Metal Forming Mechanics and Metallurgy (4th Ed.) – Study Solutions + Worked Problems

THEPROWESS STUVIA
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
 ij = −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;  – 222 -126 -119 = 0. A trial and error solution gives  -= 13.04.
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 Factoring out 13.04, 2 -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
= πsR3/2. Solving for = s, s = 2T/(πR3) = 2(400N)/(π0.0253) = 16 MPa
The axial stress is .08MN/(π0.0252) = 4.07 MPa
1,2 = 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)[ - ( 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




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,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
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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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