1. Solution Manual 3rd Ed. Metal Forming: Mechanics and Metallurgy
Chapter 1
Determine the principal stresses for the stress state
1 −3 4
0
σij = −3 5 2 .
4 2 7
Solution: I1 = 10+5+7=32, I2 = -(50+35+70) +9 +4 +16 = -126, I3 = 350 -48 -40 -80
-63 = 119; σ – 22σ2 -126σ -119 = 0. A trial and error solution gives σ -= 13.04.
3
Factoring out 13.04, σ 2 -8.96σ + 9.16 = 0. Solving; σ1 = 13.04, σ2 = 7.785, σ3 =
1.175.
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.
Solution: a. The shear stress, τ, at a radius, r, is τ = τsr/R where τsis the shear stress at the
surface R is the radius of the rod. The torque, T, is given by T = ∫2πtr2dr = (2πτs /R)∫r3dr =
πτsR3/2. Solving for = τs, τs = 2T/(πR3) = 2(400N)/(π0.0253) = 16 MPa
The axial stress is .08MN/(π0.0252) = 4.07 MPa
σ1,σ2 = 4.07/2 ± [(4.07/2)2 + (16/2)2)]1/2 = 1.029, -0.622 MPa
b. the largest shear stress is (1.229 + 0.622)/2 = 0.925 MPa
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?
Solution: Let y = hoop direction, x = axial direction, and z = radial direction. –
ex = e2 = (1/E)[σ - υ( σ3 + σ1)] = (1/E)[σ2 - υ(2σ2)] = (σ2/E)(1-2υ)
Since u < 1/2 for metals, ex = e2 is positive and the tube lengthens.
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?
Solution: The shear stresses in both are identical because a hydrostatic pressure has no
shear component.
1-5 Consider a long thin-wall, 5 cm in 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 40 N and an internal pressure of 200 kPa.
Solution: σx = PD/4t + F/(πDt) = 12.2 MPa
σy = PD/2t = 2.0 MPa
σy = 0
1

2. 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 gage, B, is at 30° to gauge A. When
the part is loaded the gauges read
Gauge A 3000x10-6
Gauge B 3500 x10-6
Gauge C 1000 x10-6
a. Find the value of γ xy.
b. Find the principal strains in the plane of the surface.
c. Sketch the Mohr’s circle diagram.
Solution: Let the B gauge be on the x’ axis, the A gauge on the x-axis and the C gauge on
y γ ′
the y-axis. e′ = l x′+ l x + l xl x , where l x′x = cosex = 30 = √3/2 and l x′y = cos
2 2
xx e
′ x ′
x x e y ′
y x x ′
y y
60 = ½. Substituting the measured strains,
3500 = 3000(√2/3)2 – 1000(1/2)2 + γ xy(√3/2)(1/2)
γ xy = (4/√3/2){3500-[3000− (1000(√3/2)2+1000(1/2)2]} = 2,309 (x10-6)
b. e1,e2 = (ex +ey)/2± [(ex-ey)2 + γ xy2]1/2/2 = (3000+1000)/2 ± [(3000-1000)2 +
23092]1/2/2 .e1 = 3530(x10-6), e2 = 470(x10-6), e3 = 0.
c)
γ/2
εx
ε2 2θ=60° ε1 ε
εx’
εy
Find the principal stresses in the part of problem 1-6 if the elastic modulus of the part is
205 GPa and Poissons’s ratio is 0.29.
Solution: e3 = 0 = (1/E)[0 - ν (σ1+σ2)], σ1 = σ2
e1 = (1/E)(σ1 - ν σ1); σ1 = Ee1/(1-ν) = 205x109(3530x10-6)/(1-.292) = 79 MPa
1
Show that the true strain after elongation may be expressed as ε=l (
n ) where r is the
1−r
1
reduction of area. ε=l (
n ).
1−r
Solution: r = (Ao-A1)/Ao =1 – A1/Ao = 1 – Lo/L1. ε = ln[1/(1-r)]
A thin sheet of steel, 1-mm thick, is bent as described in Example 1-11. Assuming that E =
is 205 GPa and ν = 0.29, ρ = 2.0 m 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.
