Question

The reactions for the frame ABCD below are: Ax = 20 kip to the left), Ay =8.89 kip (upward), Dy = 11.11 kip (upward). Determine: a) the internal forces at a section just above the 20-kip concentrated force (on member AB) b) the internal forces at a section 4.45 ft from B (on member BC) c) the internal forces at a section at midpoint between C and D (on member CD). Tip: choose the right side of the section and resolve Dy into components that are perpendicular and parallel to member CD. Make the x axis parallel to CD and the y axis perpendicular to CD and proceed. Remember Vis perpendicular to the member and N is parallel to the member. 2 кг 1, 2014 1 tront

Answer 1

2k14 Au=20k ) Ay=8.89K (1) By = 11:11K (1) - zok --- 8! Ay zok N=-8.89K (compressen) V-20-20 k 20k 8.89k (1=0 M = 2045 m = 100K-ff 2 k/d V N-20-20 B 山AN 4.45 VUM zok (N=o7 V=8.89-2x4.45 V=-0.01.20 [=0 Azok on - 20x10-20x5+8.898 4.45 8.89k -2x 4.458 4.45 [m= 119.758 k-ft

2K1ft B 10! N=-11.11 Eino N =-11 11x 2ok N- - 20k- IN=-8.68 K (compression) 889 sino = 5 - 4 COSO = 4 4 4 V = !1!1 caso = 1111x [V = 6.94 k M = 8.89x14 +20%* 5 - 2x10x9 TM = 44.46 K- ft