ENGI 1313 Mechanics I - Memorial University of … Kenny, Ph.D., P.Eng. Assistant Professor Faculty...
23
Shawn Kenny, Ph.D., P.Eng. Assistant Professor Faculty of Engineering and Applied Science Memorial University of Newfoundland [email protected] ENGI 1313 Mechanics I Lecture 26: 3D Equilibrium of a Rigid Body
Transcript of ENGI 1313 Mechanics I - Memorial University of … Kenny, Ph.D., P.Eng. Assistant Professor Faculty...
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 ClassFriday Nov. 9
Two OptionsUse review class Wednesday Nov. 28• Preferred option
Schedule time on Thursday Nov.15 or 22Please 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
AxAy
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 systemResolve forces• Scalar notation?• Vector notation?
x
y
z
TBC
TBD
AxAy
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 ForcesPosition vectors
Unit vectors
x
y
z
TBC
TBD
AxAy
Az
{ }mk2j1i2rBC +−=
( ) ( ) ( ){ }mk13j10i02rBC −+−+−=
{ }mk2j1i2rBD +−−=
( ) ( ) ( ){ }mk13j10i02rBD −+−+−−=
⎭⎬⎫
⎩⎨⎧ +−= k
32j
31i
32uBC
⎭⎬⎫
⎩⎨⎧ +−−= k
32j
31i
32uBD
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
AxAy
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?
ΣMo = 0• Why?
Find moment arm vectors
x
y
z
TBC
TBD
AxAy
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
AxAy
Az
Due to symmetry TBC = TBD
( )( ) K++× BDBDBCBCAB uTuTr
F1= 3 kNF2 = 4 kN
∑ = 0MO
( ) ( ) 0FrFr 22F11F =×+×
( )( ) K++× 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
AxAy
Az
K+− 34320
110kji
TBC
F1= 3 kNF2 = 4 kN0
40005.50kji
300040kji
=−
+−
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
AxAy
Az
F1= 3 kNF2 = 4 kN
∑ =→+ 0Fx
0AuTuT xBDxBDBCxBC =++
( ) ( ) 0A32kN17
32kN17 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
AxAy
Az
F1= 3 kNF2 = 4 kN
∑ =→+ 0Fy
0AuTuT yBDyBDBCyBC =++
( ) ( ) 0A31kN17
31kN17 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
AxAy
Az
F1= 3 kNF2 = 4 kN
0kN4kN3AuTuT ZBDzBDBCzBC =−−++
( ) ( ) 0kN7A32kN17
32kN17 z =−+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛
kN67.15Az −=
∑ =↑+ 0Fz
15 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Example 26-02The 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 SystemDraw FBD
AzBz 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
AzBz 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
AzBz 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
AzBz Cz
F = 250lb
W = 3500 lb
Example 26-02 (cont.)
∑ = 0Mx
030cosFh30sinrC30sinrBrA zzz =+++− ooo
lb250Fft15h
ft5r
===
03248C5.2B5.2A5 zzz =+++−
20 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
Moment About y-axis
AzBz Cz
F = 250lb
W = 3500 lb
Example 26-02 (cont.)
∑ = 0My
030sinFh30cosrC30cosrB zz =−− ooo
lb250Fft15h
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.)
AzBz Cz
F = 250lb
W = 3500 lb
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−
z
z
z
CBA
324818753500
5.25.2533.433.40
111
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−
z
z
z
CBA
1425218753500
5.75.7033.433.40
111
22 ENGI 1313 Statics I – Lecture 26© 2007 S. Kenny, Ph.D., P.Eng.
System of Equations
Gaussian elimination
Example 26-02 (cont.)
AzBz Cz
F = 250lb
W = 3500 lb
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−
z
z
z
CBA
1750018753500
015033.433.40
111
lb1600734
1167
ACB
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
