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    September 21, 2012

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    Ir.Muhamad Ryanto, MT.

    Jurusan Teknik Sipil

    Universitas Sangga Buana YPKP

    2

    GESER (SHEAR)

    GESER

    Geser pada balok langsing

    Geser pons (punching shear)

    Geser pada : balok tinggi

    brackets/corbels

    deep beam

    foundation

    Geser Friksi

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    Untuk kesetimbangan :

    * Jumlah tegangan geser pada penampang harus setimbang

    dengan gaya geser eksternal.

    * Tegangan geser vertikal dan horizontal besarnya pada setiap

    elemen harus seragam .

    l

    R =w l

    2

    w

    R

    dxShear Forces

    4

    lR =

    w l

    2

    w

    R

    dxShear Forces

    y bAi

    NA

    T

    V

    C

    dx

    v

    w

    V-w dx

    T + dT

    C+dC

    z

    qmax=Vz

    q = b v

    Section Beamelement

    Flexuralstresses

    Shearflow

    Shearstresses

    vmax

    v =V Ai y

    b I

    Geser pada BalokElastik IsotropikHomogen

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    Tegangan Geser Rata - Rata

    Diantara Retak - Retak

    TT +T

    jd

    C C + C

    T + TT

    V

    M

    V M + M

    V

    x

    6

    TM

    jd==

    M + Mjd

    =dan T + T

    jd = KonstanTM

    jd==

    Untuk Keseimbangan Momen

    Tegangan rata - rata geser

    Jd = 0.875d bw = lebar balok

    (ACI + PB) v =V

    bw d

    =M V x =TV

    xjd

    ;

    VT

    bw x== v

    V

    bw jd==;

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    Aksi Balok dan Aksi Busur bila Balok Prismatis dan jd

    Konstan

    Vd

    dx== (T jd)

    Vd(T)

    dx== jd +

    d(jd)

    dxT

    Keadaan Ekstrim

    Bila jd konstan d(jd)

    dx= 0

    d(T)dx

    V == jd

    bila dT

    dx= 0

    d(jd)

    dxTV ==

    jd varies

    C

    T

    C

    8

    Momen besar 1000 u Vdfc`M

    Momen kecil

    0 4 8 12 24 ~

    0.16

    0.30

    V Vdfc M fc

    = 0.14 + 17 < 0.3

    V

    fc

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    Penurunan Rumus Kekuatan Geser Dari

    Balok Tanpa Tulangan Sengkang

    Asumsi :

    Kapasitas beban dicapai ketika tegangan tarik utama

    mencapai kekuatan tarik beton yang sebanding dengan fc`

    Meskipun secara eksak distribusi dari tegangan lentur dan

    tegangan geser pada suatu potongan tidak diketahui,

    dapat diasumsikan :

    ft = Ec/Es x tegangan tarik tulangan

    V = tegangan geser rata - rata

    Ec sebanding dengan fcVu = gaya geser batas

    Mu = momen batas pada potongan tersebut

    10

    Tegangan geser v :

    v = k1Vubd

    fs = tegangan baja tulangan

    Mu

    As.d=

    dan tegangan tarik ft dalam beton :

    MufcEc.fsftEc.Mu

    Es As.d.Es As.d.Es

    Mubd2

    f

    E

    1

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    Atau dapat ditulis :

    k4ft

    fc`

    E bd2=

    Mu 2

    k4 = Konstanta

    Es = nilainya tetap

    Kekuatan tarik beton dapat

    ditulis :

    3ft max

    = k5fc`

    12

    ft max = ft + ( ft)2 + v2

    Pada persamaan 4 ada 2 variabel

    Vu/bdfc dan Mufc /Es Vud

    Dari 440 test didapatkan :

    Persamaan 1 s.d. 3 dirumuskan dalam persamaan

    tegangan utama :

    k5 fc = + { } + ks2Vu ku Mufc ku Mu fcbd Es Vud Es Vud

    = k5 + { } + ks2ku Mufc ku Mu fcEs Vud Es Vud

    Vubdfc

    .. 4

    Vu Vudbdfc Mufc

    = 0.14 + 17 + 0.3

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    Penampang tanpa tulangan geser

    V = Vcz + Vd + Vay

    A

    B

    Vcz

    C1Va

    Vax

    Vay

    VdC

    D

    EF

    T2

    Vd

    Vay

    Vax

    Va

    T1

    Penampang dengan tulangan geser

    V = Vcz + Vd + Vay + Vs

    A

    B

    Vcz

    C1Va

    Vax

    Vs

    VdC

    D

    EF

    T2

    Vd

    Vay

    Vax

    Va

    T1T2

    Vs

    Vcz

    C1

    Vc

    14

    Mengapa lebih banyak dipakai sengkang

    dibandingkan tulangan tarik miring :

    masalah kalau terjadi tegangan terbalik

    sengkang diperlukan sebagai pemegang

    tulangan

    pelaksanaan sulit

    Truss Analogy berlaku untuk balok langsing

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    GESER DALAM BALOK BETON

    BERTULANG (PB 1989)

