A third spring of constant k3and a weight W3are added to the system of Example 5.3. Determine the displacements q1, q2, and q3of weights W1, W2, and W3. Example 5.3 Consider weights W, and W, supported by the linear springs shown in Figure ES.3. The spring con- stants are k and kz. Determine the displacements q, and q, of the weights as functions of W. W2, k. and k. Assume that the weights are applied slowly so that the system is always in equilibrium as the springs are stretched from their initially unstretched lengths. to FIGURE E5.3 Under the displacements q, and q2, spring I undergoes an elongation q and spring 2 undergoes an elongation q3 – 41. Hence, by Eq. 5.12, the potential energy stored in the springs is The internal tension forces in springs I and 2 are F, = kj91. F2 = kz(qz-91) (b) But by oquilibrium, F, = W +W2, F2 - W2 Then, by Eqs. (a)-(c), the potential energy in terms of W1,W2, k, and k, is (c) 1(W,+W23 1 W (d) By Eqs. 5.4 and (d), we obtain au W,+W2 W W. (e) au 92 = W, Equation (e) agrees with a direct solution of Eqs. (b) and (c), without consideration of potential energy. www

A third spring of constant k3and a weight W3are added to the system of Example 5.3. Determine the displacements q1, q2, and q3of weights W1, W2, and W3. Example 5.3 Consider weights W, and W, supported by the linear springs shown in Figure ES.3. The spring con- stants are k and kz. Determine the displacements q, and q, of the weights as functions of W. W2, k. and k. Assume that the weights are applied slowly so that the system is always in equilibrium as the springs are stretched from their initially unstretched lengths. to FIGURE E5.3 Under the displacements q, and q2, spring I undergoes an elongation q and spring 2 undergoes an elongation q3 – 41. Hence, by Eq. 5.12, the potential energy stored in the springs is The internal tension forces in springs I and 2 are F, = kj91. F2 = kz(qz-91) (b) But by oquilibrium, F, = W +W2, F2 - W2 Then, by Eqs. (a)-(c), the potential energy in terms of W1,W2, k, and k, is (c) 1(W,+W23 1 W (d) By Eqs. 5.4 and (d), we obtain au W,+W2 W W. (e) au 92 = W, Equation (e) agrees with a direct solution of Eqs. (b) and (c), without consideration of potential energy. www

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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