Problem Set 2 Solutions

Transcription

1 M 4733: Deformation and Fracture of ngineering Materials Spring 00 Problem Set Solutions *) Prove that for the case of cubic crystals the modulus of elasticity in any given direction may be given by the following equation in terms of the three independent elastic constant s, s, s 44, and the direction cosines of the crystallographic direction under study: = s s s ( ) s ( ll + ll + ll) where l, l, l 3 are direction cosines As shown in the following figure, the axis direction is n, the three orthogonal crystal directions are e x = [ 00 ], e y = [ 00 ] and e z = [ 00 ] The three direction cosines are l i = n e i, and n can be expressed in the crystal coordinates as n= l i e i The stress state is σ = σnn = σll i j e i e j, where σ is the applied tensile stress The generated strain is ε = ε ij ee i j The normal tensile strain in axis n direction is ε= n ε n= liεijlj For cubic crystal, the elastic compliance matrix gives ε = s σ + s σ + s σ 33

2 ε = s σ + s σ + s σ ε = s σ + s σ + s σ s44σ ε = ε = γxy = s44σ 3 ε3 = ε3 = γyz = s44σ3 ε3 = ε3 = γzx = ε i j ij llε l ( sσ + sσ + sσ33) l ( sσ + sσ + sσ33) = = = + + σ σ σ σ l3 ( sσ + sσ + sσ33) s44( ll σ + ll3σ3 + l3l σ3) + σ σ = s ( l + l + l ) + ( s + s )( ll + ll + ll ) With l + l + l =, we get 3 = s s s ( ) s44 ll + ll 3 + ll 3 ( )

3 ) Hertzberg, 5 Calculate the elastic modulus for gold in <0> direction and compare your results with the 00 and values reported in Table 4 and the polycrystalline isotropic value given in Table Gold is a cubic material So we can use = s s s ( ) s ll + ll + ll where l, l, l 3 are direction cosines ( ) From Hertzberg Table 3, for gold s = 33 0 Pa, s = 07 0 Pa, = 38 0 Pa s 44 For [0] direction, l = l =, l 3 = 0 So = ( + ) 0 0 Pa = 5 0 = 8 6GPa 0 Pa From Hertzberg Table 4, for gold = 6 7GPa, 00 = 49 GPa So we have > 0 > 00 From Hertzberg Table, for isotropic polycrystalline gold isotropic = 78 0GPa 0 is closer to isotropic than and 00

4 3) Hertzberg, 0 Compute the modulus of elasticity for copper and aluminum single crystals in the <00>, <0>, and <> directions Compare these values with the modulus reported for polycrystalline samples of these two materials Both copper and aluminum are cubic materials So we can use = s s s ( ) s ( ll + ll + ll) where l, l, l 3 are direction cosines From Hertzberg Table 3, s (0 Pa - ) s (0 Pa - ) s 44 (0 Pa - ) Cu Al Copper: For <00>, <0> and <>, ll + ll + ll ( 3 3 ) are 0, 4, and 3, respectively = 50 0 Pa - 00 = 66 7GPa = Pa - 0 = 30 3GPa = Pa - = 9 GPa > 0 > 00 From Hertzberg Table, for isotropic polycrystalline copper isotropic =9 8GPa 0 is closer to isotropic than and 00 Aluminum: = 57 0 Pa - 00 = 63 7GPa = Pa = 7 6GPa = + Pa - = 76 GPa isotropic > 0 > 00 From Hertzberg Table, for isotropic polycrystalline aluminum = 70 3GPa 0 is closer to isotropic than and 00

5 4) Hertzberg, 8 For three BCC metals, tungsten, molybdenum, and iron, compute the elastic moduli in the <00> and <> directions Compare the anisotropy in these three metals From Hertzberg Table 3, s (0 Pa - ) s (0 Pa - ) s 44 (0 Pa - ) tungsten molybdenum iron For tungsten: ( s s ) s = + 066= 0 tungsten is isotropic and 00 = = = 3846 GPa s For molybdenum: 00 = = 357 GPa s = + Pa - = 9GPa 00 = 7 For iron: 00 = = 5GPa s = + Pa - = 7GPa 7 00 = 0458

