R5-1. If the roller at B can sustain a maximum load of 3 kN, determine the largest magnitude of each of the three forces F that can be supported by the truss. Ay AX 2 m By B -2 m V F C 2 m V F 45° 10 BX 2 m V F 14 Prob. R5-1 Equations of Moments SMA = 0 Roller F Break B into 2 components Bx = Bcos 45 = 3 cos 450 = 2.12 KN L By = B sin 45° = 3 sin 45 F.2+4F+6F- Bx.2m =0 1 2F = 2 (3 cos45) => F = 0.35KN
R5-2. Determine the reactions at the supports A and B for equilibrium of the beam. Diagram Redraw using forces centered at ONE point SOON Ay 400 N/m 4 m 120ON - * 1By 7 200 N/m Ax 3 B A OB A Roller 4 m 3 m Prob. R5-2 Equations of Equilibrium x- dir: Ax = 0 -80ON-1200N+Ay+ By = D -y-dir: Ay + By = 200ON A y = 343N = 0.343KN Equations of Moments -800.4m-1200.7m+By.7m = 0 EM = A Assumed they occurred at the endpoints By = 1657N = 1.66KN
R5-3. Determine the normal reaction at the roller A and horizontal and vertical components at pin B for equilibrium of the member. y x= 6cos 30° y = 6 sin 30" x Equilibrium Equations 10 kN -0.6 m - 0.6 m x- dir : Bx - 6cos 30° = 0 y-dir: N + By - 10KN - 6KN sin 30°= 0 A Equations of Moments No 0.8 m 60° 6 kN $30° 60 By 0.4 m 1 Bx Prob. R5-3 + 1.2m cos60 -> 1.2m = 1.8 m EMA = 1.8NA - 10KN. (1.2m) - 6.0.4m = 0 => NA= 8.0KN => Using Fy equation By = - NA + 10KN+ 6KN sin 30° = 5KN Bx = 6cos 30' = 5.19KN