AllChaptersCovered
z z
SOLUTIONMANUAL
z
, Solution Manual3rd Ed. MetalForming: Mechanics and Metallurgy z z z z z z z z
Chapter 1 z z
Determine the principal stresses for the stress state z z z z z z z
10 –3 4
σij = –3 5 2 .
z
z
z z
4 2 7
Solution: I1 = 10+5+7=32, I2 = -(50+35+70) +9 +4 +16 = -126, I3 = 350 -48 -40 -80 z z z z z z z z z z z z z z z z
-63 = 119; σ – 22σ2 -126σ -119 = 0. A trial and error solution gives σ -= 13.04.
3
z z z z z z z z z z z z z z z z z z z
◻ Factoring out 13.04, σ2 -8.96σ + 9.16 = 0. Solving; σ1 = 13.04, σ2 = 7.785, σ3 =
z z z z z z z z z z z z z z z z
1.175.
z
1-2 A 5-cm. diameter solid shaft is simultaneously subjected to an axial load of 80 kN and a
z z z z z z z z z z z z z z z z
torque of 400 Nm.
z z z z
a. Determine the principal stresses at the surface assuming elastic behavior. z z z z z z z z z
b. Find the largest shear stress. z z z z
Solution: a. The shear stress, τ, at a radius, r, is τ = τsr/R where τsis theshearstress at the surface
z z z z z z z z z z z z z z z z z z z z z
R is the radius of the rod. The torque, T, is given by T = ∫2πtr2dr = (2πτs /R)∫r3dr
z z z z z z z z z z z z z z z z z z z
= πτsR3/2. Solving for = τs, τs = 2T/(πR3) = 2(400N)/(π0.0253) = 16 MPa The
z z z z z z z z z z z z z z
axial stress is .08MN/(π0.0252) = 4.07 MPa
z z z z z z z
σ1,σ2 = 4.07/2 ± [(4.07/2)2 + (16/2)2)]1/2 = 1.029, -0.622 MPa
z z z z z z z z z z
b. the largest shear stress is (1.229 + 0.622)/2 = 0.925 MPa
z z z z z z z z z z z
A long thin-wall tube, capped on both ends is subjected to internal pressure. During elastic
z z z z z z z z z z z z z z
loading, does the tube length increase, decrease or remain constant?
z z z z z z z z z z
Solution: Let y= hoop direction, x = axial direction, and z = radial direction. – ex = e2
z z z z z z z z z z z z z z z z
z
z
= (1/E)[σ - v( σ3 + σ1)] = (1/E)[σ2 - v(2σ2)] = (σ2/E)(1-2v)
z
z z z z
z
z z z
z
z z z
Since u < 1/2 for metals, ex = e2 is positive and the tube lengthens.
z z z z z z
z
z
z
z z z z z
4 A solid 2-cm. diameter rod is subjected to a tensile force of 40 kN. An identical rod is
z z z z z z z z z z z z z z z z z
subjected to a fluid pressure of 35 MPa and then to a tensile force of 40 kN. Which rod
z z z z z z z z z z z z z z z z z z z
experiences the largest shear stress?
z z z z z
Solution: The shear stresses in both are identical because a hydrostatic pressure has no shear
z z z z z z z z z z z z z z
component.
z
1-5 Consider a long thin-wall, 5 cm in diameter tube, with a wall thickness of 0.25 mm z z z z z z z z z z z z z z z
that is capped on both ends. Find the three principal stresses when it is loaded under a tensile
z z z z z z z z z z z z z z z z z z
force of 40 N and an internal pressure of 200 kPa.
