Question

Can you breifly show how the values in the solutions were solved

Problem 3.2 4 kip/ft 10 kip RAx B RAY RD 20 ft 10 ft + 10 ft Σ-0 R 0 Ac 20, ,20 -Ro(40)+10(30)+4 G=0 ΣΜ-0 RD14.17 kip1 at A ΣΕ-0 -4+14.17 0 -35.83 kip T R-10-40 RAV Ay 10 kip 4 kip/ft B A f14.17kip C 35.83kip 20 ft 10 ft 10 ft 6.46 ft 14.17 4.17 V, kip 201.3 35.83 183.4 141.7 M, kip-ft 3

Answer 1

1OK RAx0 (6.46) 20 1292 t RAX 0 - 56 EK646ftB 20ft 777 7MAD (40) n20)( 20) +10 (30) KA = 35.63 kp oy om =) : a Calcalaton of shen force valuus J1 ending monant valus - 35.83 KP A 4.1 (26.40)-10(16146 12926.u6) = -33.83+1(4)(20) = 4.)9 kip max F 3 201.3 K-Ft mar -4.17+ 10 = 417 kif "BI41 ( 20) 1o (10) 4.17 E 183.4 kip-Ft -20) (x-20) SFx=0 ラ 4.17= 10+ 20 4.17(10)141 kip ft 26.46 Ft By Bolung om SF Zeno at 6.46 fE. 0