2. Objects in Equilibrium
2142111 Statics, 2011/2
© Department of Mechanical
Engineering, Chulalongkorn University
1
Objectives Students must be able to #1
Course Objective
„ Analyze rigid bodies in equilibrium
Chapter Objectives
„ State the equations of equilibrium for various bodies (particles,
2D/3D rigid bodies)
„ Draw free body diagrams (FBDs) of various bodies (select the
body, draw the isolated body, apply loads and support
reactions, add axes and dimensions, use 3 colors to
differentiate the body, loads/reactions and other information)
„ Substitute information from FBDs into equation of equilibrium in
scalar & vector forms and solve for support reactions
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Objectives Students must be able to #2
Chapter Objectives
„ State the characteristics and identify 2 and 3-force members
„ Determine from FBDs whether a body in equilibrium is statically
determinate, statically indeterminate with redundant supports
or statically indeterminate with improper supports
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Contents
„
„
Equilibrium of Objects
„ Particles
„ 2D Rigid Bodies
„ 3D Rigid Bodies
This is the
< of Statics
SD and SI Problems
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4
Equilibrium
Equilibrium Definition
An object is in equilibrium when it
is stationary or in steady
translation relative to an inertial
reference frame.
v
v v
FR = ∑ F = 0
v
v
v
MRO = ∑ MO = 0
v
When a body is in equilibrium, the resultant force FR and
v
the resultant couple MR about any point O are both zero.
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5
Equilibrium
Equilibrium Procedure
„
„
„
„
„
„
Formulate problems from physical situations. Simplify
problems by making appropriate assumptions
Draw the free body diagram (FBD) of objects under
consideration
State the condition of equilibrium
Substitute variables from the FBD into the equilibrium
equations
Substitute the numbers and solve for solutions
„ Delay substitute numbers
„ Use appropriate significant figures
Technical judgment and engineering sense
„ Try to predict the answers
„ Is the answer reasonable?
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Equilibrium
Equilibrium Free Body Diagram (FBD)
„
„
„
FBD is the sketch of the body under consideration that is
isolated from all other bodies or surroundings.
The isolation of body clearly separate cause and effects
of loads on the body.
A thorough understanding of FBD is most vital for solving
problems.
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Equilibrium
Equilibrium
„
„
„
„
„
„
„
FBD Construction
Select the body to be isolated
Draw boundary of isolated body, excluding supports
Indicate a coordinate system by drawing axes
Add all applied loads (forces and couples) on the isolated
body.
Add all to support reactions (forces and couples) represent
the supports that were removed.
Beware of loads or support reactions with specific directions due
to physical meanings
Add dimensions and other information that are required in
the equilibrium equation
No FBDs → Cannot apply equilibrium conditions
→ Stop...........
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Equilibrium
Equilibrium Helps on FBD
„
„
„
„
Establish the x, y axes in any suitable orientation.
Label all the known and unknown applied load and
support reaction magnitudes
Beware of loads or support reactions with specific
directions due to physical meanings
Otherwise, directions of unknown loads and support
reactions can be assumed.
Use different colours in FBDs
„ Body outline
„ Load (force and couple)
„ Miscellaneous (dimension, angle, etc.)
- blue
- red
- black
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Equilibrium
Equilibrium On FBD Analyses
Objective: To find support reactions
„ Apply the equations of equilibrium: state equations and
substitute in the variables from the FBD under
consideration
v v
F =0
or
Fx = 0, ∑ Fy = 0, ∑ Fz = 0
∑
∑
v
v
∑ MO = 0 or ∑ MO,x = 0, ∑ MO,y = 0, ∑ MO,z = 0
„
„
Load components are positive if they are directed along a
positive direction, and vice versa
It is possible to assume positive directions for unknown
forces and moments. If the solution yields a negative
result, the actual load direction is opposite of that shown in
the FBD.
