Nav Composites 03-beam Jacketing.pdf

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STRUKTUR KOMPOSIT DIPLOMA IV TEKNIK INFRASTRUKTUR SIPIL ITS

CHAPTER – 3 BEAM JACKETING

Dosen Pengajar : Afif Navir Refani, ST., MT. [email protected] / +628113411985

IPITS 2015

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IPITS 2015

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LECTURE CONTENTS Introduction

Composites Material Application for structural strengthening

Composites Material (concrete – concrete)

Evaluation 2

Steel – Concrete Composites

Evaluasi 1

Concrete – carbon fiber composites

Designing Of Composites Member

Evaluation 3

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COMPOSITES STRUCTURE CONCRETE EXISTING & NEW CONCRETE

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Slab Overlay

Beam Jacketing

Column Jacketing

CONCRETE COMPOSITES

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COMPOSITES STRUCTURE BEAM JACKETING (FLEXTURE COMPOSITES ELEMENT)

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CONCRETE JACKETING Jacketing a Reinforced concrete member is done to enhance it's strength. This may be necessary either in case of increase of load or to overcome deficiencies that may develop over time. This is mostly done for compression members like columns in a building, piers and abutments of bridges. The process is similar to wearing a jacket in extreme weather conditions and hence the name. The old concrete surface is thoroughly cleared of loose material or Cement plaster if any and cleaned properly. Reinforcement bars both vertical and horizontal are provided as per design and concrete is poured.

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BEAM JACKETING

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BEAM JACKETING

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BEAM JACKETING

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BEAM JACKETING

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Balanced Strain Condition

Condition where the yield of steel reinforcement and the crushing of outer concrete compressive fiber occur at the same time. Or condition when ultimate strain of concrete occur the same time as steel yield strain 13

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Balanced Strain Condition Strain linear relationship : cb εu = d εu + ε y

εu 0.003 600 = = ε u + ε y 0.003 + fy / 200000 600 + fy

From force equilibrium : Cb = Tb 0.85 f c'bab = Asb f y 0.85 f c'b(β1cb ) = ρ b bdf y

ρb =

0.85β1 f c' cb fy d

ρb =

0.85β1 f c' 600 fy 600 + f y

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Balanced Strain Condition To ensure the structure is under reinforced condition : ρ max = 0.75 ρ b = 0.75

0.85β1 f c' 600 fy 600 + f y

Minimum reinforcement for flexural member : Asmin =

3 f c' 1.4 bw d ≥ bw d fy fy

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Doubly Reinforced Concrete Section

A’s : Compression Steel Reinforcement. C’s : Compression Force in Steel Reinforcement. f’s : Compression Stress in Steel Reinforcement. 16

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Doubly Reinforced Concrete Section

A’s : Compression Steel Reinforcement. C’s : Compression Force in Steel Reinforcement. f’s : Compression Stress in Steel Reinforcement. 17

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Doubly Reinforced Concrete Section

Cc = T1

0.85 f 'c ab = ( As − As ') f y a  M n1 = ( As − As ') f y  d −  2 

Cs ' = T2

M n 2 = As ' f s (d − d ')

M n = M n1 + M n 2 a  M n = ( As − As ') f y  d −  + As ' f s (d − d ') 2 

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CONCRETE JACKETING MOMENT CAPACITY

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Doubly Reinforced Concrete Section Force Equilibrium : C =T C = Cc + C s ' T = T1 + T2

Solving the force equilibrium we can achieve that : a=

As f y − As ' f s 0.85 f 'c b

;a =

( As − As ') f y 0.85 f 'c b

All the formula above can be used only if all the steel compression is yields, otherwise the fs value must be calculate accurately.

