Some Examples

(a) Vibration of an Infinite Plate

Consider an infinite plate bounded by the planes $X_1 = 0$ and $X_1 = \ell$. The vibrations of the plate are caused by a prescribed motion of these bounding planes or by prescribed surface tractions on these planes. Since we study steady state vibration of the plate, the initial displacement and velocity fields in the plate are not required. We will neglect the effect of gravity in these problems.

We begin by assuming that the displacement field in the plate is of the form

$$u_1 = u_1 (X_1,t),\ u_2 = u_3 = 0.$$ (5.7.1)

For this displacement field, the equations of motion (5.3.5) reduce to the following equation

$$\rho_0\frac{\partial^2u_1}{\partial t^2} = (\lambda + 2\mu )\frac{\partial^2u_1}{\partial X^2_1},$$ (5.7.2)

which can be rewritten as

$$\frac{\partial^2u_1}{\partial t^2} = C^2_L\frac{\partial^2u_1}{\partial X^2_1},\ C^2_L = \frac{\lambda + 2\mu}{\rho_0}.$$ (5.7.3)

A steady state vibration solution to this equation is of the form

$$u_1 = (A\cos\lambda X_1 + B\sin \lambda X_1)(C\cos C_L\lambda t + D\sin C_L\lambda t),$$ (5.7.4)

where the constants $A,\ B,\ C,\ D$ and $\lambda$ are determined by boundary conditions. This vibration mode is sometimes termed a ``thickness stretch'' because the plate is being stretched through its thickness. It is analogous to acoustic vibration of organ pipes and to the longitudinal vibration of slender rods.

Another vibration mode can be obtained by assuming the displacement field

$$u_2 = u_2 (X_1,t),\ u_1 = u_3 = 0.$$ (5.7.5)

In this case, the displacement field must satisfy the equation

$$C^2_T\frac{\partial^2u_2}{\partial X^2_1} = \frac{\partial^2u_2}{\partial t^2},\ C^2_T = \frac{\mu}{\rho},$$ (5.7.6)

and the solution is of the same form as in the previous case.

This vibration is termed ``thickness shear'' and it is analogous to the motion of a vibrating string.

Example 5.7.1: (a) Find the thickness-stretch vibration of a plate, where the left face $(X_1 = 0)$ is subjected to a forced displacement $u_i = (\alpha \cos\omega t)\delta_{i1}$ and the right face $(X_1 = \ell )$ is fixed.
(b) Determine the values of $\omega$ that give resonance.

Solution: (a) The boundary condition on the left face $X_1 = 0$ and eqn. (5.7.4) give

$$\alpha \cos\omega t = A(C\cos C_L\lambda t + D\sin C_L\lambda t).$$

Therefore,

$$AC = \alpha,\ \lambda = \frac{\omega}{C_L},\ D = 0.$$

The second boundary condition gives

$$0 = u_1 (\ell ,t) = \left(\alpha \cos \frac{\omega \ell} {C_L} + BC\sin\frac{\omega \ell}{C_L}\right) \cos \omega t.$$

Therefore,

$$BC = -\alpha \cot \frac{\omega\ell}{C_L}$$

and the vibration is given by

$$u_1(X_1,t) = \alpha \left[\cos\frac{\omega}{C_L}X_1 - \frac{\sin\... ...omega}{C_1}X_1}{\tan\displaystyle\frac{\omega\ell }{C_L}}\right] \cos \omega t.$$

(b) Resonance is indicated by unbounded displacements. This occurs in part (a) for forcing frequencies corresponding to $\tan \displaystyle\frac{\omega\ell}{C_L} = 0$, that is, when

$$\omega = \frac{n\pi C_L}{\ell},\ n = 1,2,3,\ldots$$

Example 5.7.2 (a) Find the thickness-shear vibration of an infinite plate which has an applied surface traction $f_i = -\beta\cos\omega t\delta_{i2}$ on the plane $X_1 = 0$ and is fixed on the plane $X_1 = \ell$.

(b) Determine the resonant frequencies.