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Particle Equilibrium
Particles
Definition of Equilibrium
A particle is in equilibrium when it is
stationary or in steady translation
relative to an inertial reference
frame.
v v
∑F = 0
∑F
x
= 0,
∑F
y
= 0,
∑F
z
=0
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Particle Equilibrium
Example Hibbeler Ex 3-1 #1
The sphere has a mass of 6 kg and is supported as shown.
Draw a free-body diagram of the sphere, the cord CE, the knot at C
and the spring CD.
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12
Particle Equilibrium
Example Hibbeler Ex 3-1 #2
2
5
1
4
3
6
™ State the physical meanings
of these 9 forces
™ Do any of these forces have
specific directions?
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8
9
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Particle Equilibrium
Example Hibbeler Ex 3-2 #1
Determine the tension in cables AB
and AD for equilibrium of the 250-kg
engine shown.
™ Physical meanings of forces
™ Specific directions?
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Particle Equilibrium
Example Hibbeler Ex 3-2 #2
g = 9.807 m/s2
Particle A is in equilibrium.
⎡⎣ ∑ Fx = 0 ⎤⎦
TB cos 30° − TD = 0
(1)
⎡⎣ ∑ Fy = 0 ⎤⎦
TB sin30° − 0.25g kN = 0
(2)
Solve (2) and subst. into (1) TB = 4.9035 kN, TD = 4.2466 kN
TB = 4.90 kN, TD = 4.25 kN
Ans
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Particle Equilibrium
Example Hibbeler Ex 3-3 #1
If the sack at A has a weight of 20 lb,
determine the weight of the sack at B
and the force in each cord needed to
hold the system in the equilibrium
position shown.
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Particle Equilibrium
Example Hibbeler Ex 3-3 #2
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Particle Equilibrium
Example Hibbeler Ex 3-3 #3
Consider ring E in equilibrium
⎡⎣ ∑ Fx = 0 ⎤⎦
TEG sin30° − TEC cos 45° = 0
⎡⎣ ∑ Fy = 0 ⎤⎦
TEG cos 30° − TEC sin 45° − (20 lb) = 0 (2)
(1)
Solve (1) & (2), TEC = 38.637 lb, TEG = 54.641 lb
TEC = 38.6 lb, TEG = 54.6 lb
Ans
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Particle Equilibrium
Example Hibbeler Ex 3-3 #4
FBD of C in equilibrium
⎡⎣ ∑ Fx = 0 ⎤⎦
(TEC lb)cos 45° − (4 5)TCD = 0
(3)
⎡⎣ ∑ Fy = 0 ⎤⎦
(3 5)TCD + (TEC lb) sin 45° − WB = 0
(4)
Solve (3) & (4), TCD = 34.151 lb, WB = 47.811 lb
TCD = 34.2 lb, WB = 47.8 lb
Ans
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Particle Equilibrium
Particles
Equilibrium in 3D
z
v v
∑F = 0
v
F3
v
F2
v
ˆ
ˆ
ˆ
∑ (Fx i + Fy j + Fz k ) = 0
y
x
v
F1
∑F
∑F
∑F
x
=0
y
=0
z
=0
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Particle Equilibrium
Example Hibbeler Ex 3-5 #1
A 90-N load is suspended from the
hook. The load is supported by two
cables and a spring having a
stiffness k = 500 N/m. Determine the
force in the cables and the stretch of
the spring for equilibrium. Cable AD
lies in the x-y plane and cable AC
lies in the x-z plane.
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Particle Equilibrium
Example Hibbeler Ex 3-5 #2
Equilibrium of FBD of A
⎡⎣ ∑ Fx = 0 ⎤⎦
FD sin30° − (4 5)FC = 0
(1)
⎡⎣ ∑ Fy = 0 ⎤⎦
⎡⎣ ∑ Fz = 0 ⎤⎦
−FD cos 30° + FB = 0
(2)
(3 5)FC − (90 N) = 0
(3)
Solve (3), (1), (2) : FC = 150 N,
FD = 240 N, FB = 207.85 N = 208 N
[Fs = ks ]
Ans
FB = ksAB
sAB = FB / k = (207.85 N) /(500 N / m)
sAB = 0.41569 m = 0.416 m
Ans
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Particle Equilibrium
Example Hibbeler Ex 3-7 #1
Determine the force
developed in each cable
used to support the 40kN crate.
