EML 3004C
Chapter 4: Equilibrium
Equilibrium means balance of forces to
prevent body from translating, and balance of
moments to prevent body from rotating.
Vector analysis in 3-D is the preferred method
of solution.
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-1
EML 3004C
4.1 Conditions for Equilibrium
Equilibrium means that the object is at rest (if originally at rest), or
in constant velocity (if originally moving).
F  0
 F  ma
(Static)
(dynamic) but a  0 for static
Let Fi be the external force and f i be the
 F +  f = 0.
Since for equilibrium  f = 0, we have  F  0
internal force. Then
i
i
i
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-2
EML 3004C
4.1 Conditions of Equilibrium Con’t
Moment about any point O,
M = 0
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-3
EML 3004C
4.2 Free Body Diagrams
Need to know how to represent support and contact
conditions.
If a support prevents translation in any direction, we have a
reaction force in that direction.
If a support prevents rotation in any orientation, then we
have a couple moment exerted.
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-4
EML 3004C
4.2 Free Body Diagrams Con’t
Weight always acts at the center of gravity.
W=mg
Consider the case
of springs.
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-5
EML 3004C
4.2 Free Body Diagrams Con’t
Consider the cantilever
beam supported by a
fixed support at A.
Free body
diagram
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-6
EML 3004C
Free Body Diagrams of a Platform
Consider the
platform
Exclude all other
effects except the
platform now!
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-7
EML 3004C
4.3 Equations of Equilibrium
In two dimensions (x-y plane)
 F  F  F
x
y
z
0
Rarely though
 F   M   M 0
 M   M   M 0
a
A
A
Namas Chandra
Introduction to Mechanical engineering
B
B
C
Hibbeler
Chapter 4-8
EML 3004C
4.3 Supports and Reactions
Supports are idealized first. Reaction forces (magnitude
and direction) and moments then depends on the type of
support.
Roller allows motion along
the plane. Reaction force is
perpendicular to the
surface.
Rocker allows rotation at
that point. Reaction force is
perpendicular to the
surface.
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-9
EML 3004C
4.3 Supports and Reactions-2
Pin connected to a collar.
Reaction force is
perpendicular to the rod.
Hinge allows motion both in x
and y direction but no rotation.
Reaction force in x and y only.
F not along member
Fixed allows no rotation
and no translation.
Reaction force vector and
moment will result.
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-10
EML 3004C
4.3 Analysis Precedure
First draw the free body
diagram for the loading
shown to the right.
Apply equations of
equilibrium through force
and moment balance.
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-11
EML 3004C
Problem 4-3 (page 138, Section 4.1-4.3)
4.3 Draw the free-body diagram of the automobile, which has
a mass of 5 Mg and center of mass at G. The tires are free to
roll, so rolling resistance can be neglected. Explain the
significance of each force on the diagram.
Solution:
W  effect of gravity
(weight) on the car.
T = effect of the cable
on the car.
N A and N B = effect of the
road surface on the car.
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-12
EML 3004C
Problem 4-7 (page 139, Section 4.1-4.3)
4.7 Draw the free body diagram of the beam. The incline at B
is smooth.
Solution:
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-13
EML 3004C
Problem 4-17 (page 154, Section 4.4-4.5)
4.17 Determine the stretch of each spring for
equilibrium of 20-Kg block. The springs are
shown in their equilibrium position.
Solution:
Equilibrium:
Spring AD
+
F
y
= 0; FAD  20(9.81)  0
Spring AB and AC
4
+   Fy = 0; FAB  FAC sin 45=0
5
3
+   Fx = 0; FAB  FAC cos 45  196.2=0
4
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-14
EML 3004C
4.17 Determine the stretch of each spring
for equilibrium of 20-Kg block. The
springs are shown in their equilibrium
position.
Solution-Con’t (slide 2)
Solving Eq. 1 and  2 yields:
FAC  158.55 N
FAB  140.14N
Spring elongation:
140.14
x AB 
 0.467m
300
158.55
x AC 
 0.793m
200
196.2
x AD 
 0.490m
400
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-15
EML 3004C
4.4 Two-Force Members
Two force members are trusses that have forces (tension or
compression) but not couple moments.
