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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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ME 221 StaticsLecture #32Section 6.9

Lecture #32


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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 l.jpg

    Quiz #7

    Friday, November 21

    Analysis of Structures

    Method of Joints or

    Method of Sections

    Lecture #32


    Slide4 l.jpg

    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


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    6 kN

    A

    4

    B

    4

    5

    9

    15

    3 kN

    9

    15

    5

    C

    16

    16

    4

    4

    E

    D

    Lecture #32


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    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 l.jpg

    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 l.jpg

    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


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    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


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    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


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    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


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    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


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    Frame Example

    Lecture #32


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