Lecture 2.b.2 Analysis and Design of RC Beams (Strength Design by (JBM and RLF)
Lecture 2.b.2 Analysis and Design of RC Beams (Strength Design by (JBM and RLF)
Lecture 2.b.2 Analysis and Design of RC Beams (Strength Design by (JBM and RLF)
Capacity of RC Beams
Lecture Notes by Engr. Jerry B. Maratas and Engr. Ricardo Fornis
Design Methods
In the United States, in the early 1900s until the early 1960s, nearly
all reinforced concrete design was performed by the working-stress
design method (also called allowable-stress design or straight-line
design).
In the ultimate strength design method (USD), the service dead and
live loads are multiplied by certain load factors (equivalent to safety
factors), and the resulting values are called factored loads.
The members are then proportioned so they will theoretically just fail
under the factored loads.
Design Methods
Even with the application of USD, it is still useful to be familiar with WSD for several
reasons:
1. Some designers use WSD for proportioning fluid-containing structures (such as water
tanks and various sanitary structures). When these structures are designed by WSD,
stresses are kept at fairly low levels, with the result that there is appreciably less
cracking and less consequent leakage
2. The ACI method for calculating the moments of inertia to be used for deflection
calculations requires some knowledge of the working-stress procedure.
3. The design of prestressed concrete members is based not only on the strength method
but also on elastic stress calculations at service load conditions.
Strength Reduction Factor in Strength Design
Method
Underreinforced and overreinforced beams
A beam with a given size (b and d) and steel reinforcement of area, As, is said to have a steel ratio,
p=As/bd. A balanced steel ratio is a design concept which considers a condition where steel is
yielding when concrete reaches a strain of 0.003 (considered as strain of concrete at failure).
In cases where the actual steel ratio is less than the balanced steel ratio, the beam is said to be
underreinforced. Underreinforced beam, when overloading occurs and the reinforced beam is
“failing”, the failure is initiated by yielding of steel which is manifested by excessive deformation of
the beam. This is an observable occurrence that serves as a kind of warning to the impending failure
of the beam.
In another situation, where the steel ratio is greater than the balanced steel ratio, the reinforced
beam is said to be overreinforced. This is a condition where the steel reinforcement has not yielded
when concrete reaches its strain at failure. It should be noted that when overloading happens and
the beam is failing, the failure is initiated by the crushing of concrete, a sudden event which can
occur without warning.
Calculation of balanced steel ratio of
Rectangular RC Beams Compressive force
C = 0.85 fc '(ab)
b 0.85fc’ 𝜖𝑐 = 0.003 Tensile force at yielding
a/2 T = As f y
c a C
c
Balanced Condition
d
d-a/2 C =T
d-c 0.85 fc ' ab = As f y
As
T=Asfy 𝜖𝑠 = fy/Es As f y As
Stress Diagram Strain Diagram a= letting b =
Beam’s section 0.85 fc ' b bd
b f y d
a= Eq(1)
0.85 fc '
Calculation of balanced steel ration of From the Strain Diagram
