l. A rod of unit cross section area, elastic modulus E and density is suspended

Question

l. A rod of unit cross section area, elastic modulus E and density is suspended in a vertical plane such that it deforms under its own weight: The governing differential equation and boundary conditions for the axial displacement u(x) are dr2 dr where g is the acceleration due to gravity. By analogy with the 1-D model problem, the total potential energy functional for the unit cross-section rod can be written as E (du a) Assuming an approximate solution of the formu(x)-Ax(x-2H), determine the displacement at the end x ,, using the Rayleigh-Ritz Method. (10 points) b) If you were to solve the governing differential equation BVP in Equation to obtain the exact solution for u(r), would it be the same as your Rayleigh-Riz solution? Clearly explain why or why not. Note: you do not need to solve Equation (1) to answer this. (5 points)

Explanation / Answer

1. given cross section area = 1

elastic modulus = E

density = rho

suspended in vertical plane

deforms under own weight

governing equations

Ed^2u/dx^2 + rho*g = 0

u(0) = 0

sigma(H) = Edu/dx|x = H = 0

U = integral((E/2)(du/dx)^2 - rho*g*u(x))dx from 0 to H

a. assuming

u(x) = Ax(x - 2H)

then

U = integral((E/2)(A^2x^2(x - 2H))^2 - rho*g*Ax(x - 2H))dx from 0 to H

dU/dA = integral((E/2)(2Ax^2(x - 2H))^2- rho*gx(x - 2H))dx = 0

A = -5*rho*g/4EH^2

hence

u(x) = -5*rho*gx(x - 2H)/4EH^2

for x = H

u(H) = 5*rho*g/4E

b) for exactly solving the equation

we need dependence of u on x

which we dont have

so we solve 1 instead

E*d^2u/dx^2 = -rho*g

double integrating

E*u = -rho*g*x^2/2 + C1*x + C2

for x = 0

C2 = 0

du/dx = 0 at x = H

C1 = rho*g*H

hence

u = -rho*g*x^2/2E + rho*g*H*x/E = rho*g*x(-x/2E + H/E)

is differnt from the reyligs mehtod as its approximnate

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