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Analysis Procedure of Doubly RC Beam Step – by – Step : #1 Calculate the ρ, ρ’ and ρmin. #2 Check The Compression Steel Yields (If it yields then proceed to #4) : ρ − ρ' ≥

0.85β1 f c' d ' 600 f yd 600 − f y

#3 Calculate The Compression Steel Stress (fs) :  0.85β1 f c' d '   f s' = 6001 −  ρ − ρ ' f y d  

(

)

#4 Calculate stress block depth (a) : a=

As f y − As ' f s 0.85 f 'c b

;a =

( As − As ') f y 0.85 f 'c b

#5 Calculate φMn : a  M n = ( As − As ') f y  d −  + As ' f s (d − d ') 2  21

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Example Analysis [1] Calculate the nominal moment strength Mn of the doubly reinforced section with data specified below : f’c = 30 MPa fy = 400 Mpa d’ = 65 mm dt = 525 mm As = 3000 mm2 [6D25] As’ = 1000 mm2 [2D25] As-As’ = 2000 mm2 b = 350 mm #1 Calculate the ρ, ρ’ and ρmin. As = 0.0162 bd A' ρ ' = s = 0.0054 bd 1.4 ρ min = = 0.0035 fy

ρ=

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Example Analysis [1] #2 Check The Compression Steel Yields (If it yields then proceed to #4) : ρ − ρ' ≥

0.85β1 f c' d ' 600 f yd 600 − f y

0.85(0.85)(30)(65) 600 400(525) 600 − 400 0.0108 ≥ 0.0201 0.0108 ≥

#3 Calculate The Compression Steel Stress (fs) :  0.85β1 f c' d '   f s' = 6001 −  ρ − ρ ' f y d    0.85(0.85)(30)(65)   f s' = 6001 −  (0.0108)(400 )(525)  f s' = 227 MPa

(

)

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Example Analysis [1] #4 Calculate stress block depth (a) : a=

As f y − As ' f s 0.85 f 'c b

=

(3000)(400) − (1000)(227 ) = 109.02mm 0.85(30 )(350 )

#5 Calculate φMn : a  M n = (As f y − As ' f s ) d −  + As ' f s (d − d ') 2  109.02   M n = ((3000)(400 ) − (1000 )(227 )) 525 −  + (1000)(227 )(525 − 65) 2   M n = 480.812kNm

φMn = 0.8(480 ) = 384.649kNm

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Shear Strength of Concrete Section • Shear nominal strength (Vn) of concrete section is a combination of shear strength from concrete (Vc) and sehar strength from shear reinforcement (Vs) : Vn=Vc+Vs • Where Vc : Vc =

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f 'c b w d

• If accurate calculation were used then : 1  V d  Vc =   f 'c + 120 ρ w u bw d ≤ 0.3 f 'c bw d M u b  7 

• Where : ρw =

As Vu d ≤1 bd M u

Prepared By : Bambang Piscesa, ST, MT.

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Shear Strength of Concrete Section • If there is an axial compression load acting in the concrete section therefore Vn can be calculated as follows (Nu/Ag in MPa):  N Vc =  1 + u  14 A g 

1  6 

f 'c b w d

• If accurate calculation were used (Axial Compression Load) : 1  V d  0.3N u Vc =   f 'c + 120 ρ w u bw d ≤ 0.3 f 'c bw d 1 + 7 M Ag m   

• Where :

 4h − d  M m = M u − Nu    8 

Prepared By : Bambang Piscesa, ST, MT.

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Shear Strength of Concrete Section • If there is an axial tension load acting in the concrete section therefore Vn can be calculated as follows:  0.3N u Vc =  1 +  Ag 

1  6 

f 'c b w d ≤ 0

• Nu/Ag is in MPa. • For circular concrete section area for Vc can be calculated from diameter multiplied by effective width (d=0.8h).

Prepared By : Bambang Piscesa, ST, MT.

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Shear Strength of Concrete Section • Minimum shear reinforcement area Avmin : Av min =

bw s 3fy

• Minimum shear force acquired from shear reinforcement : 1 Vs min = bw d 3

Prepared By : Bambang Piscesa, ST, MT.

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LIMITATION FOR FLEXTURE COMPOSITE COMPONENT

Ps 21.11.6 SNI 2847-2013

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LIMITATION FOR FLEXTURE COMPOSITE COMPONENT

The length of steel headed stud anchors shall not be less than four stud diameters from the base of the steel headed stud anchor to the top of the stud head after installation.

