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DEPARTMENT OF CIVIL ENGINEERING
CE-2105
DESIGN OF CONCRETE STRUCTURE-I
DESIGN OF BEAM
(AS PER ACI CODE)
Course Teacher- Presented By-
Dr. Md. Rezaul Karim Md. Jahidur Rahman
Associate Professor S.ID- 121041
Dhaka University of Engineering &
Technology, Gazipur-17001
Content
2
 USD(Ultimate Strength Design)
 Classification of beam with respect to design
system
 Assumptions
 Evolution of design parameters
 Moment Factors Kn, 
 Balanced Reinforcement Ratio b
 Calculating Strength Reduction Factor 
 Calculating
 Design procedure for Singly Reinforced Beam
 Design procedure for Doubly Reinforced Beam
 Design procedure for T-Beam
 Appendix
Jahidur Rahman
Ultimate Strength Design(USD)
3
Assuming tensile failure condition
Additional strength of steel after yielding
ACI code emphasizes this method
Jahidur Rahman
Classification of beam with respect to design
system
Rectangular beam (reinforced at tension zone
only)
Doubly reinforced beam (reinforced at both
tension and compression zone)
T – section beam (both beam and slab are
designed together)
ASSUMPTIONS
5
 Plane sections before bending remain plane and
perpendicular to the N.A. after bending
 Strain distribution is linear both in concrete & steel and is
directly proportional to the distance from N.A.
 Strain in the steel & surrounding concrete is the same prior to
cracking of concrete or yielding of steel
 Concrete in the tension zone is neglected in the flexural
analysis & design computation
εc=0.003
εs = fy / Es
h d
c
0.85fc’
a a/2
d-a/2
b
C
T
Jahidur Rahman
6
 Concrete stress of 0.85fc’ is uniformly distributed over an
equivalent compressive zone.
fc’ = Specified compressive strength of concrete in psi.
 Maximum allowable strain of 0.003 is adopted as safe limiting
value in concrete.
 The tensile strain for the balanced section is fy/Es
 Moment redistribution is limited to tensile strain of at least
0.0075
εs
εy
fy
fs
Idealized
Actual
Es
1
Jahidur Rahman
EVALUATION OF DESIGN PARAMETERS
 Total compressive force - C = 0.85fc’ ba (Refer stress diagram)
 Total Tensile force- T = As fy
C = T
0.85fc’ ba = As fy
a = As fy / (0.85fc’ b)
= d fy / (0.85 fc’) [  = As / bd]
 Moment of Resistance/Nominal Moment-
Mn = 0.85fc’ ba (d – a/2) or,
Mn = As fy (d – a/2)
=  bd fy [ d – (dfyb / 1.7fc’) ]
=  fc’ [ 1 – 0.59 ] bd2
  =  fy / fc’
Mn = Kn bd2 Kn =  fc’ [ 1 – 0.59 ]
Ultimate Moment- Mu =  Mn
=  Kn bd2
( = Strength Reduction Factor)
7 Jahidur Rahman
 Balaced Reinforcement Ratio ( b)
From strain diagram, similar triangles
cb / d = 0.003 / (0.003 + fy / Es); Es = 29x106 psi
cb / d = 87,000 / (87,000+fy)
b = Asb / bd
= 0.85fc’ ab / (fy. d)
= β1 ( 0.85 fc’ / fy) [ 87,000 / (87,000+fy)]
Relationship b / n the depth `a’ of the equivalent rectangular stress block
& depth `c’ of the N.A. is
a = β1c
β1= 0.85 ; fc’ 4000 psi
β1= 0.85 - 0.05(fc’ – 4000) / 1000 ; 4000 < fc’ 8000
β1= 0.65 ; fc’> 8000 psi
8 Jahidur Rahman
 For beams the ACI code limits the max. amount of steel to 75% of
that required for balanced section.
  0.75  b
 Min. reinforcement is greater of the following:
Asmin = 3fc’ x bwd / fy or 200 bwd / fy
min = 3fc’ / fy or 200 / fy
 For statically determinate member, when the flange is in tension,
the bw is replaced with 2bw or bf whichever is smaller
 The above min steel requirement need not be applied, if at every
section, Ast provided is at least 1/3 greater than the analysis
9 Jahidur Rahman
Yes
N
o
Yes
N
o
Calculating
Calculating
Jahidur Rahman12
SINGLY REINFORCED BEAM
Beam is reinforced near the tensile face
Reinforcement resists the tension.