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Particle Equilibrium
Example Hibbeler Ex 3-7 #2
Position vectors
v
rAB = −3iˆ − 4 jˆ + 8kˆ m
v
rAC = −3iˆ + 4 jˆ + 8kˆ m
Force vectors
v
v
r
F
FB = FB AB = B ( −3iˆ − 4 ˆj + 8kˆ )
rAB
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v
v
rAC
F
FC = FC
= C ( −3iˆ + 4 ˆj + 8kˆ )
rAC
89
v
FD = FD iˆ
v
W = −40kˆ kN
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Particle Equilibrium
Example Hibbeler Ex 3-7 #3
Equilibrium of A
v v
⎡∑ F = 0⎤
⎣
⎦
v
v
v
v v
FB + FC + FD + W = 0
v
FC
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
( −3i − 4 j + 8k ) +
( −3i + 4 j + 8k ) + (FD i ) + ( −40k kN) = 0
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89
v
8FC
−3FB 3FC
−4FB 4FC ˆ
8FB
ˆ
ˆ
(
)j + (
−
+ FD )i + (
+
+
− 40 kN)k = 0
89
89
89
89
89
89
FB
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Particle Equilibrium
Example Hibbeler Ex 3-7 #4
x dir.:
y dir.:
z dir.:
−3FB
89
−4FB
89
8FB
−
+
+
3FC
89
4FC
89
8FC
89
89
From (2), FB = FC
From (3),
From (1),
16FC
89
−6FC
+ FD = 0
(1)
=0
(2)
− (40 kN) = 0 (3)
− (40 kN) = 0 or FC = 23.585 kN
+ FD = 0 or FD = 15.0 kN
89
FC = FB = 23.6 kN, FD = 15.0 kN
Ans
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2D Equilibrium
Equilibrium of 2D Rigid Bodies
„
„
Use similar analyses as those for particles
Additional consideration
„ Action forces in supports/constraints
„ Free-body diagram (FBD) of 2D rigid bodies
„ Equilibrium equations (scalar form) for rigid bodies
„ Two-force and three-force members
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2D Equilibrium
Scalar Form
Equilibrium 2D Equations
∑F
∑F
∑M
x
=0
y
=0
O
=0
Particle
Rigid Body
The sum of the moment about any point O is zero.
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2D Equilibrium
2D Supports Pin Joints
The pin support
„ prevents the translation by exerting a reaction force on the
bar
„ allows a free rotation so it does not exert a couple on the
bar.
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2D Equilibrium
2D Supports Roller Supports
The roller support
„ prevents the translation in y-direction by exerting a reaction
force to the bar.
„ allows a translation in x direction and a free rotation so it
does not exert and a couple to the bar.
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2D Equilibrium
2D Supports Fixed Supports
The fixed or built-in support
„ prevents all translation and rotation by exerting a reaction
force and a couple to the bar.
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2D Equilibrium
2D Supports Summary #1
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2D Equilibrium
2D Supports Summary #2
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2D Equilibrium
2D Supports Summary #3
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2D Equilibrium
Example Hibbeler Ex 5-1 #1
Draw the free-body diagram of the uniform beam shown.
The beam has the weight of 981 N.
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2D Equilibrium
Example Hibbeler Ex 5-1 #2
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2D Equilibrium
Example Hibbeler Ex 5-1 #3
Find: Ay , Ay , M A
Equilibrium of AG
⎡⎣ ∑ Fx = 0 ⎤⎦
− Ax = 0, Ax = 0
⎡⎣ ∑ Fy = 0 ⎤⎦
Ay − (1200 N) − (981 N) = 0
By = 2181 N = 2.18 kN
⎡⎣ ∑ M A = 0 ⎤⎦
Ans
Ans
M A − (1200 N)(2 m) − (981 N)(3 m) = 0
M A = 5343 N ⋅ m = 5.34 kN ⋅ m
Ans
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2D Equilibrium
Example Hibbeler Ex 5-3 #1
Two smooth pipes, each having the weight of 2943 N, are
supported by the forks of the tractor. Draw the FBDs for each
pipe and both pipes together.
