engi 1313 mechanics i

Shawn Kenny, Ph.D., P.Eng.Assistant ProfessorFaculty of Engineering and Applied ScienceMemorial University of Newfoundlandspkenny@engr.mun.ca

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