2

3. Et
Solution: a. Substituting E = 205x109, t = 0.001, ρ = 2.0 and ν = 0.29 into σ =
2 ( −ν2)
ρ1
x
νt
E
and σ = , σx = 56 MPa, , σy = 16.2 MPa
2 ( ν2)
ρ1−
y
νE t
b. Now σy = 0, so σy = = 51 MPa
2ρ
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.
2
Solution: ey = (1/E)(σy-νσy), ex = (1/E)(σx – νΕey – ν σx). Solving for σx,
2 2
σx = [E/(1-ν )]ey + νey). Similarly, σy = [E/(1-ν )](ey + νex). Substituting into
2 2
ez = (1/E)(-νσy-νσy) = (-ν /E)(E/(1-ν )[ey + νey+ ey + νex ) = [-ν(1+ ν)//(1-ν )](ey + ey) =
0.29(-1.29/0.916)(0.004) = -0.00163
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.
Solution: w = (1/2)(σ1e1 + σ2e2 + σ3e3). Substituting e1 = (1/E)[σ1 - ν(σ 2 + σ3)],
e2 = (1/E)[σ2 - ν(σ 3 + σ1)] and e3 = (1/E)[σ3 - ν(σ 1 + σ2)],
w = 1/(2E)[σ12 + σ22 + σ32 - 2ν(σ 2σ3+σ3σ1+σ1σ2)] =
(1/(2x205x109)[3002 +2502 + 2002 –(2x0.29)(-200x250 – 300x250 + 250+300)]x1012 =
400J/m3
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.
Solution: w = (1/2)(σ1e1 + σ2e2 + σ3e3) = (1/2)(σ1e1 + 0 + 0) = σ1e1/2
σ1 = 40ksi(6.89MPa/ksi) = 276 MPa
0 = e2 = (1/E)[σ2 -ν σ1], σ2 =ν σ1 = 0.29x276 = 80 MPa
e1 = (1/E)(σ1 -ν σ2) =(1/205x103)[276-.29(80)] = 0.00121
w = (276x106)(0.00121)/2 = 167 kJ/m3
Chapter 2
a) If the principal stresses on a material with a yield stress in shear of 200 MPa are σ2 =
175 MPa and σ1 = 350 MPa., what is the stress, σ3, at yielding according to the Tresca
criterion?
b) If the stresses in (a) were compressive, what tensile stress σ3 must be applied to cause
yielding according to the Tresca criterion?
3

4. Solution: a) σ1 - σ3 = 2k, σ3 = 2k – σ1 = 400 - 350 = 50 MPa.
b) σ3 = 2k – σ1 = 400 – (350) = 50 MPa
Consider a 6-cm diameter tube with 1-mm thick wall with closed ends made from a metal
with a tensile yield strength of 25 MPa. After applying a compressive load of 2000 N to the
ends. What internal pressure is required to cause yielding according to a) the Tresca
criterion. b) the von Mises criterion?
Solution: a) The ratio of the tube diameter to wall thickness is very large, so it can be treated as a thin
wall tube. The stress caused by the pressure can be found by x- and y- direction force balances.
From pressure, σx = Pd/(2t) = 60P and σy = Pd/(4t) = 30P. The stress caused by the axial load is σy =
F/(dt) = -2000N/[π(0.060)(0.001)]= -10.6 MPa, so the total stress, σy = 30P -10.6 MPa
a) σx = 60P = σmax is the largest stress, σy = 30P -10.6 MPa and σz = 0. There are two
possibilities which must be checked.
i. If σz < σy, σz = σmin, yielding will occur when 60P-0 = Y, or P=Y/60 =25/60 = 0.416 MPa
ii. If σy < σz, σy = σmin, and yielding will occur when
60P-(30P-10.6) = Y, or 30P = Y + 10.6, P = (Y+10.6)/30 = 35.6/30 = 1.1187 MPa
Yielding will occur when the smaller of the two values is reached, and therefore the smaller one is
appropriate. P = 0.415 MPa
b) Substituting into eq. 2-7 (in MPa),
2(25)2 = [60P-(30P -10.6)]2 +[(30P -10.6)-0]2 + [0-60P]2
1250 = 5400P2 + 224, p = 0.436 MPa
2-3 Consider a 0.5 m-diameter cylindrical pressure vessel with hemispherical ends
made from a metal for which k = 500 MPa. If no section of the pressure vessel is to yield
under an internal pressure of 35 MPa, what is the minimum wall thickness according to a)
the Tresca criterion? b) the von Mises criterion?
