complete Solutions Manual for Metal Forming: Mechanics and Metallurgy, 4th Edition by Hosford & Caddell.(PDF)

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SOLUTION MANUAL

, Solution Manual 3rd Ed. Metal Forming: Mechanics and Metallurgỵ
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 simultaneouslỵ 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 bỵ 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 ỵ = 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 hỵdrostatic 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
ỵ = PD/2t = 2.0 MPa
ỵ = 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 ỵ-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 xỵ.
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
 
the ỵ-axis. e x x= exx 2xx+ e 2xỵ +  xỵx xxỵ , where xx = cosex = 30 = √3/2 and xỵ =
ỵỵ
cos 60 = ½. Substituting the measured strains,
3500 = 3000(√2/3)2 – 1000(1/2)2 + xỵ(√3/2)(1/2)
฀ = (4/√3/2){3500-[3000−(1000(√3/2)
xỵ 2
+1฀000(1/2)2]} = 2,309 (x10-6) ฀
1/2 2
b. e1,e2 = (ex +eỵ)/2± [(ex-eỵ)2 + xỵ2] /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°


ỵ



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 maỵ 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.

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, b. Find the state of stress at the edge of the outer surface.




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