Solution Manual Metal Forming Mechanics and Metallurgy 4th edition_removed (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
Gauge A 3000x10-6
Gauge B 3500 x10-6
Gauge C 1000 x10-6
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 be on the x’ axis, the A gauge on the x-axis and the C gauge on
2 2
the y-axis. exx= exx xx+ e yy xy +  xy xx xy , where  xx = cosex = 30 = √3/2 and  xy =
cos 60 = ½. Substituting the measured strains,
3500 = 3000(√2/3)2 – 1000(1/2)2 + xy(√3/2)(1/2)
xy
 = (4/√3/2){3500-[3000−(1000(√3/2)
2
+1000(1/2)2]} = 2,309 (x10-6) 
1/2 2
b. e1,e2 = (ex +ey)/2± [(ex-ey)2 + xy2] /2 = (3000+1000)/2 ± [(3000-1000) +
23092]1/2/2 .e1 = 3530(x10-6), e2 = 470(x10-6), e3 = 0.
c)

x



2 1
2=60°


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
e1 = (1/E)(1 -  1); 1 = Ee1/(1-) = 205x109(3530x10-6)/(1-.292) = 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)]

A thin sheet of steel, 1-mm thick, is bent as described in Example 1-11. Assuming that E

= 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.


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, Solution: a. Substituting E = 205x109, t = 0.001,  = 2.0 and  = 0.29
Et Et
into  x = and  = , x = 56 MPa, , y = 16.2 MPa
2(1 −  2 ) y
2(1 −  2 )
Et
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.
Assuming 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 -2 (32 + 1)]2 and e3 = (1/E)[3 - (1 + 2)],
w = 1/(2E)[ +  +  - 2(  +  +  )] =
1 2 3 2 3 3 1 1 2
(1/(2x205x10 )[300 +250 + 200 –(2x0.29)(-200x250 – 300x250 + 250+300)]x1012 =
9 2 2 2

400J/m3

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 - 2) =(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 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.



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, 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)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 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
b. (x/2 - 0)2 + (0 - x)2+ (x -x/2)2 = 6k2, (3/2) x2 = 6k2, x = 2k = PD/(2t), t =
PD/(4k) which is identical to part a. t = 8.75 mm
 = 2(x + y ) /3
2 2




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
Tresca criterion predicts yielding when  /2  ( /2)2 +  2 = Y . If  /2 − ( /2)2 +  2 <
0, min = − ( /2)2 +  2 , so the Tresca criterion predicts yielding when 2 ( /2)2 +  2
b) {2[ /2 − ( /2)2 +  2 ]2 +[ 2 ( /2)2 +  2 ]2}1/2 = √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.
d) an expression in the form of y = f(K,y,n) assuming  = K n .

 Solution: a. If z = 0, y = - ex  = 2( 2 +  2) /3= = 1.154y
x y


b. x = 0, z = -(1/2)y;  = (1/2)[( y −  y /2)
2
+ ( y /2 − 0)2 + (0 −  y )2 ] = y/1.154

c. w = ∫ydy
n+1/2

d.  y = 4 /3 = K = 4 /3K( ) y ) = (4/3)
n n
ey

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




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