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Statics:Hibbeler 1 Smithfield Bridge, Pittsburgh 1882 Smithfield Other Lindenthal Projects Royal Albert Bridge at Saltash 1859 Isambard Kingdom Brunel, Engineer Saltash Brunel Balance of Forces in Equilibrium Balance of Forces in Equilibrium PULLEYS, FRAMES, AND MACHINES Compute the forces in pulleys, ropes, beam, and frame members that require multiple free-body diagrams . Compute the reactions for compound beams, frames, or similar devices. Show why it is advantageous to: Separate elements at pins and draw multiple FBDs & identify 2-force members

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  • Statics:Hibbeler 1

    Smithfield Bridge, Pittsburgh 1882

    Smithfield

    Other Lindenthal Projects

    Royal Albert Bridge at Saltash 1859 Isambard Kingdom Brunel, Engineer

    SaltashBrunel

    Balance of Forces in Equilibrium

    Balance of Forces in EquilibriumPULLEYS, FRAMES, AND MACHINES

    Compute the forces in pulleys, ropes, beam, and frame members that require multiple free-body diagrams.

    Compute the reactions for compound beams, frames, or similar devices.

    Show why it is advantageous to: Separate elements at pins and draw multiple FBDs & identify 2-force members

  • Statics:Hibbeler 2

    Pulley Systems Pulley Systems – Mechanical Advantage

    W

    T

    Pulley Systems – Mechanical Advantage

    W

    T

    Free Body Diagrams of Pulley System

    FBD Top

    T

    Reaction

    TT

    FBD A

    W

    T T

    Pulley Systems – Other Cuts and FBDs

    Fb

    FBD

    W

    TFbar

    T T

    Pulley Systems – Other Cuts and FBDs

    Fbar T

    FBD

    W

    T

    W

  • Statics:Hibbeler 3

    Pulleys – Method of Joints Approach Pulleys – Method of Joints Approach

    Pulleys – Method of Joints ApproachAy

    P

    Cy

    RR

    R

    10 lb

    FBD 2

    P P R

    P

    10 lb 10 lb

    100 lbFBD 1

    FBD 3

    Pulleys – Method of Sections Approach

    Pulleys – Method of Sections Approach

    RP

    R

    10 lb

    100 lb

    10 lb

    10 lb

  • Statics:Hibbeler 4

    QUICK PROBLEM SOLVING

    Given: A frame and loads as shown.

    Find: The reactions that the pins exert on the frame at A, B, and C.

    Plan:

    a) Draw a FBD of members AB and BC.

    b) Apply the equations of equilibrium to each FBD to solve for the six unknowns.

    QUICK PROBLEM SOLVING (continued)FBDs of members AB and BC:

    BYB

    BX

    0.4m500N

    CA X A

    B

    BY

    BX

    1000N

    45º

    + MA = BX (0.4) + BY (0.4) – 1000 (0.2) = 0

    + MC = -BX (0.4) + BY (0.6) + 500 (0.4) = 0

    BY = 0 and BX = 500 N

    Summing moments about A and C on each member, we get:

    0.2m 0.4mCY

    AY

    0.2m 0.2m

    FBDs of members AB and BC:BYBB X

    0.4m500N

    C

    0.2m 0.4mC

    A X A

    B

    BY

    BX

    1000N

    45º

    0.2m 0.2m

    QUICK PROBLEM SOLVING (continued)

    + FX = AX – 500 = 0 ; AX = 500 N

    + FY = AY – 1000 = 0 ; AY = 1,000 N

    For FBD of BC: + FX = 500 – CX = 0 ; CX = 500 N + FY = CY – 500 = 0 ; CY = 500 N

    For FBD AB: CYAY

    WORKING EXAMPLE

    Given: The wall crane supports an external load of 700 lb.

    Find: The force in the cable at the winch motor W and the horizontal and vertical components of the pin reactions at A, B, C, and D.

    Plan:

    a) Draw FBDs of the frame’s members and pulleys.

    b) Apply the equations of equilibrium and solve for the unknowns.

  • Statics:Hibbeler 5

    FBD of the Pulley E

    T T

    E

    EXAMPLE (continued)

    +↑ Fy = 2 T – 700 lb = 0

    T = 350 lb

    700 lb

    Equations of Equilibrium:

    +Fx = Cx – 350 = 0 Cx = 350 lbFBD of pulley C

    C

    350 lb

    CY

    CX

    350 lb

    +↑ Fy = Cy – 350 = 0 Cy = 350 lb +FX = – Bx + 350 – 350 sin 30° = 0

    Bx = 175 lb

    +↑ Fy = By – 350 cos 30° = 0By = 303.1 lbFBD of pulley B

    BY

    BX30°

    350 lb

    350 lb

    B

    Note that member BD is a two-force member

    FBD of member ABC

    AX

    AY

    A 45°

    TBDB

    175 lb303.11 lb 700 lb

    350 lb

    4 ft 4 ft

    +MA = TBD sin 45° (4 ft) – 303.1 (4 ft) – 700 (8 ft) = 0

    TBD = 2409 lb+FY = AY + 2409 sin 45° – 303.1 – 700 = 0

    AY = – 700 lb +FX = AX – 2409 cos 45° + 175 – 350 = 0

    AX = 1880 lb

    EXAMPLE (continued)

    A FBD of member BD

    45

    2409 lb

    B

    D

    At D, the x and y component are

    + DX = –2409 cos 45° = –1700 lb

    + DY = 2409 sin 45° = 1700 lb

    45°

    B

    2409 lb

    Why are trusses like donuts?

    Donuts are not good in compression eitherAfter the 5th one you feel kinda sick

    Both = fun!You can’t drive to school without them

    If they fall to the ground they are not any good anymore

    Analysis of Machines

  • Statics:Hibbeler 6

    Pliers, Cutters, and GripsFind vertical clamping force at E

    APPLICATIONS

    Frames are commonly used to support external loads.

    How is a frame different than a t ?truss?

    How can you determine the forces at the joints and supports of a frame?

    Find all reaction forces and forces at pinned connections in press

  • Statics:Hibbeler 7

    Elevation View of Cairo Bridge

    (Source: S. Nam, University of Illinois)

    2D Schematic of Cairo Bridge

    (Source: S. Nam, University of Illinois)

    View of Finite Element Model

    -1

    -0.5

    0

    0.5

    1

    0 10 20 30 40

    Time (sec)

    Acc

    . (g)

  • Statics:Hibbeler 8

    Animation of Bridge Deflection

    (Source: S. Nam, University of Illinois)BACK

    Graph of Maximum Deflection

    Vertical Displacement

    0

    2

    4

    6

    8

    10

    0 500 1000 1500 2000 2500 3000 3500 4000

    bridge cood. (ft)

    TEDA Stadium Tianjin, China