Rectangular RC Beams c
=
d −c
0.003 fy
0.85fc’
Es
b 𝜖𝑐 = 0.003
a/2 0.003 ( Es )( d − c )
C c=
c a c fy
using ES = 200000MPa
d
d-a/2 0.003 ( 200000 )( d − c )
d-c c=
As fy
T=Asfy 𝜖𝑠 = fy/Es
600d − 600c
Beam’s section Stress Diagram Strain Diagram c=
fy
600d
c= Eq(2)
600 + f y
Calculation of balanced steel ration of Rectangular Stress block relation
Rectangular RC Beams a = 1c
0.85fc 𝜖𝑐 = 0.003
using
b
’ a/2 b f y d 600d
a C a= and c =
c c 0.85 fc ' f y + 600
d b f y d 600d
d-a/2 = 1
As
d-c 0.85 fc ' f y + 600
T=Asfy 𝜖𝑠 = fy/Es
b =
( 0.85 fc ') 1 ( 600 )
f y ( f y + 600 )
Beam’s section Stress Diagram Strain Diagram
b =
( 0.85 fc ') 1 ( 600 )
f y ( f y + 600 )
If min b , then the design is underreinforced,
If b , then the design is Overreinforced
The choice of the value of rho, ρ
The choice of the value of
rho, ρ
That provide strain of
steel greater than or
equal to 0.005
Recommended value of rho
maximum, ρMAX for tension-
controlled design
Nominal Moment Capacity and Ultimate
Moment capacity of Rectangular RC Nominal Moment Capacity
Beams Reinforcing steel yields
a
0.85fc’ M n = As f y d −
2
a/2 b f y d
C Plug in a =
0.85 fc '
d d-a/2 1 f y d
M n = As f y d −
2 0.85 fc '
As
T=Asfy 0.59 f y
M n = As f y d 1 −
b Stress diagram, steel yields fc '
Ultimate Moment Capacity Mu
Mu = Mn
= strength reduction factor
Example
Determine the maximum permissible ultimate moment capacity of the beam shown. Use
f’c=20.7 MPa and fy=276 MPa.
1) Determination of actual steel ratio
Steel Area
(20) 2
As = ( 3)
4
As = 942.48 mm 2
Steel Ratio ( )
As 942.48
= =
bd ( 300 )( 425 )
= 0.007392
2. Compute minimum steel ratio, min
0.25 fc ' 1.4
min = or
fy fy
which ever is bigger
0.25 fc ' 0.25 20.7
= = 0.00412
fy 276
or
1.4 1.4
= = 0.00507 0.00412 The actual rho is greater than min ,
fy 276
that is, = 0.007392 min = 0.00507
min = 0.00507 governs
Okay !
3) Check the failure behavior of the beam
based on the steel ratio rho, (ρ)
3-28 mm dia
75mm
b=250 mm
Solution 1)Steel area and steel ratio
( 28 )
2
As = ( 3) = 1874.25mm2
4
As 1874.25
= = = 0.01974
bd ( 250 )( 375 )
d=375mm
Compare with min
3-28 mm dia
0.25 fc ' 1.4
75mm
min or
fy fy
b=250 mm
0.25 27.6 1.4
min = 0.003172 or = 0.003382
414 414
min =0.003382 governs
Since ( = 0.01974 ) ( min =0.003382 ) okay!
Solution
For fc’=27.6 Mpa (4000psi)and
fy=414 Mpa(60000 psi), the rho
that provides strain of 0.005 is
0.0181
c a c
Transition zone:
d t − ty
=0.65+0.25
0.005 − ty
As
d-c
𝜖𝑡 0.0043 − 0.00207
Strain Diagram =0.65+0.25
−
Beam’s section
0.005 0.00207
=0.84
Solution Nominal moment capcity, Mn
a
Mn = As fy d −
2
130.4
Mn = 1847.25(414) 375 −
2
Mn = 240386057 N.mm 240.386 kN.m
Ultimate Moment Capacity, Mu
Mu= Mn=0.84 ( 240.386 )
Mu = 201.92 kN.m
Overreinforced beams
Overreinforced beam
(steel not yielding) In an overreinforced beam,
the stress of steel is less than
b 0.85fc’ 𝜖𝑐 = 0.003
the yield strength. The stress
a/2
a C can be calculated as follows:
c c
Balanced Forces
d C =T
d-a/2
d-c 0.85 fc ' ab = As f s
As
T=Asfs As f s
𝜖𝑠 = fs/Es
a=
Stress Diagram Strain Diagram 0.85 fc ' b
Beam’s section
apply a = 1c
0.85 fc ' b1c
= f s Eq(1a)
As
Overreinforced beam Since the steel does not yield the
(steel not yielding) stress in the steel is not equal to f y .