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2D Equilibrium
Example Hibbeler Ex 5-3 #2
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2D Equilibrium
Example Hibbeler Ex 5-3 #3
Pipes in equilibrium
Find T , P, F , R
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2D Equilibrium
Example Hibbeler Ex 5-3 #4
⎡⎣ ∑ Fx = 0 ⎤⎦
R − (2943 N)sin30° = 0
R = 1471.5 N = 1.47 kN
Ans
⎡⎣ ∑ Fy = 0 ⎤⎦
P − (2943 N)cos30° = 0
P = 2548.7 N = 2.55 kN
Ans
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2D Equilibrium
Example Hibbeler Ex 5-3 #5
⎡⎣ ∑ Fx = 0 ⎤⎦
−R − (2943 N)sin30° + T = 0
T = 2943 N = 2.94 kN
⎡⎣ ∑ Fy = 0 ⎤⎦
Ans
F − (2943 N)cos30° = 0
F = 2548.7 N = 2.55 kN
Ans
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2D Equilibrium
Example Hibbeler Ex 5-6 #1
Determine the horizontal and vertical components of
reaction for the beam loaded as shown. Neglect the
weight of the beam in the calculation.
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2D Equilibrium
Example Hibbeler Ex 5-6 #2
Find: Ay , Bx , By
Equilibrium of ADB
⎡⎣ ∑ Fx = 0 ⎤⎦
(600 N)cos 45° − Bx = 0
Bx = 424.26 N = 424 N
⎡⎣ ∑ MB = 0 ⎤⎦
Ans
(100 N)(2 m) + (600 N) sin 45°(5 m) −
(600 N)cos 45°(0.2 m) − Ay (7 m) = 0
Ay = 319.50 = 319 N
⎡⎣ ∑ Fy = 0 ⎤⎦
Ans
Ay − (600 N) sin 45° − (100 N) − (200 N) + By = 0
By = 404.76 N = 405 N
Ans
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2D Equilibrium
Example Hibbeler Ex 5-8 #1
The link shown is pin-connected at A and rests against a
smooth support at B. Compute the horizontal and vertical
components of reactions at the pin A.
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2D Equilibrium
Example Hibbeler Ex 5-8 #2
FBD of link BA
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2D Equilibrium
Example Hibbeler Ex 5-8 #3
Equilibrium of FBD of link BA
⎡⎣ ∑ M A = 0 4+ ⎤⎦
−(90 N ⋅ m) − (60 N)(1 m) + NB (0.75 m) = 0
NB = 200 N
Ans
⎡⎣ ∑ Fx = 0 → + ⎤⎦
Ax − NB sin30° = 0,
⎡⎣ ∑ Fy = 0 ↑ + ⎤⎦
Ay − NB cos 30° − (60 N) = 0
Ay = 233.21 N = 233 N
Ax = 100 N
Ans
Ans
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2D Equilibrium
Definition
Special Member 2-Force Member #1
The object is a two-force member
when subjected to two equivalent forces
acting at different points.
For a two-force member in equilibrium, forces
„ are equal in magnitude.
„ are opposite in direction.
„ have the same line of action.
„ The are no other loads, e.g. weight
The ability to recognize 2-force members is
important in the analyses of structures.
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2D Equilibrium
Example
Special Member 2-Force Member #2
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49
2D Equilibrium
Definition
Special Member 3-Force Member #1
The object is a three-force member when subjected to
three equivalent forces acting at different points.
If a three force member is in equilibrium, the three
forces are:
„ coplanar and
„ either parallel or concurrent.
