lecture 4 concrete - doubly rc design - usakti 25 september 2013 sw

8/9/2019 Lecture 4 Concrete - Doubly RC Design - Usakti 25 September 2013 SW

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Term 1 :

Sept. –

Okt. 2013

Sugeng Wijanto, Ph.D.

25 September 2013

Jurusan Teknik Sipil

FTSP USAKTI

Struktur Beton Bertulang - 1

Referensi:

Chapter 4 dan 5

“Reinforced Concrete”, 4th or 6th Edition,

by J.K. Wight & J.G Mac Gregor

Badan Standarisasi Nasional,

Tata Cara Perencanaan Struktur Beton Untuk

Bangunan Gedung , SNI 03-2487-2002

hd

b

MU+

dJd

T

C

Singly Reinforced Concrete

Equilibrium :

H = 0 C = T

C = 0.85f c’ .a.b

T = As. f y

M = 0 Mn = C . (d - ½ a) = T . (d - ½ a)

Where (SNI 12.2.7.3):1 = 0.85 for fc’ 30 MPa

1 = 0.85 - 0.0071 (fc’ - 30)

for 30 < f’c 55 MPa

In all cases 1

0.65

Ref: “Reinforced Concrete” by J.K.

Wight & J.G Mac Gregor ; Chapter 4

Effect of compression

reinforcement on moment

strength

Summing moments about the centroid of the resultant

compressive force gives the following results:

For the beam without compression steel: Mn = As.f y (j1.d)

For the beam with compression steel: Mn = As.f y (j2.d)

J2 > J1

a2 < a1

The addition of compression steel has little effect on the Mn value



d’

As’

Mu

b

h

As

0.85 f c’

f s = f y

c

’cu = 0.003

s

aTc

Ts

(d-½a)

Mn

Analysis of Doubly Reinforced Section

MU+

d

Cc

’s

Strain Diagram:

ε’s= 0.003 (c-d’)/c = 0.003 (a-β1d’)/a = 0.003 {1- (β1d’)/a}εs = 0.003 (d-c)/c = 0.003 (β1d-a)/a = 0.003 { (β1d)/a -1}

Stress Diagram:

f’s = f y “IF” ε’s ≥ f y/Es OR ε’s Es ≥ f y

f’s < f y “IF” ε’s < f y/Es OR ε’s Es < f y

Neutral Axis

Strain Stress Internal Forces

Beams With Compression Reinforcement

Fungsi Tulangan Tekan adalah

1. Pada balok dengan ketinggian terbatas, dimana ρ = ρmax tapi

ØMn < Mu, maka dapat dipergunakan tulangan tekan untuk

menambah kekuatan ØMn

2. Menambah besaran daktilitas dari elemen balok

3. Untuk mengurangi besarnya lendutan balok pada bentang4. Untuk mengantisipasi kemungkinan terjadinya momen

berlawan arah (misal, akibat gaya gempa)

Perencanaan Tulangan Ganda:

Doubly Reinforced Rectangular Sections at Ultimate





Ref: “Reinforced Concrete” by

J.K. Wight & J.G Mac Gregor ;

Chapter 4

EXAMPLE 4-4 Analysis of Doubly Reinforced Rectangular Beam Section

d - a / 2

d - d ’NA



Note:

1 psi = 0.006895 MN/m2

4000 psi = 27,5 MPa

60ksi = 60000 psi60ksi = 413,6 MPa






Analysis of Flanged Sections

Office Building during

construction stage in Jakarta




Actual flow of forces on a T-beam flange.

Typical beam sections in concrete floor systems


Analysis of Mn for Flanged Sections in Positive Bending

Case 1 analysis ( β1c ≤ hf ) for M n in T-section.


Case 1 analysis ( β1c > hf ) for M n in T-section.





1. Determine beff for a beam T-section that is part of a

continuous floor system

2. Calculate Mu , Mn and As-min

Home Work ; Kumpul 2 Oktober Jam 07.40 Tepat

3.00 m

3.00 m

8.00 m

130 mm600 mm

300 mm

3-D22

3-D25

d’

3-D16 3-D16

t s l a b

300 mm

600 mm

6-D22

2-D25 t

s l a b

d’

d”

Potongan A-A

Potongan B-B

Ref: “Reinforced Concrete” by J.K.

Wight & J.G Mac Gregor ;Chapter 4 – page 160

Any Question?

Information from RC Textbook - Mac Gregor,etc. data were used in this presentation, their

contributions are gratefully acknowledged

lecture 4 concrete - doubly rc design - usakti 25 september 2013 sw

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