3. Design Assumption and Beam Analysis
Transcript of 3. Design Assumption and Beam Analysis
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REINFORCED CONCRETE
STRUCTURE I
Design AssumptionRevised : 17-September-2013
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CIVIL ENGINEERING DEPARTMENT
FACULTY OF CIVIL ENGINEERING AND PLANNING
INSTITUT TEKNOLOGI SEPULUH NOPEMBER
SURABAYA
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Design Assumptions
1. Strain in reinforcement and concrete shall be assumed directly
proportional to the distance from the neutral axis.
2. Strain in steel and surrounding concrete is the same prior to
cracking of the concrete or yielding of the steel.
3. Tensile Strength of Concrete shall be neglected in flexural
calculation of reinforced concrete.
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3
Design Assumptions
4. Maximum usable strain at extreme concrete compression fiber
shall be assumed equal to u = 0.003.
5. Stress in reinforcement fs below the yield strength fy shall be
taken as Es times the steel strain s . For strains greater than
fy/Es, stress in reinforcement shall be considered independent
of strain and equal to fy.
6. Relationship between concrete compressive stress distribution
and concrete strain shall be assumed to be rectangular.
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Design Assumption #1
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Design Assumption #1
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Design Assumption #6
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Parabolic Stress-Strain distribution in concrete may be considered satisfied by
an equivalent rectangular concrete stress distribution.
30MPa 58MPa
65.005.07
30'85.01
cf
SNI-03-2847-2002, pasal 12.2.7.3
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Design Assumption #6
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Relationship between concrete compressive stress distribution and concrete
strain shall be assumed to be rectangular, trapezoidal, parabolic, or any othershape that results in prediction of strength in substantial agreement with results
of comprehensive tests.
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REINFORCED CONCRETE
STRUCTURE
Analysis and Design of Single RC Beam
Revised : 17-September-2013
9
CIVIL ENGINEERING DEPARTMENT
FACULTY OF CIVIL ENGINEERING AND PLANNING
INSTITUT TEKNOLOGI SEPULUH NOPEMBER
SURABAYA
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Balanced Strain Condition
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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
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Balanced Strain Condition
Strain linear relationship :
From force equilibrium :
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yu
ub
d
c
fyfyyu
u
600
600
200000/003.0
003.0
ybbc
ysbbc
bb
bdfcbf
fAbaf
TC
1
'
'
85.0
85.0
yy
cb
b
y
cb
ff
f
dc
ff
600
60085.0
85.0
'
1
'
1
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Balanced Strain Condition
To ensure the structure is under reinforced condition :
Minimum reinforcement for flexural member :
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yy
cb
ff
f
600
60085.075.075.0
'
1max
db
f
db
f
fAs w
y
w
y
c 4.13 '
min
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Design of Reinforced Concrete Beam
Design of reinforced concrete beam is used to design the beam dimension with
only known moment forces. Other rules in RC design can also be based on ACI
318-99 or SNI 2847-2002.
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Design of Reinforced Concrete Beam
Force Equilibrium :
Moment Equilibrium :
A nominal strength coefficient of resistance (Rn) is obtained when both sides of
Eq. above are divided by bd2:
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''
'
85.085.0
85.0
c
y
c
ys
yysc
f
df
bf
fAa
bdffAbaf
TC
'85.0
5.0
2
c
y
yn
n
f
fddbdfM
adTorCM
'285.0
5.01
c
y
yn
nf
ff
bd
MR
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Design of Reinforced Concrete Beam
When b and d are preset (determined), is obtained by solving the quadratic
equation for Rn :
Equation above can be used to determine the steel ratio given Mu or vice-versa
if the section properties b and q are known. Substituting Mn = Mu/ into
Equation above and divided each side by fc:
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'
'
1
2
'
2
85.0
2
11
85.0
085.0
5.0
c
n
y
c
c
y
f
R
f
f
Rnfyf
f
59.01
;85.0
5.01
2
'
'''2'
bdf
M
f
f
f
f
f
f
bdf
MR
c
u
c
y
c
y
c
y
c
un
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Design of Reinforced Concrete Beam
Example of design using graphics guide of single RC beam in flexure.
