This preview shows page 1. Sign up to view the full content.
Unformatted text preview: n of the infinitesimal element abcd, denoting the r
and displacements by u and v , respectively. The general deformation experienced
by an element may be regarded as composed of (1) a change in length of the sides,
as in Figs. 3.8a and b, and (2) rotation of the sides, as in Figs. 3.8c and d.
In the analysis that follows, the small angle approximation sin L is employed, and arcs ab and cd are regarded as straight lines. Referring to Fig. 3.8a, it is
observed that a u displacement of side ab results in both radial and tangential
strain. The radial strain r, the deformation per unit length of side ad, is associated
only with the u displacement:
r 3.8 Basic Relations in Polar Coordinates = 0u
0r (3.30a) 113 ch03.qxd 12/20/02 7:20 AM Page 114 FIGURE 3.8. Deformation and displacement of an element in polar coordinates. The tangential strain owing to u, the deformation per unit length of ab, is
1 2u = 1r + u2 d  r d
u
=
r
rd (d) Clearly, a v displacement of element abcd (Fig. 3.8b) also produces a tangential
strain,
1 2v = 10v/0 2 d
1 0v
=
r0
rd (e) since the increase in length of ab is 10v/0 2d . The resultant tangential strain, combining Eqs. (d) and (e), is
= 1 0v
u
+
r0
r (3.30b) Figure 3.8c shows the angle of rotation eb ¿ f of side a ¿ b ¿ due to a u displacement. The associated strain is
1 r 10u/0 2 d
1 0u
=
r0
rd 2u = (f) The rotation of side bc associated with a v displacement alone is shown in Fig. 3.8d.
Since an initial rotation of b – through an angle v /r has occurred, the relative rotation gb – h of side bc is
1 r 2v = v
0v
r
0r (g) The sum of Eqs. (f) and (g) provides the total shearing strain
114 Chapter 3 TwoDimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 115 1 0u
v
0v
+
r0
r
0r = r (3.30c) The strain–displacement relationships in polar coordinates are thus given by Eqs.
(3.30). Hooke’s Law
To write Hooke’s law in polar coordinates, we need only replace subscripts x by r
and y by in the appropriate Cartesian equations. In the case of plane stress, from
Eqs. (3.10) we have r = 1
1
E = r 1
1
E = 1
G  2  r r2 (3.31) r For plane strain, Eqs. (3.5) lead to
1+
[11  2
E
1+
[11  2
E = r = = r 1
G   r r] (3.32) r Transformation Equations
Replacement of the subscripts x ¿ by r and y ¿ by
r =
=1
= r cos2 x x y 2 sin +
x2 sin + y y sin2 +2 cos + 2 cos xy 2 xy
y =
=1 r = r cos2
r 2 sin + sin2 2 sin cos
+ 2 cos 2
+
+2 xy
r, r
r
r sin cos 1cos2 xy We can also express x, xy, and y in terms of
placing with  in Eqs. (1.13). Thus,
x in Eqs. (1.13) results in
 sin2 2 sin cos
r , and (Problem 3.26) by re sin cos 1cos2  sin2 2 Basic Relations in Polar Coordinates (3.34) sin cos Similar transformation equations may also be written for the strains
3.8 (3.33) r, r , and .
115 ch03.qxd 12/20/02 7:20 AM Page 116 Compatibility Equation
It can be shown that Eqs. (3.30) result in the following form of the equation of compatibility:
02
1 02
+2
r0
0r2 r
2 + 20
10r
1 02 r
1 0r
=
+2
r 0r
r 0r
r 0r 0
r0 (3.35) To arrive at a compatibility equation expressed in terms of the stress function £ , it
is necessary to evaluate the partial derivatives 02 £ /0x2 and 02 £ /0y2 in terms of r and
by means of the chain rule together with Eqs. (a). These derivatives lead to the
Laplacian operator:
§2 £ = 1 0£
1 02 £
02 £
02 £
02 £
+
=
+
+2
2
2
2
r 0r
r 02
0x
0y
0r (3.36) The equation of compatibility in alternative form is thus
§4 £ = ¢ 10
1 02
02
+
+ 2 2 ≤ 1 §2 £ 2 = 0
r 0r
r0
0r2 (3.37) For the axisymmetrical, zero body force case, the compatibility equation is, from
Eq. (3.9) [referring to (3.36)],
§21 r + 2= d21 r + dr 2 2 + 1 d1 r +
r
dr 2 =0 (3.38) The remaining relationships appropriate to twodimensional elasticity are found in
a manner similar to that outlined in the foregoing discussion.
EXAMPLE 3.3
A large thin plate is subjected to uniform tensile stress o at its ends, as
shown in Fig. 3.9. Determine the field of stress existing within the plate.
Solution For purposes of this analysis, it will prove convenient to locate the origin of coordinate axes at the center of the plate as shown.
The state of stress in the plate is expressed by
x = o, y = xy =0 2 The stress function, £ = oy /2, satisfies the biharmonic equation, Eq.
(3.14). The geometry suggests polar form. The stress function £ may be
transformed by substituting y = r sin , with the following result:
FIGURE 3.9. Example 3.3. A plate in uniaxial tension. 116 Chapter 3 TwoDimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 117 £= 1
4 or 11  cos 2 2 2 (h) The stresses in the plate now follow from Eqs. (h) and (3.29):
r =
= r 1
2
1
2 = o11 o11
1
2o + cos 2 2
 cos 2 2 (3.39) sin 2 Clearly, substitution of y = xy = 0 could have led directly to the foregoing result, using the transformation expressions of stress, Eqs. (3.33). Part B—Stress Concentrations
3.9 STRESSES DUE TO CONCENTRATED LOADS Let us now consider a concentrated force P or F acting at the vertex of a very large
or semiinfinite wedge (Fig. 3.10). The load distribution alon...
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