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2D Equilibrium
Special Member 3-Force Member #2
v
F1
v
F2
O
v
F1
v
F3
v
F3
v
F2
Concurrent Forces
Parallel Forces
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2D Equilibrium
Example Hibbeler Ex 5-13 #1
The lever ABC is pin-supported
at A and connected to a short
link BD as shown. If the weight
of the members is negligible,
determine the force of the pin on
the lever at A.
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2D Equilibrium
Example Hibbeler Ex 5-13 #2
Equilibrium of FBD of ABC
⎡⎣ ∑ M A = 0 ⎤⎦
−(400 N)(0.7 m) + FBD cos 45°(0.2 m) +
FBD sin 45°(0.1 m) = 0
FBD = 1319.9 N = 1.32 kN
Ans
⎡⎣ ∑ Fx = 0 ⎤⎦
(400 N) − FBD cos 45° + Ax = 0
Ax = 533.33 N = 533 kN
Ans
⎡⎣ ∑ Fy = 0 ⎤⎦
−FBD sin 45° + Ay = 0
Ay = 933.33 N = 933 kN
Ans
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2D Equilibrium
Example Hibbeler Ex 5-13 #3
BD is a two force member
ABC is a three-force member.
Must be concurrent at O
⎛ 0.7 ⎞
= 60.255°
⎟
⎝ 0.4 ⎠
θ = tan−1 ⎜
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2D Equilibrium
Example Hibbeler Ex 5-13 #4
Equilibrium of FBD of ABC
⎡⎣ ∑ Fx = 0 → + ⎤⎦
FA cos 60.255° − F cos 45° + (400 N) = 0 (1)
⎡⎣ ∑ Fy = 0 ↑ + ⎤⎦
FA sin 60.255° − F sin 45° = 0
(2)
Solve (1) & (2)
FA = 1074.9 N = 1.07 kN
F = 1319.9 N = 1.32 kN
Ans
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55
2D Equilibrium
Example Hibbeler Ex 5-4 #1
Draw the FBD of the unloaded platform that is suspended off the edge
of the oil rig shown. The platform has the weight of 1962 N.
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2D Equilibrium
Example Hibbeler Ex 5-4 #2
Platform in equilibrium: Find Ay , Ay , T
⎡⎣ ∑ M A = 0 ⎤⎦
T cos 70°(1 m) + T sin70°(2.2 m) −
(1962 N)(1.4 m) = 0
T = 1140.1 N = 1.14 kN
Ans
⎡⎣ ∑ Fx = 0 ⎤⎦
Ax − T cos 70° = 0
Ax = 389.92 N = 390 N
Ans
⎡⎣ ∑ Fy = 0 ⎤⎦
Ay + T sin70° − (1962 N) = 0
Ay = 890.69 N = 891 N
Ans
57
57
2D Equilibrium
Example Hibbeler Ex 5-4 #3
Platform in equilibrium: Find FA , θ , T
d
, d = 2.1980 m
0.8 m
d +1m
tanθ =
, θ = 66.357°
1.4 m
⎡⎣ ∑ Fx = 0 ⎤⎦
tan70° =
FA cos θ − T cos 70° = 0
(1)
⎡⎣ ∑ Fy = 0 ⎤⎦
FA sin θ + T sin70° − (1962 N) = 0 (2)
Solve (1) & (2)
T = 1140.1 N, FA = 972.30 N
T = 1.14 kN, FA = 972 N, θ = 66.4°
Ans
58
58
3D Equilibrium
Equilibrium of 3D Rigid Bodies
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59
3D Equilibrium
3D Supports Summary #1
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60
3D Equilibrium
3D Supports Summary #2
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61
3D Equilibrium
3D Supports Summary #3
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62
3D Equilibrium
3D Supports Summary #4
Comparison with 2D supports
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3D Equilibrium
Example Hibbeler Ex 5-14 #1
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64
3D Equilibrium
Example Hibbeler Ex 5-14 #2
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65
3D Equilibrium
Example Hibbeler Ex 5-14 #3
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66
3D Equilibrium
Example Hibbeler Ex 5-14 #4
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67
3D Equilibrium
3D Equilibrium Rigid Bodies
„
„
Action forces in supports
Equilibrium conditions
v v
∑ vF = 0v
∑ MO = 0
∑F
∑F
∑F
x
=0
y
= 0 and
z
=0
∑M
∑M
∑M
Vector form
x
=0
y
=0
z
=0
Scalar form
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68
3D Equilibrium
Example Hibbeler Ex 5-15 #1
The homogeneous plate
shown has a mass of 100 kg
and is subjected to a force
and couple moment along
its edges. If it is supported
in the horizontal plane by
means of a roller at A, a
ball-and-socket joint at B,
and a cord at C, determine
the components of reaction
at the supports.
