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ME 221 Statics Lecture #32 Section 6.9

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ME 221 Statics Lecture #32 Section 6.9. Homework #11. Chapter 6 problems: 2, 3, 6 & 7 – Method of Joints 32, 36, 47 & 53 – Method of Sections 68 & 75 Due Friday, November 21. Quiz #7. Friday, November 21 Analysis of Structures Method of Joints or Method of Sections. P By. P Bx. B.

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homework 11
Homework #11

Chapter 6 problems:

    • 2, 3, 6 & 7 – Method of Joints
    • 32, 36, 47 & 53 – Method of Sections
    • 68 & 75
  • Due Friday, November 21

Lecture #32

quiz 7

Quiz #7

Friday, November 21

Analysis of Structures

Method of Joints or

Method of Sections

Lecture #32

slide4
PBy

PBx

B

PAY

PCy

PAx

a b

PCx

Ax

C

A

Ay

Cy

Analysis of Trusses Using the Method of Joints

We need to solve for:

(1) - Internal forces

FAB, FAC, and FBC

(2) - Reactions

Ax, Ay and Cy

Lecture #32

slide5
6 kN

A

4

B

4

5

9

15

3 kN

9

15

5

C

16

16

4

4

E

D

Lecture #32

slide6
Method of Sectioning

12 kN 12 kN

A C E G I

B D F H J

4 @ 2.4 m=9.6 m

1.8 m

If the question is to find internal forces in selected members of the truss, then one can alternatively use the method of sectioning.

Example: Determine the force in members FG and FH

Lecture #32

slide7
12 kN 12 kN

A C E G I

FGE

1.8 m

FGF

FHF

B D F H J

12 kN 12 kN

A C E G I

FEG

FFG

FEH

1.8 m

B D F H J

Lecture #32

slide8
Frames

Are designed to support, prevent or transmit loads (forces, no moments). At least one member is not a two-force member.

Machines

Are designed to transmit force, motion or energy (forces and moments). Will always have at least one member with multiple forces.

Lecture #32

slide9
Frames and Machines
  • Frames and machines have at least one multi-force member
  • Frames are designed to support loads
  • Machines are designed to transmit and modify loads
  • Just like trusses, frames and machines need to be disassembled to determine member forces
  • Need to draw a FBD for each member

Lecture #32

disassembly exposes internal forces
D

E

C

B

W

A

D

Cx

Cy

E

Cx

FBE

Cy

W

C

T

T

FBE

FBE

W

B

A

Ax

Ax

FBE

Ay

Ay

Disassembly Exposes Internal Forces
  • First, reactions of entire structure
  • Second, disassemble and solve for internal forces

FBE is a

two force

member

Lecture #32

notes on previous problem
Notes on Previous Problem
  • Without recognizing BE is a two force member, the problem is not tractable.
  • MUST draw a FBD for entire structure and each disassembled part
  • Each FBD may have three unknown forces which are found from SFx = 0; SFy = 0; SMz = 0

Lecture #32

general procedure
General Procedure
  • Draw FBD for entire structure and solve for reaction forces SFx = 0; SFy = 0; SMz = 0
  • Disassemble into component parts
  • Recognize two force members
  • Draw FBD and solve for reaction forces - using SFx = 0 ; SFy = 0 ; SMz = 0

Lecture #32

frame example

Frame Example

Lecture #32

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