Equilibrium of Rigid Bodies
F 0
~
~
M  0
~
~
Draw the free body diagram ( replace the supports with reactions)
Eqlb. Rigid Bodies
Jacob Y. Kazakia © 2002
1
Types of reactions - 1
A
A
B
B
Roller support : replace
with a reaction normal to
AB
Frictionless floor : again,
replace with a reaction
normal to AB
Rigid member : replace
with a reaction in the
direction of the member
Eqlb. Rigid Bodies
Jacob Y. Kazakia © 2002
2
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
3
Equilibrium in 2D
F  0 , F
x
y
0,
M
o
0
Where O is a convenient point. Try to choose it so that each
equation has only one unknown.
OR
M
A
0,
Eqlb. Rigid Bodies
M
B
0,
Jacob Y. Kazakia © 2002
M
o
0
4
Equilibrium in 2D – example 1
A
200 mm
150 mm
Determine : a) tension in cable
b) reaction at C
30 o
C
B
400 N
200 mm
D
A
200 mm
Cy
Angle ACD is equal to 90o + 30o
or 120o. The triangle ACD is
isosceles. Hence angle CAD is 30o
150 mm
Cx
30 o
C
B
T AD
D
Eqlb. Rigid Bodies
400 N
Resolve T AD into two
components one in the direction of
AB and the other normal to AB.
Take moments about C
Jacob Y. Kazakia © 2002
5
Equilibrium in 2D – example 1-cont.
A
200 mm
Cy
C
Moments about C:
(T AD sin30o) 200 – (150 sin30o) 400 = 0
150 mm
Cx
30 o
B
We used the red components of T AD
400 N
T AD
D
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
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
6
Equilibrium in 2D – example 1- better method.
200 mm
Cy
A
200 mm
Moments about C:
(T AD sin30o) 200 – (150 sin30o) 400 = 0
150 mm
Cx
30 o
C
B
We used the red components of T AD
400 N
T AD
D
Take moments about point D (use T AD itself)
400 ( 200 – 75) – C x 200 = 0 --> C x = 250
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
7
Equilibrium in 2D – example 2
6 kN
Determine the reactions at the supports
2 kN
1.5 m
2 kN
6 kN
1.5 m
2 kN
2 kN
2m
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
Eqlb. Rigid Bodies
Ay
By
Sum of forces in y dir.
B y = 4.5 kN
Jacob Y. Kazakia © 2002
8
Equilibrium in 2D – example 3
15 x 9.81 = 147.2 N
Obtain reactions
.25m
.25m
Bx
FBD
15 kg
350 mm
Ax
Ay
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
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