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69
3D Equilibrium
Example Hibbeler Ex 5-15 #2
FBD of plate
Equilibrium of plate ABC
⎡⎣ ∑ Fx = 0 ⎤⎦ Bx = 0
⎡⎣ ∑ Fy = 0 ⎤⎦ By = 0
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70
3D Equilibrium
Example Hibbeler Ex 5-15 #3
Equilibrium of plate ABC
⎡⎣ ∑ Fz = 0 ⎤⎦
Az + Bz + TC − (300 N) − (980.7 N) = 0
(1)
⎡⎣ ∑ M x = 0 ⎤⎦
⎡⎣ ∑ M y = 0 ⎤⎦
TC (2 m) − (980.7 N)(1 m) + Bz (2 m) = 0
(2)
(300 N)(1.5 m) + (980.7 N)(1.5 m) − Bz (3 m)
− Az (3 m) − (200 N ⋅ m) = 0
(3)
Solve (1), (2) & (3)
Az = 790.35 N, Bz = −216.67 N,
TC = 707.02 N
Az = 790 N, TC = 707 N,
Bx = By = 0, Bz = −217 N Ans
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71
3D Equilibrium
Example Hibbeler Ex 5-15 #4
Alternatively, for equilibrium of plate ABC
⎡⎣ ∑ Fz = 0 ⎤⎦
Az + Bz + TC − (300 N) − (980.7 N) = 0
(1)
⎡⎣ ∑ M x ′ = 0 ⎤⎦
⎡⎣ ∑ M y ′ = 0 ⎤⎦
(980.7 N)(1 m) + (300 N)(2 m) − Az (2 m) = 0
(4)
−(300 N)(1.5 m) − (980.7 N)(1.5 m) − (200 N ⋅ m)
−TC (3 m) = 0
(5)
Solve (1), (4) & (5)
Az = 790.35 N,
Bz = −216.67 N,
TC = 707.02 N
72
72
3D Equilibrium
Example Hibbeler Ex 5-16 #1
The windlass shown is
supported by a thrust
bearing at A and a smooth
journal bearing at B, which
are properly aligned on the
shaft. Determine the
magnitude of the vertical
force P that must be applied
to the handle to maintain
equilibrium of the 100 kg
bucket. Also calculate the
reactions at the bearings.
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73
3D Equilibrium
Example Hibbeler Ex 5-16 #2
FBD of windlass
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74
3D Equilibrium
Example Hibbeler Ex 5-16 #3
Equilibrium of windlass
⎡⎣ ∑ M x = 0 ⎤⎦
(981 N)(0.1 m) − P (0.3 m)cos 30° = 0
P = 377.59 N
⎡⎣ ∑ M y = 0 ⎤⎦
−(981 N)(0.5 m) + Az (0.8 m) + P (0.4 m) = 0
Az = 424.33 N
⎡⎣ ∑ M z = 0 ⎤⎦
− Ay (0.8 m) = 0
→ Ay = 0
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75
3D Equilibrium
Example Hibbeler Ex 5-16 #4
⎡⎣ ∑ Fx = 0 ⎤⎦ Ax = 0
⎡⎣ ∑ Fy = 0 ⎤⎦ Ay + By = 0
⎡⎣ ∑ Fz = 0 ⎤⎦ Az − (981 N) + Bz − P = 0
→ By = 0
(424.33 N) − (981 N) + Bz − (377.59 N) = 0
→ Bz = 934.26 N
P = 378 N, Ax = Ay = 0, Az = 424 N, By = 0, Bz = 934 N
Ans
76
76
3D Equilibrium
Example Hibbeler Ex 5-18 #1
Rod AB shown is subjected to the 200-N force. Determine the
reactions at the ball-and-socket joint A and the tension in cables BD
and BE.
