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I Made Gatot Karohika ST. MT.
Mechanical Engineering
Udayana university
4 - 2
Contents
Introduction
Free-Body Diagram
Reactions at Supports and Connections
for a Two-Dimensional Structure
Equilibrium of a Rigid Body in Two
Dimensions
Statically Indeterminate Reactions
Sample Problem 4.1
Sample Problem 4.3
Sample Problem 4.4
Equilibrium of a Two-Force Body
Equilibrium of a Three-Force Body
Sample Problem 4.6
Equilibrium of a Rigid Body in Three
Dimensions
Reactions at Supports and Connections for a
Three-Dimensional Structure
Sample Problem 4.8
4 - 3
Introduction
• The necessary and sufficient condition for the static equilibrium of a
body are that the resultant force and couple from all external forces
form a system equivalent to zero,
00 FrMF O

000
000
zyx
zyx
MMM
FFF
• Resolving each force and moment into its rectangular components
leads to 6 scalar equations which also express the conditions for static
equilibrium,
• For a rigid body in static equilibrium, the external forces and
moments are balanced and will impart no translational or rotational
motion to the body.
4 - 4
Free-Body Diagram
First step in the static equilibrium analysis of a rigid
body is identification of all forces acting on the
body with a free-body diagram.
• Select the extent of the free-body and detach it
from the ground and all other bodies.
• Include the dimensions necessary to compute
the moments of the forces.
• 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.
• Indicate point of application, magnitude, and
direction of external forces, including the rigid
body weight.
4 - 5
Reactions at Supports and Connections for a Two-
Dimensional Structure
• Reactions equivalent to a
force with known line of
action.
4 - 6
Reactions at Supports and Connections for a Two-
Dimensional Structure
• Reactions equivalent to a
force of unknown direction
and magnitude.
• Reactions equivalent to a
force of unknown
direction and magnitude
and a couple.of unknown
magnitude
4 - 7
Equilibrium of a Rigid Body in Two Dimensions
• For all forces and moments acting on a two-
dimensional structure,
Ozyxz MMMMF 00
• Equations of equilibrium become
000 Ayx MFF
where A is any point in the plane of the
structure.
• The 3 equations can be solved for no more
than 3 unknowns.
• The 3 equations can not be augmented with
additional equations, but they can be replaced
000 BAx MMF
4 - 8
Statically Indeterminate Reactions
• More unknowns than
equations
• Fewer unknowns than
equations, partially
constrained
• Equal number unknowns
and equations but
improperly constrained
4 - 9
Sample Problem 4.1
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:
• Create a free-body diagram for the crane.
• Determine B by solving the equation for
the sum of the moments of all forces
about A. Note there will be no
contribution from the unknown
reactions at A.
• Determine the reactions at A by
solving the equations for the sum of
all horizontal force components and
all vertical force components.
• Check the values obtained for the
reactions by verifying that the sum of
the moments about B of all forces is
zero.
4 - 10
Sample Problem 4.1
• Create the free-body diagram.
• Check the values obtained.
• Determine B by solving the equation for the
sum of the moments of all forces about A.
0m6kN5.23
m2kN81.9m5.1:0 BM A
kN1.107B
• Determine the reactions at A by solving the
equations for the sum of all horizontal forces
and all vertical forces.
0:0 BAF xx
kN1.107xA
0kN5.23kN81.9:0 yy AF
kN3.33yA
4 - 11
Sample Problem 4.3
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
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:
• Create a free-body diagram for the car
with the coordinate system aligned
with the track.
• Determine the reactions at the wheels
by solving equations for the sum of
moments about points above each axle.
• Determine the cable tension by
solving the equation for the sum of
force components parallel to the track.
• Check the values obtained by verifying
that the sum of force components
perpendicular to the track are zero.
4 - 12
Sample Problem 4.3
• Create a free-body diagram
lb2320
25sinlb5500
lb4980
25coslb5500


y
x
W
W
• Determine the reactions at the wheels.
00in.5
in.6lb9804in.25lb2320:0
2R
M A
lb17582R
00in.5
in.6lb9804in.25lb2320:0
1R
MB
lb5621R
• Determine the cable tension.
0Tlb4980:0xF
lb4980T
4 - 13
Sample Problem 4.4
The frame supports part of the roof of
a small building. The tension in the
cable is 150 kN.
Determine the reaction at the fixed
end E.
SOLUTION:
• Create a free-body diagram for the
frame and cable.
• Solve 3 equilibrium equations for the
reaction force components and
couple at E.
4 - 14
Sample Problem 4.4
• Create a free-body diagram for
the frame and cable.
• Solve 3 equilibrium equations for the
reaction force components and couple.
0kN150
5.7
5.4
:0 xx EF
kN0.90xE
0kN150
5.7
6
kN204:0 yy EF
kN200yE
:0EM
0m5.4kN150
5.7
6
m8.1kN20m6.3kN20
m4.5kN20m7.2kN20
EM
mkN0.180EM
4 - 15
Equilibrium of a Two-Force Body
• Consider a plate subjected to two forces F1 and F2
• 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.
• 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.
4 - 16
Equilibrium of a Three-Force Body
• Consider a rigid body subjected to forces acting at
only 3 points.
• Assuming that their lines of action intersect, the
moment of F1 and F2 about the point of intersection
represented by D is zero.
• 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.
• The lines of action of the three forces must be
concurrent or parallel.
4 - 17
Sample Problem 4.6
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:
• 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.
• 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.
• Utilize a force triangle to determine the
magnitude of the reaction force R.
4 - 18
Sample Problem 4.6
• Create a free-body diagram of the joist.
• Determine the direction of the reaction
force R.
636.1
414.1
313.2
tan
m2.313m515.0828.2
m515.020tanm414.1)2045cot(
m414.1
m828.245cosm445cos
2
1
AE
CE
BDBFCE
CDBD
AFAECD
ABAF

