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Notes for Section 5.2

Notes for Section 5.2 - = ′ ′ = y y#2#6 Consider a long...

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Differential Equations – Section 5.2 – Homework: 1, 3, 5, 9, 13, 17 First we will solve the boundary-value problem 0 ) ( , 0 ) 0 ( , 0 = = = + L y y y y λ , using eigenvalues and eigenfunctions. λ is called a parameter. Case 1: 0 = λ . The general form of the solution is 2 1 c x c y + = . What happens? Case 2: 0 < λ . The general form of the solution is x c x c y λ λ - + - = sinh cosh 2 1 . What happens? Case 3: 0 λ . The general form of the solution is x c x c y λ λ sin cos 2 1 + = . What happens?

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We will now consider structures that are constructed using beams and find the deflection of the beams. The deflection curve approximates the shape of the beam. In the following formula E = Young’s modulus of elasticity of the material of the beam and I = the moment of inertia of a cross-section of the beam. The product EI is called the flexural rigidity of the beam. The deflection y(x) satisfies the first-order differential equation ) ( 4 4 x w dx y d EI = . Boundary conditions depend on how the ends of the beam are supported. Ends of the Beam Boundary Conditions Embedded 0 , 0 = = y y Free 0 , 0 = =

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Unformatted text preview: , = ′ ′ = y y #2 #6 Consider a long slender vertical column of uniform cross-section and length L. Let y(x) denote the deflection of the column when a constant vertical compressive force, or load, P is applied to its top. Then 2 2 = + Py dx y d EI . If we let EI P / = λ we have ) ( , ) ( , = = = + ′ ′ L y y y y . As we saw earlier, = L x n c x y n π sin ) ( 2 , corresponding to the eigenvalues ,... 3 , 2 , 1 , / / 2 2 2 = = = n L n EI P n n . Physically this means that the column will buckle or deflect only when the compressive force is one of the values 2 2 2 / L EI n P n = . These different forces are called critical loads . The deflection curve corresponding to the smallest critical load 2 2 1 / L EI P = , called the Euler load , is ( 29 L x c x y / sin ) ( 2 1 = and is known as the first buckling mode. #23...
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