Solution: A force balance in the hemispherical ends gives σx ( =σy) = PD/(4t).
A force balance in the cylindrical section gives σx = PD/(2t). σy = PD/(4t) so this section
has the greatest stress.
a. σmax - σmin = 2k, PD/2t – 0 = 2k, t = PD/(4k) = 35(0.5)/(4x500) = 8.75 mm
b. (σx/2 - 0)2 + (0 - σx)2+ (σx -σx/2)2 = 6k2, (3/2)σx2 = 6k2, σx = 2k = PD/(2t), t = PD/(4k)
which is identical to part a. t = 8.75 mm
ε= 2 x2+y2 3
(ε ε )/
2-4 A thin-wall tube is subjected to combined tensile and torsional loading. Find the
relationship between the axial stress, σ, the shear stress, τ, and the tensile yield strength, Y,
to cause yielding according to a) the Tresca criterion, b) the von Mises criterion.
Solution: a) σ 2= / ±( / ) +2 If σ −( / ) + 2 > 0, σmin = 0, so the
1,σσ 2 σ 2 τ 2 /2 σ 2 τ 2
σ ±( / ) +2= . If σ −( / ) + 2<
/2 σ 2 τ Y
Tresca criterion predicts yielding when 2 /2 σ 2 τ
2
σ2 τ σ22 τ
0, σmin = −( / ) + , so the Tresca criterion predicts yielding when 2 ( / ) + 2
2 2
b) {2[ σ −( / ) + 2]2 +[ 2 ( / ) + 2]2}1/2 = √2Y+
/2 σ 2 τ
2 σ22 τ
4

5. Consider a plane-strain compression test with a compressive load, Fy, a strip width, w, an
indenter width, b, and a strip thickness, t. Using the von Mises criterion, find:
a) ε as a function of εy.
b) σ as a function of σy.
c) an expression for the work per volume in terms of εy and σy.
d) an expression in the form of σy = f(K,εy,n) assuming σ Kn= ε.
Solution: a. If εz = 0, εy = - ex ε= 2 x +y )/ = = 1.154εy
ε ε 3
2 2
(
b. σx = 0, σz = -(1/2)σy; σ 1 ) σ σ2 + y/ − 2+ −y 2 = σy/1.154
=(/ [ y−y/ ) ( 2 0 ( σ ]
2 ( 2
σ ) 0 )
c. w = ∫σydεy
d. σ y=4σ /K=4 K / ) )= (4/3)n+1/2 ey
/ =4 ε
3 3n / (4 εn
3 3y
2-6 The following yield criterion has been proposed: “Yielding will occur when the sum of
the two largest shear stresses reaches a critical value. “ Stated mathematically
(σ1- σ3) + (σ1- σ2) = C if (σ1 - σ2) > (σ2 - σ3) or (σ2 - σ3) + (σ1 - σ2) = C if (σ1- σ2) ≤ (σ2-
σ3) where σ1 > σ 2 > σ3, C = 2Y and Y = tensile yield strength.
a) Is this criterion satisfactory for an isotropic solid where Y is insensitive to pressure?
Justify your answer.
b) Plot the σz = 0 yield locus. Sketch the Tresca yield locus on the same plot
c) Where σz = 0, find the values of σx and σy for
i. plane strain, εz = 0, with εx > 0
ii. axisymmetric flow with εy = εz = εx/2 and εx > 0
Solution: a) Yes. The value of the left hand sides are not affected if each principal stress is increased
the same amount.
b) First find the constant C. Consider an x-direction tension test. At yielding, σx = σ1 = Y,
σy = σz = σ2 = σ3 = 0. Therefore (σ1 - σ2)> (σ2 - σ3) so criterion I applies, and C = (σ1 -
σ3) + (σ1 - σ2) = 2Y. Therefore C = 2Y.
