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Cambridge University Press
9780521881210  Metal Forming  Mechanics and Metallurgy  by William F. Hosford and Robert M. Caddell
Excerpt
1 Stress and Strain
An understanding of stress and strain is essential for analyzing metal forming operations. Often the words stress and strain are used synonymously by the nonscientific public. In engineering, however, stress is the intensity of force and strain is a measure of the amount of deformation.
1.1 STRESS
Stress is defined as the intensity of force, F, at a point.
where A is the area on which the force acts.
If the stress is the same everywhere,
There are nine components of stress as shown in Figure 1.1. A normal stress component is one in which the force is acting normal to the plane. It may be tensile or compressive. A shear stress component is one in which the force acts parallel to the plane.
Stress components are defined with two subscripts. The first denotes the normal to the planeon which the force acts and the second is the direction of the force.^{1} For example, σ_{xx} is a tensile stress in the xdirection. A shear stress acting on the xplane in the ydirection is denoted σ_{xy}.
Repeated subscripts (e.g., σ_{xx}, σ_{yy}, σ_{zz}) indicate normal stresses. They are tensile if both subscripts are positive or both are negative. If one is positive and the other is negative, they are compressive. Mixed subscripts (e.g., σ_{zx}, σ_{xy}, σ_{yz}) denote shear stresses. A state of stress in tensor notation is expressed as
where i and j are iterated over x, y, and z. Except where tensor notation is required, it is simpler to use a single subscript for a normal stress and denote a shear stress by τ. For example, σ_{x} ≡ σ_{xx} and τ_{xy} ≡ σ_{xy}.
1.1. Nine components of stress acting on an infinitesimal element.
1.2 STRESS TRANSFORMATION
Stress components expressed along one set of axes may be expressed along any other set of axes. Consider resolving the stress component σ_{y} = F_{y}/A_{y} onto the xʹ and yʹ axes as shown in Figure 1.2.
The force F′_{y} in the yʹ direction is F′_{y} = F_{y} cos θ and the area normal to yʹ is A_{y′} = A_{y}/cos θ, so
Similarly
Note that transformation of stresses requires two sine and/or cosine terms.
1.2. The stresses acting on a plane, A′, under a normal stress, σ_{y}.
Pairs of shear stresses with the same subscripts in reverse order are always equal (e.g., τ_{ij} = τ_{ji}). This is illustrated in Figure 1.3 by a simple moment balance on an infinitesimal element. Unless τ_{ij} = τ_{ji}, there would be an infinite rotational acceleration. Therefore
The general equation for transforming the stresses from one set of axes (e.g., n, m, p) to another set of axes (e.g., i, j, k) is
Here, the term ℓ_{im} is the cosine of the angle between the i and m axes and the term ℓ_{jn} is the cosine of the angle between the j and n axes. This is often written as
with the summation implied. Consider transforming stresses from the x, y, z axis system to the x′, y′, z′ system shown in Figure 1.4.
1.3. Unless τ_{xy} = τ_{yx}, there would not be a moment balance.
1.4. Two orthogonal coordinate systems.
Using equation 1.6,
and
These can be simplified to
and
1.3 PRINCIPAL STRESSES
It is always possible to find a set of axes along which the shear stress terms vanish. In this case σ_{1}, σ_{2}, and σ_{3} are called the principal stresses. The magnitudes of the principal stresses, σ_{p}, are the roots of
where I_{1}, I_{2}, and I_{3} are called the invariants of the stress tensor. They are
The first invariant I_{1} = −p/3 where p is the pressure. I_{1}, I_{2}, and I_{3} are independent of the orientation of the axes. Expressed in terms of the principal stresses, they are
EXAMPLE 1.1: Consider a stress state with σ_{x} = 70 MPa, σ_{y} = 35 MPa, τ_{xy} = 20, σ_{z} = τ_{zx} = τ_{yz} = 0. Find the principal stresses using equations 1.10 and 1.11.
SOLUTION: Using equations 1.11, I_{1} = 105 MPa, I_{2} = –2050 MPa, I_{3} = 0. From equation 1.10, σ^{3}_{p} – 105 σ^{2}_{p} + 2050σ_{p} + 0 = 0, so
The principal stresses are the roots σ_{1} = 79.1 MPa, σ_{2} = 25.9 MPa, and σ_{3} = σ_{z} = 0.
EXAMPLE 1.2: Repeat Example 1.1 with I_{3} = 170,700.
SOLUTION: The principal stresses are the roots of σ^{3}_{p} – 105σ^{2}_{p} + 2050σ_{p} + 170,700 = 0. Since one of the roots is σ_{z} = σ_{3} =  40, σ_{p} + 40 = 0 can be factored out. This gives σ^{2}_{p}  105σ_{p} + 2050 = 0, so the other two principal stresses are σ_{1} = 79.1 MPa, σ_{2} = 25.9 MPa. This shows that when σ_{z} is one of the principal stresses, the other two principal stresses are independent of σ_{z}.
1.4: MOHR’S CIRCLE EQUATIONS
In the special cases where two of the three shear stress terms vanish (e.g., τ_{yx} = τ_{zx} = 0), the stress σ_{z} normal to the xy plane is a principal stress and the other two principal stresses lie in the xy plane. This is illustrated in Figure 1.5.
