Unformatted text preview: chapter is subdivided into two parts. In Part A, derivations of the governing differential equations and various approaches for solution of twodimensional
problems in Cartesian and polar coordinates are considered. Part B treats stress
96 Chapter 3 TwoDimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 97 concentrations in members whose cross sections manifest pronounced changes and
cases of load application over small areas. Part A—Formulation and Methods of Solution
3.3 PLANE STRAIN PROBLEMS Consider a long prismatic member subject to lateral loading (for example, a cylinder under pressure), held between fixed, smooth, rigid planes (Fig. 3.1). Assume
the external force to be functions of the x and y coordinates only. As a consequence, we expect all cross sections to experience identical deformation, including
those sections near the ends. The frictionless nature of the end constraint permits
x, y deformation, but precludes z displacement; that is, w = 0 at z = ; L/2. Considerations of symmetry dictate that w must also be zero at midspan. Symmetry arguments can again be used to infer that w = 0 at ; L/4, and so on, until every
cross section is taken into account. For the case described, the strain depends on x
and y only:
x z = 0w
= 0,
0z = 0u
,
0x y xz = = 0v
,
0y 0w
0u
+
= 0,
0x
0z xy = 0u
0v
+
0y
0x
yz = 0w
0v
+
=0
0y
0z (3.1)
(3.2) The latter expressions depend on 0u/0z and 0v/0z vanishing, since w and its derivatives are zero. A state of plane strain has thus been described wherein each point
remains within its transverse plane, following application of the load. We next proceed to develop the equations governing the behavior of bodies under plane
strain.
Substitution of z = yz = xz = 0 into Eq. (2.30) provides the following
stress–strain relationships: FIGURE 3.1. Plane strain in a cylindrical
body. 3.3 Plane Strain Problems 97 ch03.qxd 12/20/02 7:20 AM Page 98 x = 2G x y = 2G y + xy =G = yz = 0, 1 x
x y2 +
+ y2 (3.3) xy and
xz 1 + z 1 = x + y2 =1 x + y2 (3.4) Because z is not contained in the other governing expressions for plane strain, it is
determined independently by applying Eq. (3.4). The strain–stress relations, Eqs.
(2.28), for this case become
= 1E 2 x = 1E 2 y xy = xy G ¢
¢ x  1 y y  1 x ≤
≤ (3.5) Inasmuch as these stress components are functions of x and y only, the first two
equations of (1.11) yield the following equations of equilibrium of plane strain:
0 xy
0x
+
+ Fx = 0
x
0
0y
0 xy
0y
+
+ Fy = 0
0y
0x (3.6) The third equation of (1.11) is satisfied if Fz = 0. In the case of plane strain, therefore, no body force in the axial direction can exist.
A similar restriction is imposed on the surface forces. That is, plane strain will
result in a prismatic body if the surface forces px and py are each functions of x and
y and pz = 0. On the lateral surface, n = 0 (Fig. 3.2). The boundary conditions,
from the first two equations of (1.41), are thus given by
px = xl + xym py = xyl + ym (3.7) Clearly, the last equation of (1.41) is also satisfied.
In the case of a plane strain problem, therefore, eight quantities, x, y, xy, x,
, xy, u, and v , must be determined so as to satisfy Eqs. (3.1), (3.3), and (3.6) and
y
FIGURE 3.2. Surface forces. 98 Chapter 3 TwoDimensional Problems in Elasticity ch03.qxd 12/20/02 7:20 AM Page 99 the boundary conditions (3.7). How eight governing equations, (3.1), (3.3), and
(3.6), may be reduced to three is now discussed.
Three expressions for twodimensional strain at a point [Eq. (3.1)] are functions of only two displacements, u and v , and therefore a compatibility relationship
exists among the strains [Eq. (2.8)]:
02 y
02 xy
02 x
+
=
0x 0y
0y 2
0x2 (3.8) This equation must be satisfied for the strain components to be related to the displacements as in Eqs. (3.1). The condition as expressed by Eq. (3.8) may be transformed into one involving components of stress by substituting the strain–stress
relations and employing the equations of equilibrium. Performing the operations
indicated, using Eqs. (3.5) and (3.8), we have
02
[11  2
0y2 x  y] + 02
[11  2
0x2 02
y  x] = 2 xy 0x 0y (a) Next, the first and second equations of (3.6) are differentiated with respect to x and
y, respectively, and added to yield
02
2 xy 0x 0y = ¢ 02 y
0Fy
02 x
0Fx
+
¢
+
≤
2
2≤
0x
0y
0x
0y Finally, substitution of this into Eq. (a) results in ¢ 02
02
+ 2 ≤1
2
0x
0y x + y2 = 1
1 ¢ 0Fy
0Fx
+
≤
0x
0y (3.9) This is the equation of compatibility in terms of stress.
We now have three expressions, Eqs. (3.6) and (3.9), in terms of three unknown
quantities: x, y, and xy. This set of equations, together with the boundary conditions (3.7), is used in the solution of plane strain problems. For a given situation,
after determining the stress, Eqs. (3.5) and (3.1) yield the strain and displacement,
respectively. In Sec. 3.5, Eqs. (3.6) and (3.9) will further be reduced to one equation
containing a single variable. 3.4 PLANE STRESS PROBLEMS In many problems of practical importance, the str...
View
Full Document
 Spring '08
 yok
 Force, Stress, Eqs., plane stress

Click to edit the document details