Metal Forming: Mechanics and Metallurgy / Edition 4

by William F. Hosford, Robert M. Caddell

Overview

This book helps the engineer understand the principles of metal forming and analyze forming problems—both the mechanics of forming processes and how the properties of metals interact with the processes. In this third edition, an entire chapter has been devoted to forming limit diagrams and various aspects of stamping and another on other sheet forming operations. Sheet testing is covered in a separate chapter. Coverage of sheet metal properties has been expanded. Interesting end-of-chapter notes have been added throughout, as well as references. More than 200 end-of-chapter problems are also included.

Product Details

ISBN-13:	9781107670969
Publisher:	Cambridge University Press
Publication date:	03/27/2014
Edition description:	New Edition
Pages:	344
Product dimensions:	7.01(w) x 10.00(h) x 0.71(d)

About the Author

William F. Hosford is a Professor Emeritus of Materials Science and Engineering at the University of Michigan. Professor Hosford is the author of more than 80 technical articles and numerous books, including The Mechanics of Crystals and Textured Polycrystals; Physical Metallurgy; Mechanical Behavior of Materials, 2nd edition; Materials Science: An Intermediate Text; Materials for Engineers and Reporting Results (with David VanAken).

Robert M. Caddell was a Professor of Mechanical Engineering at the University of Michigan, Ann Arbor.

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Cambridge University Press
978-0-521-88121-0 - 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.

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where A is the area on which the force acts.

   If the stress is the same everywhere,

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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 x-direction. A shear stress acting on the x-plane in the y-direction 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

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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.

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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

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Similarly

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Note that transformation of stresses requires two sine and/or cosine terms.

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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

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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

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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

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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.

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1.3. Unless τ_xy = τ_yx, there would not be a moment balance.

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1.4. Two orthogonal coordinate systems.

   Using equation 1.6,

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and

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These can be simplified to

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and

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1.3 PRINCIPAL STRESSES

It is always possible to find a set of axes along which the shear stress terms vanish. In this case σ₁, σ₂, and σ₃ are called the principal stresses. The magnitudes of the principal stresses, σ_p, are the roots of

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where I₁, I₂, and I₃ are called the invariants of the stress tensor. They are

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The first invariant I₁ = −p/3 where p is the pressure. I₁, I₂, and I₃ are independent of the orientation of the axes. Expressed in terms of the principal stresses, they are

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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, I₁ = 105 MPa, I₂ = –2050 MPa, I₃ = 0. From equation 1.10, σ³_p – 105 σ²_p + 2050σ_p + 0 = 0, so

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The principal stresses are the roots σ₁ = 79.1 MPa, σ₂ = 25.9 MPa, and σ₃ = σ_z = 0.

EXAMPLE 1.2: Repeat Example 1.1 with I₃ = 170,700.

SOLUTION: The principal stresses are the roots of σ³_p – 105σ²_p + 2050σ_p + 170,700 = 0. Since one of the roots is σ_z = σ₃ = - 40, σ_p + 40 = 0 can be factored out. This gives σ²_p - 105σ_p + 2050 = 0, so the other two principal stresses are σ₁ = 79.1 MPa, σ₂ = 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.

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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

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    These can be simplified with the trigonometric relations

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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

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The principal stresses, σ₁ and σ₂ are then the values of σ_x′ and σ_y′

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   A Mohr’s* circle diagram is a graphical representation of equations 1.16 and 1.17. They form a circle of radius (σ₁ - σ₂)/2 with the center at (σ₁ +σ₂)/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.

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1.6. Mohr’s circle diagram for stress.

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1.7. Three-dimensional Mohr’s circles for stresses.

   Using the Pythagorean theorem on the triangle in Figure 1.6,

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and

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   A three-dimensional stress state can be represented by three Mohr’s circles as shown in Figure 1.7. The three principal stresses σ₁, σ₂, and σ₃ 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 =(σ₁ - σ₃)/2 = [79.1 - (-40)]/2 = 59.6 MPa.

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1.8. Mohr’s circle for stress state in Example 1.2.

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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:

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An alternative definition* is that of true or logarithmic strain, ε defined by

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which on integrating gives

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   The true and engineering strains are almost equal when they are small. Expressing ε as ε = ln (ℓ/ℓ₀) = ln (1 + e) and expanding,

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   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 ℓ₀ is uniformly extended until its length ℓ = 2ℓ₀. Compute the values of the engineering and true strains.

(b) To what final length must a bar of length ℓ₀ be compressed if the strains are the same (except sign) as in part (a)?

SOLUTION:

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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:

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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 ℓ₀, w₀, t₀ is deformed to dimensions of ℓ, w, t.

(a) Calculate the volume strain, ε_v = ln(v/v₀) 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:

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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:

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As e gets larger the difference between ε and e become greater.

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1.10. Distortion of a two-dimensional element.

1.6 SMALL STRAINS

Figure 1.10 shows a small two-dimensional element, ABCD, deformed into AʹBʹCʹDʹ where the displacements are u and v. The normal strain, e_xx, is defined as

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Neglecting the rotation

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Similarly, e_yy = ∂v/∂y and e_zz = ∂w/∂z for a three-dimensional case.

   The shear strains are associated with the angles between AD and AʹDʹ and between AB and AʹBʹ. For small deformations

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The total shear strain is the sum of these two angles,

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Similarly,

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   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

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small shear strains form a tensor,

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   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,

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and

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The principal strains can be found from the Mohr’s circle equations for strains,

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Strains on other planes are given by

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and

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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.

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© Cambridge University Press

Table of Contents

1. Stress and strain; 2. Plasticity; 3. Strain hardening; 4. Plastic instability; 5. Temperature and strain-rate dependence; 6. Work balance; 7. Slab analysis and friction; 8. Friction and lubrication; 9. Upper-bound analysis; 10. Slip-line field analysis; 11. Deformation zone geometry; 12. Formability; 13. Bending; 14. Plastic anisotropy; 15. Cupping, redrawing and ironing; 16. Forming limit diagrams; 17. Stamping; 18. Hydroforming; 19. Other sheet forming operations; 20. Formability tests; 21. Sheet metal properties.

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"very good coverage of the principles of mechanical metallurgy...Recommended." - CHOICE

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