6.58
4 - 19
Sample Problem 4.6
• Determine the magnitude of the reaction
force R.

38.6sin
N1.98
110sin4.31sin
RT
N8.147
N9.81
R
T
4 - 20
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.
000
000
zyx
zyx
MMM
FFF
• 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,
00 FrMF O

4 - 21
Reactions at Supports and Connections for a Three-
Dimensional Structure
4 - 22
Reactions at Supports and Connections for a Three-
Dimensional Structure
4 - 23
Sample Problem 4.8
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:
• Create a free-body diagram for the sign.
• Apply the conditions for static
equilibrium to develop equations for
the unknown reactions.
4 - 24
Sample Problem 4.8
• Create a free-body diagram for the
sign.
Since there are only 5 unknowns,
the sign is partially constrain. It is
free to rotate about the x axis. It is,
however, in equilibrium for the
given loading.
kjiT
kji
T
rr
rr
TT
kjiT
kji
T
rr
rr
TT
EC
EC
EC
EC
ECEC
BD
BD
BD
BD
BDBD








7
2
7
3
7
6
3
2
3
1
3
2
7
236
12
848
4 - 25
Sample Problem 4.8
• Apply the conditions for
static equilibrium to
develop equations for the
unknown reactions.
0lb1080571.2667.2:
0714.1333.5:
0lb270ft4
0:
0lb270:
0:
0lb270
7
2
3
2
7
3
3
1
7
6
3
2
ECBD
ECBD
ECEBDBA
ECBDz
ECBDy
ECBDx
ECBD
TTk
TTj
jiTrTrM
TTAk
TTAj
TTAi
jTTAF