We can also think about an x-direction compression test. At yielding, σx = σ3 = -Y, σy = σz = σ2
= σ3 = 0 . Therefore (σ2 - σ3)>(σ1 - σ2)> so criterion II applies, and C = (σ1 - σ3) + (σ2 -
σ3) = -(-2Y) or again C = 2Y.
Now consider several loading paths:
In region A, σx = σ1, σy = σ2, σz= σ3 = 0 and σx >2σy so (σ1 - σ3) >(σ1 - σ2)
Therefore criterion I, (σx - 0) + (σx - σy) = 2Y, or σx = Y + σy/2
In region B, σx = σ1, σy = σ2, σz= σ3 = 0 but σx <2σy so (σ1 - σ3)<(σ1 - σ2)
Therefore criterion II, (σx - 0) + (σy - 0) = 2Y, or σx = 2Y - σy
In region C, σy = σ1, σx= σ2, σz= σ3 = 0 but σy <2σx so (σ1 - σ3)<(σ1 - σ2)
Therefore criterion II, (σy - 0) + (σx - 0) = 2Y, or σy = 2Y - σx
5

6. In region D, σy = σ1, σx = σ2, σz= σ3 = 0 and σy >2σx so (σ1 - σ2) >(σ2 - σ3)
Therefore criterion I, (σy - 0) + (σy - σx) = 2Y, or σy = Y + σx/2
In region E, σx = σ1, σy = σ3, σz= σ2 = 0 and (σ1 - σ2) >(σ2 - σ3)
Therefore criterion I, (σx - 0) + (σx - σy) = 2Y, or σx = Y + σy/2
In region f, σx = σ1, σy = σ3, σz= σ2 = 0 so (σ1 - σ2) >(σ2 - σ3)
Therefore criterion I, (σx - 0) + (σx - σy) = 2Y, or σx = Y + σy/2
Plotting these in the appropriate regions, and using symmetry to construct the left hand half:
c) i. For plane strain (εy = 0) and εx > 0, The normal to the locus is at the corner between A
and B regions. Both σx = Y + σy/2 and σx = 2Y - σy must be satisfied. Solving
simultaneously, σx = (4/3)Y but σy = (2/3)Y
ii. Axisymmetric flow with εy = εz = -(1/2)εx with εx > 0, is satisfied everywhere in
Region I, so σx = Y + σy/2, with (2/3)Y ≤ σx ≤ (4/3)
2-7 Consider the stress states
15 3 0 10 3 0
3 10 0 and 3 5 0 .
0 0 5 0 0 0
a) Find σm for each.
b) Find the deviatoric stress in the normal directions for each
c) What is the sum of the deviatoric stresses for each?
Solution: a) (15 + 10 + 5)/3 = 10 and (10 + 5 + 0)/3 = 5
b) 15 – 10 = 5, 10-10 = 0 5 – 10 = -5 and 10-5 = 5, 5-5 = 0, 0-5 = -5
c) The sum of the deviatoric stresses both = 0.
6

7. 2-8 A thin wall tube with closed ends is made from steel with a yield strength of 250
MPa. The tube is 2 m. long with a wall thickness of 2 mm. and a diameter of 8 cm. In
service it will experience an axial load of 8 kN and a torque of 2.7 Nm. What is the
maximum internal pressure it can withstand without yielding according to a) the Tresca
criterion, b) the von Mises criterion?
Solution: D/t = 40 so this can be regarded as a thin-wall tube. For this solution, stresses
will be expressed in ksi.
