3. Design Assumption and Beam Analysis

REINFORCED CONCRETE
STRUCTURE I
“Design Assumption”

Revised : 17-September-2013
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CIVIL ENGINEERING DEPARTMENT
FACULTY OF CIVIL ENGINEERING AND PLANNING
INSTITUT TEKNOLOGI SEPULUH NOPEMBER
SURABAYA
Prepared By : LB3
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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Design Assumptions
4. Maximum usable strain at extreme concrete compression fiber
shall be assumed equal to cu = 0.003.

5. Stress in reinforcement fs below the yield strength fy shall be
taken as Es times the steel strain cs . 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.
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 . 0 05 . 0
7
30 '
85 . 0 1 > ×
|
.
|

\
|
÷
÷ =
c f
|
SNI-03-2847-2002, pasal 12.2.7.3
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 other
shape that results in prediction of strength in substantial agreement with results
of comprehensive tests.
Force Equilibrium
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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
Prepared By : LB3
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
Balanced Strain Condition
Strain linear relationship :




From force equilibrium :
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y u
u b
d
c
c c
c
+
=
fy fy
y u
u
+
=
+
=
+ 600
600
200000 / 003 . 0
003 . 0
c c
c
( )
y b b c
y sb b c
b b
bdf c b f
f A ba f
T C
µ | =
=
=
1
'
'
85 . 0
85 . 0
y y
c
b
b
y
c
b
f f
f
d
c
f
f
+
=
=
600
600 85 . 0
85 . 0
'
1
'
1
|
µ
|
µ
Balanced Strain Condition
To ensure the structure is under reinforced condition :


Minimum reinforcement for flexural member :
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y y
c
b
f f
f
+
= =
600
600 85 . 0
75 . 0 75 . 0
'
1
max
|
µ µ
d b
f
d b
f
f
As
w
y
w
y
c
4 . 1
3
'
min
> =
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 bd
2
:
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' '
'
85 . 0 85 . 0
85 . 0
c
y
c
y s
y y s c
f
df
b f
f A
a
bdf f A ba f
T C
µ
µ
= =
= =
=
( )
(
¸
(

¸

÷ =
|
.
|

\
|
÷ =
'
85 . 0
5 . 0
2
c
y
y n
n
f
f
d
d bdf M
a
d T or C M
µ
µ
|
|
.
|


\
|
÷ = =
' 2
85 . 0
5 . 0
1
c
y
y
n
n
f
f
f
bd
M
R
µ
µ
Design of Reinforced Concrete Beam
When b and d are preset (determined), µ is obtained by solving the quadratic
equation for Rn :





Equation above can be used to determine the steel ratio µ 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 f’c :
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|
|
.
|


\
|
÷ ÷ =
= + ÷
|
|
.
|


\
|
'
'
1
2
'
2
85 . 0
2
1 1
85 . 0
0
85 . 0
5 . 0
c
n
y
c
c
y
f
R
f
f
Rn fy
f
f
µ
µ µ
( ) e e
|
µ
e
µ µ
|
59 . 0 1
;
85 . 0
5 . 0
1
2
'
' ' ' 2 '
÷ =
=
|
|
.
|


\
|
÷ = =
bd f
M
f
f
f
f
f
f
bd f
M
R
c
u
c
y
c
y
c
y
c
u
n
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 f’c and fy.
2. Use the constraint of fy in one
graphics for example fy : 240
MPa, 300 MPa, 320 MPa, 350
MPa, 400 MPa.
3. Use several f’c in one graphics
for example f’c : 20 MPa, 25
MPa, 30 MPa, 35 MPa, 40 MPa,
45 Mpa, 50 MPa.
Example Design [1]
Example Design :
Design the concrete section of the beam, which is simple supported and are
loaded as below : [f’c = 35 MPa, fy = 400 MPa].
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L=6m
Ql=1.5t/m;Qd=1t/m
kNm M M M
kNm tm l q M
kNm tm l q M
l d u
l l
d d
76 . 158 15 . 66 6 . 1 1 . 44 2 . 1 6 . 1 2 . 1
15 . 66 75 . 6 6 5 . 1
8
1
8
1
1 . 44 5 . 4 6 1
8
1
8
1
2 2
2 2
= × + × = + =
= = × × = =
= = × × = =
Example Design [1]
#1 Determine maximum reinforcement ration (µmax) for material strength f’c =
35 MPa and fy = 400 MPa.
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65 . 0 814 . 0 1
65 . 0 05 . 0
7
30 35
85 . 0 1
65 . 0 05 . 0
7
30 '
85 . 0 1
> =
> ×
|
.
|

