Chapter four
A.Lecturer Saddam K. Kwais
Equilibrium Of Rigid Bodies
4/1 Rigid body in equilibrium
A rigid body is said to be in equilibrium the resultant of all forces acting on it is zero.
Thus, the resultant force R and the resultant couple M are both zero, and we have the
equilibrium equations
R 0M 0 … … … … … … … . 4.1
4/2 Free Body Diagram
1. Select the extent of the free-body and detach it from the ground and all other
bodies.
2. Indicate point of application, magnitude, and direction of external forces, including
the rigid body weight.
3. Indicate point of application and assumed direction of unknown applied forces.
These usually consist of reactions through which the ground and other bodies oppose
the possible motion of the rigid body.
4. Include the dimensions necessary to compute the moments of the forces.
For example:
44
Chapter four
A.Lecturer Saddam K. Kwais
Equilibrium Of Rigid Bodies
4/3 Reactions at Supports and Connections for a Two-Dimensional Structure
45
Chapter four
A.Lecturer Saddam K. Kwais
Equilibrium Of Rigid Bodies
Examples Of Free – Body Diagram:
Figure 3/2 gives four examples of mechanisms and structures together with their
correct free-body diagrams.
Fig. (4-2)
46
Chapter four
A.Lecturer Saddam K. Kwais
Equilibrium Of Rigid Bodies
4/4 Equilibrium of a Rigid Body in Two Dimensions
For all forces and moments acting on a two-dimensional structure,
Fz = 0 M x = M y = 0 M z = M O
Equations of equilibrium become
∑F
x
=0
∑F
y
=0
∑M
A
=0
where A is any point in the plane of the structure.
The 3 equations can be solved for no more than 3 unknowns.
EXAMPLE 1. A fixed crane has a mass of 1000 kg
and is used to lift a 2400 kg crate. It is held in
place by a pin at A and a rocker at B.
The center of gravity of the crane is located at G.
Determine the components of the reactions at A and B.
SOLUTION:
1. Create a free-body diagram for the crane.
2. Determine B by solving the equation for the
sum of the moments of all forces about A.
∑M
A
= 0 : + B(1.5m ) − 9.81 kN(2m )
− 23.5 kN(6m ) = 0
B = +107.1 kN
3. Determine the reactions at A by solving the equations for the sum of all horizontal
forces and all vertical forces.
∑F
x
= 0 : Ax + B = 0
Ax = −107.1 kN
47
Chapter four
∑F
y
A.Lecturer Saddam K. Kwais
= 0 : Ay − 9.81 kN − 23.5 kN = 0
A y = +33.3 kN
EXAMPLE 2. A loading car is at rest on an
inclined track. The gross weight of the car
and its load is 5500 lb, and it is applied at G.
The cart is held in position by the cable.
Determine the tension in the cable and the reaction
at each pair of wheels.
SOLUTION:
1. Create a free-body diagram for the car with
the coordinate system aligned with the track.
Wx = +(5500 lb) cos 25o
= +4980 lb
Wy = −(5500 lb )sin 25o
= −2320 lb
2. Determine the reactions at the wheels.
∑M
A
= 0 : − (2320 lb) 25in. − (4980 lb) 6in.
+ R2 (50in.) = 0
R2 = 1758 lb
∑M
B
= 0 : + (2320 lb ) 25in. − (4980 lb) 6in.
− R1 (50in.) = 0
R1 = 562 lb
48
Equilibrium Of Rigid Bodies
Chapter four
A.Lecturer Saddam K. Kwais
Equilibrium Of Rigid Bodies
3. Determine the cable tension.
∑F
x
= 0 : + 4980 lb − T = 0
T = +4980 lb
4/5 Categories of Equilibrium
The categories of force systems acting on bodies in two-dimensional equilibrium are
summarized in Fig. (4-3).
Fig. (4-3)
49
Chapter four
A.Lecturer Saddam K. Kwais
Equilibrium Of Rigid Bodies
4/6 Equilibrium of a Two-Force Body
Consider a plate subjected to two forces F1 and F2 as shown in Fig. (4-4a).
For static equilibrium, the sum of moments about A must be zero. The moment of
F2 must be zero. It follows that the line of action of F2 must pass through A see
Fig. (4-4b).
