×

Close

Type:
**Note**Offline Downloads:
**475**Views:
**8116**Uploaded:
**11 months ago**

Touch here to read

Page-1

Topic:

Module2/Lesson2
Module 2: Analysis of Stress
2.2.1 PRINCIPAL STRESS IN THREE DIMENSIONS
For the three-dimensional case, for principal stresses it is required that three planes of zero
shear stress exist, that these planes are mutually perpendicular, and that on these planes the
normal stresses have maximum or minimum values. As discussed earlier, these normal
stresses are referred to as principal stresses, usually denoted by s1, s2 and s3. The largest
stress is represented by s1 and the smallest by s3.
Again considering an oblique plane
Equation (2.25).
x ¢ , the normal stress acting on this plane is given by the
s x¢ = sx l2 + sy m2 + sz n2 + 2 (txy lm + tyz mn + txz ln)
(2.27)
The problem here is to determine the extreme or stationary values of s x¢ . To accomplish
this, we examine the variation of s x¢ relative to the direction cosines. As l, m and n are not
independent, but connected by l2 + m2 + n2 = 1, only l and m may be regarded as
independent variables.
Thus,
¶s x '
= 0,
¶l
¶s x '
=0
¶m
(2.27a)
Differentiating Equation (2.27), in terms of the quantities in Equations (2.22a), (2.22b),
(2.22c), we obtain
¶n
= 0,
¶l
¶n
Ty + Tz
= 0,
¶m
T x+ T z
(2.27b)
From n2 = 1 - l2 - m2, we have
¶n
l
=¶l
n
and
¶n
m
=¶m
n
Introducing the above into Equation (2.27b), the following relationship between the
components of T and n is determined
1
Applied Elasticity for Engineers
T.G.Sitharam & L.GovindaRaju

Module2/Lesson2
T x T y Tz
=
=
l
m
n
(2.27c)
These proportionalities indicate that the stress resultant must be parallel to the unit normal
and therefore contains no shear component. Therefore from Equations (2.22a), (2.22b),
(2.22c) we can write as below denoting the principal stress by s P
Tx = s P l
Ty = s P m
T z = sP n
(2.27d)
These expressions together with Equations (2.22a), (2.22b), (2.22c) lead to
(sx - sP)l + txy m + txz n = 0
txy l+(sy - sP) m + tyz n = 0
(2.28)
txz l + tyz m + (sz - sP) n = 0
A non-trivial solution for the direction cosines requires that the characteristic determinant
should vanish.
(s x - s P )
t xy
t xz
t xy
t xz
(s y - s P )
t yz
t yz
(s z - s P )
=0
(2.29)
Expanding (2.29) leads to s P3 - I 1s P2 + I 2s P - I 3 = 0
(2.30)
where I1 = sx + sy + sz
(2.30a)
I2 = sx sy + sy sz + szsx - t 2xy - t 2yz -t 2xz
(2.30b)
s x t xy t xz
I3 = t xy s y t yz
t xz t yz s z
(2.30c)
The three roots of Equation (2.30) are the principal stresses, corresponding to which are
three sets of direction cosines that establish the relationship of the principal planes to the
origin of the non-principal axes.
2.2.2 STRESS INVARIANTS
Invariants mean those quantities that are unexchangeable and do not vary under different
conditions. In the context of stress tensor, invariants are such quantities that do not change
with rotation of axes or which remain unaffected under transformation, from one set of axes
2
Applied Elasticity for Engineers
T.G.Sitharam & L.GovindaRaju

Module2/Lesson2
to another. Therefore, the combination of stresses at a point that do not change with the
orientation of co-ordinate axes is called stress-invariants. Hence, from Equation (2.30)
sx + sy + sz = I1 = First invariant of stress
sxsy + sysz + szsx - t 2xy - t 2yz - t 2zx = I2 = Second invariant of stress
sxsysz - sxt 2yz - syt 2xz - szt 2xy + 2txy tyz txz = I3 = Third invariant of stress
2.2.3 EQUILIBRIUM OF A DIFFERENTIAL ELEMENT
Figure 2.11(a) Stress components acting on a plane element
When a body is in equilibrium, any isolated part of the body is acted upon by an equilibrium
set of forces. The small element with unit thickness shown in Figure 2.11(a) represents part
3
Applied Elasticity for Engineers
T.G.Sitharam & L.GovindaRaju

Module2/Lesson2
of a body and therefore must be in equilibrium if the entire body is to be in equilibrium. It is
to be noted that the components of stress generally vary from point to point in a stressed
body. These variations are governed by the conditions of equilibrium of statics. Fulfillment
of these conditions establishes certain relationships, known as the differential equations of
equilibrium. These involve the derivatives of the stress components.
Assume that sx, sy, txy, tyx are functions of X, Y but do not vary throughout the thickness
(are independent of Z) and that the other stress components are zero.
Also assume that the X and Y components of the body forces per unit volume, Fx and Fy,
are independent of Z, and that the Z component of the body force Fz = 0. As the element is
very small, the stress components may be considered to be distributed uniformly over each
face.
Now, taking moments of force about the lower left corner and equating to zero,
¶s y ö Dx æ
¶t
ö
Dy
1 æ
+ (t xy Dy ) - çç s y +
Dy ÷÷Dx
+ ççt yx + yx Dy ÷÷DxDy
2
2 è
¶y
2 è
¶y
ø
ø
¶t
æ
ö
¶s x ö Dy
Dx
1
æ
- ççt xy + xy Dx ÷÷DxDy + ç s x +
Dx ÷ Dy
+ s y Dx
- t yx Dx +
¶x
¶x
2
2
2
è
ø
è
ø
(Fx DyDx ) Dy - ( Fy DxDy ) Dx = 0
2
2
- (s x Dy )
Neglecting the higher terms involving Dx, and Dy and simplifying, the above expression is
reduced to
txy Dx Dy = tyx Dx Dy
or
txy = tyx
In a like manner, it may be shown that
tyz = tzy and txz = tzx
Now, from the equilibrium of forces in x-direction, we obtain
¶t yx ö
æ
¶s x ö
æ
-sx Dy + ç s x +
Dx ÷Dy + ççt yx +
Dy ÷÷Dx - t yx Dx + Fx DxDy = 0
¶x
¶y
è
ø
è
ø
Simplifying, we get
or
¶s x ¶t yx
+
+ Fx = 0
¶x
¶y
¶s x ¶t xy
+
+ Fx = 0
¶x
¶y
4
Applied Elasticity for Engineers
T.G.Sitharam & L.GovindaRaju

## Leave your Comments

## GARIKAPATI RAMBABU

1 month ago02## Venni Sadhana

1 month ago00