eim is the cosine of the angle between the i and the m axes and the term eim is the cosine of the angle between the j and n axes.This is often written more simply as ij =EinEinOmn, (1.7) with the summation implied.Consider transforming stresses from the x,y,z axis system to the x',y,z'system shown in Figure 1.4. Using equation 1.6, Ox'x'=Ex'xex'xOrx ex'xeryOxy ex'xex'zOxz +Ex'xer'-Oxz ex'yex=Oy:+Ex'zex=Ozz (1.8a) and Oxy=Ex'xeyxOxx +Ex'yeyxOxy ex'zey=Oxz +Ex'xey'yoxx +er'yeyyoyy +ex'zeyyoy: ex'xey=Oxz ex'yeyzoyz+exzeyzozz (1.8b) Figure 1.4.Two orthogonal coordinate systems
1.2 STRESS TRANSFORMATION 3 τyx τyx τxy τxy x y Figure 1.3. Unless τxy = τyx, there would not be a moment balance. on an infinitesimal element. Unless τi j = τ ji , there would be an infinite rotational acceleration. Therefore τi j = τ ji. (1.5) The general equation for transforming the stresses from one set of orthogonal axes (e.g., n, m, p) to another set of axes (e.g., i, j, k), is σi j = 3 n=1 3 m=1 imjnσmn. (1.6) Here, the term im is the cosine of the angle between the i and the m axes and the term jn is the cosine of the angle between the j and n axes. This is often written more simply as σi j = injnσmn, (1.7) with the summation implied. Consider transforming stresses from the x, y,z axis system to the x , y , z system shown in Figure 1.4. Using equation 1.6, σx x = x xx xσx x + x xx yσx y + xxx zσx z + x xx zσx z + x yx zσyz + x zx zσzz (1.8a) and σx y = x xy xσx x + x yy xσx y + x zy zσx z + xxy yσx x + x yy yσyy + xzy yσyz + xxy zσx z + x yy zσyz + xzy zσzz. (1.8b) x y z x y z Figure 1.4. Two orthogonal coordinate systems.
STRESS AND STRAIN These can be simplified to Ox=Eixox +eiyoy+eo:+2exyer-ty:+2ex'-er'xtsx +2Ex'xex'ytxy (1.9a) and tx'y =Ex'xey'xOx +exyeyyoy +exzly:o:+(ex'yey=+exzeyy)tyz +(erzeyx +ex'xey=)tzx +(exxeyy +ex'yeyx)txy (1.9b) 1.3 PRINCIPAL STRESSES It is always possible to find a set of axes along which the shear stress terms vanish.In this case o1,o2 and o3 are called the principal stresses.The magnitudes of the principal stresses,op,are the roots of o2-4o3-20p-1=0, (1.10) where I1,12 and I3 are called the inariants of the stress tensor.They are I1=Oxx+Oyy十Oz, 12=++diy -dyyd::-d:Oxx -axxoyy and (1.11) h=0x0y0e:十20z020y-axx0.-an0-02:0 The first invariant,I=-p/3 where p is the pressure.I1,12 and I3 are independent of the orientation of the axes.Expressed in terms of the principal stresses they are 1=01+02十03， I2=-0203-0301-0102and (1.12) 3=010203. EXAMPLE 1.1:Consider a stress state withox=70 MPa,oy=35 MPa,t.xy=20,0:= =ty=0.Find the principal stresses using equations 1.10 and 1.11. SOLUTION:Using equations 1.11,/1 105 MPa,12 =-2,050 MPa,/3 =0.From equation1.10,o2-105o3+2,050op+0=0,o3-105op+2,050=0. The principal stresses are the roots,o1=79.1 MPa,o2=25.9 MPa and o3=o:=0. EXAMPLE 1.2:Repeat Example 1.1,with /3 170,700. SOLUTION:The principal stresses are the roots of-65+1750p+170,700=0. Since one of the roots is o==o3 =-40,op+40 =0 can be factored out.This gives o2-1050p+2,050 =0,so the other two principal stresses are o1=79.1 MPa, 02 =25.9 MPa.This shows that when o:is one of the principal stresses,the other two principal stresses are independent of o=
