engi 1313 mechanics i
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ENGI 1313 Mechanics I. Lecture 26:3D Equilibrium of a Rigid Body. Schedule Change. Postponed Class Friday Nov. 9 Two Options Use review class Wednesday Nov. 28 Preferred option Schedule time on Thursday Nov.15 or 22 Please Advise Class Representative of Preference. Lecture 26 Objective. - PowerPoint PPT PresentationTRANSCRIPT
Shawn Kenny, Ph.D., P.Eng.Assistant ProfessorFaculty of Engineering and Applied ScienceMemorial University of [email protected]
ENGI 1313 Mechanics I
Lecture 26: 3D Equilibrium of a Rigid Body
2 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Schedule Change
Postponed Class Friday Nov. 9
Two Options Use review class Wednesday Nov. 28
• Preferred option Schedule time on Thursday Nov.15 or 22
Please Advise Class Representative of Preference
3 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Lecture 26 Objective
to illustrate application of scalar and vector analysis for 3D rigid body equilibrium problems
4 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-01
The pipe assembly supports the vertical loads shown. Determine the components of reaction at the ball-and-socket joint A and the tension in the supporting cables BC and BD.
5 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-01 (cont.)
Draw FBD
x
y
z
TBC
TBD
Ax
Ay
Az
Due to symmetryTBC = TBD
F1= 3 kNF2 = 4 kN
6 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-01 (cont.)
What are the First Steps? Define Cartesian coordinate
system Resolve forces
• Scalar notation?
• Vector notation?
x
y
z
TBC
TBD
Ax
Ay
Az
F1= 3 kNF2 = 4 kN
7 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-01 (cont.)
Cable Tension Forces Position vectors
Unit vectors
x
y
z
TBC
TBD
Ax
Ay
Az
mk2j1i2rBC
mk13j10i02rBC
mk2j1i2rBD
mk13j10i02rBD
k
3
2j
3
1i
3
2uBC
k
3
2j
3
1i
3
2uBD
F1= 3 kNF2 = 4 kN
8 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-01 (cont.)
Ball-and-Socket Reaction Forces Unit vectors
x
y
z
TBC
TBD
Ax
Ay
Az
iuAx
juAy
kuAz F1= 3 kN
F2 = 4 kN
9 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-01 (cont.)
What Equilibrium Equation Should be Used? Mo = 0
• Why? Find moment arm vectors
x
y
z
TBC
TBD
Ax
Ay
Az
mk01j01i00rAB
mk1j1i0rAB
mk0j4i0r 1F
mk0j5.5i0rAB
F1= 3 kNF2 = 4 kN
10 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-01 (cont.)
Moment Equation
x
y
z
TBC
TBD
Ax
Ay
Az
Due to symmetry TBC = TBD
BDBDBCBCAB uTuTr
F1= 3 kNF2 = 4 kN
0MO
0FrFr 22F11F
BDBCABBC uurT
0FrFr 22F11F
11 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-01 (cont.)
Moment Equation
x
y
z
TBC
TBD
Ax
Ay
Az
34320
110
kji
TBC
F1= 3 kNF2 = 4 kN0
400
05.50
kji
300
040
kji
0mkN22mkN12T2 BD
kN17TT BCBD
0MO
12 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-01 (cont.)
Force Equilibrium
x
y
z
TBC
TBD
Ax
Ay
Az
F1= 3 kNF2 = 4 kN
0Fx
0AuTuT xBDxBDBCxBC
0A3
2kN17
3
2kN17 x
0Ax
13 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-01 (cont.)
Force Equilibrium
x
y
z
TBC
TBD
Ax
Ay
Az
F1= 3 kNF2 = 4 kN
0Fy
0AuTuT yBDyBDBCyBC
0A3
1kN17
3
1kN17 y
kN333.11Ay
14 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-01 (cont.)
Force Equilibrium
x
y
z
TBC
TBD
Ax
Ay
Az
F1= 3 kNF2 = 4 kN
0kN4kN3AuTuT ZBDzBDBCzBC
0kN7A3
2kN17
3
2kN17 z
kN67.15Az
0Fz
15 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
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.
16 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Establish Cartesian Coordinate System
Draw FBD
Az
Bz Cz
F = 250lb
W = 3500 lb
Example 26-02 (cont.)
17 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
What Equilibrium Equation Should be Used? Three equations to
solve for three unknown vertical support reactions
Az
Bz Cz
F = 250lb
W = 3500 lb
Example 26-02 (cont.)
0Fz
0Mx
0My
18 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Vertical Forces
Az
Bz Cz
F = 250lb
W = 3500 lb
Example 26-02 (cont.)
0Fz
lb3500CBA zzz
19 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Moment About x-axis
Az
Bz Cz
F = 250lb
W = 3500 lb
Example 26-02 (cont.)
0Mx
030cosFh30sinrC30sinrBrA zzz
lb250F
ft15h
ft5r
03248C5.2B5.2A5 zzz
20 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Moment About y-axis
Az
Bz Cz
F = 250lb
W = 3500 lb
Example 26-02 (cont.)
0My
030sinFh30cosrC30cosrB zz
lb250F
ft15h
ft5r
01875C33.4B33.4 zz
21 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
System of Equations Gaussian
elimination
Example 26-02 (cont.)
Az
Bz Cz
F = 250lb
W = 3500 lb
z
z
z
C
B
A
3248
1875
3500
5.25.25
33.433.40
111
z
z
z
C
B
A
14252
1875
3500
5.75.70
33.433.40
111
22 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
System of Equations Gaussian
elimination
Example 26-02 (cont.)
Az
Bz Cz
F = 250lb
W = 3500 lb
z
z
z
C
B
A
17500
1875
3500
0150
33.433.40
111
lb
1600
734
1167
A
C
B
z
z
z
23 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
References
Hibbeler (2007) http://wps.prenhall.com/
esm_hibbeler_engmech_1