0.85fc’ 𝜖𝑐 = 0.003
From the Strain Diagram
b
a/2 c d −c
C =
a c 0.003 fs
c
Es
d
d-a/2 0.003 ( Es )( d − c )
fs =
As d-c c
T=Asfs 𝜖𝑠 = fs/Es using ES = 200000MPa
0.003 ( 200000 )( d − c )
Stress Diagram Strain Diagram
Beam’s section
fs =
c
600d − 600c
fs = Eq(1b)
c
Eq1a = Eq1b After obtaining the value
0.85 fc ' b1c 600d − 600c of c, the value of a and fs
=
As c can be calculated using the
c 2
=
( 600d − 600c ) As following expressions
0.85 fc ' b1 a = 1c
600 As c 600d ( As ) and
c + 2
=
0.85 fc ' b1 0.85 fc ' b1 600d − 600c
fs =
600 d 600 d 2
c
c2 + c =
0.85 fc ' 1 0.85 fc ' 1 The nominal moment capacity
In the above equation, the value a
M n = As f s d −
of c can be obtained 2
Ultimate Moment; Mu = Mn
Overreinforced beam
(Illustrative Example
Determine the ultimate moment capacity
Given
of the beam shown, given that
b=300mm
fc ' = 20.7 MPa and fy = 414 MPa
Solution
d=415mm
(25)2 625
6-25 mm Ab = = 490.87mm2
bars 4 4
625 1875
As = NAb = 6 = 2945.2 mm 2
4 4
As 2945.2
= = = 0.023656
bd 300 ( 415 )
check for min
Overreinforced beam
0.25 fc ' 0.25 20.7
min = = = 0.002747
fy 414
1.4 1.4
Given or min = = = 0.003382
fy 414
b=300mm
min = 0.003382 governs
( = 0.023656 ) ( min = 0.003382 ) okay
d=415mm 0.85 fc ' 1 600
check for b =
6-25 mm fy fy + 600
bars
0.85 ( 20.7 )( 0.85 ) 600
b = = 0.021376
( 414 ) 414 + 600
Note that
( = 0.023656 ) ( b =0.021376 ) Overreinforced beam
Overreinforced beam
(Illustrative Example
b=300mm
Solve for c
600 d 600 d 2
c +
2
c =
d=415mm 0.85 fc ' 1 0.85 fc ' 1
600 ( 0.023656 )( 415 ) 600 ( 0.023656 )( 415 )
2
6-25 mm c + c=
0.85 ( 20.7 )( 0.85 )
2
bars
0.85 ( 20.7 )( 0.85 )
c 2 + ( 393.851) c = 163448.3567
c = 252.77
a = 0.85 ( 252.77 ) = 214.856mm
Overreinforced beam
(Illustrative Example
Knowing a = 214.86mm
b=300mm 0.85fc’
a
a/2 Mn = C d −
a C 2
a
d=415mm Mn = 0.85 f c(ab) d −
'
d-a/2 2
6-25 mm
214.856
bars Mn = 0.85 ( 20.7 ) (214.856) ( 300 ) 415 −
T=Asfs 2
Stress Diagram Mn = 348822756 N .mm = 348.8kN .m
Mu = Mn
Mu = 0.65 ( 348.8 ) = 226.7kN .m
Overreinforced beam
Knowing c = 252.77
c + c=
0.85 ( 20.7 )( 0.85 )
2
wL 116.8 ( 6 )
2 2
0.25 fc ' 1.4
Mu = = min or
8 8 fy fy
Mu = 525.6 kN .m (At midspan) min
0.25 27.6
= 0.004759
276
3) Calculate the value of rho,
1.4
fc ' or min = = 0.005072
= 0.18 276
fy min =0.005072 governs
27.6 Since ( = 0.018 ) ( min =0.005072 ) okay!
= 0.18 = 0.018
276
In Table A.7, the value of rho for fc’=27.6MPa (4000 psi) and
fy=276MPa, that produces strain of εt=0.005 in the steel is
ρTC=0.0271.
