3 Pinned Portal Frames

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    PORTAL FRAME STRUCTURES

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    PORTAL FRAME STRUCTURES

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    PORTAL FRAME STRUCTURES - ANALYSIS

    PINNED BASE

    PORTAL FRAME

    A

    HAHE

    B

    E

    D

    VA VE

    3 equations :

    M=0, V=0 & H=0

    BUT

    4 UNKNOWNS :-

    HA, VA, VE& HE !!!!

    SOLUTION:- Introduce an INTERNAL PIN at, say, position C ......

    Hence an extra equation is generated, MC= 0 (4 unknowns & 4 equations)

    C

    internal pin,

    BM = 0

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    PORTAL FRAME STRUCTURESSIGN CONVENTION FOR BENDING

    MOMENT AND SHEAR FORCE

    NEGATIVEBENDING MOMENT

    AND

    SHEAR FORCE

    POSITIVE

    BENDING MOMENTAND

    SHEAR FORCE

    A

    B

    E

    DC

    A

    B

    E

    DC

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    WORKED EXAMPLE3 PINNED FRAME (SUPPORT REACTIONS)

    HE

    VE

    5 kN/m

    15 kN/m

    1m1m 2m2m

    4m

    25 kN

    C

    25 kN

    AHA

    B

    E

    D

    VA

    M@A, M = 0:-

    (15 x 4 x 2) + (25 x 2) + (5 x 6 x 3)

    + (25 x 4)6VE= 0

    360 = 6VE or VE= 60 kN ()

    M@C (pin), for RHS of frame, MC=0 :-

    (25 x 1) + (5 x 3 x 1.5) + 4HE3VE = 0

    4HE= 18047.5 or HE= 33.1 kN ()

    V = 0 :-

    VA+ VE= 80 VA = 20 kN (

    )

    H = 0 :-

    HA+ HE= (15 x 4) or HA = 26.9 kN ()

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    NBD (33.1)

    MBD= 12.4 (-ve BM)

    SBD(20)

    Beam BC - D

    WORKED EXAMPLE3 PINNED FRAME ( COLUMN AB )

    COLUMN A - B

    15 kN/m

    4m

    A26.9 kN

    20 kN

    B

    MBA

    MBA= (26.9 x 4)(15 x 4 x 2) =12.4 kNm

    SBA

    SBA= 26.9 15 x 4 =33.1 kN

    NBA

    NBA= 20 kN

    BM at mid-span AB

    = (26.9 x 2)(15 x 2 x 1) = 23.8 kNm

    SBA

    MBA

    NBA

    Forces (shown +ve)

    at top of column,

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    WORKED EXAMPLE3 PINNED FRAME (BM DISTRIBUTION BEAM B - C - D)

    M at x = 0m = - 12.4 kNm

    M at x = 2m = - 12.4 + ( 20 x 2 )( 5 x 2 x 1 ) = 17.6 kNm

    x

    1m

    NBD (33.1)

    MBD= 12.4 (-ve BM)

    SBD(20)

    Beam BC - D

    25 kN 25 kN

    2m1m

    2m

    5 kN/m

    M at C (pin) = - 12.4 + ( 20 x 3 )( 5 x 3 x 1.5 )( 25 x 1 ) = 0 kNm

    (OK re pin)

    M at x = 4m = - 12.4 + ( 20 x 4 )( 5 x 4 x 2 )( 25 x 2) = - 22.4 kNm

    Actually =

    0.1 but OK

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    WORKED EXAMPLECOLUMN D - E

    MDE= (33.1 x 4) = 132.4 kNm ( -ve sign)

    MDB

    Beam BC - D

    MDE

    COLUMN D - E

    4m

    E33.1 kN

    60 kN

    D

    33.1 kN

    (SDE)

    MDB= 132.4 kNm ( -ve sign)

    NDE = 60 kN

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    3 PINNED FRAME (BENDING MOMENT DIAGRAM)

    +

    +--

    C

    B

    E

    D

    A

    23.8

    12.4

    17.6

    132.4

    132.4

    2m

    2m

    -

    -

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    3 PINNED FRAME (SHEAR FORCE DIAGRAM)

    +

    +

    +-

    33.1 CB

    E

    D

    A

    20

    26.9

    33.110

    15

    25

    50

    60

    33.1

    -

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    CLASS EXAMPLE/TUTORIAL3 PINNED FRAME

    (SUPPORT REACTIONS)

    (i) Calculate the values of the support reactions

    (ii) Draw BMD and SFD

    VE

    HE

    15 kN/m

    6 kN/m

    1m1m

    6m

    6m

    C

    26 kN

    A

    HA

    B

    E

    D

    VA