Solutions Manual for Metal Forming: Mechanics and Metallurgy (4th Edition) by Hosford and Caddell

All Chapters Covered




SOLUTION MANUAL

, Solution Manual 3rd Ed. Metal Forming: Mechanics and Metallurgy
Chapter 1

Determine the principal stresses for the stress state 10
3 4
3 5 2.
ij
4 2 7
Solution: I1 = 10+5+7=32, I2 = -(50+35+70) +9 +4 +16 = -126, I3 = 350 -48 -40 -80
3 2
-63 = 119; – 22 -126 -119 = 0. A trial and error solution gives -= 13.04.
2
Factoring out 13.04, -8.96 + 9.16 = 0. Solving; = 13.04, = 7.785, = 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, , at a radius, r, is = sr/R where sis the shear stress at the surface
R is the radius of the rod. The torque, T, is given by T = ∫2πtr2dr = (2π s /R)∫r3dr
3
= π sR3/2. Solving for = s, s = 2T/(πR3) = 2(400N)/(π0.025 ) = 16 MPa The axial
2
stress is .08MN/(π0.025 ) = 4.07 MPa
2 2 1/2
1, 2 = 4.07/2 ± [(4.07/2) + (16/2) )] = 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)[ - ( 3 + 1)] = (1/E)[ 2 - (2 2)] = ( 2/E)(1-2 )
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: x = PD/4t + F/(πDt) = 12.2 MPa
y = PD/2t = 2.0 MPa
y=0




1

,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
6
Gauge A 3000x10-
6
Gauge B 3500 x10-
6
Gauge C 1000 x10-
a. Find the value of xy.
b. Find the principal strains in the plane of the surface.
c. Sketch the Mohr’s circle diagram.
Solution: Let the B gauge 2
be on the
2
x’ axis, the A gauge on the x-axis and the C gauge on
the y-axis. e x x e xx x x e yy xy xy x x xy , where x x = cosex = 30 = √3/2 and xy =
cos 60 = ½. Substituting the measured strains, 3500 =
2 2
3000(√2/3) – 1000(1/2) + xy(√3/2)(1/2) 2 2 -6
xy = (4/√3/2){3500-[3000 (1000(√3/2) +1000(1/2) ]} = 2,309 (x10 )
1/2 2 2 1/2
b. e1,e2 = (ex +ey)/2± [(ex-ey)2 + xy2] /2 = (3000+1000)/2 ± [(3000-1000) + 2309 ] /2 .e1 =
-6 -6
3530(x10 ), e2 = 470(x10 ), e3 = 0.
c)

x



2 1
2=60°

x’

y



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 - ( 1+ 2)], 1 = 2
9 -6 2
e1 = (1/E)( 1 - 1); 1 = Ee1/(1- ) = 205x10 (3530x10 )/(1-.29 ) = 79 MPa
1
Show that the true strain after elongation may be expressed as ln(
) where r is the
1 r
1
reduction of area. ln( ).
1 r
Solution: r = (Ao-A1)/Ao =1 – A1/Ao = 1 – Lo/L1. ln[1/(1-r)]=
heet of steel, 1-mm thick, is bent as described in Example 1-11. Assuming that E
A thin s
= is 205 GPa and = 0.29, = 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

, 9
Solution: a. Substituting E = 205x10 , t = 0.001, = 2.0 and = 0.29
Et Et
into x and y , x = 56 MPa, , y = 16.2 MPa
2 (1 2 ) 2 (1Et 2 )
b. Now y = 0, so y = 51 MPa
2

1-10 For an aluminum sheet, under plane stress loading x = 0.003 and y = 0.001.
Assuming that E = is 68 GPa and = 0.30, find z.
Solution: ey = (1/E)( y- y), ex = (1/E)( x – ey – x). Solving for x,
x = [E/(1- )]ey + ey). Similarly, y = [E/(1- )](ey + ex). Substituting into
ez = (1/E)(- y- y) = (- /E)(E/(1- )[ey + ey+ ey + ex ) = [- /(1- )](ey + ey) = 0.29(-
1.29/0.916)(0.004) = -0.00163