The required rho is 0.018<0.0271, tension-controlled condition
0.59 f y
M n = f y bd 1 −
2
4) Nominal Moment Capacity fc '
0.59 f y 2
0.59 f y
M n = As f y d 1 − Mu = Mn = f y bd 1 −
fc ' fc '
where As = bd where = 0.90(Tension controlled condition)
0.59 f y 2 0.59(0.18)(276)
M n = f y bd 1 −
2
525.6 x10 6
= 0.90(0.018)276bd 1 −
fc ' 27.6
0.59 f y bd 2
= 131519730.3 mm 3
2
Mu = Mn = f y bd 1 −
fc ' 5)Using b = 350mm
( 350 ) d 2 = 131519730.3
d = 613mm say d = 625mm
6) Calculate A s 7) Number of bars
using As = bd using 25 mm bars
As = 0.018 ( 350 )( 625 ) ( 25 )
2
A1 = = 490.87mm 2
As = 3937.5mm 2 4
0.59 f y As 3937.5
Mu = Mn = f y bd 1 − N= = = 8.02 say 8.0 pcs
2
fc ' A1 490.87
8) Clear spacing of bars
b − 50(2) − 4(db )
S=
n −1
350 − 50(2) − 4(25)
S= = 50mm 25mm
4 −1
Design of RC Beams (Proportioning Beams for Flexure)
fc '
0.59 ( 0.037115 )( 276 )
6
(
525 x10 = 0.75 ( 0.037115 )( 276 ) bd 2
) 1 −
27.6
bd 2 = 87493643.8mm3
using d = 1.5b
b (1.5b ) = 87493643.8
2
350mm 350mm
fy
M n = bd f y 1 −
2 ( 0.85 fc ')
2
Mn Mu fy fy =
= = f y 1 −
bd 2 ( 0.85 fc ')
bd 2 2 2
fy = ( 0.85 fc ') − ( 0.85 fc ') − 2 ( 0.85 fc ') Rm
2
Mu fy
= Rm = f y 1 −
2 ( 0.85 fc ')
Let
bd 2
2 Rm
fy = ( 0.85 fc ') − 0.85 fc ' 1 −
( 0.85 fc ')
0.85 fc ' 2 Rm
= 1− 1−
f y
( 0.85 fc ')
Illustrative example
The size of the beam shown has been selected for architectural considerations. The factored moment of
200 kN.m is the estimated moment for the design. Using fc’=20.7MPa, and fy=414 MPa design the steel
reinforcement.
1) Compute Ru
Mu 200 x106
Ru = =
bd 0.9 ( 350 )( 525 )
2 2
Ru = 2.3036 MPa
2) Compute rho
0.85 fc ' 2 Ru
= 1 − 1 −
fy 0.85 fc
0.85(20.7) 2(2.3036)
= 1 − 1 − = 0.005986
(414) 0.85(20.7)
4) Compute the rho balanced, b
b =
( 0.85 fc ') 1 ( 600 ) ( 0.85 )( 20.7 )( 0.85 )( 600 )
= = 0.0213757
f y ( f y + 600 ) 414 ( 414 + 600 )
Since
min = 0.003382 = 0.005986 b = 0.0213757
Underreinforced beam
In Table A.7, the value of rho for fc’=20.7MPa (3000 psi) and fy=414
MPa, that produces strain of εt=0.005 in the steel is ρTC=0.0136.
1) Factored Moment
Mu = 1.2 ( 24 ) + 1.6 ( 32 ) = 80 kN .m
2) Compute R m
Mn Mu
Rm = =
bd 2 bd 2
Rm =
(
80 1x106 ) = 3.168
( 0.9 )( 250 )( 335)
2
3) Compute
0.85 fc ' 2 Rm
= 1− 1−
f y
( 0.85 fc ')
0.85 ( 34.5 ) 2 ( 3.168 )
= 1 − 1 − = 0.008117
414 ( 0.85)( 34.5)