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x = c2, y = a2, xy =  b2 All three stress components are constant throughout the body. For a rectangular
plate (Fig. 3.4a), it is apparent that the foregoing may be adapted to represent
simple tension 1c2 Z 02, double tension 1c2 Z 0, a2 Z 02, or pure shear 1b2 Z 02.
A polynomial of the third degree
£3 = a3 3
b3
c3
d3
x + x2y + xy 2 + y3
6
2
2
6 (3.18) fulfills Eq. (3.14). It leads to stresses
x = c3x + d3y, y = a3x + b3y, xy =  b3x  c3y For a3 = b3 = c3 = 0, these expressions reduce to
x = d3y, y = xy =0 representing the case of pure bending of the rectangular plate (Fig. 3.4b). FIGURE 3.4. Stress fields of (a) Eq. (3.17) and (b) Eq. (3.18). 104 Chapter 3 TwoDimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 105 A polynomial of the fourth degree,
£4 = a4 4
b4
c4
d4
e4 4
x + x3y + x2y2 + xy3 +
y
12
6
2
6
12 (3.19) satisfies Eq. (3.14) if e4 =  12c4 + a42. The corresponding stresses are
x
y xy = c4x2 + d4 xy  12c4 + a42y2
= a4 x2 + b4xy + c4y2
= b4 2
d4
x  2c4 xy  y2
2
2 A polynomial of the fifth degree
£5 = f5 5
a5 5
b5 4
c5
d5
e5 4
x+
x y + x3y 2 + x2y3 +
xy +
y
20
12
6
6
12
20 (3.20) fulfills Eq. (3.14) provided that 13a 5 + 2c5 + e52x + 1b5 + 2d5 + 3f52y = 0 It follows that
e5 =  3a5  2c5, b5 =  2d5  3f5 The components of stress are then
x y xy = c5 3
x + d5x2y  13a5 + 2c52xy2 + f5y3
3 = a5x3  13f5 + 2d52x2y + c5xy 2 + d5 3
y
3 = 113f5 + 2d52x3  c5x2y  d5xy2 + 113d5 + 2c52y3
3
3 Problems of practical importance may be solved by combining functions (3.17)
through (3.20), as required. With experience, the analyst begins to understand the
types of stress distributions arising from a variety of polynomials.
EXAMPLE 3.1
A narrow cantilever of rectangular cross section is loaded by a concentrated force at its free end of such magnitude that the beam weight may
be neglected (Fig. 3.5a). Determine the stress distribution in the beam.
Solution The situation described may be regarded as a case of plane
stress provided that the beam thickness t is small relative to the beam
depth 2h.
The following boundary conditions are consistent with the coordinate system in Fig. 3.5a:
1 xy2y = ; h = 0, 1 y2y = ; h = 0 (a) These conditions simply express the fact that the top and bottom edges
of the beam are not loaded. In addition to Eq. (a) it is necessary, on the
3.6 Solution of Elasticity Problems 105 ch03.qxd 12/20/02 7:20 AM Page 106 FIGURE 3.5. Example 3.1. Endloaded cantilever beam. basis of zero external loading in the x direction at x = 0, that x = 0
along the vertical surface at x = 0. Finally, the applied load P must be
equal to the resultant of the shearing forces distributed across the free
end:
+h P= Lh xyt dy (b) The negative sign agrees with the convention for stress discussed in
Sec. 1.4.
For purposes of illustration, three approaches will be employed to
determine the distribution of stress within the beam.
Method 1. Inasmuch as the bending moment varies linearly with x and
x at any section depends on y, it is reasonable to assume a general expression of the form
x = 02 £
= c1xy
0y2 (c) in which c1 represents a constant. Integrating twice with respect to y,
£ = 1c1xy3 + yf11x2 + f21x2
6 (d) where f11x2 and f21x2 are functions of x to be determined. Introducing
the £ thus obtained into Eq. (3.14), we have
y d4f1
dx4 + d4f2
dx4 =0 Since the second term is independent of y, a solution exists for all x and
y provided that d4f1/dx4 = 0 and d4f2/dx4 = 0, which, upon integrating,
leads to
f11x2 = c2x3 + c3x2 + c4x + c5
f21x2 = c6x3 + c7x2 + c8x + c9 where c2, c3, Á , are constants of integration. Substitution of f11x2 and
f21x2 into Eq. (d) gives
106 Chapter 3 TwoDimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 107 £ = 1 c1xy3 + 1c2x3 + c3x2 + c4x + c52y
6
+ c6x3 + c7x2 + c8x + c9 Expressions for
y xy = y and follow from Eq. (3.13): xy 02 £
= 61c2y + c62x + 21c3y + c72
0x2 (e) 02 £
==  1c1y2  3c2x2  2c3x  c4
2
0x 0y At this point, we are prepared to apply the boundary conditions. Substituting Eqs. (a) into (e), we obtain c2 = c3 = c6 = c7 = 0 and c4 =  1c1h2.
2
The final condition, Eq. (b), may now be written as
h  Lh
 h
xyt dy = Lh
 1
2
2 c1t1y  h22dy = P from which
c1 =  3P
P
=3
I
2th where I = 2th3 is the moment of inertia of the cross section about the
3
neutral axis. From Eqs. (c) and (e), together with the values of the constants, the stresses are found to be
x = Pxy
,
I y = 0, xy = P2
1h  y22
2I (3.21) The distribution of these stresses at sections away from the ends is
shown in Fig. 3.5b.
Method 2. Beginning with bending moments Mz = Px, we may assume a stress field similar to that for the case of pure bending:
x = ¢ Px
≤ y,
I xy = xy1x, y2, y = z = xz = yz =0 (f) Equation of compatibility (3.12) is satisfied by these stresses. On the
basis of Eqs. (f), the equations of equilibrium lead to
0 xy
0x
+
= 0,
0x
0y 0 xy 0x =0 (g) From the second expression, xy can depend only on y. The first equation of (g) together with Eqs. (f) yields
d xy dy 3.6 Solution of Elasticity Problems = Py
I 107 ch03.qxd 12/20/02 7:20 AM Page 108 from which
xy = Py2
+c
2I Here c is determined on the basis of 1 xy2y + ; h = 0: c =  Ph2/2I. The
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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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