STRESS STATE 2- D - PowerPoint PPT Presentation

1 / 19
About This Presentation
Title:

STRESS STATE 2- D

Description:

Taking the unstressed plane to be parallel to the x-y plane gives Figure 3-4. ... The transformation of stress equations now become It is important to note that . – PowerPoint PPT presentation

Number of Views:52
Avg rating:3.0/5.0
Slides: 20
Provided by: kalu
Learn more at: https://eng.fsu.edu
Category:

less

Transcript and Presenter's Notes

Title: STRESS STATE 2- D


1
STRESS STATE 2- D
2
Stresses acting on an elemental cube
  • It is convenient to resolve the stresses at a
    point into normal and shear components.
  • The stresses used to describe the state of a
    tri-dimensional body have two indices or
    subscripts. The first subscript indicates the
    plane (or normal to) in which the stress acts and
    the second indicates the direction in which the
    stress is pointing.

Figure 3-1 Stresses acting on an elemental
cube.
3
Description of Stress at a Point
  • As described in Sec.1-8, it is often convenient
    to resolve the stresses at a point into normal
    and shear components. In the general case the
    shear components are at arbitrary angles to the
    coordinates axes, so that it is convenient to
    resolve each shear stress further into two
    components. The general case shown in Fig. 3-1.
  • A shear stress is positive if it points in the
    positive direction on the positive face of a unit
    cube. (It is also positive if it points in the
    negative direction on the negative face of a unit
    cube.)

4
Figure 3-2. Sign convention for shear stress (a)
Positive (b) negative
5
Two Dimensions
Figure 3-3 Forces and Stresses related to
different sets of axes.
6
  • With the coordinate system shown, the stress s,
    acting in a direction parallel to F across area A
    is simply F/A. Because F has no component
    parallel to A, there is no shear stress acting on
    that plane. Now consider a plane located at
    angle which defines new coordinates axes in
    relation to the original x-y system. The forces
    F has components Fy and Fx acting on the plane
    whose area A equals A/cos ?. Thus, the stresses
    acting on the inclined plane are
  • and
  • The developments leading to Eqs. (3.3) and (3.4)
    have, in effect, transformed the stress ?y to a
    new set of coordinates axes.

(3.1)
(3.2)
7
Figure 3-4. The six components needed to
completely describe the state of stress
at a point.
  • If the three components of stress acting on one
    face of the element are all zero, then a state of
    plane stress exists. Taking the unstressed plane
    to be parallel to the x-y plane gives

(3.3)
8
  • Rotation of Coordinate Axes
  • The same state of plane stress may be described
    on any other coordinate system, such as
    in Fig. 3.5 (b).
  • This system is related to the original one by an
    angle of rotation q, and the values of the stress
    components change to , , and in
    the new coordinate system.
  • It is important to recognize that the new
    quantities do not represent a new state of
    stress, but rather an equivalent representation
    of the original one.
  • The values of the stress components in the new
    coordinate system may be obtained by considering
    the freebody diagram of a portion of the element
    as indicated by the dashed line Fig. 3.5 (a).

9
(a)
(b)
Figure 3-5. The three components needed to
describe a state of plane stress (a), and an
equivalent representation of the same state of
stress for a rotated coordinate system (b)
10
  • The resulting freebody is shown in Fig. 3.6.
    Equilibrium of forces in both the x and y
    directions provides two equations, which are
    sufficient to evaluate the unknown normal and
    shear stress components ? and ? on the inclined
    plane.
  • The stresses must first be multiplied by the
    unequal areas of the sides of the triangular
    element to obtain forces.
  • For convenience, the hypotenuse is taken to be of
    unit length, as is the thickness of the element
    normal to the diagram.

11
Figure 3-6. Stresses on an oblique plane
Summing the forces in the x- direction, and then
in the y-direction, gives two equations.
(3.4)
(3.5)
12
Solving for the unknowns s and t, and also
invoking some basic trigonometric identities,
yields The desired complete state of
stress in the new coordinate system may now be
obtained. Equations 3.6 and 3.7 give and
directly, and substitution of q 90o gives
.
(3.6)
(3.7)
13
Figure 3-7. Stress on oblique plane (two
dimensional)
Let Sx and Sy denote the x and y components of
the total stress acting on the inclined face. By
taking the summation of the forces in the x
direction and the y direction, we obtain
or
14
The components of Sx and Sy in the direction of
the normal stress s are and so that the
normal stress acting on the oblique plane is
given by
(3.8)
15
The shearing stress on the oblique plane is given
by
(3.9)
The stress may be found by substituting
qp/2 for q in Eq. (2-2), since is
orthogonal to
(3.10)
Equation (3.10) to (3.12) are the transformation
of stress equations which give the stresses in an
coordinate system if the stresses in the
xy coordinate system and the angle q are known.
16
To aid in computation, it is often convenient to
express Eqs. (3.10) to (3.12) in terms of the
double angle 2q. This can be done with the
following identities.
The transformation of stress equations now become
(3.11) (3.12) (3.13)
17
It is important to note that
. Thus the sum of the normal stresses on
two perpendicular planes is an invariant
quantity, that is, it is independent of
orientation or angle q. 1. The maximum and
minimum values of normal stress on the oblique
plane through point O occur when the shear
stresses is zero. 2. The maximum and minimum
values of both normal stress and shear stress
occur at angles which are 90 degrees apart. 3.
The maximum shear stress occurs at an angle
halfway between the maximum and minimum normal
stresses. 4. The variation of normal stress
and shear stress occurs in the form of a sine
wave, with a period of q 180o. These
relationships are valid for any state of stresses.
18
  • Poissons ratio is defined as the ratio between
    the lateral and the longitudinal strains. Both
    ?11 and ?22 are negative (decrease in length) and
    ?33 is positive. In order for Poissons ratio to
    be positive, the negative sign is used. Hence,
  • ? - ?11 - ?22
  • ?33 ?33
    (3.1)

Figure 3-2 Unit cube being extended in direction
Ox3
19
Since VVo
(3.2)
  • Substituting Eq. 3.2 into Eq. 3.1, we arrive at
  • ? 0.5
  • For the case in which there is no lateral
    contraction, ? is equal to zero. Poissons
    ratio for most metals is usually around 0.3
Write a Comment
User Comments (0)
About PowerShow.com