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Engineering 36
Chp 5: Mech
Equilibrium
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Vector Mechanics - Statics
1
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
ReCall Equilibrium Conditions
 Rigid Body in Static Equilibrium
Characterized by
• Balanced external forces and moments
– Will impart no Tendency toward Translational
or Rotational motion to the body
 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
Engineering-36: Vector Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Equilibrium cont.
 Rigid Body Equilibrium Mathematically

 F  0 and



 
 MO   r  F  0
 Resolving into Rectangular
Components the Resultant Forces &
Moments Leads to an Equivalent
Definition of Rigid Body Equilibrium
F  0 F  0 F  0
M  0 M  0 M  0
x
y
x
Engineering-36: Vector Mechanics - Statics
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z
y
z
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
2D Planar System
 In 2D systems it is assumed that
• The System Geometry resides
completely the XY Plane
• There is NO Tendency to
– Translate in the Z-Direction
– Rotate about the X or Y Axes
 These Conditions Simplify The
Equilibrium Equations
F
x
0
Engineering-36: Vector Mechanics - Statics
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F
y
0
M
z
0
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
2D Planar System:
F
x
0
F
y
0
M
z
0
 No Z-Translation → NO Z-Directed Force:
F
x
0
F
y
0
F
z
0
 No X or Y Rotation → NO X or Y
Applied Moments
M
x
0
Engineering-36: Vector Mechanics - Statics
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M
y
0
M
z
0
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
2D Planar System:
F
x
0
F
y
0
M
z
0
 If r Lies in the XY Plane, then rz = 0.
With Fz = 0 the rxF Determinant:
iˆ
ˆj
r  F  rx
Fx
ry
Fy
kˆ
0  ry 0  0 Fy iˆ  rx 0  Fx 0 ˆj  rx Fy  ry Fx kˆ
0
Mx  0
My  0
Mz  0
 So in this case M due to rxF is confined
to the Z-Direction:
r  F 
XY
 M z  M z kˆ  rx Fy  ry Fx kˆ
Engineering-36: Vector Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Example  Crane problem

Solution Plan
•
•
Create a free-body diagram
for the crane
Determine Rcns at B by
solving the equation for the
sum of the moments of all
forces about Pin-A
–

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.

Engineering-36: Vector Mechanics - Statics
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•
Note there will be no
Moment 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.
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Example  Crane problem

Reaction Analysis
•
PIN at A
–
•
X & Y Reactions
ROCKER at B
–
NORMAL Reaction
Only
 +X Rcn in This Case

Solution Plan cont.
•
Check the values obtained
for the reactions by
verifying that the moments
about B of all forces Sum
to zero.
Engineering-36: Vector Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Example  Crane problem

 M A  0 :  B1.5m   9.81 kN2m 
Draw the Free Body
Diagram (FBD)
 23.5 kN6m   0
B  107.1 kN


Determine Pt-B Rcn by
solving the equation for
the sum of the
moments of all forces
about Pt-A.
Engineering-36: Vector Mechanics - Statics
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Determine the reactions at
Pt-A by solving the eqns
for the sum of all
horizontal & vertical forces
 Fx  0 : Ax  B  0
Ax  107.1kN
 Fy  0 : Ay  9.81kN  23.5 kN  0
Ay  33.3 kN
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Example  Coal Car

Solution Plan
•
•


A loaded coal car is at rest on an
inclined track. The car has a gross
weight of 5500 lb, for the car and
its load as applied at G. The car is
held in position by the cable.
Determine the TENSION in the
cable and the REACTION at each
pair of wheels.
Engineering-36: Vector Mechanics - Statics
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•
Create a free-body diagram
for the car with the coord
system ALIGNED WITH
the CABLE
Determine the reactions at
the wheels by solving
equations for the sum of
moments about points
above each axle, in the
LoA of the Pull Cable
Determine the cable tension
by solving the equation for
the sum of force
components parallel
to the cable
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Example  Coal Car

Create Free Body Dia.

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

Wx  5500lb cos 25  4980lb
Wy  5500lbsin 25  2320lb
Engineering-36: Vector Mechanics - Statics
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Determine cable tension
 Fx  0 :  4980 lb  T  0
T  4980 lb
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Example  Cable Braced Beam

Solution Plan
•
•

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.
Engineering-36: Vector Mechanics - Statics
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
Reaction Analysis
•

Create a free-body diagram
for the frame and cable.
Solve 3 equilibrium
equations for the reaction
force components and
Moment at E.
The Support at E is a
CANTILEVER; can Resist
–
Planar Forces
–
Planar Moment
Also
DF  4.52  62  7.5m
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Triangle Trig Review
adj
cos 
hyp
Engineering-36: Vector Mechanics - Statics
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opp
sin  
hyp
opp
tan 
adj
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Example  Cable Braced Beam

