Problems for Chapter 5

 

Analytical Techniques and Solutions for Linear Elastic Solids

 

 

 

5.7.  Energy Methods

 

 

5.7.1.      A shaft with length L and square cross section is fixed at one end, and subjected to a twisting moment T at the other.  The shaft is made from a linear elastic solid with Young’s modulus E and Poisson’s ratio $\nu$.  The torque causes the top end of the shaft to rotate through an angle $\varphi$.

5.7.1.1.            Consider the following displacement field

$v_1=-\frac{\varphi }{L}{x}_1{x}_3{v}_2=\frac{\varphi }{L}{x}_2{x}_3{v}_3=0$

Show that this is a kinematically admissible displacement field for the twisted shaft.

5.7.1.2.            Calculate the strains associated with this kinematically admissible displacement field

5.7.1.3.            Hence, show that the potential energy of the shaft is

You may assume that the potential energy of the torsional load is $-T\varphi$

5.7.1.4.            Find the value of $\varphi$ that minimizes the potential energy, and hence estimate the torsional stiffness of the shaft.

 

 

 

 

5.7.2.      In this problem you will use the principle of minimum potential energy to find an approximate solution to the displacement in a pressurized cylinder.  Assume that the cylinder is an isotropic, linear elastic solid with Young’s modulus $E$ and Poisson’s ratio $\nu$, and subjected to internal pressure p at r=a.

5.7.2.1.            Approximate the radial displacement field as $u_r=C_1+C_{2}r$, where $C_1,C_2$ are constants to be determined.  Assume all other components of displacement are zero.  Calculate the strains in the solid ${\epsilon }_{rr},{\epsilon }_{\theta \theta }$

5.7.2.2.            Find an expression for the total strain energy of the cylinder per unit length, in terms of $C_1,C_2$ and relevant geometric and material parameters

5.7.2.3.            Hence, write down the potential energy (per unit length) of the cylinder.

5.7.2.4.            Find the values of $C_1,C_2$ that minimize the potential energy

5.7.2.5.            Plot a graph showing the normalized radial displacement field $E\left(b^2-a^2\right)u_r(r)/(pa{b}^2)$ as a function of the normalized position $(r-a)/(b-a)$ in the cylinder, for $\nu =0.3$, and $b/a=1.1$ and $b/a=2$.   On the same graph, plot the exact solution, given in Section 4.1.9.

 

 

 

 

5.7.3.      A bi-metallic strip is made by welding together two materials with identical Young’s modulus and Poisson’s ratio $E,\nu$, but with different thermal expansion coefficients ${\alpha }_1,{\alpha }_2$, as shown in the picture.  At some arbitrary temperature $T_0$ the strip is straight and free of stress.   The temperature is then increased to a new value $T$, causing the strip to bend.  Assume that, after heating, the displacement field in the strip can be approximated as $u_1=\lambda x_1+x_1{x}_2/R{u}_2=-x_1^2/(2R)-\beta x_2{u}_3=0$, where $\lambda ,R,\beta$ are constants to be determined.

5.7.3.1.            Briefly describe the physical significance of the shape changes associated with $\lambda ,R,\beta$.

5.7.3.2.            Calculate the distribution of (infinitesimal) strain associated with the kinematically admissible displacement field

5.7.3.3.            Hence, calculate the strain energy density distribution in the solid.  Don’t forget to account for the effects of thermal expansion

5.7.3.4.            Minimize the potential energy to determine values for $\lambda ,R,\beta$ in terms of relevant geometric and material parameters.

 

 

 

 

5.7.4.      By guessing the deflected shape, estimate the stiffness of a clamped$—$clamped beam subjected to a point force at mid-span.   Note that your guess for the deflected shape must satisfy $w(x_1)=\frac{dw}{d{x}_1}=0$, so you can’t assume that it bends into a circular shape as done in class.  Instead, try a deflection of the form $w(x_1)=1-\cos \left(2\pi x_1/L\right)$, or a similar function of your choice (you could try a suitable polynomial, for example).  If you try more than one guess and want to know which one gives the best result, remember that energy minimization always overestimates stiffness.  The best guess is the one that gives the lowest stiffness.

 

 

 

 

5.7.5.      A slender rod with length L and cross sectional area A is subjected to an axial body force $b=b(x_2)e_2$.  Our objective is to determine an approximate solution to the displacement field in the rod.

5.7.5.1.            Assume that the displacement field has the form

$u_2=w(x_2)u_1=u_3=0$

where the function w is to be determined.  Find an expression for the strains in terms of w and hence deduce the strain energy density.

 

5.7.5.2.            Show that the potential energy of the rod is

$V(w)=EA\frac{1-\nu }{\left(1-2\nu \right)\left(1+\nu \right)}{\displaystyle \underset{0}{\overset{L}{\int }}\frac{1}{2}{\left\{\frac{dw}{d{x}_2^{}}\right\}}^2}d{x}_2-A{\displaystyle \underset{0}{\overset{L}{\int }}b(x_2)w(x_2)d{x}_2}$

 

5.7.5.3.            To minimize the potential energy, suppose that w is perturbed from the value the minimizes V  to a value $w+\delta w$.  Assume that $\delta w$ is kinematically admissible, which requires that $\delta w=0$ at any point on the bar where the value of w is prescribed.  Calculate the potential energy $V(w+\delta w)$ and show that it can be expressed in the form

$V(w+\delta w)=V(w)+\delta V+\frac{1}{2}{\delta }^{2}V$

where $V$ is a function of w only, $\delta V$ is a function of w and $\delta w$, and ${\delta }^{2}V$ is a function of $\delta w$ only.

 

5.7.5.4.            As discussed in Section 8 of the online notes (or in class), if $V$ is stationary at $\delta w=0$, then $\delta V=0$.  Show that, to satisfy $\delta V=0$, we must choose w to satisfy

$\frac{EA(1-\nu )}{(1-2\nu )(1+\nu )}{\displaystyle \underset{0}{\overset{L}{\int }}\frac{dw}{d{x}_2}\frac{d\delta w}{d{x}_2}d{x}_2-A{\displaystyle \underset{0}{\overset{L}{\int }}b\delta w}}d{x}_2=0$

 

5.7.5.5.            Integrate the first term by parts to deduce that, to minimize, V, w must satisfy

$\frac{d^{2}w}{d{x}_2^2}+b(x_2)=0$

Show that this is equivalent to the equilibrium condition

$\frac{d{\sigma }_{22}}{d{x}_2}+b_2=0$

Furthermore, deduce that if $w$ is not prescribed at either $x_2=0,x_2=L$ or both, then the boundary conditions on the end(s) of the rod must be

$\frac{dw}{d{x}_2}=0$

Show that this corresponds to the condition that ${\sigma }_{22}=0$ at a free end.

 

5.7.5.6.            Use your results in (2.5) to estimate the displacement field in a bar with mass density $\rho$, which is  attached to a rigid wall at $x_2=L$, is free at $x_2=0$, and subjected to the force of gravity (acting vertically downwards…)