Question

Problem 3130 points Recall from lecture the "work equivalence method" is based on the assumption distributed load w(x) in going through displacement field v(x) is that the work of the equal to the work done by nodal loads fry and mu Cl 12) in going through nodal displacements dy and rotations mathematically as: Walstr Wnodal ure below: wx) is wi at node 1 (x ) equivalent loads f ms ay and ma. You may employ one of Either (1) the method we have used in lecture by finding Waistr through requires you to express wr) as a linear function of x or (2) by using the principle of out the proper combination of linear and wniforn loading cases given in the formula Now consider the linecarly distributed load wx) shown in the and w2 at node 2 (x L). Find the nodal the following two methods: the the superposition and work w(a) W1 W2 .끅분(Xy.ny)-박 Page 6 of 8

Answer 1

レ w| | 나レ 4 -ST 2 レ 珂실L)