Solution: On the plane $X_1 = 0$, $\mathbf{n} = (-1,0,0)$. Therefore

$$f_i = T_{ij} n_j = -T_{i1}$$

gives

$$T_{21}\big\vert _{X_1 = 0} = \beta \cos\omega t.$$

This shearing stress forces a vibration of the form

$$u_2 = (A\cos\lambda X_1 + B\sin\lambda X_1)(C\cos C_T \lambda t + D\sin C_T \lambda t).$$

Using Hooke's law (5.2.5), we have

$$T_{12} = 2 \mu e_{12} = \mu \frac{\partial u_2}{\partial X_1}\ .$$

Therefore

$$\mu \frac{\partial u_2}{\partial X_1}\bigg\vert _{X_1 = 0} = \beta \cos \omega t$$

$$\mu(B\lambda )(C\cos C_T \lambda t + D\sin C_T \lambda t) = \beta \cos \omega t.$$

Thus

$$D = 0,\ \lambda = \frac{\omega}{C_T},\ BC = \frac{\beta C_T}{\mu \omega}\ .$$

The boundary condition at $X_1 = \ell$ gives $u_2(\ell ,t) = 0$.

$$\therefore \hspace*{.5in} (A\cos\lambda\ell + B\sin\lambda\ell ) C\cos C_T\lambda t = 0.$$

Hence

$$A = -B\tan \lambda \ell .$$

Thus

$$u_2(X_1,t) = \frac{\beta C_T}{\omega \mu}\left(\sin\frac{\omega}{... ...X_1 - \tan \frac{\omega \ell}{C_T}\cos\frac{\omega}{C_T}X_1\right)\cos\omega t.$$

(b) Resonance occurs for

$$\tan\frac{\omega \ell}{C_T} = \infty$$

or

$$\omega = \frac{n\pi C_T}{2\ell}\ , \ \ n = 1,3,5,\ldots$$

(B) Torsion of a Circular Shaft

Consider elastic deformations of a cylindrical bar of circular cross-section of radius $a$ and length $L$, that is being twisted by an end moment $M_t$ at the right end and is fixed at the left end. We choose the $X_3$-axis to coincide with the axis of the cylinder and the left-hand and right-hand faces to correspond to the planes $X_3 = 0$ and $X_3 = L$ respectively.

This problem involves the solution of equilibrium equations

$$(\lambda + \mu )u_{j,ji} + \mu u_{i,jj} = 0$$ (5.7.7)

subject to the boundary conditions

$$\left.\begin{array}{l} T_{ij}n_j = 0 \ \mbox{on the lateral sur... ... .1in} u_i = 0 \ \mbox{on the plane}\ X_3 = 0.\end{array}\right\}$$ (5.7.8)

Note that if the problem is formulated in terms of displacements, then compatibility conditions (5.6.6) are not needed. We use St. Venant's semi-inverse method to solve the problem in an inverse way. That is, we make a kinematic assumption about the displacement field and then ensure that eqns. (5.7.7) and (5.7.8) are satisfied.

Because of the symmetry of the problem, it is reasonable to assume that the motion of each cross-sectional plane induced by the end moments is a rigid body motion about the $X_3$-axis. This motion is similar to that of a stack of coins in which each coin is rotated by a slightly different angle than the previous coin. We will see that this assumption results in a displacement field that satisfies (5.7.7) and (5.7.8). To ensure that the deformations are small, we will assume that the angle $\theta$ of rotation of any section with respect to the left-end is very small as compared to one.

Under the preceding assumptions, the displacement of any point can be calculated as follows.

$$\begin{array}{l} u_1 = r\cos (\beta + \theta ) - r\cos\beta\\... ... \end{array} \ \ \ \mbox{\includegraphics{continuumfig5.2.eps}}$$

For this displacement field to satisfy the equations of equilibrium (5.7.7), we must have

$$\frac{d^2\theta}{dX^2_3} = 0.$$

Thus

$$\frac{d\theta}{dX_3} = \alpha = .$$ (5.7.9)

That is, the angle of twist per unit length is the same over the entire length of the shaft. From (5.7.9) and recalling that $\theta = 0$ at $X_3 = 0$, we get

$$\theta = \alpha X_3$$

and, therefore,

$$u_1 = -\alpha X_2X_3,\ u_2 = \alpha X_1X_3,\ u_3 = 0.$$ (5.7.10)