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77
3D Equilibrium
Example Hibbeler Ex 5-18 #2
FBD of rod AB
v
A = ( Ax iˆ + Ay jˆ + Az kˆ ) N
v
v
v
TE = TE iˆ N
rC = 0.5rB
v
v
TD = TD ˆj N
rC = (0.5iˆ + 1ˆj − kˆ ) m
v
v
F = −200kˆ N
rB = (iˆ + 2 ˆj − 2kˆ ) m
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78
3D Equilibrium
Example Hibbeler Ex 5-18 #3
Equilibrium of body ACB
v v
⎡∑ F = 0⎤
⎣
⎦
v v
v
v v
A + TE + TD + F = 0
v
ˆ
ˆ
ˆ
( Ax + TE )i + ( Ay + TD ) j + ( Az + F )k = 0
⎡⎣ ∑ Fx = 0 ⎤⎦
⎡⎣ ∑ Fy = 0 ⎤⎦
Ax + TE = 0
(1)
Ay + TD = 0
(2)
⎡⎣ ∑ Fz = 0 ⎤⎦
Az − (200 N) = 0
(3)
79
79
3D Equilibrium
Example Hibbeler Ex 5-18 #4
Equilibrium of body ACB
v
v
⎡∑ M A = 0⎤
⎣
⎦
v
v
v
v v v
(rC × F ) + rB × (TE + TD ) = 0
v
v
v
v v v
(0.5rB × F ) + rB × (TE + TD ) = 0
v v
v
v
v
rB × (0.5F + TE + TD ) = 0
v v
v
v
ˆ
ˆ
ˆ
(i + 2 j − 2k ) m × (0.5F + TE + TD ) = 0
⎡⎣ ∑ M x = 0 ⎤⎦
⎡⎣ ∑ M y = 0 ⎤⎦
⎡⎣ ∑ M z = 0 ⎤⎦
(2 m)TD − (200 N ⋅ m) = 0
(1)
( −2 m)TE + (100 N ⋅ m) = 0
(2)
(1 m)TD − (2 m)TE = 0
(3)
80
80
3D Equilibrium
Example Hibbeler Ex 5-18 #5
Solve eqns (1) through (6)
TD = 100 N
TE = 50 N
Ax = −50 N
Ay = −100 N
Az = 200 N
Ans
The negative sign indicates
that Ax and Ay have a direction
which is opposite to those
shown in the FBD.
81
81
3D Equilibrium
Example Hibbeler Ex 5-19 #1
The bent rod is supported at A
by a journal bearing, at D by a
ball-and-socket joint,
at B by means of cable BC.
Using only one equilibrium
equation, obtain a direct
solution for the tension in cable
BC. The bearing at A is
capable of exerting force
components only in the z and y
directions since it is properly
aligned on the shaft.