F/A = 2/(πdt) = 2/(πx3x0.05) = 4.244 ksi
T = τ(πdt)(d/2); τ = 2T/(πd2t) = 2x2./(π320.05) = 2.829 ksi
σx = Pd/(2t), σy = Pd/(4t) + 4.244 = σx/2 + A, where A = 4.244 ksi
a) For Mises, substituting σz = τxy = τyz = τzx = 0 into the yield criterion, Eq. (2-12)
2Y2 = (σy - σz)2 + (σz - σx)2 + (σx - σy)2 + 6txy2
2Y2 = σ y2 + σ x2 + (σ x − σ y) 2 + 6τ xy2 = 2[σ y2 − σ xσy+ σ x2 ] + 6τ xy2
Y2 = σ y2 − σ xσy + σ x2 + 3τ xy2
Substituting σy = σx/2 + A,
σx/2 + A2 - σx/2 + Aσx +σ+ 3τxy2 - Y2 = 0
  
σx2(1/4 -1/2 + 1) + σx(A - A) + (A2 + 3τxy2- Y2) = 0
(3/4)σx2 + (A2 + 3τxy2- Y2) = 0
σx2 + B = 0 where B = (4/3)(A2 + 3τxy2- Y2)
Substituting B = (4/3)(4.2442 + 3x2.8292- 202) = -2.098
σx2 = 2.098, σx = 45.8, σx = Pd/(2τ),
P = (2τ/d)sx = 45.8(2x0.050/3) = 1.528 ksi
b) For Tresca, we must find the principal stresses.
σ1,2 = (σx + σy)/2 ± (1/2)[(σx - σy)2 + 4τxy2]1/2
Substituting σy = σx/2 +A,
σ1,2 = (3/4)σx + A/2 ± (1/2)[(σx/2 - A)2 + 4τxy2]1/2 = (3/4)σx + A/2 ± (1/2)[σx2/4 - Asσ + A2 +
4τxy2]1/2
σ1 = (3/4)σx + A/2 + (1/2)[σx2/4 - Aσx + A2 + 4τxy2]1/2
σ2 = (3/4)σx + A/2 - (1/2)[σx2/4 - Aσx + A2 + 4τxy2]1/2
There are two possibilities: σ2 > 0, and σ2 < 0.
1st assume that s2 > 0. Then σ1 - 0 = Y,
(3/4)σx + A/2 + (1/2)[σx2/4 - Aσx + A2 + 4τxy2]1/2 = Y
(3/4)σx + A/2 + (1/2)[σx2/4 - Aσx + A2 + 4τxy2]1/2 = Y
(3/4)σx + 4.244/2 + (1/2)[σx2/4 - 4.244σx + 4.2442 + 4x2.8292]1/2 = 40
7

8. (3/2)σx + 4.244 + [σx2/4 - 4.244σx + 49.855]1/2 = 80
[σx2/4 - 4.244σx + 49.855]1/2 = 75.756 -(3/2)σx2
σx2/4 - 4.244σx + 49.855] = [75.756 -(3/2)σx]2 = 5738.9 -227.26σx +2.25σx2
σx2[0.25 - 2.25] + [-4.244 +227.26]sx + 49.855 - 5738.9 = 0
2σx2 -223.3sx + 5689 = 0; σx = {223.3 ± [223.32 -4x2x 5689]1/2}/(2x2) =
55.85 ± 16.49,
σx = 72.34 or 39.36, The smaller value is correct
Then P = (2t/d)σx = 39.36(2x0.050/3) = 1.312 ksi
Now we must check to see whether σ2 > 0. Substituting A = 4.244, τ = 2.829 and σx = 39.36 into
σ2 = (3/4)σx + A/2 - (1/2)[σx2/4 - Aσx + A2 + 4τxy2]1/2
σ2 = (3/4)x39.36 + 4.244/2 - (1/2)[39.362/4 - 4.244x39.36 + 4.2442 +
4x2.829 2]1/2 = 31.48. Therefore the solution for σ2 > 0 is appropriate.
2-9 Calculate the ratio of σ τa for a) pure shear. b) uniaxial tension, and c) plane
/m x
strain tension. Assume the von Mises criterion.
Solution:
a) σ1 = τ, σ2 = 0, σ3 = -τ, σ = {[τ2 + (2τ) 2 + τ2]/2}1/2, σ /τ = √3
b) σ /τ = 2
c) σ1 = τ, σ2 = τ/2, σ3 = 0, σ = {[(τ/2)2 + τ2 +(τ/2)2]/2}1/2, σ /τ = √(3/2)
A material yields under a biaxial stress state, σ3 = -(1/2)σ1, σ2 = 0.
a) Assuming the von Mises criterion, find dε1/ dε2.