4 STRESS AND STRAIN These can be simplified to σx = 2 xxσx + 2 x yσy + 2 xzσz + 2x yx zτyz + 2x zxx τzx + 2x xx y τx y (1.9a) and τx y = x x y xσx + x yy yσy + x zyzσz + (x yy z + x zy y )τyz + (x zy x + x xy z)τzx + (x xy y + x yy x )τx y (1.9b) 1.3 PRINCIPAL STRESSES It is always possible to find a set of axes along which the shear stress terms vanish. In this case σ1, σ2 and σ3 are called the principal stresses. The magnitudes of the principal stresses, σp, are the roots of σ 3 p − I1σ 2 p − I2σp − I3 = 0, (1.10) where I1, I2 and I3 are called the invariants of the stress tensor. They are I1 = σx x + σyy + σzz, I2 = σ2 yz + σ2 zx + σ 2 x y − σyyσzz − σzzσx x − σx xσyy and (1.11) I3 = σx xσyyσzz + 2σyzσzxσx y − σx xσ 2 yz − σyyσ2 zx − σzzσ 2 x y . The first invariant, I1 = −p/3 where p is the pressure. I1, I2 and I3 are independent of the orientation of the axes. Expressed in terms of the principal stresses they are I1 = σ1 + σ2 + σ3, I2 = −σ2σ3 − σ3σ1 − σ1σ2 and (1.12) I3 = σ1σ2σ3. EXAMPLE 1.1: Consider a stress state with σx = 70 MPa, σy = 35 MPa, τ x y = 20, σz = τzx = τyz = 0. Find the principal stresses using equations 1.10 and 1.11. SOLUTION: Using equations 1.11, I1 = 105 MPa, I2 = −2,050 MPa, I3 = 0. From equation 1.10, σ 3 p − 105σ 2 p + 2,050σp + 0 = 0, σ 2 p − 105σp + 2,050 = 0. The principal stresses are the roots, σ1 = 79.1 MPa, σ2 = 25.9 MPa and σ3 = σz = 0. EXAMPLE 1.2: Repeat Example 1.1, with I3 = 170,700. SOLUTION: The principal stresses are the roots of σ3 p − 65σ2 p + 1750σp + 170,700 = 0. Since one of the roots is σz = σ3 = −40, σp + 40 = 0 can be factored out. This gives σ2 p −105σp + 2,050 = 0, so the other two principal stresses are σ1 = 79.1 MPa, σ2 = 25.9 MPa. This shows that when σz is one of the principal stresses, the other two principal stresses are independent of σz
1.4 MOHR'S CIRCLE EQUATIONS 5 1.4 MOHR'S CIRCLE EQUATIONS In the special cases where two of the three shear stress terms vanish (e.g.,x==0), the stress,o=,normal to the xy plane is a principal stress and the other two principal stresses lie in the xy plane.This is illustrated in Figure 1.5. For these conditions ex'==y=0,tyz=tzx=0,ex'x =ey'y cos and r'=-yx=sin.Substituting these relations into equations 1.9 results in tx'y =cos sin(-0x +y)+(cos2o-sin2)tsy. ox=cos2pox+sin2poy+2cos中sinφtxy,and (1.13) oy=sin2por+cos2pay+2cosφsinφtxw. These can be simplified with the trigonometric relations, sin2φ=2 sin cosφand cos2φ=cos2φ-sin2φto obtain tx'y =-sin 2(ax-dy)/2+cos 2otxy, (1.14a) ox=(ox+ov)/2+cos2o(ox-o)+txysin2φ，and (1.14b) oy=(ox+o,)/2-cos2φ(ox-o)+Uxy sin2中 (1.14c) If is set to zero in equation 1.14a,becomes the angle between the principal axes and the x and y axes.Then tan 20 txy/[(ox -ay)/2]. (1.15) The principal stresses,o and o2,are then the values of ox and oy, 01.2 =(ox +oy)/2 [(ox -d)/cos 20]+txy sin 20 or 2=a,+a,/2±2[a,-a,P+4n. (1.16) yx Figure 1.5.Stress state for which the Mohr's circle equations apply