    Vu < Vn

    Vu = beban geser batas (terfaktor)

    = faktor reduksi = 0.60 PB`89Vn = Vc + VsVc = geser yang ditahan beton

    Vs = geser yang ditahan baja tulangan

    Vc = 1/6 fc bw.d= 1/7 ( fc + 120 w ) bw.d

    Vu d

    Mu(rumus untuk geser yang lebih teliti)

    16

    1. KERUNTUHAN GESER BALOK

    AKIBAT PELELEHAN SENGKANG

    Sengkang tidak dapat menahan geser bila tidak memotong

    retak miring. PB`89 pasal 11.5.4.1.

    menentukan bahwa jarak sengkang maksimum (vertikal)

    adalah :

    smax = d/2 atau 600 mm

    Bila Vs > 1/3 fc bw.d maka jarak sengkang :smax = d/4 atau 300 mm

    Jarak Maksimum Sengkang

    45

    Max S = d/2

    d/245

    Max S = d

    d/2

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    2. KERUNTUHAN GESER AKIBAT

    KEGAGALAN PENJANGKARAN

    Keruntuhan akibat sengkang yang mencapai tegangan

    lelehnya dapat terjadi bila sengkang-sengkang tersebut

    dijangkarkan dengan sem-purna. Sebaiknya sengkang -

    sengkang diterus-kan masuk ke daerah tekan dan tarik

    beton dan didetail sesuai dengan syarat-syarat PB`89 pasal

    12.12.2 pasal 12.13.5.

    fy sengkang sebaiknya BJTP/24

    18

    3. RETAK YANG BERLEBIHAN PADA

    WAKTU PEMBEBANAN LAYAN

    Lebar retak pada tulangan miring >

    lebar retak pada sengkang.

    Lebar retak besar - jarak besar > kecil - jarak dekat

    PB`89 : Vs < 2/3 fc bw.d(syarat lebar retak pada geser)

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    4. KERUNTUHAN GESER AKIBAT HANCURNYA

    BAGIAN BADAN BALOK YANG TERTEKAN

    Pada bagian balok yang tipis keruntuhan

    akibat

    hancurnya diagonal compression member

    dibatasi agar tegangan geser yang terjadi (0.5 Vc)

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    Jumlah tulangan minimum

    Av = bw. s3 fy

    Vc = 1/6 fc` bw.d Vc = 2/3 bw.d

    Av.fy.dVs = = 1/3 bw.ds

    Dimana :

    Av = Luas tulangan sengkang (2 kaki)

    bw = Lebar balok

    s = jarak tulangan sengkang

    fy = mutu tulanganVc = kapasitas beton terhadap geser

    fc = mutu beton

    Vs = kapasitas tulangan terhadap geser

    d = tinggi efektif balok

    24

    LETAK PENAMPANG KRITIS PADA BALOK

    BETON DENGAN BEBAN GESER

    Tumpuan yang menghasilkan tegangan tekan pada

    balok, letak penampang kritis dapat dievaluasi pada jarak

    d dari perlelatakan dan tidak ada beban terpusat yang

    bekerja dalam daerah d dari perle-takan. Sedang untuk

    penampang yang menerima tumpuan tarik penampangkritis dievaluasi pada muka kolom.

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    d

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    KAPASITAS GESER BILA DISERTAI

    BEBAN AKSIAL

    Kapasitas Geser tanpa Beban Aksial

    Vc = [(fc + 120 w ) : 7] bw.d < 0.3 fc.bw.dVu.d

    Mu

    Kapasitas Geser dengan Beban Aksial TekanVu.d

    MmVc = [(fc + 120 w ) : 7] bw.d

    < (0.3 fc.bw.d) x 1 +0.3 Nu

    Ag

    Mm = Mu - Nu ( )4k-d

    8

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    Kapasitas Geser dengan Beban Aksial Tarik

    Vc = (1/6.fc.bw.d) ( 1 + )0.3 Nu

    Ag

    28

    Langkah Perencanaan terhadap

    BEBAN GESER

    1. Tentukan besar gaya geser terfaktor Vu pada

    penampang kritis (mis : Vu = 1.2 VD + 1.6 VL)

    2. Untuk penampang kritis tersebut, hitung Vc;

    bila :

    [ Vc + fc bw . d ]Vu 23

    Vu

    bw . s

    3 . fy

    penampang harus diperbesar

    3. Bila : Vc > Vc

    gunakan tulangan minimum :

    As min =

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    4. Jika : > VcVu

    gunakan tul. geser sehingga memenuhi :Vu Vc + Vs

    30

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    Stirrups can not sustain the shear force

    if it does not intersect the diagonal

    crack. SKSNI - sec 3.4.5 : define that

    maximum space of each vertical

    stirrup is :

    smax = d/2 or 600 mm

    If the shear force in the reinforcement

    Vs

    > (fc

    /3) bw

    d ; then the maximumspace of each vertical stirrup is :

    smax = d/4 or 300 mm

    45

    smax= d/2

    d/2

    45

    smax= d

    d/2

    1. Shear Failure in Beams due to Yielding in Stirrup

    32

    2. Shear Failure due to the Loss of Development Length

    * Shear failure due to yielding in the stirrup only could be reached if

    the stirrups have sufficient development length.