The strains and stresses associated with these displacements can be calculated from eqns. (3.14.3) and (5.2.5). The non-zero components of strains are

$$\begin{split}e_{23} =\ & e_{32} = \alpha x_1/2,\\ e_{31}=\ & e_{13} = -\alpha X_2/2, \end{split}$$ (5.7.11)

and the non-zero components of stress are

$$\begin{split}T_{13} =\ & T_{31} = -\mu\alpha X_2,\\ T_{23} =\ & T_{32} = \mu\alpha X_1. \end{split}$$ (5.7.12)

To see whether the boundary conditions are satisfied or not, we note that on the lateral surface a unit outer normal $\mathbf{n} = \displaystyle\frac{1}{a} (X_1,X_2,0)$. Therefore, on the lateral surface,

$$T_{ij}n_j = T_{i1}\frac{X_1}{a} + T_{i2}\frac{X_2}{a}\ ,$$

and the calculated stress field does satisfy (5.7.8)$_1$. On the end plane $X_3 = L$, $\mathbf{n} = (0,0,1)$ and in order that (5.7.8)$_2$ be satisfied, $\int T_{i3}dA = 0$ which is true because $\int X_1dA = 0$, $\int X_2dA = 0$. The boundary condition (5.7.8)$_3$ requires that, on the end plane $X_3 = L$,

$$\int\varepsilon_{ijk}X_jT_{k3}dA = M_t\delta_{i3}\ .$$

This gives

$$\mu\alpha\int (X^2_1 + X^2_2)dA = M_t\ ,$$

or

$$\mu\alpha J_p = M_t,$$ (5.7.13)

in which $J_p$ is the polar moment of inertia and equals $\pi a^4/2$. The boundary condition (5.7.8)$_4$ is clearly satisfied by the displacement field (5.7.10).

Substituting for $\alpha$ from (5.7.13) into (5.7.10), (5.7.11) and (5.7.12) we obtain the displacement components, non-zero strain components and non-zero stress components at any point in the cylindrical bar. In terms of the twisting moment $M_t$, the stress tensor becomes

$$[T_{ij}]= \frac{M_t}{J_p}\left[\begin{array}{ccc} 0 & 0 & -X_2\... ...0 & X_1\\ \noalign{\vskip .1in} -X_2 & X_1 & 0\end{array}\right]$$ (5.7.14)

At any point $(X_1,X_2,b)$ of a cross-section $X_3 = b,\ \mathbf{n} = (0,0,1)$ and the stress vector $\mathbf{f}$ is given by

$$f_i = T_{ij}n_j = \mu\alpha (-X_2\delta_{1i} + X_1\delta_{2i})\ .$$

Note that $\mathbf{f}$ lies in the plane $X_3 = b$ implying thereby that there is no normal stress at any point on the plane $X_3 = b$. Also $\mathbf{f}$ is perpendicular to the radius vector joining the point $(X_1,X_2,b)$ with the center $(0,0,b)$ of the cross-section. The magnitude of $\mathbf{f}$ is

$$\vert\mathbf{f}\vert = \mu\alpha r = \frac{M_t}{J_p} r,\ r = \sqrt{X^2_1 + X^2_2}\ .$$

From this we see that the maximum stress is a tangential stress that acts on the boundary of the cylinder and has the magnitude $$\frac{M_t}{J_p}a$$.

(c) Torsion of Non-Circular Cylinders

For cross-sections other than circular, the stress field (5.7.14) does not satisfy the boundary conditions of zero tractions on the mantle of the cylinder. We will see that in order for this boundary condition to be satisfied, the cross-sections will not remain plane.

We begin by assuming a displacement field that still rotates each cross-section by a small angle $\theta$, but in addition there may be a displacement in the axial direction. This warping of a cross-sectional plane will be defined by $u_3 = \hat\phi\ (X_1,X_2)$. Thus we assume that each cross-section is warped in the same way. For this to be true the left end of the cylinder can not be fixed to a rigid flat support but is subjected to a torque equal and opposite to that applied to the right end. The rigid motion is removed by applying suitable constraints.