z z z z z z z z z z z
Solution: σx = PD/4t + F/(πDt) = 12.2 MPa z z z z z z z z
σy = PD/2t = 2.0 MPa z z z z z
σy = 0 z z
1
,1-6 Three strain gauges are mounted on the surface of a part. Gauge A is parallel to the x- z z z z z z z z z z z z z z z z z
axis and gauge C is parallel to the y-axis. The third gage, B, is at 30° to gauge A. When the
z z z z z z z z z z z z z z z z z z z z
part is loaded the gauges read
z z z z z z
Gauge A 3000x10-6 z
Gauge B 3500 x10-6 z z
Gauge C 1000 x10-6 z z
a. Find the value of γxy. z z z z
b. Find the principal strains in the plane of the surface. z z z z z z z z z
c. Sketch the Mohr’s circle diagram. z z z z
Solution: Let the B gauge be on the x’ axis, the A gauge on the x-axis and the C gauge on
z z z z z z z z z z z z z z z z z z z z
2 2
they-axis. ex x =exxℓ x x +e ℓ x y +γxyℓx yyxℓx y, where ℓx x =cosex = 30=√3/2and ℓx y=
z z z z z z z z z z z z z z z z
z
z z z z z z z z z
cos 60 = ½. Substituting the measured strains, 3500 =
z z z z z z z z z
3000(√2/3)2 – 1000(1/2)2 + γxy(√3/2)(1/2)
z z z z z
2 2 -6
γ◻xy=(4/√3/2){3500-[3000–(1000(√3/2) +1 000(1/2) ]} =2,309(x10 )
z z z z z
1/2 2
b. e1,e2 = (ex +ey)/2± [(ex-ey)2 + γxy2]
z z /2 = (3000+1000)/2 ± [(3000-1000) +
z z z z z z z z z z
23092]1/2/2.e1 = 3530(x10-6), e2 = 470(x10-6), e3 = 0. z z z z z z z z z
c)
γ/2
sx
s2 s1
s
sx’
sy
Find the principal stresses in the part of problem 1-6 if the elastic modulus ofthe part is 205
z z z z z z z z z z z z z z z z z z
GPa and Poissons’s ratio is 0.29.
z z z z z z
Solution: e3 = 0 = (1/E)[0- v(σ1+σ2)], σ1= σ2 z z z z z z z z z z z
e1 = (1/E)(σ1 - v σ1); σ1 = Ee1/(1-v) = 205x109(3530x10-6)/(1-.292) = 79 MPa
z z z z z z z z z z z z z
1
Show that the true strain after elongation may be expressed as s =ln(
z z z z z z z z z z z z z z
) where r is the
1–r
z z z z
z z
1
reduction of area. s =ln( ).
z
z z z z z z
1–r z z
Solution: r = (Ao-A1)/Ao =1 – A1/Ao = 1 – Lo/L1. s = ln[1/(1-r)]
◻
z z z z z z z z z z z z z
A thin sheet of steel, 1-mm thick, is bent as described in Example 1-11. Assuming that E
z z z z z z z z z z z z z z z z
=is205 GPaand v=0.29,p=2.0 mandthattheneutralaxis doesn’t shift.
z z z z z z z z z z z z z z z z z
a. Find the state of stress on most of the outer surface. z z z z z z z z z z
b. Find the state of stress at the edge of the outer surface. z z z z z z z z z z z
2
, Solution: a. Substituting E = 205x109, t = 0.001, p = 2.0 and v = 0.29 z z z z z z z z z z z z z z z
Et vEt
intoσx = and σy = ,σx=56MPa,,σy=16.2MPa
2p(1–v )2
z z
2p(1–v2)
z
z z z z z z z z z z
z
z z z z z
z z z z
vEt
b. Now σy = 0, so σy =
z = 51 MPa z z z z z z z z
2p
1-10 For an aluminum sheet, under plane stress loading sx = 0.003 and sy = 0.001. z z z z z z z z z z z z z z
Assumin gthatE=is68GPaand v=0.30,find sz. z z z z z z z z z z z z
Solution: ey = (1/E)(σy-vσy), ex = (1/E)(σx – vEey – v2σx). Solving for σx, z z z z z z z z z z z z z
σx = [E/(1-v2)]ey+ vey). Similarly, σy = [E/(1-v2)](ey+ vex). Substituting into
z z z z z z z z z z z z
ez = (1/E)(-vσy-vσy) = (-v /E)(E/(1-v2)[ey+ vey+ ey+ vex ) = [-v(1+v)//(1-v2)](ey+ey) = 0.29(-
z z z z z z z z z z z z z z z z z
1.29/0.916)(0.004) = -0.00163 z z
1-11 A piece of steel is elastically loaded under principal stresses, σ1 = 300 MPa, σ2 = 250z z z z z z z z z z z z z z z z
MPa and σ3 = -200 MPa. Assuming that E = is 205 GPa and v = 0.29 find the stored elastic
z z z z z z z z z z z z z z z z z z z z z
energy per volume.