kjiA
TT ECBD

lb22.5lb101.2lb338
lb315lb3.101
Solve the 5 equations for the 5 unknowns,

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equilibrium-of-rigid-body

  • 1. I Made Gatot Karohika ST. MT. Mechanical Engineering Udayana university
  • 2. 4 - 2 Contents Introduction Free-Body Diagram Reactions at Supports and Connections for a Two-Dimensional Structure Equilibrium of a Rigid Body in Two Dimensions Statically Indeterminate Reactions Sample Problem 4.1 Sample Problem 4.3 Sample Problem 4.4 Equilibrium of a Two-Force Body Equilibrium of a Three-Force Body Sample Problem 4.6 Equilibrium of a Rigid Body in Three Dimensions Reactions at Supports and Connections for a Three-Dimensional Structure Sample Problem 4.8
  • 3. 4 - 3 Introduction • The necessary and sufficient condition for the static equilibrium of a body are that the resultant force and couple from all external forces form a system equivalent to zero, 00 FrMF O  000 000 zyx zyx MMM FFF • Resolving each force and moment into its rectangular components leads to 6 scalar equations which also express the conditions for static equilibrium, • For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body.
  • 4. 4 - 4 Free-Body Diagram First step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-body diagram. • Select the extent of the free-body and detach it from the ground and all other bodies. • Include the dimensions necessary to compute the moments of the forces. • 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. • Indicate point of application, magnitude, and direction of external forces, including the rigid body weight.
  • 5. 4 - 5 Reactions at Supports and Connections for a Two- Dimensional Structure • Reactions equivalent to a force with known line of action.
  • 6. 4 - 6 Reactions at Supports and Connections for a Two- Dimensional Structure • Reactions equivalent to a force of unknown direction and magnitude. • Reactions equivalent to a force of unknown direction and magnitude and a couple.of unknown magnitude
  • 7. 4 - 7 Equilibrium of a Rigid Body in Two Dimensions • For all forces and moments acting on a two- dimensional structure, Ozyxz MMMMF 00 • Equations of equilibrium become 000 Ayx MFF where A is any point in the plane of the structure. • The 3 equations can be solved for no more than 3 unknowns. • The 3 equations can not be augmented with additional equations, but they can be replaced 000 BAx MMF
  • 8. 4 - 8 Statically Indeterminate Reactions • More unknowns than equations • Fewer unknowns than equations, partially constrained • Equal number unknowns and equations but improperly constrained
  • 9. 4 - 9 Sample Problem 4.1 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: • Create a free-body diagram for the crane. • Determine B by solving the equation for the sum of the moments of all forces about A. Note there will be no contribution from the unknown reactions at A. • Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components. • Check the values obtained for the reactions by verifying that the sum of the moments about B of all forces is zero.
  • 10. 4 - 10 Sample Problem 4.1 • Create the free-body diagram. • Check the values obtained. • Determine B by solving the equation for the sum of the moments of all forces about A. 0m6kN5.23 m2kN81.9m5.1:0 BM A kN1.107B • Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces. 0:0 BAF xx kN1.107xA 0kN5.23kN81.9:0 yy AF kN3.33yA
  • 11. 4 - 11 Sample Problem 4.3 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 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: • Create a free-body diagram for the car with the coordinate system aligned with the track. • Determine the reactions at the wheels by solving equations for the sum of moments about points above each axle. • Determine the cable tension by solving the equation for the sum of force components parallel to the track. • Check the values obtained by verifying that the sum of force components perpendicular to the track are zero.
  • 12. 4 - 12 Sample Problem 4.3 • Create a free-body diagram lb2320 25sinlb5500 lb4980 25coslb5500   y x W W • Determine the reactions at the wheels. 00in.5 in.6lb9804in.25lb2320:0 2R M A lb17582R 00in.5 in.6lb9804in.25lb2320:0 1R MB lb5621R • Determine the cable tension. 0Tlb4980:0xF lb4980T
  • 13. 4 - 13 Sample Problem 4.4 The frame supports part of the roof of a small building. The tension in the cable is 150 kN. Determine the reaction at the fixed end E. SOLUTION: • Create a free-body diagram for the frame and cable. • Solve 3 equilibrium equations for the reaction force components and couple at E.
  • 14. 4 - 14 Sample Problem 4.4 • Create a free-body diagram for the frame and cable. • Solve 3 equilibrium equations for the reaction force components and couple. 0kN150 5.7 5.4 :0 xx EF kN0.90xE 0kN150 5.7 6 kN204:0 yy EF kN200yE :0EM 0m5.4kN150 5.7 6 m8.1kN20m6.3kN20 m4.5kN20m7.2kN20 EM mkN0.180EM
  • 15. 4 - 15 Equilibrium of a Two-Force Body • Consider a plate subjected to two forces F1 and F2 • 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. • 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.
  • 16. 4 - 16 Equilibrium of a Three-Force Body • Consider a rigid body subjected to forces acting at only 3 points. • Assuming that their lines of action intersect, the moment of F1 and F2 about the point of intersection represented by D is zero. • 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. • The lines of action of the three forces must be concurrent or parallel.
  • 17. 4 - 17 Sample Problem 4.6 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: • 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. • 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. • Utilize a force triangle to determine the magnitude of the reaction force R.
  • 18. 4 - 18 Sample Problem 4.6 • Create a free-body diagram of the joist. • Determine the direction of the reaction force R. 636.1 414.1 313.2 tan m2.313m515.0828.2 m515.020tanm414.1)2045cot( m414.1 m828.245cosm445cos 2 1 AE CE BDBFCE CDBD AFAECD ABAF  6.58
  • 19. 4 - 19 Sample Problem 4.6 • Determine the magnitude of the reaction force R.  38.6sin N1.98 110sin4.31sin RT N8.147 N9.81 R T
  • 20. 4 - 20 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. 000 000 zyx zyx MMM FFF • 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, 00 FrMF O 
  • 21. 4 - 21 Reactions at Supports and Connections for a Three- Dimensional Structure
  • 22. 4 - 22 Reactions at Supports and Connections for a Three- Dimensional Structure
  • 23. 4 - 23 Sample Problem 4.8 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: • Create a free-body diagram for the sign. • Apply the conditions for static equilibrium to develop equations for the unknown reactions.
  • 24. 4 - 24 Sample Problem 4.8 • Create a free-body diagram for the sign. Since there are only 5 unknowns, the sign is partially constrain. It is free to rotate about the x axis. It is, however, in equilibrium for the given loading. kjiT kji T rr rr TT kjiT kji T rr rr TT EC EC EC EC ECEC BD BD BD BD BDBD         7 2 7 3 7 6 3 2 3 1 3 2 7 236 12 848
  • 25. 4 - 25 Sample Problem 4.8 • Apply the conditions for static equilibrium to develop equations for the unknown reactions. 0lb1080571.2667.2: 0714.1333.5: 0lb270ft4 0: 0lb270: 0: 0lb270 7 2 3 2 7 3 3 1 7 6 3 2 ECBD ECBD ECEBDBA ECBDz ECBDy ECBDx ECBD TTk TTj jiTrTrM TTAk TTAj TTAi jTTAF        kjiA TT ECBD  lb22.5lb101.2lb338 lb315lb3.101 Solve the 5 equations for the 5 unknowns,