Transcript Stresses acting on an elemental cube

STRESS STATE 2- D
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Stresses acting on an elemental cube
It is convenient to resolve the
stresses at a point into normal and
shear components.
Figure 3-1 Stresses acting on
an elemental cube.
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.
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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.)
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Figure 3-2. Sign convention for shear stress (a) Positive; (b)
negative
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Two Dimensions
Figure 3-3 Forces and Stresses related to different sets of axes.
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• 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:
Fy '
s y'
F
 '  cos 2   s y cos 2 
A
A
(3.1)
 x' 
Fx ' F
 sin  cos  s y sin  cos
'
A
A
(3.2)
and
• The developments leading to Eqs. (3.3) and (3.4) have, in
effect, transformed the stress sy to a new set of coordinates
axes.
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 s x  xy  xz 


s y  yz 



sz 

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
s z   zx   yz  0
(3.3)
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Rotation of Coordinate Axes
• The same state of plane stress may be described on any other
coordinate system, such as x  y in Fig. 3.5 (b).
• This system is related to the original one by an angle of rotation ,
and the values of the stress components change to s x , s y , and  xy
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).
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(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)
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• 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 s 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.
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Figure 3-6. Stresses on an oblique plane
Summing the forces in the x- direction, and then in the y-direction,
gives two equations.
s cos   sin   s x cos   xy sin   0
(3.4)
s sin    cos  s y sin    xy cos  0
(3.5)
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Solving for the unknowns s and , and also invoking some basic
trigonometric identities, yields  xy
s x s y s x s y
s

cos 2   xy sin 2
2
2
 
s x s y
2
sin 2   xy cos 2
(3.6)
(3.7)
The desired complete state of stress in the new coordinate system may
now be obtained. Equations 3.6 and 3.7 give s x and  xy directly, and
substitution of  + 90o gives s y .
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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
S x A  s x Al   xy Am
S y  s y sin    xy cos 
or
Sy A  s
y
Am   xy Al
S x  s x cos    xy sin 
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The components of Sx and Sy in the direction of the normal stress s
are S xN  S x cos and S yN  S y sin 
so that the normal stress acting on the oblique plane is given by
s x '  S x cos   S y sin 
(3.8)
s x'  s x cos 2   s y sin 2   2 xy sin  cos
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The shearing stress on the oblique plane is given by
 xy  S y cos  S x sin 

(3.9)

 xy   xy cos 2   sin 2   s y  s x sin  cos
The stress s y may be found by substituting p/2 for 
in Eq. (2-2), since s y is orthogonal to s x
p
p
p
p
s y  s x cos 2     s y sin 2     2 xy sin    cos  

2
and since sin   p   cos

2

2

2

2
p

and cos     sin 
2

s y  s x sin 2   s y cos 2   2 xy sin  cos
(3.10)
Equation (3.10) to (3.12) are the transformation of stress equations
which give the stresses in an xy coordinate system if the stresses
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in the xy coordinate system and the angle  are known.
To aid in computation, it is often convenient to express Eqs. (3.10) to
(3.12) in terms of the double angle 2. This can be done with the
following identities.
cos 2  1
2
cos  
2
1  cos 2
sin 2  
2
2 sin  cos  sin 2
cos 2   sin 2   cos 2
The transformation of stress equations now become
s x 
s y 
s x s y s x s y
2
s x s y
 xy 
2
s y s x
2


2
s x s y
2
cos 2   xy sin 2
(3.11)
cos 2   xy sin 2
(3.12)
sin 2   xy cos 2
(3.13)
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It is important to note that s x '  s y '  s x  s y . Thus the sum of the
normal stresses on two perpendicular planes is an invariant quantity,
that is, it is independent of orientation or angle .
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  = 180o. These relationships are
valid for any state of stresses.
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Poisson’s 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 Poisson’s 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
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V 1
V  (1  11 )(1   22 )(1   33 )
V  1  11   22   33
Since V=Vo

 11   22   33  0
2 11   33
(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. Poisson’s ratio for most metals is usually around 0.3
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