12 STRESS AND STRAIN For elastic loading,substituting o,=Ee,into equation 1.38 and integrating w axex/2 Ee:/2. (1.39) For multiaxial loading dw oxdex +oydey+o-de:ty-dyy:+tzxdy=x txydyxy. (1.40) and if the deformation is elastic, dw =(1/2)(o de +o2de2 +o3de3). (1.41) 1.10 FORCE AND MOMENT BALANCES Many analyses of metal forming operations involve force or moment balances.The net force acting on any portion of a body must be zero.External forces on a portion of a body are balanced by internal forces acting on the arbitrary cut through the body. As an example,find the stresses in the walls of thin wall tube under internal pressure (Figure 1.11).Let the tube length be L,its diameter D and its wall thickness t and let the pressure be P(Figure 1.1la).The axial stress,oy,can be found from a force balance on a cross section of the tube.Since in Px D2/4=x Dtoy, oy=PD/(41) (1.42) The hoop stress,ox,can be found from a force balance on a longitudinal section of the tube (Figure 1.11b).PDL 20tL or 1 ax=PD/t ay PD/(21). (1.43) A moment balance can be made about any axis through a material.The internal moment must balance the external moment.Consider a cylindrical rod under torsion. A moment balance relates the torque,T,to the distribution of shear stress,x(Fig- ure 1.12).Consider an annular element of thickness dr at a distance r from the axis.The shear force on this element is the shear stress times the area of the element,(2r)txdr. The moment caused by this element is the shear force times the distance,r,from the axis so dT=(2xr)txy(r)dr so R T=2π (1.44) Figure 1.11.Forces acting on cuts through a tube under pressure. (a) (b)
12 STRESS AND STRAIN For elastic loading, substituting σx = Eex into equation 1.38 and integrating w = σx ex/2 = Ee2 x/2. (1.39) For multiaxial loading dw = σxdex + σydey + σzdez + τyzdγyz + τzxdγzx + τx ydγx y . (1.40) and if the deformation is elastic, dw = (1/2)(σ1de1 + σ2de2 + σ3de3). (1.41) 1.10 FORCE AND MOMENT BALANCES Many analyses of metal forming operations involve force or moment balances. The net force acting on any portion of a body must be zero. External forces on a portion of a body are balanced by internal forces acting on the arbitrary cut through the body. As an example, find the stresses in the walls of thin wall tube under internal pressure (Figure 1.11). Let the tube length be L, its diameter D and its wall thickness t and let the pressure be P (Figure 1.11a). The axial stress, σy, can be found from a force balance on a cross section of the tube. Since in Pπ D2/4 = π Dtσy , σy = P D/(4t). (1.42) The hoop stress, σx, can be found from a force balance on a longitudinal section of the tube (Figure 1.11b). PDL = 2σx t L or σx = 1 2 P D/t σy = P D/(2t). (1.43) A moment balance can be made about any axis through a material. The internal moment must balance the external moment. Consider a cylindrical rod under torsion. A moment balance relates the torque, T, to the distribution of shear stress, τ x y (Figure 1.12). Consider an annular element of thickness dr at a distance rfrom the axis. The shear force on this element is the shear stress times the area of the element, (2πr)τ x ydr. The moment caused by this element is the shear force times the distance, r, from the axis so dT = (2πr)τ x y (r)dr so T = 2π R 0 τx yr 2 dr. (1.44) P σy σy t D D P σx σx t (a) (b) Figure 1.11. Forces acting on cuts through a tube under pressure
1.11 BOUNDARY CONDITIONS 13 Figure 1.12.Moment balance on an annular element. An explicit solution requires knowledge of how txy varies with r for inte- gration. 1.11 BOUNDARY CONDITIONS In analyzing metal forming problems,it is important to be able to recognize boundary conditions.Often these are not stated explicitly.Some of these are listed below: 1.A stress,o.,normal to a free surface and the two shear stresses in the surface are zero. 2.Likewise there are no shear stresses in surfaces that are assumed to be frictionless. 