4.Member end forces -Nodal displacements P)=IklA) -Member end forces {Fy=[kδ 同源大学 土本工程学院 6kN.m> 3kN.m Example 3kN.m Solution: 1=1 i2=2 1)discretization 2 2)Global stiffness/loading ② 1 4.Member end forces = 42 1 「84N2 242 k9 区3 -%-{2 4-116j1-712 42 0 -6 [= 212 4 P)= 3 811/24 0 48 3 712 3)displacements -17/12 P)=kHA) A= -1/6 6 11/24