6 5) What s the angle between fracture surface and axial direction for a ductile round rod under the pure torsion at two ends, why? The round rod gets the maximum shear stress at the surface side of the circular cross section According to Tresca s criterion for the failure of ductile material, the fracture of the round rod under pure torsion goes from the surface to inside along the cross-section perpendicular to the axis

7 6) Consider a face-centered cubic material with a tensile stress parallel to <00>, <0>, and <> types of directions How many initial slip systems operate for these respective cases? For FCC, the slip system is {}<0>, which includes 4 slip planes and 3 <0> directions in each plane The following picture shows the 4 slip plane, three slip directions are located on the edges of the triangles How many initial slip systems operate means that for the slip systems, how many of them will operate when the resolved shearing stress first reaches the critical value This number depends on how many slip systems provide the maximum Schmid factor cosφ cosλ, where φ is the angle between the rod axis and the normal to the slip plane, and λ is the angle between load axis and slip direction A) Tensile stress in [00] direction, cosφ = 3 for ( ), ( ), ( ), ( ) planes cosλ = for [ 0 ], and [ 0 ] directions, cosλ = 0 for [ 0 ] direction

8 There are 4 = 8 initial slip systems, that gives the maximum Schmid factor of cosφ cosλ = 6 B) When tensile stress in [0] direction, cosφ = 6 for ( ), ( ) planes; cosφ = 0 for ( ), ( ) planes Consider () plane, cosλ = for [ 0 ] and [ 0 ] directions; cosλ = 0 for [ 0 ] direction The situation is similar to ( ) plane So there are = 4 initial slip systems that gives the maximum Schmid factor of cosφ cosλ = 6 C) When tensile stress in [] direction, cosφ = for ( ) plane; cosφ = 3 for ( ), ( ), ( ) planes Consider () plane, cosλ = 0 for [ 0 ], [ 0 ], and [ 0 ] directions Consider ( ) plane, cosλ = 3 for [ 0 ] and [ 0 ] directions cosλ = 0 for [ 0 ] direction The situation is similar to ( ) and ( ) planes So, there are 3 = 6 initial slip systems, that gives the maximum Schmid factor of cosφ cosλ = 3 3

9 7) The reversibility nature of elastic deformation behavior is that if the loads are removed from the specimen, the corresponding strain will retrace itself along the same plot back to zero For a general linear material, the six independent stress components determine the strains of the body The strains in each direction can be given by ε x s s s s s s εy s s s s s s ε z s s s s s s = γ xy s s s s s s γ yz s s s s s s γ zx s s s s s s σ x σ y σ z τ xy τ yz τ zx The reversibility of elastic strains leads to the fact that s,,3,4,5,6) Prove this ij = s, ( i j, i,j = ji The reversibility means that only there exists only one energy-state for a given stress state non-reversible deformation reversible deformation In the left pictures, after a path from O A P B O, the strain energy left in the body is the shade of the figure So, for a stress level, there may be several associated strain energy state In a reversible deformation, the strain-energy state and the stress level have a one to one correspondence Apply {σ x, σ y, 0, 0, 0, 0} to a elastic body, Method I, first apply σ x, then applyσ y ; Method II, first apply σ y, then apply σ x

10 There strain energy should be the same, because the final stress states are the same For method I, after applied σ x, the stored strain energy is I = sσ x after applied σ y, the increased strain energy is I = ( sσy) σx + sσy For method II, after applied σ y, the stored strain energy is I = sσ y after applied σ x, the increased strain energy is I = ( sσx) σy + sσx Reversibility requires that I + I = II + II This gives s = s With the similarly approach, we can prove s = s, ( i j, i,j =,,3,4,5,6) ij ji