Transcript Lecture 26 - Faculty of Engineering and Applied Science

ENGI 1313 Mechanics I
Lecture 26:
3D Equilibrium of a Rigid Body
Shawn Kenny, Ph.D., P.Eng.
Assistant Professor
Faculty of Engineering and Applied Science
Memorial University of Newfoundland
[email protected]
Schedule Change

Postponed Class


Friday Nov. 9
Two Options

Use review class Wednesday Nov. 28
• Preferred option


2
Schedule time on Thursday Nov.15 or 22
Please Advise Class Representative of
Preference
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
Lecture 26 Objective

3
to illustrate application of scalar and vector
analysis for 3D rigid body equilibrium
problems
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
Example 26-01

4
The pipe assembly
supports the
vertical loads
shown. Determine
the components of
reaction at the balland-socket joint A
and the tension in
the supporting
cables BC and BD.
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
Example 26-01 (cont.)

Draw FBD
z
Due to symmetry
TBC = TBD
TBD
F1= 3 kN
F2 = 4 kN
TBC
Az
Ax
Ay
x
5
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
y
Example 26-01 (cont.)

What are the First Steps?
Define Cartesian
coordinate system
 Resolve forces

z
• Scalar notation?
• Vector notation?
TBD
F1= 3 kN
F2 = 4 kN
TBC
Az
Ax
Ay
x
6
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
y
Example 26-01 (cont.)

Cable Tension Forces
Position vectors
rBC  2  0 ˆi  0 1 ˆj  3 1kˆ m




  2 ˆi 1 ˆj  2 kˆ m
rBC  2 ˆi 1 ˆj  2 kˆ m
rBD   2  0 ˆi  0 1 ˆj  3 1kˆ m
rBD

7

© 2007 S. Kenny, Ph.D., P.Eng.
TBD
F1= 3 kN
F2 = 4 kN
TBC
Az
Unit vectors
2 ˆ 1 ˆ 2 ˆ
ˆuBC  
 i  j  k
3
3 
3
1
2 
 2
uˆ BD    ˆi  ˆj  kˆ 
3
3 
 3
z
Ax
Ay
x
ENGI 1313 Statics I – Lecture 26
y
Example 26-01 (cont.)

Ball-and-Socket Reaction
Forces

Unit vectors
z

  ˆj 
 kˆ 
uˆ Ax  ˆi
uˆ Ay
uˆ Az
TBD
F1= 3 kN
F2 = 4 kN
TBC
Az
Ax
Ay
x
8
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
y
Example 26-01 (cont.)

What Equilibrium Equation
Should be Used?

Mo = 0
z
• Why?
Find moment arm vectors
rAB  0  0 ˆi  1  0  ˆj  1  0 kˆ m



 0 ˆi  4 ˆj  0 kˆ m
 0 ˆi  5.5 ˆj  0 kˆ m
rAB
9
© 2007 S. Kenny, Ph.D., P.Eng.
F1= 3 kN
F2 = 4 kN
TBC
Az
rAB  0 ˆi  1 ˆj  1 kˆ m
rF1
TBD
Ax
Ay
x
ENGI 1313 Statics I – Lecture 26
y
Example 26-01 (cont.)

Moment Equation
MO  0
rAB  TBC uˆ BC  TBD uˆ BD   
rF1  F1   rF 2  F2   0
z
TBD
Due to symmetry TBC = TBD
TBC rAB  uˆ BC  uˆ BD   
F1= 3 kN
F2 = 4 kN
TBC
Az
rF1  F1   rF 2  F2   0
Ax
Ay
x
10
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
y
Example 26-01 (cont.)

Moment Equation
MO  0
i
j
k
TBC 0
1
1 
0
i
j
0 4
0 0
z
2 3 4 3
k
i
j
0  0 5.5
3 0 0
TBD
k
0 0
4
TBC
Az
2TBD 12kN  m  22kN  m  0
TBD  TBC  17 kN
11
© 2007 S. Kenny, Ph.D., P.Eng.
F1= 3 kN
F2 = 4 kN
Ax
Ay
x
ENGI 1313 Statics I – Lecture 26
y
Example 26-01 (cont.)