What is the ratio of τa / at yielding?
m Y
x
Solution: dε1/dε2 = [σ1 - (σ2 + σ3)/2]/[σ2 - (σ3 + σ1)/2] =
[σ1 - (0-σ1/2)/2]/[0 - (-σ1/2 -σ1)/2] = (5/4)/(3/4) = 5/3
2-11 A material is subjected to stresses in the ratio, σ1 , σ2 = 0.3σ1 and σ3 = -0.5σ1. Find
the ratio of σ1/Y at yielding using the a) Tresca and b) von Mises criteria.
Solution: a) For Tresca, σ1 – (-0.5σ1) = Y, σ1 /Y= 2/3
b) For von Mises, {[(.3+.5)2 + (-.5 – 1)2 + (1-.3)2]/2}1/2σ1 = Y, σ1 /Y= 0.77
2-12 A proposed yield criterion is that yielding will occur when the diameter of the
largest Mohr’s circle plus half the diameter of the second largest Mohr’s circle reaches a
critical value. Plot the yield locus in σ1 vs. σ2 in σ3 = 0 space.
Solution: Divide stress space into regions with different conditions for yielding.
To evaluate C, consider an x-direction tension test. At yielding σx = Y, σy = 0,
The diameters of the two largest Mohr’s circle are Y. Y = Y/2 = C. C = 3/2Y
8

9. σy + (1/2)(σ y-σx) = (3/2)Y σy + (1/2)(σ x) = (3/2)Y
σy - (1/3)σ x =Y (2 /3)σ + (1/3)(σ x) = Y
y
σy
σx + (1/2)(σ y) = (3/2)Y
(2 /3)σ + (1/3)(σ y) = Y
x
σx + (1/2)(σ x-σy) = (3/2)Y
σx - (1/3)σ y =Y
σx
(σx+σy)+(1/2)σx = (3/2)Y
σx + (2/3)σ y =Y
(σx+σy)+(1/2)(-σy) = (3/2)Y
(2 /3)σ + σy =Y
x
2.13 Make plot of ε 1 versus ε2 for a constant level of ε = 0.10 according to
a. von Mises.
b. Tresca.
Solution: Taking ε ε 4 ) ++2 −2so ε ε4 ) ++2 1 and ε = ε
=( / ( ρρ 1
[ 3
1 1 )/
] 1= / ( ρρ /
[ 3
( 1 )2
] 2 ρ 1
for von Mises and 1ε−2= for Tresca,, ε /ε and ε2 /ε can be calculated for various
ε ε 1
values of ρ. ε2 /ε
ε 1 /ε
ε1
10
1
Tresca
von Mises
-10 1
ε2 /ε
-1 0 10
CHAPTER 3
When a brass tensile specimen, initially 0.505 in. in diameter, is tested, the maximum load
of 15,000 lbs was recorded at an elongation of 40%. What would the load be on an
identical tensile specimen when the elongation is 20%?
Solution: n = εmax load = ln(1+emax load) = ln(1.4) = 0.365.
σmax load = smax load (1+emax load) = (12,000)/0.2)(1.4) = 84x103. But also σmax load =
K(.365).365 = 0.6932K. Equating and solving for K, K = 84x103/0.6932 = 121,000.
At 20% elongation, ε = ln(1.2) = 0.1823. σ = 121,000(0.1823).365 = 65,000. s = 65,000/1.2
= 54,180. F = 54,000(0.2) = 10.
8 lbs.
9

10. 3-2 During a tension test the tensile strength was found to be 340 MPa. This was
recorded at an elongation of 30%. Determine n and K if the approximation σ Kn =ε
applies.
Solution: n = εmax load = ln(1+emax load) = ln(1.3) = 0.262.
σmax load = smax load (1+emax load) = 340(1.3) = 442 MPa. But also σmax load = K(.0.262)0.262 =
0.704K. K = 442/0.704 = 627 MPa.
*
3.3 Show that the plastic work per volume is σ /n 1for a metal stretched in tension to
ε( +)
11
ε1 if σ k n
=ε .