Similarly, the line of action of F1 must pass through B for the sum of moments
about B to be zero.
Requiring that the sum of forces in any direction be zero leads to the conclusion
that F1 and F2 must have equal magnitude but opposite sense see Fig. (4-4c).
Fig. (4-4)
4/7 Equilibrium of a Three-Force Body
Consider a rigid body subjected to forces acting at only 3 points as shown in Fig.
(4-5a).
Assuming that their lines of action intersect, the moment of F1 and F2 about the
point of intersection represented by D is zero see Fig. (4-5b).
Since the rigid body is in equilibrium, the sum of the moments of F1, F2, and F3
about any axis must be zero. It follows that the moment of F3 about D must be zero
as well and that the line of action of F3 must pass through D see Fig. (4-5c).
The lines of action of the three forces must be concurrent or parallel.
50
Chapter four
A.Lecturer Saddam K. Kwais
Fig. (4-5)
EXAMPLE 1. A man raises a 10 kg joist,
of length 4 m, by pulling on a rope. Find
the tension in the rope and the reaction at A.
SOLUTION:
1. Create a free-body diagram of the joist.
Note that the joist is a 3 force body acted upon
by the rope, its weight, and the reaction at A.
2. The three forces must be concurrent for static
equilibrium. Therefore, the reaction R must
pass through the intersection of the lines of action
of the weight and rope forces.
Determine the direction of the reaction force R.
AF = AB cos 45 = (4 m ) cos 45 = 2.828 m
CD = AE = 12 AF = 1.414 m
BD = CD cot(45 + 20) = (1.414 m ) tan 20 = 0.515 m
CE = BF − BD = (2.828 − 0.515) m = 2.313 m
CE 2.313
tanα =
=
= 1.636
AE 1.414
α = 58.6o
51
Equilibrium Of Rigid Bodies
Chapter four
A.Lecturer Saddam K. Kwais
Equilibrium Of Rigid Bodies
3. Utilize a force triangle to determine the magnitude of the reaction force R.
T
R
98.1 N
=
=
o
o
sin 31.4
sin 110
sin 38.6 o
T = 81.9 N
R = 147.8 N
4/8 Equilibrium of a Rigid Body in Three Dimensions
Six scalar equations are required to express the conditions for the equilibrium of a
rigid body in the general three dimensional case.
∑F = 0 ∑F = 0 ∑F = 0
∑M = 0 ∑M = 0 ∑M = 0
x
y
x
z
y
z
These equations can be solved for no more than 6 unknowns which generally
represent reactions at supports or connections.
The scalar equations are conveniently obtained by applying the vector forms of the
conditions for equilibrium,
r
r
∑F = 0 ∑M
O
(
)
r r
=∑ r ×F =0
4/9 Reactions at Supports and Connections for a Three-Dimensional Structure
52
Chapter four
A.Lecturer Saddam K. Kwais
Equilibrium Of Rigid Bodies
Fig. (4-6): Reactions at Supports and Connections.
53
Chapter four
A.Lecturer Saddam K. Kwais
Equilibrium Of Rigid Bodies
EXAMPLE 1. A sign of uniform density weighs
270 lb and is supported by a ball-and-socket joint
at A and by two cables. Determine the tension in
each cable and the reaction at A.
SOLUTION:
1.Create a free-body diagram for the sign.
v
v
v
rD = 0 i + 4 j − 8 k
v
v
v
rB = 8 i + 0 j + 0 k
r r
r
rD − rB
TBD = TBD r r
rD − rB
r
r
r
− 8i + 4 j − 8k
= TBD
12
 2 r 1 r 2 r
= TBD  − i + j − k 
3
3 
 3
v
v
v
rC = 0 i + 3 j + 2 k
v
v
v
rE = 6 i + 0 j + 0 k
r r
r
rC − rE
TEC = TEC r r
rC − rE
r
r
r
− 6i + 3 j + 2 k
= TEC
7
 6 r 3 r 2 r
= TEC  − i + j + k 
7
7 
 7
2.Apply the conditions for static equilibrium to develop equations for the unknown
reactions.
