This document discusses how to calculate strain components in cylindrical coordinates. It shows that the normal strains in the r, θ, and z directions are given by the partial derivatives of the displacement components ur, uθ, and uz with respect to r, θ, and z, respectively. Additionally, it explains that the shear strain γrθ contains terms for the partial derivatives of both ur and uθ due to the effects of radial and angular displacements on elements. Figures are provided to illustrate these effects.
This document discusses how to calculate strain components in cylindrical coordinates. It shows that the normal strains in the r, θ, and z directions are given by the partial derivatives of the displacement components ur, uθ, and uz with respect to r, θ, and z, respectively. Additionally, it explains that the shear strain γrθ contains terms for the partial derivatives of both ur and uθ due to the effects of radial and angular displacements on elements. Figures are provided to illustrate these effects.
This document discusses how to calculate strain components in cylindrical coordinates. It shows that the normal strains in the r, θ, and z directions are given by the partial derivatives of the displacement components ur, uθ, and uz with respect to r, θ, and z, respectively. Additionally, it explains that the shear strain γrθ contains terms for the partial derivatives of both ur and uθ due to the effects of radial and angular displacements on elements. Figures are provided to illustrate these effects.
This document discusses how to calculate strain components in cylindrical coordinates. It shows that the normal strains in the r, θ, and z directions are given by the partial derivatives of the displacement components ur, uθ, and uz with respect to r, θ, and z, respectively. Additionally, it explains that the shear strain γrθ contains terms for the partial derivatives of both ur and uθ due to the effects of radial and angular displacements on elements. Figures are provided to illustrate these effects.
In class, we showed that the strain components relative to a set of (n, t, s) directions that were mutually orthogonal in the undeformed body could be expressed in terms of derivatives of the displacement vector, u, as:
u nn = n sn u tt = t st u ss = s s s u u nt = n +t , etc. st sn
Before, we chose n, t, and s along a set of Cartesian axes to obtain the strain components in rectangular coordinates. Similarly, we can use these expressions to find the strain components in any other coordinate system. Consider a set of cylindrical coordinate, for example:
ez e e z ey er ex y r z
x Figure 1. Cylindrical coordinates
In this case we can take
EM424 : Strain in Cylindrical Coordinates
sn = r st = r sz = z and
n = e r = cos e x + sin e y t = e = sin e x + cos e y s = ez
and where the displacement, u, is given by
u = ur er + u e + uz e z
Consider, now the normal strain in the r direction. We have
u rr = e r r u u u = e r r e r + e + z e z r r r ur = r
In the direction, however,
1 u 1 u u u e e = e = e r e r + e + z e z + ur r + u r r 1 u ur = + r r
where we have used the fact that
e r e = e , = e r
The additional strain term coming from the radial displacement can be understood if we consider what such a displacement does in terms of the elongation of an element initially in the direction: EM424 : Strain in Cylindrical Coordinates
(r+ur )d rd
d ur r
Figure 2. Effects of a radial displacement on an element in the direction
From Fig. 2 we see that a radial displacement causes an strain in the direction of an element initially in that direction given by
(r + ur ) r ( )u r = r ur = r
Also, in the z direction we find
u u u u zz = e z = e z r e r + e + z e z z z z z uz = z
Finally, consider one if the shear strains, namely
1 u u r = e r + e r r 1 u u u e e = e r r e r + e + z e z + ur r + u r u u u +e r e r + e + z e z r r r 1 ur u u = + r r r
The first two terms in this shear strain expression are analogous to the terms that appear in Cartesian coordinates. The last term can be understood by the fact that the displacement in the direction can itself cause a shear strain as shown in Figure 3: EM424 : Strain in Cylindrical Coordinates
Figure 3. Shear strain caused by a displacement in the direction
where
u ( r )u = =
r
(the minus sign exists because a constant displacement causes the angle between two lines initially along the r and directions to be greater than ninety degrees as shown in Fig. 3.)
