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Unformatted text preview: y 2.1 Determine an approximate value for the critical load of a propped column. The column is hinged at the top loaded end and fixed at its base. Use the energy method. Assume the deflected shape of the column by the deflection curve of a uniformly loaded propped beam whose boundary conditions are the same as those of the column. Assuming the coordinate origin at the propped end, the deflection curve is given by Taking successive derivatives yields For stability, the variation of with respect to the natural coordinate must vanish. Hence, 1 Since cannot be equal to zero for nontrivial solution (or the system must be in equilibrium in a slightly deformed configuration), the quantity inside the parentheses must vanish. Hence, This compares very well with the exact value of (less than 2.5% error) 2.2 Find the critical load for a rigid bar system loaded as shown in Fig. P22. Assume the two rigid bars of length are connected by a hinge and displacements remain small. Fig. P22 Spring supported rigid bar Let the displacements at the spring location be. Then The strain energy stored in the elastic supports during buckling is 2 The rotation of the left side bar is and the rotation of the right side bar is. The distance moved by the force P is found to be . The work done by the force P is Equating these two expressions yields The correct critical load is the minimum. To find the critical value of P , one must adjust the deflections, which are unknown, so as to make P a minimum value. This is accomplished by setting the determinant for the unknown equal to zero after taking . Similarly, 3 and Substituting these values, one obtains For nontrivial solution, the coefficient determinant must vanish. Hence, The critical value is of course the smaller one, and the corresponding eigenvector is as assumed in the sketch. 2.3 Use the principle of minimum potential energy to derive the governing differential equation of equilibrium and the natural boundary conditions for a prismatic 4 column resting on an elastic foundation with a foundation modulus. Then, compute the critical load for the pinned column shown in Fig. P23. Fig. P23 Pinned column resting on elastic foundation The strain energy stored in the elastic body is The loss of the potential energy of the applied load is and the total potential energy functional is For equilibrium the first variation of the total potential energy is zero, so that Note that. The first and third terms of the integrand in the above equation may be integrated by parts to obtain 5 Hence, the first variation of the total potential energy is...
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This note was uploaded on 09/24/2011 for the course CIVL 3110 taught by Professor Melville during the Spring '08 term at Auburn University.
 Spring '08
 MELVILLE

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