Quiz Chapter 08.07: Finite Difference Method

1. The exact solution to the boundary value problem

• • • \dfrac{d^{2}y}{dx^{2}} = 6x - 0.5x^{2}, \, y(0) = 0, \, y(12) = 0

for y(4) is

2. Given

• • • \dfrac{d^{2}y}{dx^{2}} = 6x - 0.5x^{2}, \, y(0) = 0, \, y(12) = 0

the value of \dfrac{d^{2}y}{dx^{2}} at y(4) using the Finite Difference Method and a step size of h=4 can be approximated by

3. Given

• • • \dfrac{d^{2}y}{dx^{2}} = 6x - 0.5x^{2}, \, y(0) = 0, \, y(12) = 0

The value of y(4) using the Finite Difference Method with a second order accurate central divided difference method and a step size of h=4 is

4. The transverse deflection, u of a cable of length, L, fixed at bothe ends is given as a solution to

• • • \dfrac{d^{2}u}{dx^{2}} = \dfrac{Tu}{R} + \dfrac{qx \left( x - L \right)}{2R}

where

• • • T= tension in cable
    • R= flexural stiffness
    • q= distributed transverse load

Given values are L=50", \, T=2000lbs, q=75 lbs/in, R=75 \times 10^{6} lbs⋅in², using Finite Difference Method modeling with second order central divided difference accuracy and a step size of h=12.5", the value of the deflection in inches at the center of the cable is most nearly

5. The radial displacement, u is a pressurized hollow, thick cylinder (inner radius=5", outer radius=8") is given at different radial locations.

The maximum normal stress, in psi, on the cylinder is given by

• • • \sigma_{max} = 3.2967 \times 10^{6} \left( \dfrac{u(5)}{5} + 0.3 \dfrac{du}{dr}(5) \right)

The maximum stress, in psi, with second order accuracy is

Hint: f(x_{i+1}) = f(x_{i}) + {f}'(x_{i})h + \dfrac{{f}''(x_{i})h^{2}}{2} + \dfrac{{f}'''(x_{i})h^{3}}{6} + \, ... \,,

and f(x_{i+2}) = f(x_{i}) + {f}'(x_{i})(2h) + \dfrac{{f}''(x_{i})(2h)^{2}}{2} + \dfrac{{f}'''(x_{i})(2h)^{3}}{6} + \, ... \,

where x_{i+1} = x_{i} + h

6. For a simply supported beam (at x=0 and x=L) with a uniform load q, the deflection v(x) is described by the boundary value ordinary differential equation as

• • • \dfrac{d^{2}v}{dx^{2}} = \dfrac{qx(L-x)^{2}}{2EI}, \, 0 \leq x \leq L

where

• • • E= Young’s modulus of elasticity of beam
    • I= second moment of cross-sectional area

This is based on assuming that \dfrac{dv}{dx} is small; if \dfrac{dv}{dx} is not small, then the ordinary differential equation is