\
|
÷
÷ =
> ×
|
.
|

\
|
÷
÷ =
|
|
|
c f
Calculating |1
Calculating µmax :
0272 . 0
400 600
600
400
35 814 . 0 85 . 0
75 . 0
600
600 85 . 0
75 . 0 75 . 0
max
max
'
1
max
=
(
¸
(

¸

+
×
× ×
=
(
¸
(

¸

+
× = =
µ
µ
|
µ µ
fy fy
f
c
bal
Example Design [1]
#2 Compute bd
2
required :






#3 Determine Size member so that bd
2
> bd
2
required :



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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3 2
' 2 2
23008696
625 . 8 8 . 0
1000000 76 . 158
625 . 8
35 85 . 0
400 0272 . 0 5 . 0
1 400 0272 . 0
85 . 0
5 . 0
1
mm
R
M
bd
MPa R
f
f
f
bd
M
bd
M
R
n
u
n
c
y
y
u n
n
=
×
×
= =
=
|
.
|

\
|
×
× ×
÷ × =
|
|
.
|


\
|
÷ = = =
|
µ
µ
|
mm d
mm b
37 . 303
250
23008696
250
= =
=
Example Design [1]
#4 Using the 400 mm beam depth (h), compute a revised value of µ :
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MPa
bd
M
R
bd
M
R
u
n
u
n
786 . 6
342 250 8 . 0
1000000 76 . 158
2 2
2
=
× ×
×
= =
=
|
|
0035 . 0 0196 . 0 0272 . 0
400
4 . 1 4 . 1
35 85 . 0
786 . 6 2
1 1
400
35 85 . 0
0272 . 0
4 . 1
85 . 0
2
1 1
85 . 0
min max
min max
min
'
'
max
= > = > =
= = >
|
|
.
|


\
|
×
×
÷ ÷
×
= > =
= >
|
|
.
|


\
|
÷ ÷ = >
µ µ µ
µ µ µ
µ µ µ
fy
fy f
R
f
f
c
n
y
c
Example Design [1]
#5 Compute As Required :
As = µ x b x d
As = 0.0196 x 250 x 342
As = 1675.80 mm
2
Use 6 D19 (As=1701 mm
2
)

#6 CrossCheck The Moment Nominal with Moment Ultimate :
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kNm M kNm M
kNm
a
d f A M
n u
y s n
26 . 161 76 . 158
26 . 161
2
48 . 91
342 400 1701 8 . 0
2
= s =
=
|
.
|

\
|
÷ × × × =
|
.
|

\
|
÷ =
|
| |
( )
( )
mm
b f
f A
a
c
y s
48 . 91
250 35 85 . 0
400 1701
85 . 0
'
= = =
Example Analysis [2]
Example Analysis :
Analyze the Moment Nominal Capacity (Mn) of the beam below : [f’c = 35
MPa, fy = 400 MPa].
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d=350mm h=400mm
b=250mm
Cover+dh+db/2=50mm
3D19
Example Analysis [2]
Example Analysis :
Analyze the Moment Nominal Capacity (Mn) of the beam below : [f’c = 35
MPa, fy = 400 MPa].
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T
C
d
a
d-a/2
2 2
850 19 25 . 0 3 mm As = × × × = t
mm
b f
f A
a
c
y s
71 . 45
250 35 85 . 0
400 850
85 . 0
'
=
× ×
×
= =
( ) kNm M
M M
kNm M
M
a
d f A M
n
n u
n
n
y s n
983 . 88 229 . 111 8 . 0
229 . 111
2
71 . 45
350 400 850
2
= =
=
=
|
.
|

\
|
÷ × × =
|
.
|

\
|
÷ =
|
|
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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
M
n
/
b
d
2

µ = 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 . 5
350 . 150
10 . 110
2
6
2
= =
bd
Mn
023 . 0 = µ
2
2
1256.6 4D20 use
5 . 1207
350 * 150 * 023 . 0
mm
mm As
bd As
=
=
= = µ
Home Work:
2. Solve Problem 5.1
3. By using your own graph, Solve Problem 5.5.a