Concrete resists the compression.
DESIGN PROCEDURE FOR
SINGLY REINFORCED BEAM
13
1. Determine the
service loads
3. Calculate d= h – Effective cover
2. Assume `h` as per
the support conditions
[As per ACI code in
table 9.5(a)]
4. Assume the value of `b` by the
rule of thumb
5. Estimate self weight
6. Primary elastic analysis
and derive B.M (M), Shear
force (V) values
7. Compute min and b 8.Choose  between min and b
Note Below:
 Design the steel reinforcement arrangement with appropriate cover and
spacing stipulated in code. Bar size and corresponding no. of bars based on
the bar size #n.
 Check crack width as per codal provisions.
14 Jahidur Rahman
9. Calculate , Kn
10. From Kn & M
calculate `d’ required
11. Check the required `d’ with
assumed `d’
12. With the final values of , b, d
determine the Total As required
OK
DESIGN PROCEDURE FOR
SINGLY REINFORCED BEAM BY FLOWCHART
15
Actual steel ratio Maximum steel ratio
Over
reinforced
beam
Yes Under
reinforced
beam
N
o
Ultimate moment
Jahidur Rahman16
DOUBLY REINFORCED BEAM
Beam is fixed for Architectural purposes.
Reinforcement are provided both in tension and
compression zone.
Concrete has limitation to resist the total compression
so extra reinforcement is required.
Actual steel ratio Maximum steel ratio
rectangular
beam
Yes Doubly
reinforced
beam
N
o
DESIGN
PROCEDURE
FOR
DOUBLY
REINFORCED
BEAM
Concrete
section,
Area of
steel are
known
YesN
o
Compression
and tension bar
reach in yielding
Compression bar
does not reach in
yielding
DESIGN
PROCEDURE
FOR
DOUBLY
REINFORCED
BEAM
Load or
ultimate
moment is
given
Maximum steel ratio
Reinforcement
required in
comp. zone
too
rectangular
beam
YesDoubly
reinforced
beam
N
o
Jahidur Rahman19
T- REINFORCED BEAM
A part of slab acts as the upper part of beam.
Resulting cross section is T – shaped.
The slab portion of the beam is flange.
The beam projecting bellow is web or stem.
DESIGN
PROCEDURE
FOR
T-REINFORCED
BEAM
Are of concrete
section, Area of
steel are known
Additional moment
Nominal moment
calculation
Rectangula
r beam
analysis
Yes
T -
beam
analysis
N
o
modification
Total moment
DESIGN
PROCEDURE
FOR
T-
REINFORCED
BEAM
Moment or loading is
given
Total area of steel
Nominal moment
calculation
Rectangula
r beam
analysis
YesT -
beam
analysis
N
o
Area of steel in flange
Area of steel in web
Additional moment
T section:
L section:
Isolated beam:
Check for
Jahidur Rahman22
APPENDIX
AS PER TABLE 9.5 (a)
Simply
Supported
One End
Continuous
Both End
Continuous
Cantilever
L / 16 L / 18.5 L / 21 L/8
Values given shall be used directly for members with normal
weight concrete (Wc = 145 lb/ft3) and Grade 60 reinforcement
 For structural light weight concrete having unit wt. In range
90-120 lb/ft3 the values shall be multiplied by
(1.65 – 0.005Wc) but not less than 1.09
 For fy other than 60,000 psi the values shall be multiplied by
(0.4 + fy/100,000)
 `h` should be rounded to the nearest whole number
23 Jahidur Rahman
RULE OF THUMB
 d/b = 1.5 to 2.0 for beam spans of 15 to 25 ft.
 d/b = 3.0 to 4.0 for beam spans > 25 ft.
 `b` is taken as an even number
 Larger the d/b, the more efficient is the section due to less
deflection
CLEAR COVER
 Not less than 1.5 in. when there is no exposure to weather or
contact with the ground
 For exposure to aggressive weather 2 in.
 Clear distance between parallel bars in a layer must not be
less than the bar diameter or 1 in.
24 Jahidur Rahman
BAR SIZE
 #n = n/8 in. diameter for n 8.
Ex. #1 = 1/8 in.
….