3.Constraints from neighboring regions:The strains in a region are often controlled by the deformation in a neighboring region.Consider a long narrow groove in a plate (Figure 1.13.)The strain ex,in the groove must be the same as the strain in the region outside the groove.However,the strains g,and s=need not be the same. 4. Saint-Venant's principle states that the constraint from discontinuity will disappear within one characteristic distance of the discontinuity.For example,the shoulder on a tensile bar tends to suppress the contraction of the adjacent region of the gauge section.However this effect is very small at a distance equal to the diameter away from the shoulder.Figure 1.14 illustrates this on sheet specimen. Bending ofa sheet(Figure 1.15)illustrates another example of Saint-Venant's principle. The plane-strain condition g=0 prevails over most of the material because the bottom and top surfaces are so close.However,the edges are not in plane strain because o= 0.However,there is appreciable deviation from plane strain only in a region within a distance equal to the sheet thickness from the edge. EXAMPLE 1.9:A metal sheet,1 m wide,3 m long and 1 mm thick is bent as shown in Figure 1.15.Find the state of stress in the surface in terms of the elastic constants and the bend radius,p. Figure 1.13.Grooved plate.The material outside the groove affects the material inside the groove,sxA =sxB
1.11 BOUNDARY CONDITIONS 13 R dr τxy x y z r Figure 1.12. Moment balance on an annular element. An explicit solution requires knowledge of how τ x y varies with r for integration. 1.11 BOUNDARY CONDITIONS In analyzing metal forming problems, it is important to be able to recognize boundary conditions. Often these are not stated explicitly. Some of these are listed below: 1. A stress, σz, normal to a free surface and the two shear stresses in the surface are zero. 2. Likewise there are no shear stresses in surfaces that are assumed to be frictionless. 3. Constraints from neighboring regions: The strains in a region are often controlled by the deformation in a neighboring region. Consider a long narrow groove in a plate (Figure 1.13.) The strain εx, in the groove must be the same as the strain in the region outside the groove. However, the strains εy and εz need not be the same. 4. Saint-Venant’s principle states that the constraint from discontinuity will disappear within one characteristic distance of the discontinuity. For example, the shoulder on a tensile bar tends to suppress the contraction of the adjacent region of the gauge section. However this effect is very small at a distance equal to the diameter away from the shoulder. Figure 1.14 illustrates this on sheet specimen. Bending of a sheet (Figure 1.15) illustrates another example of Saint-Venant’s principle. The plane-strain condition εy = 0 prevails over most of the material because the bottom and top surfaces are so close. However, the edges are not in plane strain because σy = 0. However, there is appreciable deviation from plane strain only in a region within a distance equal to the sheet thickness from the edge. EXAMPLE 1.9: A metal sheet, 1 m wide, 3 m long and 1 mm thick is bent as shown in Figure 1.15. Find the state of stress in the surface in terms of the elastic constants and the bend radius, ρ. A x y z B Figure 1.13. Grooved plate. The material outside the groove affects the material inside the groove, εxA = εxB
14 STRESS AND STRAIN 0.1 0.05 一X 0 0.5 1.5 Normalized distance from end,x/w Figure 1.14.The lateral-contraction strain of a sheet tensile specimen of copper as a function to the distance from the shoulder.The strain was measured when the elongation was 27.6%. Figure 1.15.In bending of a sheet,plane-strain(ey =0)prevails except within a distance equal to the thickness from the edges where ay=0. SOLUTION:e =(1/E)[oy-v(a:+ax)]=0 and o:=0,so ay vox.Neglecting any shift of the neutral plane,ex=t/(2p).Substituting into Hooke's law, ex =1/(2p)=(1/E)[ax-v(ay +a:)]or t/(2p)=(ax/E)(1-v2). Solving for o Et vEt Ox= 2p(1-u2) amdy=2p1-u° NOTES OF INTEREST Otto Mohr(1835-1918)made popular