Create a free-body
diagram for the frame
and cable
 Fx  0 : E x 
4.5
150 kN   0
7.5
E x  90.0 kN
 Fy  0 : E y  420 kN  
sin
6
150 kN   0
7.5
E y  200 kN
cos
 M E  0 :  20 kN7.2 m   20 kN5.4 m 
 20 kN3.6 m   20 kN1.8 m 


6
150 kN 4.5 m  M E  0
7.5
Solve 3 equilibrium eqns
for the reaction force
M E  180.0 kN  m
components and couple
Engineering-36: Vector Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Rigid Body Equilibrium in 3D
 SIX scalar equations are required to express
the conditions for the equilibrium of a rigid
body in the general THREE dimensional case
 Fx  0  Fy  0  Fz  0
Mx  0 My  0 Mz  0
 These equations can be solved for
NO MORE than 6 unknowns
• The Unknowns generally represent
REACTIONS at Supports or Connections.
Engineering-36: Vector Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Rigid Body Equilibrium in 3D
 The scalar equations are often conveniently
obtained by applying the vector forms of the
conditions for equilibrium

F  0



 
 MO   r  F  0
 Solve the Above Eqns with the Usual
Techniques; e.g., Determinant Operations
• A Clever Choice for the Pivot Point, O,
Can Eliminate from the Calculation as
Many as 3 Unknowns (a PoC somewhere)
Engineering-36: Vector Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Example  3D Ball-n-Socket

Solution Plan
•
•


A sign of Uniform Density
weighs 270 lb and is
supported by a BALL-ANDSOCKET joint at A and
by two Cables.
Determine the tension in each
cable and the reaction at A.

Engineering-36: Vector Mechanics - Statics
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Create a free-body diagram
of the Sign
Apply the conditions for
static equilibrium to develop
equations for the unknown
reactions.
Reaction Analysis
•
Ball-n-Socket at A can
Resist Translation in 3D
–
•
Can NOT Resist TWIST
(moment) in Any Direction
Cables Pull on Their
Geometric Axis
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
3D Ball-n-Socket Prob cont.

Create a free-body
diagram of the Sign

The x-Axis rotation Means
that the Sign is Only Partially
Constrained. Mathematically
M
•
Ax
 Anything
But If the Sign is NOT Swinging
(i.e., it’s hanging STILL) The
Following Condition Must Exist
M


18
0
With The Absence of the Mx
Constraint This Problem
Generates only 5 Unknowns
Note: The Sign can
SWING about the x-Axis •
(Wind Accommodation?)
Engineering-36: Vector Mechanics - Statics
Ax
MX = 0 by X-Axis P.O.A.
for W, TEC & TBC
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
3D Ball-n-Socket Prob cont.2



rAD  rAB
TBD  TBD 
  TBD
rAD  rAB
 TBD
 TBD

TEC  TEC

State in ComponentForm the Two CableTension Vectors
 TEC
 TEC
Engineering-36: Vector Mechanics - Statics
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
rBD

rBD
 8iˆ  4 ˆj  8kˆ
12
 23 iˆ  13 ˆj  23 kˆ


rAC  rAE
EC

  TEC
rAC  rAE
EC


 6iˆ  3 ˆj  2kˆ
7
 6 iˆ  3 ˆj  2 kˆ

7
7
7

Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
3D Ball-n-Socket Prob cont.3
  

 F  A  TBD  TEC  270lbˆj  0
iˆ : Ax  23 TBD  76 TEC  0
ˆj : Ay  1 TBD  3 TEC  270 lb  0
3
7
kˆ : Az  23 TBD  72 TEC  0





 M A  rAB  TBD  rAE  TEC 
4 ftiˆ   270 lb ˆj  0
ˆj : 5.333TBD  1.714TEC  0
kˆ : 2.667TBD  2.571TEC  1080ft  lb  0

Apply the conditions
for static equilibrium
 Solving 5 Eqns in 5 Unkwns
to develop
TBD  101.3 lb TEC  315lb
equations for the

unknown reactions. A  338lbiˆ  101.2lb ˆj  22.5lbkˆ
Engineering-36: Vector Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
WhiteBoard Work
Solve 3D
Problem by
Mechanics
& MATLAB
>> r = [5 -3 -7]
r =
5
-3
-7
>> F = [-13 8 11]
F =
-13
8
11
>> rxF = cross(r,F)
rxF =
23
36
1
 Determine the tensions in the cables and the
components of reaction acting on the smooth collar at G
necessary to hold the 2000 N sign in equilibrium. The
sign weight may concentrated at the center of gravity.
Engineering-36: Vector Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Engineering-36: Vector Mechanics - Statics
22
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Engineering-36: Vector Mechanics - Statics
23
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Engineering-36: Vector Mechanics - Statics
24
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
Engineering 36
Appendix
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-36: Vector Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
F
y
0
R  T1 kˆ R  T2  kˆ M A  kˆ  0
Engineering-36: Vector Mechanics - Statics
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OR
RkˆT  T   M kˆ
1
2
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx
A

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