Resolve all the forces at A and B so that
FA  F1  F2  F3 and FB  F4  F5  F6
If FA and FB are collinear then FA  FB and M  0.
Then the body is a two-force member.
It can be either in compression FA < 0
or tension FA > 0
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-16
EML 3004C
4.4 Two and Three-Force Members
If a member is subjected to three coplanar
forces, then the forces should either be
concurrent or coplanar for equilibrium.
Hydraulic
cylinder is a
two-force
member
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-17
EML 3004C
Problem 4-26 (page 155, Section 4.4-4.5)
4.26 Determine the horizontal and vertical
components of reaction at the pin A and the
force in the short link BD.
Solution:
CCW +
M
A
 0;
8 (1.5 cos 30) - FBD (0.5 sin 30) = 0
FBD  41.57 kN = 41.6 kN
+ Fx  0;
41.57 - Ax  0
Ax  41.6 kN
   Fy  0;
Ay 8  0
A y  8 kN
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-18
EML 3004C
4.5 Equilibrium in 3-D
The concept of equilibrium in 3-D is similar. Here we need to solve all
the known and unknown in 3-D. Once again we need to know the
reaction forces and moments for each type of support, see Table 4-2
Ball- Only 3 forces
Bearing- 2 forces+2 moments
Namas Chandra
Introduction to Mechanical engineering
Pin- All except 1 moment
Fixed- All 6 forces & moments
Hibbeler
Chapter 4-19
EML 3004C
4.6 Equations of Equilibrium
In 3-D we apply all the equations of motion
F  0
M  0
The same equations can be written as a set of six scalar equations.
F  F  F  0
M  M  M  0
x
y
x
z
y
z
Procedure for Analysis:
1. Draw Free body diagram for the body under analysis
2. Mark all the reaction and external forces/moments.
3. Use the above equations to solve.
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-20
EML 3004C
Problem 4-68 (page 174, Section 4.6-4.7)
4.68 Member AB is supported by a cable BC and at A by
a smooth fixed square rod which fits loosely through the
square hole of the collar. If the force F = - 45k lb,
determine the tension in cable BC and the x, y, z
components of reaction at A.
Solution: Force Vector:
 12i + 4j + 6k 
FBC  FBC 

 (12) 2  42  62 


6
2
3
= - FBC i + FBC j + FBC k
7
7
7
Equilibrium:
 Fz  0;
Namas Chandra
Introduction to Mechanical engineering
3
FBC - 45 = 0
7
FBC  105 lb
Hibbeler
Chapter 4-21
EML 3004C
4.68 Member AB is supported by a cable BC
and at A by a smooth fixed square rod which
fits loosely through the square hole of the
collar. If the force F = - 45k lb, determine the
tension in cable BC and the x, y, z components
of reaction at A.
Solution-Con’t (slide 2)
F
x
 0;
6
Ax  (105)  0
7
 Fy  0;
Ax  90 lb
2
Ay  (105)  0
7
 M x  0;
Ay  30 lb
3
M Ax  (105)(4)  0
7
Namas Chandra
Introduction to Mechanical engineering
M Ax  180 lb  ft
Hibbeler
Chapter 4-22
EML 3004C
4.68 Member AB is supported by a cable BC
and at A by a smooth fixed square rod which
fits loosely through the square hole of the
collar. If the force F = - 45k lb, determine the
tension in cable BC and the x, y, z components
of reaction at A.
Solution-Con’t (slide 3)
 M y  0;
3
M Ay  45 (12) 
(105) (12)  0
7
M Ay  0
M
z
 0;
2
6
M Az  (105) (12) + (105) (4) = 0
7
7
M Az  720 lb  ft
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-23
EML 3004C
4.7 Friction
Friction is the force of resistance offered by
a body that prevents or retards a body from
motion relative to the first.
Friction always acts tangent to the surface
and opposing any possible motion.
Friction is caused by small asperities as
shown here.
We should consider all the
minor surface asperities to get
a distributed load .
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-24
EML 3004C
4.7 Friction-2
Frictional coefficient changes from
static to kinetic when the value
reduces..
Consider the motion of the
following structure.