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Home Work :
1. Create a graphics for single
reinforced concrete beam with
various fcand fy.
2. Use the constraint of fy in onegraphics for example fy : 240
MPa, 300 MPa, 320 MPa, 350
MPa, 400 MPa.
3. Use several fc in one graphics
for example fc : 20 MPa, 25MPa, 30 MPa, 35 MPa, 40 MPa,
45 Mpa, 50 MPa.
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Example Design [1]
Example Design :
Design the concrete section of the beam, which is simple supported and are
loaded as below : [fc= 35 MPa, fy = 400 MPa].
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L=6m
Ql=1.5t/m;Qd=1t/m
kNmMMM
kNmtmlqM
kNmtmlqM
ldu
ll
dd
76.15815.666.11.442.16.12.1
15.6675.665.18
1
8
1
1.445.4618
1
8
1
22
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Example Design [1]
#1 Determine maximum reinforcement ration (max) for material strength fc=
35 MPa and fy = 400 MPa.
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65.0814.01
65.005.07
303585.01
65.005.07
30'85.01
cf
Calculating 1
Calculating max:
0272.0
400600
600
400
35814.085.075.0
600
60085.075.075.0
max
max
'
1max
fyfy
fcba l
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Example Design [1]
#2 Compute bd2required :
#3 Determine Size member so that bd2> bd2required :
Minimum Beam Depth (h) = d + cover + dh + db/2
Minimum Beam Depth (h) = 303.37 + 40 +10 + 8 = 361.37 mm 400 mm
d = h cover dh db/2 = 342 mm
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32
'22
23008696625.88.0
100000076.158
625.83585.0
4000272.05.014000272.0
85.0
5.01
mmRMbd
MPaR
f
ff
bd
M
bd
MR
n
u
n
c
y
yun
n
mmd
mmb
37.303250
23008696
250
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Example Design [1]
#4 Using the 400 mm beam depth (h), compute a revised value of :
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MPabd
MR
bd
MR
un
un
786.63422508.0
100000076.15822
2
0035.00196.00272.0
400
4.14.1
3585.0
786.6211
400
3585.00272.0
4.185.021185.0
minmax
minmax
min'
'
max
fy
fyfR
ff
c
n
y
c
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Example Design [1]
#5 Compute As Required :
As = x b x d
As = 0.0196 x 250 x 342
As = 1675.80 mm2
Use 6 D19 (As=1701 mm2)
#6 CrossCheck The Moment Nominal with Moment Ultimate :
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kNmMkNmM
kNm
a
dfAM
nu
ysn
26.16176.158
26.1612
48.91
34240017018.02
mmbf
fAa
c
ys48.91
2503585.0
4001701
85.0 '
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Example Analysis [2]
Example Analysis :
Analyze the Moment Nominal Capacity (Mn) of the beam below : [fc = 35MPa, fy = 400 MPa].
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d=350mm h=400mm
b=250mm
Cover+dh+db/2=50mm
3D19
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Example Analysis [2]
Example Analysis :
Analyze the Moment Nominal Capacity (Mn) of the beam below : [fc = 35MPa, fy = 400 MPa].
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T
C
d
a
d-a/2
228501925.03 mmAs
mmbf
fAa
c
ys71.45
2503585.0
400850
85.0 '
kNmM
MM
kNmM
M
adfAM
n
nu
n
n
ysn
983.88229.1118.0
229.1112
71.45350400850
2
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24
0
1
2
3
4
5
6
7
8
9
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
Mn/bd2
= As/bd
Example Design 1 :
Calculate the lateral reinforcement requirement for beam size of
150x350mm which supported nominal moment load of 110kN-m,with fc=30 Mpa, fy=300 Mpa.
Solution:
From Rn chart we get:
98.5350.150
10.1102
6
2
bd
Mn
023.0
2
2
1256.64D20use
5.1207
350*150*023.0
mm
mmAs
bdAs
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Home Work:
2. Solve Problem 5.1
3. By using your own graph, Solve Problem 5.5.a