1-11 A piece of steel is elastically loaded under principal stresses, 1 = 300 MPa, 2 = 250 MPa
and 3 = -200 MPa. Assuming that E = is 205 GPa and = 0.29 find the stored elastic energy per
volume.
Solution: w = (1/2)( 1e1 + 2e2 + 3e3). Substituting e1 = (1/E)[ 1 - 2 + 3)], e2 = (1/E)[ 2 -
3 + 1)] and e3 = (1/E)[ 3 - 1 + 2)],
2 2 2
w = 1/(2E)[ 1 + + - 2 2 3+ 3 1+ 1 2)] =
9 2 2 2 3 2 12 3
(1/(2x205x10 )[300 +250 + 200 –(2x0.29)(-200x250 – 300x250 + 250+300)]x10 = 400J/m

1-12 A slab of metal is subjected to plane-strain deformation (e2=0) such that 1 = 40 ksi and 3
= 0. Assume that the loading is elastic and
that E = is 205 GPa and = 0.29 (Note the mixed units.) Find
a. the three normal strains.
b. the strain energy per volume.
Solution: w = (1/2)( 1e1 + 2e2 + 3e3) = (1/2)( 1e1 + 0 + 0) = 1e1/2
1 = 40ksi(6.89MPa/ksi) = 276 MPa
0 = e2 = (1/E)[ 2 - 1], 2 = 1 = 0.29x276 = 80 MPa e1 = (1/E)( 1 -
3
2) =(1/205x10 )[276-.29(80)] = 0.00121 w =
6 3
(276x10 )(0.00121)/2 = 167 kJ/m

Chapter 2

a) If the principal stresses on a material with a yield stress in shear of 200 MPa are 2
= 175 MPa and 1 = 350 MPa., what is the stress, 3, at yielding according to the Tresca criterion?
b) If the stresses in (a) were compressive, what tensile stress 3 must be applied to cause yielding
according to the Tresca criterion?
Solution: a) 1 - 3 = 2k, 3 = 2k – 1 = 400 - 350 = 50 MPa.




3

, b) 3 = 2k – 1 = 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, x = Pd/(2t) = 60P and y = Pd/(4t) = 30P. The stress caused by the axial load is y =
F/(dt) = -2000N/[π(0.060)(0.001)]= -10.6 MPa, so the total stress, y = 30P -10.6 MPa
a) x = 60P = max is the largest stress, y = 30P -10.6 MPa and z = 0. There are two possibilities
which must be checked.
i. If z < y, z = min, yielding will occur when 60P-0 = Y, or P=Y/60 =25/60 = 0.416 MPa
ii. If y < z, y = min, 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) = [60P-(30P -10.6)]2 +[(30P -10.6)-0]2 + [0-60P]2
2
2
1250 = 5400P + 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 x ( = y) = PD/(4t).
A force balance in the cylindrical section gives x = PD/(2t). y = PD/(4t) so this section has the
greatest stress.
a. max - min = 2k, PD/2t – 0 = 2k, t = PD/(4k) = 35(0.5)/(4x500) = 8.75 mm
2 2 2 2 2 2
b. ( x/2 - 0) + (0 - x) + ( x - x/2) = 6k , (3/2) x = 6k , x = 2k = PD/(2t), t =
PD/(4k) which is identical to part a. t = 8.75 mm
2( 2x 2 ) /3
 y


2-4 A thin-wall tube is subjected to combined tensile and torsional loading. Find the
relationship between the axial stress, , the shear stress, , and the tensile yield strength,
Y, to cause yielding according to a) the Tresca criterion, b) the von Mises criterion.
Solution: a) 1, 2 /2 ( /2) 2 2 If /2 ( /2)2 2
> 0, min = 0, so the
2 2
Tresca criterion predicts yielding when /2 ( /2) Y . If /2 ( /2)2 2 <
0, min =  ( /2)2 2 , so the Tresca criterion predicts yielding when 2 ( /2)2 2
2 2 1/2
b) {2[ /2 ( /2) 2 2 ] +[ 2 ( /2) 2 2 ] } = √2Y+




4

, 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) as a function of y.
b) as a function of y.
c) an expression for the work per volume in terms of y and y.
n
d) an expression in the form of y = f(K, y,n) assuming K .
2
Solution: a. If z = 0, y = - ex 2( 2x )
 y
/3= 2 2
=
1.1
54 y
b. x = 0, z = -(1/2) y; (1/2)[( y y /2) 2 ( y /2 0) (0 y ) ] = y/1.154
c. w = ∫ yd y 4/ 3K n n+1/2
ey
d. y 4 /3 4 /3K( ) y ) = (4/3)
n