82
82
3D Equilibrium
Example Hibbeler Ex 5-19 #2
FBD of ABED
83
83
3D Equilibrium
Example Hibbeler Ex 5-19 #3
Summing moments about an axis passing
through points D and A.
v v
∑ MDA = uˆDA ⋅ ∑ (r × F ) = 0
v
Reminder: r is a position vector from any point on the DA axis to
any point in the line of action of the force.
v
rDA
1 ˆ 1 ˆ
=−
uˆDA =
i−
j
rDA
2
2
84
84
3D Equilibrium
Example Hibbeler Ex 5-19 #4
v
v v
v v v
∑ (r × F ) = rB × TB + rE × W
0.2 ˆ 0.3 ˆ 0.6 v
ˆ
= ( −1j ) m × (
TB i −
TB j +
TB k ) N +
0.7
0.7
0.7
v
= ( −0.5 ˆj ) m × ( −981k ) N
v
ˆ
= [( −0.85714T + 490.5)i + 0.28571T k ] N ⋅ m
B
B
Equilibrium of body ABED
v v
∑ MDA = uˆDA ⋅ ∑ (r × F ) = 0
v
1 ˆ 1 ˆ
ˆ
i −
j ] ⋅ [( −0.85714TB + 490.5)i + 0.28571TB k ] N ⋅ m
= [−
2
2
1
= (−
)( −0.85714TB + 490.5) N ⋅ m = 0
2
TB = 572.25 N = 572 N
Ans
85
85
SD/SI
Statically Determinate Objects
Statically determinate (SD) problems can be analyzed
using the equilibrium condition alone.
v
v v
FR = ∑ F = 0
v
v
v
MRO = ∑ MO = 0
86
86
SD/SI
Statically Indeterminate Objects
Statically indeterminate (SI) problems can not be analyze
with equilibrium equations.
„
An object with redundant supports
„
An object with improper supports
87
87
SD/SI
SI Objects Redundant Supports #1
An object has redundant supports, when it has more
supports than the minimum number necessary to maintain
it in equilibrium.
Result
More unknown forces or couples than the number of
independent equilibrium equations.
88
88
SD/SI
SI Objects Redundant Supports #2
5 Unknowns
3 Equilibrium Equations
89
89
SD/SI
SI Objects Redundant Supports #3
8 Unknowns
6 Equilibrium Equations
90
90
SD/SI
Example Redundant Supports #1
Equilibrium of beam AB
⎡⎣ ∑ Fx = 0 ⎤⎦
Ax = 0
⎡⎣ ∑ Fy = 0 ⎤⎦
Ay − F + B = 0
⎡⎣ ∑ M A = 0 ⎤⎦
−M − FL + B(2L ) = 0
3 equations vs 4 unknown reactions
91
91
SD/SI
Example Redundant Supports #2
Equilibrium of beam AB
⎡⎣ ∑ Fx = 0 ⎤⎦
Ax + Bx = 0
⎡⎣ ∑ Fy = 0 ⎤⎦
Ay + By − F = 0
⎡⎣ ∑ M A = 0 ⎤⎦
−FL + By (2L ) = 0
Three equations and four unknown reactions:
„ Ay and By can be found.
„ Ax = −Bx
92
92
SD/SI
SI Objects Redundant Supports #4
„
„
Degree of redundancy =
# of unknown reactions − # of independent equations
The unknown reactions can be determined by supplementing the
equilibrium equations with equations relating forces/couples with
deformation.
See you in Mechanics of Materials I
93
93
SD/SI
SI Objects Improper Supports #1
An object has improper supports when it has inadequate
(not enough) supports to maintain it in equilibrium.
Result
The object will move when loaded.
See you in Dynamics
94
94
SD/SI
SI Objects Improper Supports #2
45
The supports exert
parallel forces only.
The supports exert only
concurrent forces.
95
95
SD/SI
SI Objects Improper Supports #3
96
96
SD/SI
SI Objects Improper Supports #4
97
97
SD/SI
SI Objects Improper Supports #5
98
98
Equilibrium
Here ends the most important
chapter of the subject.
99
99
Review
Concepts
„
„
„
When a body is in equilibrium, the resultant force and
couple about any point O are both zero. Problems can be
analyzed by drawing free body diagrams (FBDs) and
substitute information in the FBD under consideration into
the equilibrium equations.
Statically determinate (SD) problems can be solved
using the equilibrium conditions alone.
The statically indeterminate (SI) problems cannot due to
too few or too many supports or constraints.
100
100