z z z
Solution: w = (1/2)(σ1e1 + σ2e2 + σ3e3). Substituting e1 = (1/E)[σ1 - v(σ2 + σ3)], e2 = z z z z z z z z z z z z z z z z z
(1/E)[σ2 - v(σ3 + σ1)] and e3 = (1/E)[σ3 - v(σ1 + σ2)],
w = 1/(2E)[σ 2 + σ 2 + σ 2 - 2v(σ σ +σ σ +σ σ )] =
z z z z z z z z z z z z z
z z z z z z z z z z z z z z z z z z
1 2 3 2 z 3 3 z 1 1 z 2
(1/(2x205x10 )[300 +250 + 200 –(2x0.29)(-200x250 – 300x250 + 250+300)]x1012 = 9 2
z
2
z z
2
z z z z z z
400J/m3
z
1-12 A slab of metal is subjected to plane-strain deformation (e2=0) such that σ1= 40 ksi z z z z z z z z z z z z z z z
and σ3 = 0. Assume that the loading is elastic and z z z z z z z z z z z
that E = is 205 GPa and v = 0.29 (Note the mixed units.) Find
z z z z z z z z z z z z z z
a. the three normal strains. z z z
b. the strain energy per volume. z z z z
Solution: w = (1/2)(σ1e1 + σ2e2 + σ3e3) = (1/2)(σ1e1 + 0 + 0) = σ1e1/2 σ1 = z z z z z z z z z z z z z z z z z
40ksi(6.89MPa/ksi) = 276 MPa
z z z z
0 = e2 = (1/E)[σ2 -v σ1], σ2 =v σ1= 0.29x276 = 80 MPa
z z z z z z z z z z z z z z
e1 = (1/E)(σ1 -v σ2) =(1/205x103)[276-.29(80)] = 0.00121 w
z z z z z z z z
= (276x106)(0.00121)/2 = 167 kJ/m3
z z z z z
Chapter 2 z
a) If the principal stresses on a material with a yield stress in shear of 200 MPa are σ2
z z z z z z z z z z z z z z z z z
= 175 MPa and σ1 = 350 MPa., what is the stress, σ3, at yielding according to the Tresca
z z z z z z z z z z z z z z z z z z
criterion?
z
b) If the stresses in (a) were compressive, what tensile stress σ3 must be applied to cause
z z z z z z z z z z z z z z z
yielding according to the Tresca criterion?
z z z z z z
Solution: a) σ1 - σ3 = 2k, σ3 = 2k – σ1 = 400 - 350 = 50 MPa. z z z z z z z z z z z z z z z z z z
3
z z
SOLUTIONMANUAL
z
, Solution Manual3rd Ed. MetalForming: Mechanics and Metallurgy z z z z z z z z
Chapter 1 z z
Determine the principal stresses for the stress state z z z z z z z
10 –3 4
σij = –3 5 2 .
z
z
z z
4 2 7
Solution: I1 = 10+5+7=32, I2 = -(50+35+70) +9 +4 +16 = -126, I3 = 350 -48 -40 -80 z z z z z z z z z z z z z z z z
-63 = 119; σ – 22σ2 -126σ -119 = 0. A trial and error solution gives σ -= 13.04.
3
z z z z z z z z z z z z z z z z z z z
◻ Factoring out 13.04, σ2 -8.96σ + 9.16 = 0. Solving; σ1 = 13.04, σ2 = 7.785, σ3 =
z z z z z z z z z z z z z z z z
1.175.
z
1-2 A 5-cm. diameter solid shaft is simultaneously subjected to an axial load of 80 kN and a
z z z z z z z z z z z z z z z z
torque of 400 Nm.
z z z z
a. Determine the principal stresses at the surface assuming elastic behavior. z z z z z z z z z
b. Find the largest shear stress. z z z z
Solution: a. The shear stress, τ, at a radius, r, is τ = τsr/R where τsis theshearstress at the surface
z z z z z z z z z z z z z z z z z z z z z
R is the radius of the rod. The torque, T, is given by T = ∫2πtr2dr = (2πτs /R)∫r3dr
z z z z z z z z z z z z z z z z z z z
= πτsR3/2. Solving for = τs, τs = 2T/(πR3) = 2(400N)/(π0.0253) = 16 MPa The
z z z z z z z z z z z z z z
axial stress is .08MN/(π0.0252) = 4.07 MPa
z z z z z z z
σ1,σ2 = 4.07/2 ± [(4.07/2)2 + (16/2)2)]1/2 = 1.029, -0.622 MPa
z z z z z z z z z z
b. the largest shear stress is (1.229 + 0.622)/2 = 0.925 MPa
z z z z z z z z z z z
A long thin-wall tube, capped on both ends is subjected to internal pressure. During elastic
z z z z z z z z z z z z z z
loading, does the tube length increase, decrease or remain constant?