#8 = 8/8 i.e., I in.
Weight, Area and Perimeter of individual bars
inch mm
3 0.376 0.375 9 0.11 1.178
4 0.668 0.500 13 0.20 1.571
5 1.043 0.625 16 0.31 1.963
6 1.502 0.750 19 0.44 2.356
7 2.044 0.875 22 0.60 2.749
8 2.670 1.000 25 0.79 3.142
9 3.400 1.128 28 1.00 3.544
10 4.303 1.270 31 1.27 3.990
11 5.313 1.410 33 1.56 4.430
14 7.650 1.693 43 2.25 5.319
18 13.600 2.257 56 4.00 7.091
Perimeter
(in.)
Stamdard Nominal Dimensions
Bar
No
Wt.per
Foot (lb)
Diameter db C/S Area,
Ab (in2
)
25 Jahidur Rahman
26 Jahidur Rahman

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Design Procedure of Singly,Doubly & T-Beam(As Per ACI code)

  • 1. DEPARTMENT OF CIVIL ENGINEERING CE-2105 DESIGN OF CONCRETE STRUCTURE-I DESIGN OF BEAM (AS PER ACI CODE) Course Teacher- Presented By- Dr. Md. Rezaul Karim Md. Jahidur Rahman Associate Professor S.ID- 121041 Dhaka University of Engineering & Technology, Gazipur-17001
  • 2. Content 2  USD(Ultimate Strength Design)  Classification of beam with respect to design system  Assumptions  Evolution of design parameters  Moment Factors Kn,   Balanced Reinforcement Ratio b  Calculating Strength Reduction Factor   Calculating  Design procedure for Singly Reinforced Beam  Design procedure for Doubly Reinforced Beam  Design procedure for T-Beam  Appendix Jahidur Rahman
  • 3. Ultimate Strength Design(USD) 3 Assuming tensile failure condition Additional strength of steel after yielding ACI code emphasizes this method Jahidur Rahman
  • 4. Classification of beam with respect to design system Rectangular beam (reinforced at tension zone only) Doubly reinforced beam (reinforced at both tension and compression zone) T – section beam (both beam and slab are designed together)
  • 5. ASSUMPTIONS 5  Plane sections before bending remain plane and perpendicular to the N.A. after bending  Strain distribution is linear both in concrete & steel and is directly proportional to the distance from N.A.  Strain in the steel & surrounding concrete is the same prior to cracking of concrete or yielding of steel  Concrete in the tension zone is neglected in the flexural analysis & design computation εc=0.003 εs = fy / Es h d c 0.85fc’ a a/2 d-a/2 b C T Jahidur Rahman
  • 6. 6  Concrete stress of 0.85fc’ is uniformly distributed over an equivalent compressive zone. fc’ = Specified compressive strength of concrete in psi.  Maximum allowable strain of 0.003 is adopted as safe limiting value in concrete.  The tensile strain for the balanced section is fy/Es  Moment redistribution is limited to tensile strain of at least 0.0075 εs εy fy fs Idealized Actual Es 1 Jahidur Rahman
  • 7. EVALUATION OF DESIGN PARAMETERS  Total compressive force - C = 0.85fc’ ba (Refer stress diagram)  Total Tensile force- T = As fy C = T 0.85fc’ ba = As fy a = As fy / (0.85fc’ b) = d fy / (0.85 fc’) [  = As / bd]  Moment of Resistance/Nominal Moment- Mn = 0.85fc’ ba (d – a/2) or, Mn = As fy (d – a/2) =  bd fy [ d – (dfyb / 1.7fc’) ] =  fc’ [ 1 – 0.59 ] bd2   =  fy / fc’ Mn = Kn bd2 Kn =  fc’ [ 1 – 0.59 ] Ultimate Moment- Mu =  Mn =  Kn bd2 ( = Strength Reduction Factor) 7 Jahidur Rahman
  • 8.  Balaced Reinforcement Ratio ( b) From strain diagram, similar triangles cb / d = 0.003 / (0.003 + fy / Es); Es = 29x106 psi cb / d = 87,000 / (87,000+fy) b = Asb / bd = 0.85fc’ ab / (fy. d) = β1 ( 0.85 fc’ / fy) [ 87,000 / (87,000+fy)] Relationship b / n the depth `a’ of the equivalent rectangular stress block & depth `c’ of the N.A. is a = β1c β1= 0.85 ; fc’ 4000 psi β1= 0.85 - 0.05(fc’ – 4000) / 1000 ; 4000 < fc’ 8000 β1= 0.65 ; fc’> 8000 psi 8 Jahidur Rahman
  • 9.  For beams the ACI code limits the max. amount of steel to 75% of that required for balanced section.   0.75  b  Min. reinforcement is greater of the following: Asmin = 3fc’ x bwd / fy or 200 bwd / fy min = 3fc’ / fy or 200 / fy  For statically determinate member, when the flange is in tension, the bw is replaced with 2bw or bf whichever is smaller  The above min steel requirement need not be applied, if at every section, Ast provided is at least 1/3 greater than the analysis 9 Jahidur Rahman
  • 12. Jahidur Rahman12 SINGLY REINFORCED BEAM Beam is reinforced near the tensile face Reinforcement resists the tension. Concrete resists the compression.
