Stress and Strain

Lecture No. 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. STRESS
Stress is defined as the intensity of force, F, at a point.
? = ?F/? A as ? A ? 0, …………………………………………………(1)
where A is the area on which the force acts.
If the stress is the same everywhere,
? = F/A. ………………………………………………………………...(2)
There are nine components of stress as shown in Figure 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 plane on which the force acts and the second is the direction of the force. For example, ?xx is a tensile stress in the x-direction. A shear stress acting on the x-plane in the y-direction is denoted ?xy.
Repeated subscripts (e.g., ?xx , ?yy, ?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., ?zx, ?xy, ?yz) denote shear stresses. A state of stress in tensor notation is expressed as

?_ij=|?( ?_xx&?_yx&?_zx@?_xy&?_yy&?_zy@?_xz&?_yz&?_zz )| ………………………………………..(3)


Fig. 1: Nine components of stress acting on an infinitesimal element.
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 ? . For example, ?x ? ?xx and ?xy ? ?xy.
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 ?y = Fy/Ay onto the x? and y? axes as shown in Figure 2.
The force F? y in the y? direction is F? y = Fy cos ? and the area normal to y? is Ay? = Ay/ cos ?, so
?y? = Fy?/Ay? = Fy cos ?/(Ay/ cos ?) = ?y cos2 ?. ………………………..(4a)
Similarly
?y?x? = Fx?/Ay? = Fy sin ?/(Ay/ cos ?) = ?y cos ? sin ?. …………………….(4b)
Note that transformation of stresses requires two sine and/or cosine terms.
Pairs of shear stresses with the same subscripts in reverse order are always equal (e.g., ?i j = ?ji ). This is illustrated in Figure 3 by a simple moment balance on an infinitesimal element. Unless ?i j = ?ji , there would be an infinite rotational acceleration. Therefore
?i j = ? ji . ……………………………………………………………….(5)


Fig.2: The stresses acting on a plane, A?, under a normal stress, ?y.


Fig. 3: Unless ?xy = ?yx, there would not be a moment balance.
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
? i j =?_(n=1)^3??_(m=1)^3??l_im l_jn ?_mn ?……………………………………………..(6)
Here, the term lim is the cosine of the angle between the i and m axes and the term ljn is the cosine of the angle between the j and n axes. This is often written as
?i j = lim ljn ?mn, …………………………………………………………(7)
with the summation implied. Consider transforming stresses from the x, y, z axis system to the x?, y?, z? system shown in Figure 4.
Using equation 6,
.........................................(8a)
and
…………………………..(8b)


Fig 4: Two orthogonal coordinate systems.


These can be simplified to
……(9a)

and
………..(9b)
3. PRINCIPAL STRESSES
It is always possible to find a set of axes along which the shear stress terms vanish. In this case ?1, ?2, and ?3 are called the principal stresses. The magnitudes of the principal stresses, ?p, are the roots of
……………………………………………..(10)
where I1, I2, and I3 are called the invariants of the stress tensor. They are
………………..(11)

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
I1 = ?1 + ?2 + ?3,
I2 = ??2?3 ? ?3?1 ? ?1?2
I3 = ?1?2?3. ………………...…………………………………………..(12)









EXAMPLE 1:
Consider a stress state with ?x = 70MPa, ?y = 35MPa, ?xy = 20, ?z = ?zx = ?yz = 0. Find the principal stresses.
SOLUTION:
Using equations 11, I1 = 105 MPa, I2 = –2050 MPa, I3 = 0. From equation 10,
The principal stresses are the roots ?1 = 79.1MPa, ?2 = 25.9MPa, and ?3 = ?z = 0.