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
( 1- 3) + ( 1- 2) = C if ( 1 - 2) > ( 2 - 3) or ( 2 - 3) + ( 1 - 2) = C if ( 1- 2) ≤
( 2- 3) where 1 > 2 > 3, 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 z = 0 yield locus. Sketch the Tresca yield locus on the same plot
c) Where z = 0, find the values of x and y for
i. plane strain, z = 0, with x > 0
ii. axisymmetric flow with y = z = x/2 and x > 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, x = 1 = Y,
y z 2 3 = 0. Therefore ( 1 - 2)> ( 2 - 3) so criterion I applies, and C = ( 1 - 3) + ( 1 - 2) = 2Y. Therefore
= = =
C = 2Y.
We can also think about an x-direction compression test. At yielding, x = 3 = -Y, y = z = 2 =
. Therefore ( 2 - 3)>( 1 - 2)> so criterion II applies, and C = ( 1 - 3) + ( 2 - 3) = -(- 2Y) or again C = 2Y.
Now consider several loading paths:
In region A, x = 1, y = 2, z= 3 = 0 and x >2 y so ( 1 - 3) >( 1 - 2)
Therefore criterion I, ( x - ) + ( x - y) = 2Y, or x = Y + y/2
In region B, x = 1, y = 2, z= 3 = 0 but x <2 y so ( 1 - 3)<( 1 - 2)
Therefore criterion II, ( x - ) + ( y - ) = 2Y, or x = 2Y - y
In region C, y = 1, x= 2, z= 3 = 0 but y <2 x so ( 1 - 3)<( 1 - 2)
Therefore criterion II, ( y - ) + ( x - ) = 2Y, or y = 2Y - x
In region D, y = 1, x = 2, z= 3 = 0 and y >2 x so ( 1 - 2) >( 2 - 3)
Therefore criterion I, ( y - ) + ( y - x) = 2Y, or y = Y + x/2



5

, In region E, x = 1, y = 3, z= 2 = 0 and ( 1 - 2) >( 2 - 3)
Therefore criterion I, ( x - ) + ( x - y) = 2Y, or x = Y + y/2
In region f, x = 1, y = 3, z= 2 = 0 so ( 1 - 2) >( 2 - 3)
Therefore criterion I, ( x - ) + ( x - y) = 2Y, or x = Y + y/2
Plotting these in the appropriate regions, and using symmetry to construct the left hand half:




c) i. For plane strain ( y = 0) and x > 0, The normal to the locus is at the corner between A and B
regions. Both x = Y + y/2 and x = 2Y - y must be satisfied. Solving simultaneously, x = (4/3)Y but y =
(2/3)Y
ii. Axisymmetric flow with y = z = -(1/2) x with x > 0, is satisfied everywhere in Region I, so x = Y +
y/2, with (2/3)Y ≤ x ≤ (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 m 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.

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?




