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Jurusan Teknik Sipil

Universitas Sangga Buana YPKP

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GESER (SHEAR)

GESER

Geser pons (punching shear)

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

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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

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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

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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

Bila jd konstan d(jd)

dx= 0

d(T)dx

V == jd

bila dT

dx= 0

d(jd)

dxTV ==

jd varies

C

T

C

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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

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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`

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ft max = ft + ( ft)2 + v2

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

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Mengapa lebih banyak dipakai sengkang

dibandingkan tulangan tarik miring :

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)

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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)

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

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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

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WAKTU PEMBEBANAN LAYAN

Lebar retak pada tulangan miring >

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

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

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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

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Kapasitas Geser dengan Beban Aksial Tarik

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

Ag

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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

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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

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smax= d/2

d/2

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smax= d

d/2

1. Shear Failure in Beams due to Yielding in Stirrup

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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

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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.

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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

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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

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1

Check :

spacing of shear reinforcement, s

Vs > (fc/3) bw d

s

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