Problems for Chapter 5

 

Analytical Techniques and Solutions for Linear Elastic Solids

 

 

 

5.2.  Airy Function to Plane Static Problems

 

 

5.2.1.      A rectangular dam is subjected to pressure $p(x_2)={\rho }_w{x}_2$ on one face, where ${\rho }_w$ is the weight density of water.  The dam is made from concrete, with weight density ${\rho }_c$ (and is therefore subjected to a body force ${\rho }_c{e}_2$ per unit volume).  The goal is to calculate formulas for a and L to avoid failure.

5.2.1.1.            Write down the boundary conditions on all four sides of the dam.

5.2.1.2.            Consider the following approximate state of stress in the dam

$\begin{array}{l}{\sigma }_{22}=\frac{{\rho }_w{x}_2^3{x}_1}{4a^3}+\frac{{\rho }_w{x}_2{x}_1}{20a^3}\left(-10x_1^2+6a^2\right)-{\rho }_c{x}_2\\ {\sigma }_{11}=-\frac{{\rho }_w{x}_2}{2}+\frac{{\rho }_w{x}_2{x}_1}{4a^3}\left(x_1^2-3a^2\right)\\ {\sigma }_{12}=\frac{3{\rho }_w{x}_2^2}{8a^3}(a^2-x_1^2)-\frac{{\rho }_w}{8a^3}\left(a^4-x_1^4\right)+\frac{3{\rho }_w}{20a}(a^2-x_1^2)\end{array}$

Show that (i) The stress state satisfies the equilibrium equations (ii) the stress state exactly satisfies boundary conditions on the sides $x_1=\pm a$, (iii) The stress does not satisfy the boundary condition on $x_2=0$ exactly.

5.2.1.3.            Show, however, that the resultant force acting on $x_2=0$ is zero, so by Saint Venant’s principle the stress state will be accurate away from the top of the dam.

5.2.1.4.            The concrete cannot withstand any tension.  Assuming that the greatest principal tensile stress is located at point A $(x_1=a,x_2=L)$, show that the dam width must satisfy

5.2.1.5.            The concrete fails by crushing when the minimum principal stress reaches ${\sigma }_{1\min }=-{\sigma }_c$.  Assuming the greatest principal compressive stress is located at point B, $(x_1=-a,x_2=L)$ show that the height of the dam cannot exceed

 

 

 

 

5.2.2.      The stress due to a line load magnitude P per unit out-of-plane length acting tangent to the surface of a homogeneous, isotropic half-space can be generated from the Airy function

$\varphi =-\frac{P}{\pi }r\theta \cos \theta$

Calculate the displacement field in the solid, following the procedure in Section 5.2.6

 

 

 

 

 

5.2.3.      The figure shows a simple design for a dam.

 

5.2.3.1.            Write down an expression for the hydrostatic pressure in the fluid at a depth $x_2$ below the surface

5.2.3.2.            Hence, write down an expression for the traction vector acting on face OA of the dam.

5.2.3.3.            Write down an expression for the traction acting on face OB

5.2.3.4.            Write down the components of the unit vector normal to face OB in the basis shown

5.2.3.5.            Hence write down the boundary conditions for the stress state in the dam on faces OA and OB

5.2.3.6.            Consider the candidate Airy function

$\varphi =\frac{C_1}{6}{x}_1^3+\frac{C_2}{2}{x}_1^2{x}_2+\frac{C_3}{2}{x}_1{x}_2^2+\frac{C_4}{6}{x}_2^3$

Is this a valid Airy function?  Why?

5.2.3.7.            Calculate the stresses generated by the Airy function given in 5.2.2.6

5.2.3.8.            Use 5.2.2.5 and 5.2.2.7 to find values for the coefficients in the Airy function, and hence show that the stress field in the dam is

$\begin{array}{l}{\sigma }_{11}=-\rho x_2\\ {\sigma }_{22}=\frac{-2\rho }{{\tan }^3\beta }{x}_1+\frac{\rho }{{\tan }^2\beta }{x}_2\\ {\sigma }_{12}=\frac{-\rho }{{\tan }^2\beta }{x}_1\end{array}$

 

 

 

5.2.4.      Consider the Airy function

$\varphi =-\frac{{\sigma }_0}{2}\log (r)+\frac{{\sigma }_0}{4}{r}^2+\frac{{\sigma }_0}{4}\left(2a^2-r^2-\frac{a^4}{r^2}\right)\cos 2\theta$

Verify that the Airy function satisfies the appropriate governing equation. Show that this stress state represents the solution to a large plate containing a circular hole with radius a at the origin, which is loaded by a tensile stress ${\sigma }_0$ acting parallel to the $e_1$ direction.  To do this,

5.2.4.1.            Show that the surface of the hole is traction free $–$ i.e. ${\sigma }_{rr}={\sigma }_{r\theta }=0$ on r=a

5.2.4.2.            Show that the stress at $r/a\to \infty$ is ${\sigma }_{rr}={\sigma }_0(1+\cos 2\theta )/2={\sigma }_0{\cos }^2\theta$, ${\sigma }_{\theta \theta }=0$${\sigma }_{r\theta }=-{\sigma }_0\sin 2\theta$.

5.2.4.3.            Show that the stresses in 5.2.3.2 are equivalent to a stress ${\sigma }_{11}={\sigma }_0,{\sigma }_{22}={\sigma }_{12}=0$.  It is easiest to work backwards $–$ start with the stress components in the $\left\{e_1,e_2,e_3\right\}$ basis and use the basis change formulas to find the stresses in the $\left\{e_r,e_{\theta },e_z\right\}$ basis

5.2.4.4.            Plot a graph showing the variation of hoop stress ${\sigma }_{\theta \theta }/{\sigma }_0$ with $\theta$ at $r=a$ (the surface of the hole).  What is the value of the maximum stress, and where does it occur?

 

 

 

 

 

5.2.5.      Find an expression for the vertical displacement $u_1(x_2)$ of the surface of a half-space that is subjected to a distribution of pressure p(s) as shown in the picture.  Show that the slope of the surface can be calculated as

$\frac{\partial u_2}{\partial x_2}=\frac{2\left(1-{\nu }^2\right)}{\pi E}{\displaystyle \underset{L}{\int }\frac{p(s)ds}{x_2-s}}$