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Force Transfer Mechanisms

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SHEAR FORCE CAPACITY For Non Seismic Beam :

▰ Vu ≤ Ø Vnh ▰ Vnh = 0,55 bv . d Dimana : Vu

= Gaya Geser Horizontal Ultimate

Vnh

= Gaya Geser Nominal

Ø

= 0,75 (geser)

bv

= lebar pelat komposit

D

= tebal pelat overlay 41

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For Seismic Beam : a. Daerah Tumpuan Mpr−+Mpr+ 820,36+504,86 +Vu +159,753=343,81 kN ln 7,2 + − 820,36+504,86 Mpr +Mpr −Vu= −159,753=24,305 kN Vka = 7,2 ln Gaya geser maksimum yang ditimbulkan oleh beban gempa adalah: Mpr+ +Mpr− 820,36+504,86 = =184,058 kN ln 7,2

Vki =

Jika dipakai sengkang tertutup dengan diameter 13 mm (2 kaki), maka jarak antar sengkang, s, adalah: As×fy×d 2(132,73)×400×672,5 = =230,74 mm s= Vs 309,48×103 Jarak maksimum sengkang tertutup sepanjang 2h (2 x 750 = 1500 mm) tidak boleh melebihi nilai terkecil dari: d/4 = 672,5/4 = 168,13 mm 6db = 6(22) = 132 mm 150 mm Sehingga tulangan sengkang terpasang D13-130 mm hingga sepanjang 1500 mm dari muka tumpuan mencukupi untuk menahan gaya geser.

b. Daerah Lapangan Jika digunakan db = 13 mm, maka jarak ditentukan dari nilai yang terkecil antara: 672,5 d =545,63 mm s1 =Av fyt =2 132,73 400 Vs 130,874 × 103 𝑑 672,5 𝑠 336,25 𝑚𝑚 2 2 fyt 400 s3 =Av = 2 132,73 × =758,46 mm 0,35bw 0,35×400 s4 =600 mm Jadi tulangan sengkang terpasang D13-180 mampu menahan gaya geser. 42

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CASE STUDY 1 FLEXTURE ELEMENT (BEAM)

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CEK : • Kapasitas Momen Balok Eksisting (300x400) terhadap Mu • Kapasitas Momen Balok Setelah jacketing (400x500) terhadap Mu

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Case study 1

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Tulangan Tambahan tumpuan 4D16 (atas) & 2D16 (bawah) Tulangan Tambahan lapangan 2D16 (atas) & 4D16 (bawah) Tulangan Sengkang Tambahan D10-150 (tumpuan) dan D10200 (lapangan) Mu Tump = 19.127 Kg.m Mu Lap = 14.345 Kg.m Vu Tump = 97.873 Kg fc’ = 30 Mpa Fy = 390 Mpa Fys = 240 MPa

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MOMENT CAPACITY EXISTING (30x40) ENDSPAN

Mn(tump) = 92.80 kNm = 9.280 Kg.m Mutump ≤ Ø Mn 19.127 Kgm ≤ 0.9 x 9.280 Kgm 19.127 Kgm ≤ 8.352 Kgm (NOT OK)

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MOMENT CAPACITY EXISTING (30x40) MIDSPAN

Mn(lap) = 70.11 kNm = 7.011 Kg.m Mulap ≤ Ø Mn 14.345 Kgm ≤ 0.9 x 7.011 Kgm 14.345 Kgm ≤ 6.309 Kgm (NOT OK)

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MOMENT CAPACITY BEAM JACKETING (400x500mm) ENDSPAN

Mn(tump) = 221.49 kNm = 22.149 Kgm Mutump ≤ Ø Mn 19.127 Kgm ≤ 0.9 x 22.149 Kgm 19.127 Kgm ≤ 19.934 Kgm (OK)

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MOMENT CAPACITY BEAM JACKETING (400x500mm) MIDSPAN

Mn(lap) = 196.72 kNm = 19.672Kg.m Mulap ≤ Ø Mn 14.345 Kgm ≤ 0.9 x 19.672 Kgm 14.345 Kgm ≤ 17.704 Kgm (OK)

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END OF LECTURES 3

IPITS 2015

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