The assumed displacement field has the form

$$u_1 = -X_2\theta (X_3),\ u_2 = X_1\theta (X_3),\ u_3 = \hat \phi (X_1,X_2).$$ (5.7.15)

For this displacement field to satisfy the equations of equilibrium (5.7.7), we must have

$$\frac{d^2\theta}{dX^2_3} =\ 0,$$

$$\frac{\partial^2\hat\phi}{\partial X^2_1} + \frac{\partial^2\hat\phi}{\partial X^2_2} =\ 0\ .$$ (5.7.16)

Therefore, the angle $\alpha$ of twist per unit length is the same over the entire length of the cylinder. Setting $\theta = 0$ at $X_3 = 0$, we get $\theta = \alpha X_3$ and therefore the displacement field (5.7.15) can be rewritten as

$$u_1 = -\alpha X_2X_3,\ u_2 = \alpha X_1X_3,\ u_3 = \alpha\phi (X_1,X_2),$$ (5.7.17)

where $\hat\phi = \alpha \phi$. For equilibrium, $\phi$ must satisfy

$$\frac{\partial^2\phi}{\partial X^2_1} + \frac{\partial^2\phi}{\partial X^2_2} = \nabla^2\phi = 0.$$ (5.7.18)

Stresses corresponding to the displacements (5.7.17) are

$$\begin{split}T_{11} =\ & T_{22} = T_{33} = T_{12} = 0,\\ T_{2... ...eft(\frac{\partial\phi}{\partial X_1} - X_2\right). \end{split}$$ (5.7.19)

Since the bar is cylindrical, the unit normal to the lateral surface has the form $\mathbf{n} = n_1\mathbf{e}_1 + n_2\mathbf{e}_2$ and the associated surface traction is given by

$$f_i = \mu\alpha \left[\left(\frac{\partial\phi}{\partial X_1} - X... ...1 + \left(\frac{\partial\phi}{\partial X_2} + X_1\right) n_2\right]\delta_{i3}.$$

For the lateral surface to be traction free, $\phi$ must satisfy

$$\frac{d\phi}{dn} = \frac{\partial\phi}{\partial X_1} n_1 + \frac{\partial\phi}{\partial X_2} n_2 = X_2n_1 - X_1n_2,$$ (5.7.20)

on the boundary.

In order that the boundary condition (5.7.8)$_2$ be satisfied,

$$\int T_{13}dA = 0,\ \int T_{23}dA = 0.$$

Now

$$\int T_{13} dA =\ \mu\alpha\int\left(\frac{\partial\phi}{\partial X_1} - X_2\right) dA,$$

$$=\ \mu\alpha\int\left[\frac{\partial}{\partial X_1} \left(X_1\left(\... ...\left(X_1\left(\frac{\partial\phi}{\partial X_2} + X_1\right)\right)\right] dA,$$

$$=\ \mu\alpha\oint X_1\left[\left(\frac{\partial\phi}{\partial X_1} -... ...right) n_1 + \left(\frac{\partial\phi}{\partial X_2} + X_1\right)n_2\right] ds,$$

$$=\$$ 0

because of (5.7.20). Using a similar argument one can show that $\int T_{23} dA = 0$. The boundary condition (5.7.8)$_3$ gives

$$M_t =\ \int (X_1T_{23} - X_2T_{13})dA,$$

$$=\ \mu\alpha \int \left(X^2_1 + X^2_2 + X_1\frac{\partial\phi}{\partial X_2} - X_2\frac{\partial\phi}{\partial X_1}\right)dA,$$

$$=\ D\alpha ,$$ (5.7.21)

where

$$D = \mu \int \left(X^2_1 + X^2_2 + X_1\frac{\partial\phi}{\partial X_2} - X_2\frac{\partial\phi}{\partial X_1}\right) dA.$$ (5.7.22)

The formula (5.7.21) shows that the twisting moment or torque $M_t$ is proportional to the angle of twist per unit length, so that the constant $D$ provides a measure of the rigidity of a bar subjected to torsion. $D$ is called the torsional rigidity of the bar.

Hence the torsion problem is solved once $\phi$ is determined by solving (5.7.18) and (5.7.20).

