Receive a 20% Discount on All Purchases Directly Through IGI Global's Online Bookstore.

Additionally, libraries can receive an extra 5% discount. Learn More

Additionally, libraries can receive an extra 5% discount. Learn More

DOI: 10.4018/978-1-4666-6505-7.ch005

Chapter Preview

TopStress is defined as force per unit area. As was explained in chapter 4, the stress at a point is uniquely defined by specifying the magnitude and direction of a single vector having the units of force per unit elemental area, Chapter 4, Figure 1(a). In engineering, it is found more useful to define stress at a point in terms of the direct and shear components of stress acting normally and tangentially to the six faces of an elemental prism. The components of stress are then three normal stresses σ_{xx}, σ_{yy}, σ_{zz} and three pairs of reciprocally equal tangential stresses, σ_{xy} = σ_{yx}; σ_{xz} = σ_{zx}; σ_{yz} = σ_{zy}, Chapter 4, Figure 1(b).

As stated in Chapter 4, it is possible to have a prism which is oriented in such a way that the shear components of stress are zero. The normal stresses so obtained are called principal stresses. The relationship between the principal stresses and the corresponding conjugate stresses can be obtained by considering the equilibrium of forces acting on an oblique plane which is arbitrarily inclined to the faces of the parallelopiped as in Figure1 in Chapter 4 and with direction cosines l, m and n. This can be derived more easily for a two-dimensional state of stress.

Let us consider a 2-dimensional state of stress as shown in Figure 1. It is assumed that the stresses normal to the plane of this paper are zero i.e. σ_{z} = 0; τ_{xz} = 0; τ_{zy} = 0. Consider the equilibrium of forces on the plane which makes an angle with the vertical, Figure 2.

If we divide by *ΔA _{n}*, Equation 5.1a becomes

From trigonometry, we have the following identities

;Substituting the above identifies in Equation 5.1b and rearranging we have

Considering forces parallel to the inclined plane we have

Therefore

Dividing by *ΔA _{n}* gives

Using trigonometric identities we have

Differentiate Equation 5.1c with respect to θ

i.e.

This is the plane at which σ_{n} is maximum or minimum i.e. the principal plane. The principal stresses represent the maximum and minimum values of all the stresses at the given point: the axes are commonly designated so that σ_{1} ≥ σ_{2} ≥ σ_{3}.

For the plane stress loading, the Principal Planes are the two planes where the normal stress (σ) is the maximum or minimum. Usually

*•*There are no shear stresses on principal planes.

*•*The two planes are mutually perpendicular.

*•*The orientations of the planes (θ

_{p}) are given by:*5.3*

The Principal stresses are the normal stresses (σ) acting on the principal planes and are given as

Similarly it is possible to have three planes on which the shear rather than the normal stresses are at maximum. Such stresses are called the maximum shear stresses. For plane stress loading, the maximum shear stress occurs on two mutually perpendicular planes and the orientations of the two planes (θ_{s}) are given by:

The planes are related by

The values of τ_{max} can also be obtained in terms of principal stresses using equation 5.2

For a 3-D case, the principal shear stresses are given by

; ;The angular distortions on the planes of maximum shear are

; ;**Problem 5.1:**Stress at a Point.

The principal stresses on a prism of soil are 20 kN/m^{2} and 12 kN/m^{2}. Determine the normal and shear stresses on a plane which makes an angle of 30^{o} with the horizontal plane.

Solution:

Using Equation 5.1c we have

Similarly Equation 5.2 gives

Search this Book:

Reset

Copyright © 1988-2019, IGI Global - All Rights Reserved