This preview shows page 1. Sign up to view the full content.
Unformatted text preview: in a relatively simple manner through the
use of finite difference methods.
EXAMPLE 3.2
A rectangular beam of small thickness t, depth 2h, and length 2L is subjected to an arbitrary variation of temperature throughout its depth,
T = T1y2. Determine the distribution of stress and strain for the case in
which (a) the beam is entirely free of surface forces (Fig. 3.6a), and (b)
the beam is held by rigid walls that prevent the xdirected displacement
only (Fig. 3.6b).
Solution The beam geometry indicates a problem of plane stress. We
begin with the assumptions
x = x1y2, y = xy =0 (a) Direct substitution of Eqs. (a) into Eqs. (3.6) indicates that the equations
of equilibrium are satisfied. Equations (a) reduce the compatibility
equation (3.24) to the form
d2
1
dy2 x + ET2 = 0 (b) FIGURE 3.6. Example 3.2. Rectangular beam in plane thermal stress: (a) unsupported; (b) placed between two rigid walls. 110 Chapter 3 TwoDimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 111 from which
x =  ET + c1y + c2 (c) where c1 and c2 are constants of integration. The requirement that faces
y = ; h be free of surface forces is obviously fulfilled by Eq. (b).
a. The boundary conditions at the end faces are satisfied by determining the constants that assume zero resultant force and moment at
x = ; L:
h h
xt Lh
 dy = 0, Lh
 xyt dy = 0 (d) Substituting Eq. (c) into Eqs. (d), it is found that
h
h
c1 = 13/2h32 1h ETy dy and c2 = 11/2h2 1h ET dy. The normal
stress, upon substituting the values of the constants obtained, together with the moment of inertia I = 2h3t/3 and area A = 2ht, into
Eq. (c) is thus
x h
yt h
t
= E B T +
T dy +
Ty dy R
A Lh
I Lh
 (3.26) The corresponding strains are
x = x E T, + y = x E + T, xy =0 (e) The displacements can readily be determined from Eqs. (3.1).
From Eq. (3.26), observe that the temperature distribution for
T = constant results in zero stress, as expected. Of course, the strains
(e) and the displacements will, in this case, not be zero. It is also noted
that, when the temperature is symmetrical about the midsurface
1y = 02, that is, T1y2 = T1  y2, the final integral in Eq. (3.26) vanishes. For an antisymmetrical temperature distribution about the midsurface, T1y2 =  T1  y2, and the first integral in Eq. (3.26) is zero.
b. For the situation described, x = 0 for all y. With y = xy = 0 and
Eq. (c), Eqs. (3.23a) lead to c1 = c2 = 0, regardless of how T varies
with y. Thus,
x = E T (3.27) and
x = xy = 0, y = 11 + 2 T (f) Note that the axial stress obtained here can be large even for modest
temperature changes, as can be verified by substituting properties of
a given material.
3.7 Thermal Stresses 111 ch03.qxd 12/20/02 3.8 7:20 AM Page 112 BASIC RELATIONS IN POLAR COORDINATES Geometrical considerations related either to the loading or to the boundary of a
loaded system often make it preferable to employ polar coordinates, rather than
the Cartesian system used exclusively thus far. In general, polar coordinates are
used advantageously where a degree of axial symmetry exists. Examples include a
cylinder, a disk, a wedge, a curved beam, and a large thin plate containing a circular
hole.
The polar coordinate system 1r, 2 and the Cartesian system (x, y) are related
by the following expressions (Fig. 3.7a):
x = r cos , r2 = x2 + y2 y = r sin , = tan1 (a) y
x These equations yield
x
0r
=
= cos ,
r
0x
y
sin
0
= 2=,
r
r
0x y
0r
=
= sin
r
0y
(b) x
cos
0
= 2=
r
r
0y Any derivatives with respect to x and y in the Cartesian system may be transformed
into derivatives with respect to r and by applying the chain rule:
0r 0
00
0
=
+
= cos
0x
0x 0r
0x 0 sin
0
r
0r 0
0 0r 0
00
0
=
+
= sin
0y
0y 0r
0y 0 cos
0
+
r
0r 0
0 (c) Relations governing properties at a point not containing any derivatives are not affected by the curvilinear nature of the coordinates, as is observed next. FIGURE 3.7. (a) Polar coordinates; (b) stress element in polar coordinates. 112 Chapter 3 TwoDimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 113 Equations of Equilibrium
Consider the state of stress on an infinitesimal element abcd of unit thickness described by polar coordinates (Fig. 3.7b). The r and directed body forces are denoted by Fr and F . Equilibrium of radial forces requires that  ¢ r + dr sin 0r
dr ≤ 1r + dr2d 0r d
+¢
2 r + rrd ¢ 0r
d
d ≤ dr cos
2
0 + r 0
0 d ≤ dr sin dr cos d
2 d
+ Frrdr d = 0
2 Inasmuch as d is small, sin1d /22 may be replaced by d /2 and cos1d /22 by 1. Additional simplication is achieved by dropping terms containing higherorder infinitesimals. A similar analysis may be performed for the tangential direction. When
both equilibrium equations are divided by r dr d , the results are
0r
1 0r
+
+
r0
0r
10
r0 r r + Fr = 0 2r
0r
+F =0
+
+
r
0r (3.28) In the absence of body forces, Eqs. (3.28) are satisfied by a stress function
£ 1r, 2 for which the stress components in the radial and tangential directions are
given by
r r 1 0£
1 02 £
+2
r 0r
r 02
2
0£
=
0r2
= = 1 0£
1 02 £
0 1 0£
= ¢
≤
2
r 0r 0
r0
0r r 0 (3.29) Strain–Displacement Relations Consider now the deformatio...
View
Full
Document
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

Click to edit the document details