Seven unknowns:
N A , FA , Bx , By , P, N c , Fc
Two sets of 3 equations
and
Friction  F = s N
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-25
EML 3004C
4.7 Friction-3
But we use simple law suggested by Coloumb for two possible cases.
Impending (or possible) motion-statics  Fs   s N
Actual motion-kinetic  Fk   k N
First we still need to establish equilibirum to find N
For static conditions, we use Ffriction  s N
For dynamics conditions, we use Ffriction  k N
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-26
EML 3004C
4.7 Tipping or impending motion
Tipping during motion or sliding depends if the
clockwise moment at the bottom corner is CW or CCW.
Evaluate the location of N with respect to W.
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-27
EML 3004C
Example of pipes stacked
The concrete pipes are stacked. Determine the
minimum coefficient of static friction so that
the pile does not collapse.
Solution:
Draw the Free body diagrams first.
Top
pipe
 M  0;  F  r  F  r  0  F  F
 F  0;N sin 30  F cos30  N sin 30  F cos30  0
 F  0;2 N cos30  2 F sin 30  W  0
0
x
A
A
B
A
B
B
y
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-28
EML 3004C
Example of pipes stacked
Solution: Use F  F and N  N
A
A
 M  0;  F  r  F .r  0  F  F
 F  0;  N sin 30  F cos30  F  0 (2)
 F  0;N  W  N cos30  F sin 30  0 (3)
0
C
C
x
y
C
From (2),
Bottom pipe
Namas Chandra
Introduction to Mechanical engineering
  s min
F
  0.268
N
Hibbeler
Chapter 4-29
EML 3004C
Example of man on a plank
Determine how far d the man can walk up the plank without causing
the plank to slip. s between A and B is 0.3. Man weighs 200 lb.
Solution:
 F  0; F  F cos 30  N cos 60  0
 F  0;2 N  F sin 30  N sin 60  200  0
 M  0; N cos10(15)  F sin10(15) 
x
A
y
B
A
A
B
B
B
B
(2)
B
 200sin 20(3)  200 cos 20( d )  0
Namas Chandra
Introduction to Mechanical engineering
(1)
(3)
Hibbeler
Chapter 4-30
EML 3004C
Example of man on a plank-2
Solution continued
If the plank is on the verge of moving,
slipping would occur at A.
Hence FA   N A and FB   N B  0.3N B
Substituting in eqns (1), (2) and (3),
d  10.2 ft
N A  110.09 lb and N B  88.14 lb
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-31
EML 3004C
Problem 4-100 (page 192, Section 4.6-4.7)
4.100 Two boys, each weighing 60 lb, sit at the
ends of a uniform board, which has a weight of
30 lb. If the board rests at its center on a post
having a coefficient of static friction of
(  sthe
) board,
0.6 determine the greatest angle of
with
tilt before slipping occurs. Neglect the size
of post and the thickness of the board in the
calculations.
Solution:
  Fx  0;
0.6 N cos  - N sin  =0
tan  = 0.6
 = 31.0
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-32
EML 3004C
Problem 4-112 (page 195, Review Problems)
4.112 The horizontal beam is supported by springs at its ends.
If the stiffness of the spring at A is k A  5 kN/m , determine the
required stiffness of the spring at B so that if the beam is
loaded with the 800-N force it remains in the horizontal
position both before and after loading.
Solution:
Equilibrium:
CCW +
M
A
 0;
FB (3) - 800 (1) = 0
FB  266.67 N
CCW+ M B  0;
800(2) - FA (3)=0
FA  533.33 N
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-33
EML 3004C
4.112 The horizontal beam is supported by
springs at its ends. If the stiffness of the
spring at A is k A  5 kN/m, determine the
required stiffness of the spring at B so that
if the beam is loaded with the 800-N force
it remains in the horizontal position both
before and after loading.
Solution-Con’t (slide 2)
F
Spring Force Formula: x 
k
x A  xB
533.33 266.67

5000
kB
k B  2500 N/m=2.50 kN/m
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-34
EML 3004C
Chapter 4: Equilibrium.. concludes
Namas Chandra
Introduction to Mechanical engineering
Hibbeler
Chapter 4-35