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Unformatted text preview: esulting expression for xy satisfies Eq. (b) and is identical with the result previously obtained.
Method 3. The problem may be treated by superimposing the polynomials £ 2 and £ 4,
a2 = c2 = a4 = b4 = c4 = e4 = 0
Thus,
d4 3
xy
6 £ = £ 2 + £ 4 = b2xy +
The corresponding stress components are
x = d4xy, y = 0, xy =  b2  d4 2
y
2 It is seen that the foregoing satisfies the second condition of Eqs. (a).
The first of Eqs. (a) leads to d4 =  2b2/h2. We then obtain
xy =  b2 ¢ 1  y2
h2 ≤ which
when
substituted
into
condition
(b)
results
in
b2 =  3P/4ht = Ph2/2I. As before, xy is as given in Eqs. (3.21).
Observe that the stress distribution obtained is the same as that
found by employing the elementary theory. If the boundary forces result
in a stress distribution as indicated in Fig. 3.5b, the solution is exact.
Otherwise, the solution is not exact. In any case, recall, however, that
SaintVenant’s principle permits us to regard the result as quite accurate
for sections away from the ends.
Section 5.4 illustrates the determination of the displacement field
after derivation of the curvature–moment relation. 3.7 THERMAL STRESSES Consider the consequences of increasing or decreasing the uniform temperature of
an entirely unconstrained elastic body. The resultant expansion or contraction occurs in such a way as to cause a cubic element of the solid to remain cubic, while experiencing changes of length on each of its sides. Normal strains occur in each
direction unaccompanied by normal stresses. In addition, there are neither shear
strains nor shear stresses. If the body is heated in such a way as to produce a
nonuniform temperature field, or if the thermal expansions are prohibited from
108 Chapter 3 TwoDimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 109 taking place freely because of restrictions placed on the boundary even if the temperature is uniform, or if the material exhibits anisotropy in a uniform temperature
field, thermal stresses will occur. The effects of such stresses can be severe, especially since the most adverse thermal environments are often associated with design
requirements involving unusually stringent constraints as to weight and volume.
This is especially true in aerospace applications, but is of considerable importance,
too, in many everyday machine design applications.
Solution of thermal stress problems requires reformulation of the stress–strain
relationships accomplished by superposition of the strain attributable to stress and
that due to temperature. For a change in temperature T(x, y), the change of length,
L, of a small linear element of length L in an unconstrained body is L = LT.
Here , usually a positive number, is termed the coefficient of linear thermal expansion. The thermal strain t associated with the free expansion at a point is then
t T = (3.22) The total x and y strains, x and y, are obtained by adding to the thermal strains of
the type described, the strains due to stress resulting from external forces:
x = 1
1
E y = 1
1
E xy = x  y2 + T y  x2 + T (3.23a) xy G In terms of strain components, these expressions become
x = E
1 y = E
1 xy =G 2 2 1
1 x + y2  ET
1 y + x2  ET
1 (3.23b) xy Because free thermal expansion results in no angular distortion in an isotropic
material, the shearing strain is unaffected, as indicated. Equations (3.23) represent
modified strain–stress relations for plane stress. Similar expressions may be written
for the case of plane strain. The differential equations of equilibrium (3.6) are
based on purely mechanical considerations and are unchanged for thermoelasticity.
The same is true of the strain–displacement relations (2.3) and the compatibility
equation (3.8), which are geometrical in character. Thus, for given boundary conditions (expressed either as surface forces or displacements) and temperature distribution, thermoelasticity and ordinary elasticity differ only to the extent of the
strain–stress relationship.
By substituting the strains given by Eq. (3.23a) into the equation of compatibility (3.8), employing Eq. (3.6) as well, and neglecting body forces, a compatibility
equation is derived in terms of stress:
3.7 Thermal Stresses 109 ch03.qxd 12/20/02 7:20 AM Page 110 ¢ 02
02
+ 2 ≤1
0x2
0y + x y + ET2 = 0 (3.24) Introducing Eq. (3.13), we now have §4 £ + E §2 T = 0 (3.25) This expression is valid for plane strain or plane stress provided that the body
forces are negligible.
It has been implicit in treating the matter of thermoelasticity as a superposition
problem that the distribution of stress or strain plays a negligible role in influencing
the temperature field [Refs. 3.4 and 3.5]. This lack of coupling enables the temperature field to be determined independently of any consideration of stress or strain. If
the effect of the temperature distribution on material properties cannot be disregarded, the equations become coupled and analytical solutions are significantly
more complex, occupying an area of considerable interest and importance. Numerical solutions can, however, be obtained...
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This note was uploaded on 01/09/2011 for the course ME 201 taught by Professor Yok during the Spring '08 term at Boğaziçi University.
 Spring '08
 yok

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