    * The stirrups should be anchorage into the compression and tension

    zone of concrete and should be detailed regarding SNI provision

    especially sec. 3.5.13.2.

    * The bar of BJTP 24 should be used for minimum fy requirement.

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    Crack width sloping web reinforcement > crack width at verticalstirrups

    Crack width at bigger bar diameter and bigger spaced >

    small bar diameter and close spaced.

    Vs < 2/3 . fc bw.d (crack width requirement for shear)

    At thin part of beams, failure due to diagonal compression member must

    be restricted so that the generated shear stress is less than (0.2 to 0.25)times the concrete strength, then :

    Vs < 2/3. fc bw.d

    3. Excessive Crack at Service Loads

    4. Shear Failure due to Web Compression

    34

    Beams without web reinforcement will cause a sudden and brittlefailure due to so many shear mode failure. Then SNI define :

    if Vu > (0.5 Vc) minimum shear reinforcement

    5. Shear Failure due to Yielding at Tension Steel

    Av min = bw s

    3 fy

    Where :

    Av min = minimum shear reinforcement

    bw = beam width

    s = space between each stirrup

    fy = steel strength

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    Critical section = the location of the first inclined crack.

    The critical section must be taken at the face of support when one of the

    following occurs :1. Factored shear Vu does gradually decrease

    from the face of support but the support isitself a beam or girder and therefore does notintroduce compression into the end region ofthe member.

    2. When a concentrated load occurs between theface of support and the distance d therefrom.

    3. When any loading may cause a potentialinclined crack to occur at the face of supportor extend into instead of away from the support.

    Vu

    Critical section

    d

    Critical section

    Critical section

    Instead of these three condition, critical section could be taken

    as the location at spacing d f rom suppor t surface.

    36

    In The SNI strength design method for shear, it is required that :

    Where :Vu = factored shear force

    Vn = shear strength capacityVn = nominal shear strength

    = reduction factor for shear = 0.6 (SNI 3.2.3.2.(3))Vc = portions of shear strength from concreteVs = portions of shear strength from reinforcement

    Vu Vn

    Vn = Vc + Vs

    SNI eq. 3.4-1

    SNI eq. 3.4-2

    =30 35 40 45

    Contribution ofconcrete neglected

    Contribution of stirrups

    vs

    Shearresistance,vu

    vc

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    For element with bending & shear :

    Vc = (f c/6) bw d SNI eq. 3.4-3

    For element with bending & shear & axial compression :

    Vc = 2 {1 + (Nu/14Ag)} (f c/6) bw d SNI eq. 3.4-4

    Where :Vc = shear force from contribution of concrete (N)fc = strength of concrete (MPa)

    bw = width of beam (mm)d = effective depth of beam(mm)Nu = axial compression ultimate load (N)Ag = section area of element (mm

    2)

    Concrete Contribut ion to Shear Strength

    bw

    d

    38

    Reinforcement Contribut ion to Shear Strength

    If using Vertical Stirrups : SNI eq. 3.4-17

    Vs =

    If using Sloping Stirrups : SNI eq. 3.4-18

    Vs =

    Minimum Shear Reinforcement :

    Av min = SNI eq. 3.4-14

    Maximum Shear Strength : SNI 3.4.5.6).(8)

    Vs max = (2fc/3) bw d

    Av fy d

    s

    Av fy (sin + cos ) d

    s

    bw s

    3 fy

    Where :Vs = shear force from contribution

    of shear reinforcement (N)Av = area of shear reinf. (mm

    2)fy = steel yield strength(MPa)d = effective depth of beam(mm)

    s = spacing of stirrups (mm)bw = width of beam(mm)

    s

    bw

    d

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    Start

    Find : Vu

    Vu > VcVu < Vc

    Vc Vu Vc

    define : and

    use Av min

    Av min =

    1

    bw s

    3 fy

    * No shear -reinforcement

    no

    yesVs = Vu - Vc

    Vs >(2fc/3) bw d

    define :

    Vs =

    1

    Av fy d

    s

    Change thedimension of

    beam

    yes

    no

    Finish 2

    noyes

    40

    1

    Check :

    spacing of shear reinforcement, s

    Vs > (fc/3) bw d

    s

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