Structural Mechanics 5 REACTIONS SFD BMD with multiple

Structural Mechanics 5 REACTIONS, SFD, BMD – with multiple loads 10 k. N 20 k. N 30 k. N RA 1. 50 m RB 0. 50 m 2. 00 m 1. 00 m

Sum of moments about RA = 0 10 x (1. 50) + 20 x(1. 5 +0. 5) + 30 x( 1. 5 +0. 5 +2. 0) - RB x (1. 5 +0. 5 +2. 0 +1. 0) = 10 x (1. 50) + 20 x(1. 5 +0. 5) + 30 x( 1. 5 +0. 5 +2. 0) RB x (5. 0) = 10 x (1. 5) + 20 x(2. 0) + 30 x(4) RB = {10 x (1. 5) + 20 x(2. 0) +30 x(4)} / 5. 0 RB = {15 + 40 + 120 } / 5. 0 RB = {175} / 5. 0 RB = 35 k. N 10 k. N 1. 50 m RA 20 k. N 0. 50 m 30 k. N 2. 00 m 1. 00 m RB =0

Sum of moments about RB = 0 RA x (1. 5 +0. 5 +2. 0 +1. 0) - 10 x (0. 5+2. 0+1. 0) - 20 x(2. 0 +1. 0) - 30 x( 1. 0) RA x (5. 0) = 10 x (3. 5) + 20 x(3. 0) + 30 x(1. 0) RA = {35+60+30} / 5. 0 RA = {125} / 5. 0 RA = 25 k. N

CHECK VERTICAL FORCES • UPWARD REACTION = RB + RA = 35 + 25 = 60 k. N • DOWNWARD FORCE = 10+20+30 =60 k. N • So likely that calculations are correct!

SHEAR FORCE DIAGRAM Space diagram 10 k. N RA = 20 k. N 30 k. N RB = 25 k. N 35 k. N SHEAR FORCE DIAGRAM 25 k. N 25 -10 =15 k. N 0 -35 +35 =0 15 -20 = -5 k. N -5 -30 = -35 k. N

BENDING MOMENT DIAGRAM • Space diagram 10 k. N 20 k. N 30 k. N RA = 25 k. N 1. 50 m RB = 35 k. N 0. 50 m 2. 00 m 1. 00 m 45. 0 k. Nm 37. 5 k. Nm 35. 0 k. Nm Section 1 Section 2 Section 3

BMD CALCULATIONS • • Section 1 25 k. N [1. 5]m = 37. 5 k. Nm Section 2 25 k. N [1. 5+0. 5]m -10 k. N[0. 5]m = 45. 0 k. Nm (NB max BM where shear force crosses the zero axis) Section 3 25 k. N [1. 5+0. 5+2. 0]m -10 k. N [2. 0+0. 5]m-20 k. N[2. 0]m = 35. 0 k. Nm • Check: (taking BM from the RHS) • 35 k. N [1. 0]m = 35. 0 k. Nm