This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: AUBURN UNIVERSITY Department of Civil Engineering CIVL 7690 Analysis of Plate and Shell Systems Final, Take Home, Due August 4, 2009 1. A simply supported semiinfinite plate shown below is free at the cutoff edge. A vertical line loading of a sinusoidal distribution is applied at the free edge. Derive a general expression for the deflection and the maximum bending moments. If 0.1 k/in, q = 100 in, a = 1 in, E=29,000 ksi, and 0.3 t μ = = , determine max max max max , , , and x y x w M M R and plot the variation of max max max , , and y x w M M versus x . 2. A square plate of dimension “a” is simply supported on two side boundaries and fixed at the top and bottom. The plate is subjected to a linearly varying compressive load, x N , resulting from a pure bending plus axial compression as shown in the sketch. Using the energy method discussed in the class, determine the critical load. Assume the displacement function as single waves for both x and ydirections reflecting the fixed condition in the y direction. y y x ( 29 sin y p y q a π = free edge x a ∞ a a N s.s. s.s. 1 2 N fixed fixed 1 1. M. Lévy suggested a solution of the following form for a rectangular plate that has two opposite edges simply supported: 1 sin sin m m m y y w X X a a π π ∞ = = = ∑ (a) where X m is a function of y only. As the line loading, ( 29 p y , is given by a sinusoidal function, m is taken to be equal to one. Since the semiinfinite plate is subjected to a sinusoidal line loading at its left edge ( x =0) and there is no distributed loading on the plate, the governing differential equation is a homogeneous forth order partial differential equation. 4 4 4 4 2 2 4 2 w w w x x y y ∂ ∂ ∂ + + = ∂ ∂ ∂ ∂ (b) Taking partial derivatives of (a) and substituting into (b), gives cos w y X y a a π π ∂ = ∂ sin w y X x a π ∂ ′ = ∂ 2 2 2 sin w y X y a a π π ∂ =  ∂ 2 2 sin w y X x a π ∂ ′′ = ∂ 3 3 3 cos w y X y a a π π ∂ =  ∂ 3 3 sin w y X x a π ∂ ′′′ = ∂ 4 4 4 sin w y X y a a π π ∂ = ∂ 4 4 sin iv w y X x a π ∂ = ∂ 2 cos w y X x y a a π π ∂ ′ = ∂ ∂ 2 4 2 2 sin w y X x y a a π π ∂ ′′ =  ∂ ∂ 2 4 2 sin iv y X X X a a a π π π ′′ + = (c) 2 Equation (c) can be satisfied for all...
View
Full
Document
This note was uploaded on 09/24/2011 for the course CIVL 7690 taught by Professor Staff during the Summer '10 term at Auburn University.
 Summer '10
 Staff
 Civil Engineering

Click to edit the document details