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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 28: Method of Joints

ENGI 1313 Mechanics I

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ENGI 1313 Mechanics I. Lecture 28:Method of Joints. Lecture 28 Objective. to understand the method of joints for establishing forces in truss members. Recall 2D Rigid Body Equilibrium. Support Reactions. A x. A y. C y. Method of Joints. Joint Equilibrium FBD at a joint - PowerPoint PPT Presentation

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Page 1: ENGI 1313 Mechanics I

Shawn Kenny, Ph.D., P.Eng.Assistant ProfessorFaculty of Engineering and Applied ScienceMemorial University of [email protected]

ENGI 1313 Mechanics I

Lecture 28: Method of Joints

Page 2: ENGI 1313 Mechanics I

2 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Lecture 28 Objective

to understand the method of joints for establishing forces in truss members

Page 3: ENGI 1313 Mechanics I

3 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Recall 2D Rigid Body Equilibrium

Support Reactions

0Fx

0Fy

0Mx

Ay

Ax

Cy

Page 4: ENGI 1313 Mechanics I

4 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Method of Joints

Joint Equilibrium FBD at a joint Particle equilibrium

concepts Solve for member

forces

Page 5: ENGI 1313 Mechanics I

5 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Method of Joints (cont.)

Joint Forces Tension pulls

on joint• + convention

Compression pushes on joint• - convention

Newton’s 3rd Law• T pull on member• C push on member

CBBC FFC ABBA FFT

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6 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Method of Joints Equilibrium Equations

Two-Force Member Coplanar and

concurrent force system What does this mean?

0Fx

0Fy 0Mx

Necessary for Equilibrium

Automatically Satisfied

Page 7: ENGI 1313 Mechanics I

7 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Procedure for Method of Joints

1. Find Support Reactions Typically required but not always necessary

2. Draw FBD at Truss Joint Select joint with 1 known force and at most 2

unknowns Assume forces are tensile (positive scalar)

unless obvious by inspection

3. Apply Equations of Equilibrium4. Repeat for all Joints

Page 8: ENGI 1313 Mechanics I

8 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Joint Free Body Diagrams

CBF

ABF

ACF

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9 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Coordinate Axes Orientation

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10 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Coordinate Axes Orientation (cont.)Resolve FCB

Find Support Reactions 045sinF30cosF0F CBCDx

045cosF30sinFkN5.10F CBCDy

015sinF30coskN5.10F CBy

kN02.5FCB

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11 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Coordinate Axes Orientation (cont.)Resolve FCB

Find Support Reactions Resolve FCD

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12 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Example 28-01

Determine the force in each member. Indicate whether the member is in tension (T) or compression (C).

Page 13: ENGI 1313 Mechanics I

13 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Example 28-01 (cont.)

Where to Start? Examine joints

• # Known Forces?

• # Unknown Forces? 0Fx

045sinFN500 BC

)C(N707FBC

0Fx

045sinFN500 BC

)C(N707N707FBC

FBA

FBC

500N

Page 14: ENGI 1313 Mechanics I

14 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Example 28-01 (cont.)

Joint B 0Fx

0Fy

045sinFN500 BC

)C(N707FBC

0F45cosF BABC

)T(N500FBA

Page 15: ENGI 1313 Mechanics I

15 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Example 28-01 (cont.)

Joint C 0Fx

0Fy

045cosN1.707FCA

)T(N500FCA

045sinN1.707Cy

N500Cy CBF

Page 16: ENGI 1313 Mechanics I

16 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Example 28-01 (cont.)

Joint A 0Fx

0Fy

0N500Ax

N500Ax

0N500Ay

N500Ay ABF

ACF

Page 17: ENGI 1313 Mechanics I

17 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Example 28-01 (cont.)

Support Reactions More than 2 unknowns at

each joint then determine reactions first

For this case not necessary but to show equivalence

0Fx 0Fy

0N500Ax

N500Ax

0CA yy

N500Ay

Ay

Ax

Cy

0MA

0m2Cm2N500 y

N500Cy

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18 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

Example 28-01 (cont.)

Results Summary

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19 ENGI 1313 Statics I – Lecture 28© 2007 S. Kenny, Ph.D., P.Eng.

References

Hibbeler (2007) http://wps.prenhall.com/

esm_hibbeler_engmech_1