r
F
∑ =
r
i:
r
j:
r r
r
r
A + TBD + TEC − (270 lb ) j = 0
Ax − 23 TBD − 67 TEC = 0 ......................................(1)
Ay + 13 TBD + 73 TEC − 270 lb = 0 ........................(2)
54
Chapter four
A.Lecturer Saddam K. Kwais
Equilibrium Of Rigid Bodies
r
k : Az − 23 TBD + 72 TEC = 0 ......................................(3)
r
r r
r
= rB × TBD + rE
v
v
i
j
v
MA = 8
0
∑M
A
r
r
r
× TEC + (4 ft )i × (− 270 lb ) j = 0
v
v
v
v
k
i
j
k
v
v
0 TBD + 6
0
0 TEC + (4 i ) × (−270 j )
−2 3 13 −2 3
−6 7 3 7 2 7
r
j : 5.333TBD − 1.714TEC = 0 ⇒ TBD = 0.3124TEC ......................(4)
r
k : 2.667 TBD + 2.571TEC − 1080 lb = 0 .....................................(5)
Substitution Eq.(4) in Eq.(5) we get
TBD = 101.3 lb TEC = 315 lb Substituti on these values in Eqs.(1,2,3) we get
r
r
r
v
A = (338 lb )i + (101.2 lb ) j − (22.5 lb )k
EXAMPLE 2. A 48-in. boom is held by
a ball-and-socket joint at C and by two cables
BF and DAE; cable DAE passes around
a frictionless pulley at A. For the loading shown,
determine the tension in each cable
and the reaction at C.
SOLUTION:
1.Create a free-body diagram for the sign.
rD
rE
rA
r
TAD
v
v
v
= −20 i + 0 j − 0 k
v
v
v
= 0 i + 20 j + 0 k
v
v
v
= 0 i + 0 j + 48 k
r r
rD − rA
= TAD r r
rD − rA
r
r
r
− 20i + 0 j − 48k
= TAD
52
y
20 in.
16 in.
E
D
Czk
Cyj
C
48 in.
TAE
TAD
F
TBF
A
z
55
W =- (320 Ib) j
20 in.
Cxi
B
30 in.
x
Chapter four
A.Lecturer Saddam K. Kwais
Equilibrium Of Rigid Bodies
r 12 r 
 5r
= TAD  − i + 0 j − k 
13 
 13
r
r
r
r
r
r
rE − rA
0 i + 20 j − 48 k
 r 5 r 12 r 
TAE = TAE r r = TAE
= TAE  0 i +
j − k
rE − rA
52
13
13 

v
v
v
rF = 16 i + 0 j + 0 k
v
v
v
rB = 0 i + 0 j + 30 k
r
r
r
r r
r
r 15 r 
rF − rB
16 i + 0 j − 30 k
8 r
TBF = TBF r r = TBF
= TBF  i + 0 j − k 
rF − rB
34
17 
 17
2.Apply the conditions for static equilibrium to develop equations for the unknown
reactions.
r v r
r
r
r
F
=
C
+
T
+
T
+
T
−
320
lb
j =0
(
)
∑
AD
AE
BF
r
5
8
i : C x − TAD + TBF = 0 ..........................................(1)
13
17
r
5
j : C y + TAE − 320 lb = 0 .........................................(2 )
13
r
12
12
15
k : C z − TAD − TAE − TBF = 0 .............................(3)
13
13
17
r
v
r
r r
r r
r r
M
=
r
×
T
+
r
×
T
+
r
∑ C A AD A AE B × TBF + (30 in.)k × (− 320 lb) j = 0
v
v
v
v
v
v
v
v
v
i
j
k
i
j
k
i
j
k
v
MA = 0
0
48 TAD + 0
0
48 TAE + 0
0
30 TBF
− 5 13 0 − 12 13
0 5 13 − 12 13
8 17 0 − 15 17
v
v
+ (30 k ) × (−320 j )
v
i : − 18.46 TAE + 9600 = 0 ⇒ TAE = 520 Ib = TAD sub.in eq.(2) ⇒ C y = 120 Ib
v
j : 18.46 TAD − 14.11TBF = 0 ⇒ TBF = 680 Ib sub.in eq.(1) ⇒ C x = −120 Ib
Sub. the values of TAD, TAE, and TBF in eq.(3) we get
v
v
v
C z = 1560 Ib ⇒ C = −120 i + 120 j + 1560 k
56
Ans.