1.4 MOHR’S CIRCLE EQUATIONS 5 1.4 MOHR’S CIRCLE EQUATIONS In the special cases where two of the three shear stress terms vanish (e.g., τyx = τzx = 0), the stress, σz, normal to the x y plane is a principal stress and the other two principal stresses lie in the x y plane. This is illustrated in Figure 1.5. For these conditions x z = y z = 0, τyz = τzx= 0, x x = y y = cos φ and x y = −y x = sin φ. Substituting these relations into equations 1.9 results in τx y = cos φ sin φ(−σx + σy ) + (cos2 φ − sin2 φ)τx y , σx = cos2 φσx + sin2 φσy + 2 cos φ sin φτx y , and (1.13) σy = sin2 φσx + cos2 φσy + 2 cos φ sin φτx y . These can be simplified with the trigonometric relations, sin 2φ = 2 sin φ cos φ and cos2 φ = cos2 φ − sin2 φ to obtain τx y = − sin 2φ(σx − σy )/2 + cos 2φτ x y , (1.14a) σx = (σx + σy )/2 + cos 2φ(σx − σy ) + τx y sin 2φ, and (1.14b) σy = (σx + σy )/2 − cos 2φ(σx − σy ) + τ x y sin 2φ. (1.14c) If τx y is set to zero in equation 1.14a, φ becomes the angle θ between the principal axes and the x and y axes. Then tan 2θ = τx y/[(σx − σy )/2]. (1.15) The principal stresses, σ1 and σ2, are then the values of σx and σy , σ1,2 = (σx + σy )/2 ± [(σx − σy )/ cos 2θ] + τx y sin 2θ or σ1,2 = (σx + σy )/2 ± (1/2) (σx − σy ) 2 + 4τ 2 x y1/2 . (1.16) z x y y x′ y′ σx σy σy σx τxy τyx τyx τxy x φ Figure 1.5. Stress state for which the Mohr’s circle equations apply
6 STRESS AND STRAIN y (01-022 29 (Gx-Gy2 01 ox好 y Figure 1.6.Mohr's circle diagram for stress. 03 02 y Figure 1.7.Three-dimensional Mohr's circles for stresses. A Mohr's*circle diagram is a graphical representation ofequations 1.15 and 1.16. They form a circle of radius (o-02)/2 and with the center at (o+02)/2 as shown in Figure 1.6.The normal stress components are plotted on the ordinate and the shear stress components are plotted on the abscissa. Using the Pythagorean theorem on the triangle in Figure 1.6, (a-2/2={ax+a,/2+} (1.17) and tan(28)=tx/[(ox+o)/2]. (1.18) A three-dimensional stress state can be represented by three Mohr's circles as shown in Figure 1.7.The three principal stresses o1,o2 and o3 are plotted on the ordinate.The circles represent the stress state in the 1-2,2-3 and 3-1 planes. EXAMPLE 1.3:Construct the Mohr's circle for the stress state in Example 1.2 and determine the largest shear stress. O.Mohr,Zivilingeneur(1882),p.113
6 STRESS AND STRAIN (σx-σy)/2 (σx+σy)/2 σy 2θ σx τxy σ2 σ1 (σ1-σ2)/2 σ τ x y x′ θ Figure 1.6. Mohr’s circle diagram for stress. σ τ σ3 σ2 σ1 σy σx τxy 2θ Figure 1.7. Three-dimensional Mohr’s circles for stresses. A Mohr’s∗ circle diagram is a graphical representation of equations 1.15 and 1.16. They form a circle of radius (σ1 − σ2)/2 and with the center at (σ1 + σ2)/2 as shown in Figure 1.6. The normal stress components are plotted on the ordinate and the shear stress components are plotted on the abscissa. Using the Pythagorean theorem on the triangle in Figure 1.6, (σ1 − σ2)/2 = [(σx + σy )/2]2 + τ 2 x y1/2 (1.17) and tan(2θ) = τx y/[(σx + σy )/2]. (1.18) A three-dimensional stress state can be represented by three Mohr’s circles as shown in Figure 1.7. The three principal stresses σ1, σ2 and σ3 are plotted on the ordinate. The circles represent the stress state in the 1–2, 2–3 and 3–1 planes. EXAMPLE 1.3: Construct the Mohr’s circle for the stress state in Example 1.2 and determine the largest shear stress. ∗ O. Mohr, Zivilingeneur (1882), p. 113