6

,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 = (πdt)(d/2); = 2T/(πd2t) = 2x2./(π320.05) = 2.829 ksi
x = Pd/(2t), y = Pd/(4t) + 4.244 = x/2 + A, where A = 4.244 ksi
a) For Mises, substituting z = xy = yz = zx = 0 into the yield criterion, Eq. (2-12) 2Y2 = ( y -
z)2 + ( z - 2x)2 + ( x - y)2 + 6t xy2
Y y x y xy x xy
 x y
 x
y
Y y x y x xy
Substituting y = x/2 + A, 2 2
/2 + 2 - x /2 + x + + 3 xy - Y = 0
x 2 2 2
2
x (1/4 -1/2 + 1) + x(A - A) + (A + 3 xy - Y ) = 0
2 2 2 2
(3/4) x + (A + 3 xy - Y ) = 02 2 2
2
x + B = 0 where B = (4/3)(A + 3 xy - Y )
Substituting B = (4/3)(4.244 + 3x2.8292
2 - 202) = -2.098
x = 2.098, x = 45.8, x = Pd/(2 ),
P = (2 /d)sx = 45.8(2x0.050/3) = 1.528 ksi
b) For Tresca, we must find the principal stresses. 2 1/2
2 ]
1,2 = ( x + y)/2 ± (1/2)[( x - y) + 4 xy
Substituting y = x/2 +A, 2 1/2 2 2
2 = (3/4) x + A/2 ± (1/2)[ x /4 - As + A +
1,2 =2(3/4)
1/2 x
+ A/2 ± (1/2)[( x/2 - A) + 4 xy ]
4 xy ]
2 2 2 1/2
1 = (3/4) x + A/2 + (1/2)[ x /4 - A x + A + 4 ]
xy 2 1/2
2 2
2 = (3/4) x + A/2 - (1/2)[ x /4 - A x + A + 4 xy ]
There are two possibilities: 2 > 0, and 2 < 0. 1st
assume that s2 > 0. Then 1 - 0 = Y,
2 2 2 1/2
(3/4) x + A/2 + (1/2)[ x /4 - A x + A + 4 xy ] =Y
2 2 2 1/2
(3/4) x + A/2 + (1/2)[ x /4 - A x + A + 4 xy ] = Y
2 2 2 1/2
(3/4) x + 4.244/2 + (1/2)[ x /4 - 4.244 x + 4.244 + ] = 40
2 1/2
(3/2) x + 4.244 + [ x /4 - 4.244 x + 49.855] = 80
2 1/2 2
[ x /4 - 4.244 x + 49.855] = 75.756 -(3/2) x
2 2 2
x /42
- 4.244 x + 49.855] = [75.756 -(3/2) x] = 5738.9 -227.26 x +2.25 x
x [0.25 - 2.25] + [-4.244 +227.26]sx + 49.855 - 5738.9 = 0
2 x2 -223.3sx + 5689 = 0; x = {223.3 ± [223.32 -4x2x 5689] 1/2}/(2x2) =
7

,55.85 ± 16.49,
x = 72.34 or 39.36, The smaller value is correct Then P =
(2t/d) x = 39.36(2x0.050/3) = 1.312 ksi
Now we must check to see whether 2 > 0. Substituting A = 4.244, = 2.829 and x = 39.36 into
2 2 2 1/2
2 = (3/4) x + A/2 - (1/2)[ x /4 - A x + A + 4 xy ]
2 = (3/4)x39.36 + 4.244/2 - (1/2)[39.36 /4 - 4.244x39.36 + 4.2442 +
2
4x2.829 2]1/2 = 31.48. Therefore the solution for 2 > 0 is appropriate.
2-9 Calculate the ratio of / max for a) pure shear. b) uniaxial tension, and c) plane strain
tension. Assume the von Mises criterion.
Solution:
2 2 2 1/2 / 1/2
= √3
a) 1 = , 2 = , 3 = - , = {[ + (2 ]/2} , b) / 2= 2 2
2 +( /2) ]/2} , / = √(3/2)
c) 1 = , 2 = /2, 3 = 0, = {[( /2) +

A material yields under a biaxial stress state, 3 = -(1/2) 1, 2 = 0.
a) Assumingthe von Mises criterion, find d 1/ d 2.
b) What is the ratio of max /Y at yielding?
Solution: d 1/d 2 = [ 1 - ( 2 + 3)/2]/[ 2 - ( 3 + 1)/2] = [ 1 - ( - 1/2)/2]/[0
- (- 1/2 - 1)/2] = (5/4)/(3/4) = 5/3

2-11 A material is subjected to stresses in the ratio, 1 , 2 = 0.3 1 and 3 = -0.5 1. Find the
ratio of 1/Y at yielding using the a) Tresca and b) von Mises criteria.
Solution: a) For Tresca, 1 – (-0.5 1) = Y, 1 /Y= 2/3
2 2 2 1/2
b) For von Mises, {[(.3+.5) + (-.5 – 1) + (1-.3) ]/2} 1 = Y, 1 /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 1 vs. 2 in 3 = 0 space.
Solution: Divide stress space into regions with different conditions for yielding.
To evaluate C, consider an x-direction tension test. At yielding x = Y, y = 0, The
diameters of the two largest Mohr’s circle are Y. Y = Y/2 = C. C = 3/2Y




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