For an elliptic cylindrical bar with the cross-section given by

$$\frac{X^2_1}{a^2} + \frac{X^2_2}{b^2} = 1,$$ (5.7.23)

we see that on the boundary,

$$\mathbf{n} = \frac{1}{F}\left(\frac{X_1}{a^2}\mathbf{e}_1 + \frac{X_2}{b^2}\mathbf{e}_2\right),\ F^2 = \frac{X^2_1}{a^4} + \frac{X^2_2}{b^4},$$ (5.7.24)

and, therefore, eqn. (5.7.20) becomes

$$\frac{\partial\phi}{\partial X_1}X_1b^2 + \frac{\partial\phi}{\partial X_2}X_2a^2 = X_1X_2(b^2 - a^2).$$ (5.7.25)

This suggests that

$$\phi = \frac{b^2-a^2}{b^2 + a^2}X_1X_2.$$ (5.7.26)

This choice of $\phi$ satisfies (5.7.25) and (5.7.18). Substituting for $\phi$ in (5.7.22) and then the result in (5.7.21) we arrive at

$$M_t = \frac{2\mu}{b^2 + a^2}[b^2I_{22} + a^2I_{11}]\alpha$$

in which $I_{11}$ and $I_{22}$ are, respectively, the second moments of area about $X_1$ and $X_2$ axes. Recalling that for an ellipse $I_{11} = \pi ab^3/4,\ I_{22} = \pi a^3b/4$, we obtain

$$\alpha = M_t\frac{b^2 + a^2}{\pi a^3b^3\mu}\ .$$ (5.7.27)

Substitution for $\phi$ into (5.7.17) and (5.7.19) yields

$$u_1 = -\alpha X_2X_3,\ u_2 = \alpha X_1 X_3,\ u_3 = \alpha\frac{b^2 - a^2}{b^2 + a^2}X_1X_2,$$ (5.7.28)

and

$$T_{11} =\ T_{22} = T_{33} = T_{12} = 0,$$

$$T_{23} =\ \mu\alpha\left(\frac{2b^2}{b^2 + a^2}\right) X_1,\ T_{13} = \mu\alpha\left(\frac{-2a^2}{b^2 + a^2}\right) X_2.$$ (5.7.29)

At any point $(X_1,X_2,c)$ of a cross-section $X_3=c,\ \mathbf{n} = (0,0,1)$ and the stress vector $\mathbf{f}$ is given by

$$f_i = T_{ij} n_j = \frac{2\mu\alpha}{b^2 + a^2} (-a^2X_2\delta_{1i} + b^2X_1\delta_{2i}).$$

Since $\mathbf{f}$ lies in the plane $X_3 = c$, therefore, there is no normal stress at the point on the plane $X_3 = c$. The magnitude of $\mathbf{f}$ gives the shear stress $\tau$ at the point.

$$\tau = \frac{2\mu\alpha}{a^2 + b^2} (a^4X^2_2 + b^4 X^2_1)^{1/2}.$$ (5.7.30)

A point in the interior of the cross-section where $\tau$ takes on extremum values is given by

$$\frac{\partial \tau}{\partial X_1} = \frac{\partial\tau}{\partial X_2} = 0.$$

This gives $X_1 = X_2 = 0$ and at this point $\tau = 0$. To find points on the boundary where $\tau$ may have extremum values we first write (5.7.30) as

$$\tau = \frac{2\mu\alpha}{a^2 + b^2}(a^2(a^2 - b^2) X^2_2 + b^4 a^2)^{1/2}.$$ (5.7.31)

Since $a^2>b^2$, it is obvious that $\tau$ is minimum at $X_2 = 0$ and has the value $$\frac{2\mu\alpha ab^2}{a^2 + b^2}$$. $\tau$ is maximum at $X_2 = \pm b$ and has the value $$\frac{2\mu\alpha a^2b}{a^2 + b^2}$$. Thus, the maximum shear stress occurs at the extremities of the minor axis of the ellipse, contrary to the intuitive expectation that the maximum shear stress would occur at points of maximum curvature.

It is clear from (5.7.28)$_3$ that the axial displacement of points in the first and third quadrant will be along the $-X_3$ axis and that of points in the second and fourth quadrant will be along the $X_3$-axis. The points which have the same displacement in $X_3$-direction lie on a rectangular hyperbola.