z z z z z z z z z z
Solution: Let y= hoop direction, x = axial direction, and z = radial direction. – ex = e2
z z z z z z z z z z z z z z z z
z
z
= (1/E)[σ - v( σ3 + σ1)] = (1/E)[σ2 - v(2σ2)] = (σ2/E)(1-2v)
z
z z z z
z
z z z
z
z z z
Since u < 1/2 for metals, ex = e2 is positive and the tube lengthens.
z z z z z z
z
z
z
z z z z z
4 A solid 2-cm. diameter rod is subjected to a tensile force of 40 kN. An identical rod is
z z z z z z z z z z z z z z z z z
subjected to a fluid pressure of 35 MPa and then to a tensile force of 40 kN. Which rod
z z z z z z z z z z z z z z z z z z z
experiences the largest shear stress?
z z z z z
Solution: The shear stresses in both are identical because a hydrostatic pressure has no shear
z z z z z z z z z z z z z z
component.
z
1-5 Consider a long thin-wall, 5 cm in diameter tube, with a wall thickness of 0.25 mm z z z z z z z z z z z z z z z
that is capped on both ends. Find the three principal stresses when it is loaded under a tensile
z z z z z z z z z z z z z z z z z z
force of 40 N and an internal pressure of 200 kPa.
z z z z z z z z z z z
Solution: σx = PD/4t + F/(πDt) = 12.2 MPa z z z z z z z z
σy = PD/2t = 2.0 MPa z z z z z
σy = 0 z z
1
,1-6 Three strain gauges are mounted on the surface of a part. Gauge A is parallel to the x- z z z z z z z z z z z z z z z z z
axis and gauge C is parallel to the y-axis. The third gage, B, is at 30° to gauge A. When the
z z z z z z z z z z z z z z z z z z z z
part is loaded the gauges read
z z z z z z
Gauge A 3000x10-6 z
Gauge B 3500 x10-6 z z
Gauge C 1000 x10-6 z z
a. Find the value of γxy. z z z z
b. Find the principal strains in the plane of the surface. z z z z z z z z z
c. Sketch the Mohr’s circle diagram. z z z z
Solution: Let the B gauge be on the x’ axis, the A gauge on the x-axis and the C gauge on
z z z z z z z z z z z z z z z z z z z z
2 2
they-axis. ex x =exxℓ x x +e ℓ x y +γxyℓx yyxℓx y, where ℓx x =cosex = 30=√3/2and ℓx y=
z z z z z z z z z z z z z z z z
z
z z z z z z z z z
cos 60 = ½. Substituting the measured strains, 3500 =
z z z z z z z z z
3000(√2/3)2 – 1000(1/2)2 + γxy(√3/2)(1/2)
z z z z z
2 2 -6
γ◻xy=(4/√3/2){3500-[3000–(1000(√3/2) +1 000(1/2) ]} =2,309(x10 )
z z z z z
1/2 2
b. e1,e2 = (ex +ey)/2± [(ex-ey)2 + γxy2]
z z /2 = (3000+1000)/2 ± [(3000-1000) +
z z z z z z z z z z
23092]1/2/2.e1 = 3530(x10-6), e2 = 470(x10-6), e3 = 0. z z z z z z z z z
c)
γ/2
sx
s2 s1
s
sx’
sy
Find the principal stresses in the part of problem 1-6 if the elastic modulus ofthe part is 205
z z z z z z z z z z z z z z z z z z
GPa and Poissons’s ratio is 0.29.
z z z z z z
Solution: e3 = 0 = (1/E)[0- v(σ1+σ2)], σ1= σ2 z z z z z z z z z z z
e1 = (1/E)(σ1 - v σ1); σ1 = Ee1/(1-v) = 205x109(3530x10-6)/(1-.292) = 79 MPa
z z z z z z z z z z z z z
1
Show that the true strain after elongation may be expressed as s =ln(
z z z z z z z z z z z z z z
) where r is the
1–r
z z z z
z z
1
reduction of area. s =ln( ).
z
z z z z z z
1–r z z
Solution: r = (Ao-A1)/Ao =1 – A1/Ao = 1 – Lo/L1. s = ln[1/(1-r)]
◻
z z z z z z z z z z z z z
A thin sheet of steel, 1-mm thick, is bent as described in Example 1-11. Assuming that E
z z z z z z z z z z z z z z z z
=is205 GPaand v=0.29,p=2.0 mandthattheneutralaxis doesn’t shift.
