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
8/9/2019 Lecture 4 Concrete - Doubly RC Design - Usakti 25 September 2013 SW
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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
Analysis of Doubly Reinforced Section
8/9/2019 Lecture 4 Concrete - Doubly RC Design - Usakti 25 September 2013 SW
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Analysis of Doubly Reinforced Section
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
EXAMPLE 4-4 Analysis of Doubly Reinforced Rectangular Beam Section
EXAMPLE 4-4 Analysis of Doubly Reinforced Rectangular Beam Section
Note:
1 psi = 0.006895 MN/m2
4000 psi = 27,5 MPa
60ksi = 60000 psi60ksi = 413,6 MPa
8/9/2019 Lecture 4 Concrete - Doubly RC Design - Usakti 25 September 2013 SW
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EXAMPLE 4-4 Analysis of Doubly Reinforced Rectangular Beam Section
EXAMPLE 4-4 Analysis of Doubly Reinforced Rectangular Beam Section
Analysis of Doubly Reinforced Section
Analysis of Flanged Sections
Office Building during
construction stage in Jakarta
8/9/2019 Lecture 4 Concrete - Doubly RC Design - Usakti 25 September 2013 SW
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Analysis of Flanged Sections
Actual flow of forces on a T-beam flange.
Typical beam sections in concrete floor systems
Analysis of Flanged Sections
Analysis of Mn for Flanged Sections in Positive Bending
Case 1 analysis ( β1c ≤ hf ) for M n in T-section.
Analysis of Mn for Flanged Sections in Positive Bending
Case 1 analysis ( β1c > hf ) for M n in T-section.
8/9/2019 Lecture 4 Concrete - Doubly RC Design - Usakti 25 September 2013 SW
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Analysis of Mn for Flanged Sections in Positive Bending
Analysis of Mn for Flanged Sections in Positive Bending
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