lineld - AUBURN UNIVERSITY Department of Civil Engineering...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the 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 semi-infinite plate shown below is free at the cut-off 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 w M M R 2. A square plate of dimension “a” is simply supported on all four boundaries. The plate is subjected to a linearly varying compressive load, x N , resulting from a pure bending as shown in the sketch. Using the energy method discussed in the class, determine the critical load. Assume the displacement function (a) single half sine waves for both x- and y-directions and (b) a single half sine wave in the x-direction and two half sine waves in the y-direction. y y x ( 29 sin y p y q a π = free edge x a a a N . . ss . . ss . . ss . . ss ∞ 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 semi-infinite 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...
View Full Document

{[ snackBarMessage ]}

Page1 / 7

lineld - AUBURN UNIVERSITY Department of Civil Engineering...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online