1.5. Stress state for which the Mohr’s circle equations apply.
For these conditions ℓ_{x′z} = ℓ_{y′z} = 0, τ_{yz} = τ_{zx} = 0, ℓ_{x′x} = ℓ_{y′y} = cos ɸ, and ℓ_{x′y} =  ℓ_{y′x} = sin ɸ. Substituting these relations into equations 1.9 results in
These can be simplified with the trigonometric relations
If τ_{x′y′} is set to zero in equation 1.14a, ɸ becomes the angle θ between the principal axes and the x and y axes. Then
The principal stresses, σ_{1} and σ_{2} are then the values of σ_{x′} and σ_{y′}
A Mohr’s* circle diagram is a graphical representation of equations 1.16 and 1.17. They form a circle of radius (σ_{1}  σ_{2})/2 with the center at (σ_{1} +σ_{2})/2 as shown in Figure 1.6. The normal stress components are plotted on the ordinate and the shear stress components are plotted on the abscissa.
1.6. Mohr’s circle diagram for stress.
1.7. Threedimensional Mohr’s circles for stresses.
Using the Pythagorean theorem on the triangle in Figure 1.6,
and
A threedimensional stress state can be represented by three Mohr’s circles as shown in Figure 1.7. The three principal stresses σ_{1}, σ_{2}, and σ_{3} are plotted on the ordinate. The circles represent the stress state in the 1–2, 2–3, and 3–1 planes.
EXAMPLE 1.3: Construct the Mohr’s circle for the stress state in Example 1.2 and determine the largest shear stress.
SOLUTION: The Mohr’s circle is plotted in Figure 1.8. The largest shear stress is τ_{max} =(σ_{1}  σ_{3})/2 = [79.1  (40)]/2 = 59.6 MPa.
1.8. Mohr’s circle for stress state in Example 1.2.
1.9. Deformation, translation, and rotation of a line in a material.
1.5 STRAIN
Strain describes the amount of deformation in a body. When a body is deformed, points in that body are displaced. Strain must be defined in such a way that it excludes effects of rotation and translation. Figure 1.9 shows a line in a material that has been that has been deformed. The line has been translated, rotated, and deformed. The deformation is characterized by the engineering or nominal strain, e:
An alternative definition* is that of true or logarithmic strain, ε defined by
which on integrating gives
The true and engineering strains are almost equal when they are small. Expressing ε as ε = ln (ℓ/ℓ_{0}) = ln (1 + e) and expanding,
There are several reasons why true strains are more convenient than engineering strains. The following examples indicate why.
EXAMPLE 1.4:
(a) A bar of length ℓ_{0} is uniformly extended until its length ℓ = 2ℓ_{0}. Compute the values of the engineering and true strains.
(b) To what final length must a bar of length ℓ_{0} be compressed if the strains are the same (except sign) as in part (a)?
SOLUTION:
EXAMPLE 1.5: A bar 10 cm long is elongated to 20 cm by rolling in three steps: 10 cm to 12 cm, 12 cm to 15 cm, and 15 cm to 20 cm.
(a) Calculate the engineering strain for each step and compare the sum of these with the overall engineering strain.
(b) Repeat for true strains.
SOLUTION:
With true strains, the sum of the increments equals the overall strain, but this is not so with engineering strains.
EXAMPLE 1.6: A block of initial dimensions ℓ_{0}, w_{0}, t_{0} is deformed to dimensions of ℓ, w, t.
(a) Calculate the volume strain, ε_{v} = ln(v/v_{0}) in terms of the three normal strains, ε_{ℓ}, ε_{w} and ε_{t}.
(b) Plastic deformation causes no volume change. With no volume change, what is the sum of the three normal strains?
SOLUTION:
Examples 1.4, 1.5, and 1.6 illustrate why true strains are more convenient than engineering strains.
1. True strains for an equivalent amount of tensile and compressive deformation are equal except for sign.
2. True strains are additive.
3. The volume strain is the sum of the three normal strains.
EXAMPLE 1.7: Calculate the ratio of ε/e for e = 0.001, 0.01, 0.02, 0.05, 0.1, and 0.2.
SOLUTION:
As e gets larger the difference between ε and e become greater.
1.10. Distortion of a twodimensional element.
1.6 SMALL STRAINS
Figure 1.10 shows a small twodimensional element, ABCD, deformed into AʹBʹCʹDʹ where the displacements are u and v. The normal strain, e_{xx}, is defined as
Neglecting the rotation
Similarly, e_{yy} = ∂v/∂y and e_{zz} = ∂w/∂z for a threedimensional case.
The shear strains are associated with the angles between AD and AʹDʹ and between AB and AʹBʹ. For small deformations
The total shear strain is the sum of these two angles,
Similarly,
This definition of shear strain, γ, is equivalent to the simple shear measured in a torsion of shear test.
1.7 THE STRAIN TENSOR
If tensor shear strains ε_{ij} are defined as
small shear strains form a tensor,
Because small strains form a tensor, they can be transformed from one set of axes to another in a way identical to the transformation of stresses. Mohr’s circle relations can be used. It must be remembered, however, that ε_{ij} = γ_{ij}/2 and that the transformations hold only for small strains. If γ_{yz} = γ_{zx} = 0,
and
The principal strains can be found from the Mohr’s circle equations for strains,
Strains on other planes are given by
and
1.8 ISOTROPIC ELASTICITY
Although the thrust of this book is on plastic deformation, a short treatment of elasticity is necessary to understand springback and residual stresses in forming processes.
© Cambridge University Press
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