Force Equilibrium

F
x
0
TBC uBCx  TBD uBDx  Ax  0
z
17 kN 2   17 kN  2   Ax  0
3
TBD
 3
F1= 3 kN
F2 = 4 kN
TBC
Az
Ax  0
Ax
Ay
x
12
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
y
Example 26-01 (cont.)

Force Equilibrium

F
y
0
TBC uBCy  TBD uBDy  Ay  0
17 kN  1   17 kN  1   Ay
 3
 3
z
0
Ay  11.333 kN
TBD
F1= 3 kN
F2 = 4 kN
TBC
Az
Ax
Ay
x
13
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
y
Example 26-01 (cont.)

Force Equilibrium

F
z
0
TBC uBCz  TBD uBDz  AZ  3kN  4kN  0 z
17 kN 2   17 kN 2   Az  7 kN  0
3
TBD
3
F1= 3 kN
F2 = 4 kN
TBC
Az
Az   15.67 kN
Ax
Ay
x
14
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
y
Example 26-02

The silo has a weight
of 3500 lb and a
center of gravity at G.
Determine the vertical
component of force
that each of the three
struts at A, B, and C
exerts on the silo if it is
subjected to a
resultant wind loading
of 250 lb which acts in
the direction shown.
15
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
Example 26-02 (cont.)
Establish Cartesian
Coordinate System
 Draw FBD

W = 3500 lb
F = 250lb
Az
16
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
Bz
Cz
Example 26-02 (cont.)

What Equilibrium
Equation Should be
Used?

Three equations to
solve for three
unknown vertical
support reactions

F  0
M  0
M  0
W = 3500 lb
F = 250lb
z
x
Az
y
17
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
Bz
Cz
Example 26-02 (cont.)

Vertical Forces
 Fz  0
Az  Bz  Cz  3500 lb
W = 3500 lb
F = 250lb
Az
18
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
Bz
Cz
Example 26-02 (cont.)

Moment About
x-axis
 Mx  0
r  5 ft
h  15 ft
F  250lb
W = 3500 lb
 Az r  Bz r sin30  Cz r sin30  Fhcos 30  0
F = 250lb
 5 Az  2.5Bz  2.5Cz  3248  0
Az
19
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
Bz
Cz
Example 26-02 (cont.)

Moment About
y-axis
 My  0
r  5 ft
h  15 ft
F  250lb
W = 3500 lb
Bz r cos 30   Cz r cos 30  Fhsin30   0
F = 250lb
4.33Bz  4.33Cz 1875  0
Az
20
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26
Bz
Cz
Example 26-02 (cont.)

System of
Equations

Gaussian
elimination
1
1

 0 4.33
 5 2.5
3500   Az 
 
 4.33 1875  Bz 

2.5  3248  
C
z
 
1
1
1
1
3500   Az 

 
0 4.33  4.33 1875  Bz 

0 7.5
7.5 14252  
Cz 
21
© 2007 S. Kenny, Ph.D., P.Eng.
W = 3500 lb
F = 250lb
Az
ENGI 1313 Statics I – Lecture 26
Bz
Cz
Example 26-02 (cont.)

System of
Equations

Gaussian
elimination
1
1
1
3500   Az 

 
0
4
.
33

4
.
33
1875

 Bz 

0 15
0
17500  
C
z
 
Bz  1167 
  

Cz    734  lb
 A  1600 
 z 

22
© 2007 S. Kenny, Ph.D., P.Eng.
W = 3500 lb
F = 250lb
Az
ENGI 1313 Statics I – Lecture 26
Bz
Cz
References
Hibbeler (2007)
 http://wps.prenhall.com/esm_hibbeler_eng
mech_1

23
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 26