the graphical representation of stress at a point (Civiling,1882,p.113)even though it had previously been suggested by Culman (Graphische Statik,1866,p.226). Barre de Saint-Venant was born 1797.In 1813 at the age of 16 he entered L'Ecole Polytechnique.He was a sergeant on a student detachment as the allies were attacking Paris in 1814.Because he stepped out of ranks and declared that he could not in good conscience fight for a usurper(Napoleon),he was prevented from taking further classes at L'Ecole Polytechnique.He later graduated from L'Ecole des Ponts et Chaussees where he taught through his career
14 STRESS AND STRAIN Figure 1.14. The lateral-contraction strain of a sheet tensile specimen of copper as a function to the distance from the shoulder. The strain was measured when the elongation was 27.6%. y x z Figure 1.15. In bending of a sheet, plane-strain (εy = 0) prevails except within a distance equal to the thickness from the edges where σy = 0. SOLUTION: ey = (1/E)[σy − υ(σz + σx)] = 0 and σz = 0, so σy = υσx. Neglecting any shift of the neutral plane, ex = t/(2ρ). Substituting into Hooke’s law, ex = t/(2ρ) = (1/E)[σx − υ(σy + σz)] or t/(2ρ) = (σx/E)(1 − υ2 ). Solving for σ σx = Et 2ρ(1 − υ2) and σy = υEt 2ρ(1 − υ2) . NOTES OF INTEREST Otto Mohr (1835–1918) made popular the graphical representation of stress at a point (Civiling, 1882, p. 113) even though it had previously been suggested by Culman (Graphische Statik, 1866, p. 226). Barre de Saint-Venant was born 1797. In 1813 at the age of 16 he entered L’ ´ Ecole ´ Polytechnique. He was a sergeant on a student detachment as the allies were attacking Paris in 1814. Because he stepped out of ranks and declared that he could not in good conscience fight for a usurper (Napoleon), he was prevented from taking further classes at L’Ecole Polytechnique. He later graduated from L’ ´ Ecole des Ponts et Chauss ´ ees ´ where he taught through his career
PROBLEMS 15 Gy+(Oayl ay)dy tyx+(otyxloy)dy txy+(otxyl ax)dx Figure 1.16.Variation of a stress state in space. x x+(Oo ox)dx Gy REFERENCES R.M.Caddell,Deformation and Fracture of Solids,Prentice-Hall,1980. H.Ford,Advanced Mechanics of Materials,Wiley,1963. W.F.Hosford,Mechanical Behavior of Materials,Cambridge University Press,2004. W.Johnson and P.B.Mellor,Engineering Plasticity,Van Nostrand-Reinhold,1973. N.H.Polokowski and E.J.Ripling,Strength and Stucture of Engineering Materials, Prentice-Hall,1966. APPENDIX-EQUILIBRIUM EQUATIONS As the stress state varies from one place to another,there are equilibrium conditions, which must be met.Consider Figure 1.16. An x-direction force balance gives Ox txy =0x+80x/ax +txy +8txy/8y or simply dox/8x+txy/ay =0. (1.45) In three dimensions aox/ax+atxy/8y+ty:/0z=0 atxv/ox +00v/0y+atv/0z=0 (1.46) atx=/0x +atyz/0y+00:/0z=0. These equations are used in Chapter 9. PROBLEMS 1.1.Determine the principal stresses for the stress state 10-34 -3 52 27
PROBLEMS 15 σx σy σy + (∂ σy/ ∂y)dy σx + (∂ σx/ ∂x)dx τyx τxy τyx + (∂ τyx/ ∂ y)dy τxy + (∂ τxy/ ∂ x)dx dx Figure 1.16. Variation of a stress state in space. dy REFERENCES R. M. Caddell, Deformation and Fracture of Solids, Prentice-Hall, 1980. H. Ford, Advanced Mechanics of Materials, Wiley, 1963. W. F. Hosford, Mechanical Behavior of Materials, Cambridge University Press, 2004. W. Johnson and P. B. Mellor, Engineering Plasticity, Van Nostrand-Reinhold, 1973. N. H. Polokowski and E. J. Ripling, Strength and Stucture of Engineering Materials, Prentice-Hall, 1966. APPENDIX – EQUILIBRIUM EQUATIONS As the stress state varies from one place to another, there are equilibrium conditions, which must be met. Consider Figure 1.16. An x-direction force balance gives σx + τx y = σx + ∂σx/∂x + τx y + ∂τx y/∂y or simply ∂σx/∂x + ∂τx y/∂y = 0. (1.45) In three dimensions ∂σx/∂x + ∂τx y/∂y + ∂τyz/∂z = 0 ∂τx y/∂x + ∂σy/∂y + ∂τyz/∂z = 0 (1.46) ∂τx z/∂x + ∂τyz/∂y + ∂σz/∂z = 0. These equations are used in Chapter 9. PROBLEMS 1.1. Determine the principal stresses for the stress state σi j = 10 −3 4 −3 52 4 27 .