z z z z z z z z z z z z z z z z z
a. Find the state of stress on most of the outer surface. z z z z z z z z z z
b. Find the state of stress at the edge of the outer surface. z z z z z z z z z z z
2
, Solution: a. Substituting E = 205x109, t = 0.001, p = 2.0 and v = 0.29 z z z z z z z z z z z z z z z
Et vEt
intoσx = and σy = ,σx=56MPa,,σy=16.2MPa
2p(1–v )2
z z
2p(1–v2)
z
z z z z z z z z z z
z
z z z z z
z z z z
vEt
b. Now σy = 0, so σy =
z = 51 MPa z z z z z z z z
2p
1-10 For an aluminum sheet, under plane stress loading sx = 0.003 and sy = 0.001. z z z z z z z z z z z z z z
Assumin gthatE=is68GPaand v=0.30,find sz. z z z z z z z z z z z z
Solution: ey = (1/E)(σy-vσy), ex = (1/E)(σx – vEey – v2σx). Solving for σx, z z z z z z z z z z z z z
σx = [E/(1-v2)]ey+ vey). Similarly, σy = [E/(1-v2)](ey+ vex). Substituting into
z z z z z z z z z z z z
ez = (1/E)(-vσy-vσy) = (-v /E)(E/(1-v2)[ey+ vey+ ey+ vex ) = [-v(1+v)//(1-v2)](ey+ey) = 0.29(-
z z z z z z z z z z z z z z z z z
1.29/0.916)(0.004) = -0.00163 z z
1-11 A piece of steel is elastically loaded under principal stresses, σ1 = 300 MPa, σ2 = 250z z z z z z z z z z z z z z z z
MPa and σ3 = -200 MPa. Assuming that E = is 205 GPa and v = 0.29 find the stored elastic
z z z z z z z z z z z z z z z z z z z z z
energy per volume.
z z z
Solution: w = (1/2)(σ1e1 + σ2e2 + σ3e3). Substituting e1 = (1/E)[σ1 - v(σ2 + σ3)], e2 = z z z z z z z z z z z z z z z z z
(1/E)[σ2 - v(σ3 + σ1)] and e3 = (1/E)[σ3 - v(σ1 + σ2)],
w = 1/(2E)[σ 2 + σ 2 + σ 2 - 2v(σ σ +σ σ +σ σ )] =
z z z z z z z z z z z z z
z z z z z z z z z z z z z z z z z z
1 2 3 2 z 3 3 z 1 1 z 2
(1/(2x205x10 )[300 +250 + 200 –(2x0.29)(-200x250 – 300x250 + 250+300)]x1012 = 9 2
z
2
z z
2
z z z z z z
400J/m3
z
1-12 A slab of metal is subjected to plane-strain deformation (e2=0) such that σ1= 40 ksi z z z z z z z z z z z z z z z
and σ3 = 0. Assume that the loading is elastic and z z z z z z z z z z z
that E = is 205 GPa and v = 0.29 (Note the mixed units.) Find
z z z z z z z z z z z z z z
a. the three normal strains. z z z
b. the strain energy per volume. z z z z
Solution: w = (1/2)(σ1e1 + σ2e2 + σ3e3) = (1/2)(σ1e1 + 0 + 0) = σ1e1/2 σ1 = z z z z z z z z z z z z z z z z z
40ksi(6.89MPa/ksi) = 276 MPa
z z z z
0 = e2 = (1/E)[σ2 -v σ1], σ2 =v σ1= 0.29x276 = 80 MPa
z z z z z z z z z z z z z z
e1 = (1/E)(σ1 -v σ2) =(1/205x103)[276-.29(80)] = 0.00121 w
z z z z z z z z
= (276x106)(0.00121)/2 = 167 kJ/m3
z z z z z
Chapter 2 z
a) If the principal stresses on a material with a yield stress in shear of 200 MPa are σ2
z z z z z z z z z z z z z z z z z
= 175 MPa and σ1 = 350 MPa., what is the stress, σ3, at yielding according to the Tresca
z z z z z z z z z z z z z z z z z z
criterion?
z
b) If the stresses in (a) were compressive, what tensile stress σ3 must be applied to cause
z z z z z z z z z z z z z z z
yielding according to the Tresca criterion?
z z z z z z
Solution: a) σ1 - σ3 = 2k, σ3 = 2k – σ1 = 400 - 350 = 50 MPa. z z z z z z z z z z z z z z z z z z
3
