3. (40 points) A four-node, three-element planar frame structure is shown below: Z X 2 m 3 m Fa 2 m B Th Them 4 m The global coordinate system is X - Z, as shown. The force Fa is horizontal, and is acting at the midpoint of element 1. The distributed force fb is acting vertically as shown, and is specified as force per unit length (this length being along the element 3, i.e., this is force per unit length along BD). For each of the frame elements 1 and 3, obtain the elemental force vector {F} in the global coordi- nate system. cos Be sin 20 - cos Be -sin 20 E.A sin 20 sin e -sin 20 sin 0. [K] == he - cos20 sin 20 cos Be sin 20 sin 20 sin 0 sin 20 sin e 3 =1-3 (A)*+2(A)--(1-)* 3 45=3(*) - 2 (A).4 [AA] 6 -3he -6 -3he 2Eele -3h 2h2 3he h [Ke] = h -6 3he 6 3he -3he h 3he 3he 2h2 6 gehe -he Q2 {Fe} = + 12 6 23 he Q4 S ==== 142 W 52 cos sina 0 sin a cos 0 001 141 WI S (5.4.6a) Cosa sina 0 142 0 - sin a COS Wa 00 S