16 STRESS AND STRAIN 1.2.A 5-cm-diameter solid shaft is simultaneously subjected to an axial load of 80 kN and a torque of 400 Nm. a)Determine the principal stresses at the surface assuming elastic behavior. b)Find the largest shear stress. 1.3.A long thin-wall tube,capped on both ends,is subjected to internal pressure. During elastic loading,does the tube length increase,decrease,or remain con- stant? 1.4.A solid 2-cm-diameter rod is subjected to a tensile force of 40 kN.An identical rod is subjected to a fluid pressure of 35 MPa and then to a tensile force of 40 kN.Which rod experiences the largest shear stress? 1.5.Consider a long,thin-wall,5-cm-diameter tube,with a wall thickness of 0.25 mm that is capped on both ends.Find the three principal stresses when it is loaded under a tensile force of 400 kN and an internal pressure of 200 kPa. 1.6.Three strain gauges are mounted on the surface of a part.Gauge A is parallel to the x-axis,and gauge C is parallel to the y-axis.The third gauge,B,is at 30 to gauge A.When the part is loaded,the gauges read Gauge A 3,000×10-6 Gauge B 3,500×10-6 Gauge C 1,000×10-6 a)Find the value of yxy. b)Find the principal strains in the plane of the surface. c)Sketch the Mohr's circle diagram. 1.7.Find the principal stresses in the part of problem 1.6 if the elastic modulus of the part is 205 GPa and Poisson's ratio is 0.29. 1.8.Show that the true strain after elongation may be expressed as s=In(), where r is the reduction of area. 1.9.A thin sheet of steel,1-mm thick,is bent as described in Example 1.9.Assuming that E=is 205 GPa and v =0.29,and that the neutral axis doesn't shift, a)Find the state of stress on most of the outer surface. b)Find the state of stress at the edge of the outer surface. 1.10.For an aluminum sheet,under plane stress loading &r =0.003 and y =0.001. Assuming that E=is 68 GPa and v=0.30,find z. 1.11.A piece of steel is elastically loaded under principal stresses,o1=300 MPa, o2 250 MPa,and o3 =-200 MPa.Assuming that E=is 205 GPa and v 0.29,find the stored elastic energy per volume. 1.12.A slab of metal is subjected to plane-strain deformation (e2 =0)such that o1 40 ksi and o3 =0.Assume that the loading is elastic,and that E=is 205 GPa, and v =0.29 (note the mixed units).Find a)the three normal strains. b)the strain energy per volume
16 STRESS AND STRAIN 1.2. A 5-cm-diameter solid shaft is simultaneously subjected to an axial load of 80 kN and a torque of 400 Nm. a) Determine the principal stresses at the surface assuming elastic behavior. b) Find the largest shear stress. 1.3. A long thin-wall tube, capped on both ends, is subjected to internal pressure. During elastic loading, does the tube length increase, decrease, or remain constant? 1.4. A solid 2-cm-diameter rod is subjected to a tensile force of 40 kN. An identical rod is subjected to a fluid pressure of 35 MPa and then to a tensile force of 40 kN. Which rod experiences the largest shear stress? 1.5. Consider a long, thin-wall, 5-cm-diameter tube, with a wall thickness of 0.25 mm that is capped on both ends. Find the three principal stresses when it is loaded under a tensile force of 400 kN and an internal pressure of 200 kPa. 1.6. Three strain gauges are mounted on the surface of a part. Gauge A is parallel to the x-axis, and gauge C is parallel to the y-axis. The third gauge, B, is at 30◦ to gauge A. When the part is loaded, the gauges read Gauge A 3,000 × 10−6 Gauge B 3,500 × 10−6 Gauge C 1,000 × 10−6 a) Find the value of γx y . b) Find the principal strains in the plane of the surface. c) Sketch the Mohr’s circle diagram. 1.7. Find the principal stresses in the part of problem 1.6 if the elastic modulus of the part is 205 GPa and Poisson’s ratio is 0.29. 1.8. Show that the true strain after elongation may be expressed as ε = ln( 1 1−r ), where r is the reduction of area. 1.9. A thin sheet of steel, 1-mm thick, is bent as described in Example 1.9. Assuming that E = is 205 GPa and ν = 0.29, and that the neutral axis doesn’t shift, a) Find the state of stress on most of the outer surface. b) Find the state of stress at the edge of the outer surface. 1.10. For an aluminum sheet, under plane stress loading εx = 0.003 and εy = 0.001. Assuming that E = is 68 GPa and ν = 0.30, find εz. 1.11. A piece of steel is elastically loaded under principal stresses, σ1 = 300 MPa, σ2 = 250 MPa, and σ3 = −200 MPa. Assuming that E = is 205 GPa and ν = 0.29, find the stored elastic energy per volume. 1.12. A slab of metal is subjected to plane-strain deformation (e2 = 0) such that σ1 = 40 ksi and σ3 = 0. Assume that the loading is elastic, and that E = is 205 GPa, and ν = 0.29 (note the mixed units). Find a) the three normal strains. b) the strain energy per volume