Solution: w = ∫σ1dε1 = ∫kε1ndε1 = kε1n+1/(n+1) = kε1ε1n/(n+1) = σ /n 1
ε( +)
11
3.4 For plane-strain compression (Figure 3.11)
a. Express the incremental work per volume, dw, in terms of σ and dε and
compare it with dw = σ1dε1 + σ2dε2 + σ3dε3.
b. If σ k n, express the compressive stress, as a function of σ1, K and n.
=ε
Solution: a. With εy = 0 and σx = 0, dw = σ3dεz. σy = σz/2, σx =0,
σ = {[(σy - σz)2 +(σz – σx)2 +(σx – σy)2]/2}1/2 = {[(-σz/2)2 +(-σz)2 + (-σz/2)2]/2}1/2 = (3/4)σz
d [ /) ε+ y + z)/ =2)−x2 0d ] / = (4/3)1/2dεz
e ( 3 x d d ] 2 {/ [ d + ε}
= ( ε ε 1 ( 3 ε +z 1
2 2 2 2 2
2 d ( )
σ ε = (3/4)σz(4/3)1/2dεz = (σzdεz
d
b. σ (/) σ 4) k = /) k /) ε (4/3)(n+1)/2e .
= 3 = / 1 ε ( 3 ( 32n
41 /2
( 3 n 41 4 n =
/
2 /2 / n
z
3.5 The following data were obtained from a tension test:
Load Min. Neck true true corrected
dia. radius strain stress true stress
(kN) (mm) (mm) σ (MPa) σ (MPa)
0 8.69 ∞ 0 0 0
27.0 8.13 ∞ 0.133 520 520
34.5 7.62 ∞
40.6 6.86 ∞
38.3 5.55 10.3
29.2 3.81 1.8
a. Compute the missing values
b. Plot both σ and σ vs. ε on a logarithmic scale and determine K and n.
c. Calculate the strain energy per volume when ε = 0.35.
Solution: a)
Load Min. Neck true true a/R corrected
dia. radius strain stress true stress
(kN) (mm) (mm) σ (MPa) σ (MPa)
0 8.69 ∞ 0 0 0 0
27.0 8.13 ∞ 0.133 520 0 520
34.5 7.62 ∞ 0.263 754 0 654
40.6 6.86 ∞ 0.473 1099 0 1099
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11. 38.3 5.55 10.3 0.978 1717 0.26 1631
29.2 3.81 1.8 1.65 2561 1.06 2100
3.6 Consider a steel plate with a yield strength of 40 ksi, Young’s modulus of 30x106 psi
and a Poisson’s ratio of 0.30 loaded under balanced biaxial tension. What is the volume
change, ∆V/V, just before yielding?
Solution: At yielding σ1 = σ2 = 40,000 psi, σ3 = 0. e1 = e2 = (1/E)[σ1 – υσ1], e3 = (1/E)[-
2υσ1]; ∆v/v = e1 + e2 + e3 = (σ1 /E)[2-–4 υ] = 0.107x10-3.
3.7 The strain-hardening of a certain alloy is better approximated by
σ = A[1--exp(-Bε)] than by σ k n. Determine the true strain at necking in terms of A
=ε
and B.
Solution: σ = A[1--exp(-Bε)] =dσ/dε = ABexp(-Bε); A = A(B+1)exp(-Bε); ε =
ln(1+B)/B
3-8 Express the tensile strength, in terms of A and B for the material in Problem 3-7.
Solution:
σmax load = A{1—exp[-B(ln(1+B)/B)]} = A[1+(1+B)] = A(2+B);
Tensile strength = σmax load exp(ε) =A(2+B)exp[ln(1+B)/B] = A(2+B)(1+B)1/B
3-9 A metal sheet undergoing plane-strain tensile deformation is loaded to a tensile
stress of 300 MPa. What is the major strain if the effective stress-strain relationship is
σ6 (.5+ 0 MPa?
5 0 ε.2
= 00 1 )2
Solution:
σ =√(4/3) σ =650(0.015+ ε )0.22; ε =[√(3/4)(300)/650 -0.015]1/0.22 = 0.155; ε = √(4/3) ε =
0.179
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