Equilibrium of Rigid Bodies Elementary Engineering Mechanics Equilibrium of Rigid Bodies Draw the free body diagram ( replace the supports with reactions) Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002

Elementary Engineering Mechanics Types of reactions - 1 Roller support : replace with a reaction normal to AB A B Frictionless floor : again, replace with a reaction normal to AB A B Rigid member : replace with a reaction in the direction of the member Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002

Elementary Engineering Mechanics Types of reactions - 2 A Frictionless slider : replace with a reaction normal to AB B Pin support : replace with two reactions, horizontal & vertical fixed support : replace with two reactions and a couple Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002

Elementary Engineering Mechanics Equilibrium in 2D Where O is a convenient point. Try to choose it so that each equation has only one unknown. OR Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002

Equilibrium in 2D – example 1 Elementary Engineering Mechanics Equilibrium in 2D – example 1 200 mm A 150 mm Determine : a) tension in cable b) reaction at C 30 o C B 400 N 200 mm D Angle ACD is equal to 90o + 30o or 120o. The triangle ACD is isosceles. Hence angle CAD is 30o 200 mm A C y 150 mm C x 30 o C Resolve T AD into two components one in the direction of AB and the other normal to AB. Take moments about C B 400 N T AD D Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002

Equilibrium in 2D – example 1-cont. Elementary Engineering Mechanics Equilibrium in 2D – example 1-cont. Moments about C: (T AD sin30o) 200 – (150 sin30o) 400 = 0 We used the red components of T AD 200 mm A C y 150 mm C x 30 o C B 400 N T AD Sum of forces in the x , y direction: ( use the green components of T AD ) T AD cos60o + C x – 400 = 0 -T AD sin60o + C y = 0 D We solve to obtain T AD = 300 N, C x = 400 - 150 = 250 N, and C y = 300 sin60o = 259.8 N Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002

Equilibrium in 2D – example 1- better method. Elementary Engineering Mechanics Equilibrium in 2D – example 1- better method. Moments about C: (T AD sin30o) 200 – (150 sin30o) 400 = 0 We used the red components of T AD 200 mm A C y 150 mm C x 30 o C 200 mm B 400 N T AD Take moments about point D (use T AD itself) 400 ( 200 – 75) – C x 200 = 0 --> C x = 250 D Take moments about point A C x 100 – 400 ( 100 + 75) + C y 200 cos30o = 0 --> C y = 259.8 N Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002

Equilibrium in 2D – example 2 Elementary Engineering Mechanics Equilibrium in 2D – example 2 6 kN Determine the reactions at the supports 2 kN 1.5 m 6 kN 2 kN 1.5 m 2 kN 2 kN 2 m Bx Sum of moments about B: 2 x 6 – 3 x 2 – 1.5 x 2 – 2 x A y = 0, A y = 1.5 kN Sum of forces in x – dir : B x = 4 kN A y B y Sum of forces in y dir. B y = 4.5 kN Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002

Equilibrium in 2D – example 3 Elementary Engineering Mechanics Equilibrium in 2D – example 3 Obtain reactions 15 x 9.81 = 147.2 N .25m .25m B x FBD 15 kg 350 mm A x A y From sum of forces in x dir.  A x = B x = T From sum of forces in y dir.  A y = 147.2 N Sum of moments ( about A )  .35 T - .25 147.2 = 0 T = 105.1 N Eqlb. Rigid Bodies Jacob Y. Kazakia © 2002