  • 13. DESIGN PROCEDURE FOR SINGLY REINFORCED BEAM 13 1. Determine the service loads 3. Calculate d= h – Effective cover 2. Assume `h` as per the support conditions [As per ACI code in table 9.5(a)] 4. Assume the value of `b` by the rule of thumb 5. Estimate self weight 6. Primary elastic analysis and derive B.M (M), Shear force (V) values 7. Compute min and b 8.Choose  between min and b
  • 14. Note Below:  Design the steel reinforcement arrangement with appropriate cover and spacing stipulated in code. Bar size and corresponding no. of bars based on the bar size #n.  Check crack width as per codal provisions. 14 Jahidur Rahman 9. Calculate , Kn 10. From Kn & M calculate `d’ required 11. Check the required `d’ with assumed `d’ 12. With the final values of , b, d determine the Total As required OK
  • 15. DESIGN PROCEDURE FOR SINGLY REINFORCED BEAM BY FLOWCHART 15 Actual steel ratio Maximum steel ratio Over reinforced beam Yes Under reinforced beam N o Ultimate moment
  • 16. Jahidur Rahman16 DOUBLY REINFORCED BEAM Beam is fixed for Architectural purposes. Reinforcement are provided both in tension and compression zone. Concrete has limitation to resist the total compression so extra reinforcement is required.
  • 17. Actual steel ratio Maximum steel ratio rectangular beam Yes Doubly reinforced beam N o DESIGN PROCEDURE FOR DOUBLY REINFORCED BEAM Concrete section, Area of steel are known YesN o Compression and tension bar reach in yielding Compression bar does not reach in yielding
  • 18. DESIGN PROCEDURE FOR DOUBLY REINFORCED BEAM Load or ultimate moment is given Maximum steel ratio Reinforcement required in comp. zone too rectangular beam YesDoubly reinforced beam N o
  • 19. Jahidur Rahman19 T- REINFORCED BEAM A part of slab acts as the upper part of beam. Resulting cross section is T – shaped. The slab portion of the beam is flange. The beam projecting bellow is web or stem.
  • 20. DESIGN PROCEDURE FOR T-REINFORCED BEAM Are of concrete section, Area of steel are known Additional moment Nominal moment calculation Rectangula r beam analysis Yes T - beam analysis N o modification Total moment
  • 21. DESIGN PROCEDURE FOR T- REINFORCED BEAM Moment or loading is given Total area of steel Nominal moment calculation Rectangula r beam analysis YesT - beam analysis N o Area of steel in flange Area of steel in web Additional moment T section: L section: Isolated beam: Check for
  • 23. AS PER TABLE 9.5 (a) Simply Supported One End Continuous Both End Continuous Cantilever L / 16 L / 18.5 L / 21 L/8 Values given shall be used directly for members with normal weight concrete (Wc = 145 lb/ft3) and Grade 60 reinforcement  For structural light weight concrete having unit wt. In range 90-120 lb/ft3 the values shall be multiplied by (1.65 – 0.005Wc) but not less than 1.09  For fy other than 60,000 psi the values shall be multiplied by (0.4 + fy/100,000)  `h` should be rounded to the nearest whole number 23 Jahidur Rahman
  • 24. RULE OF THUMB  d/b = 1.5 to 2.0 for beam spans of 15 to 25 ft.  d/b = 3.0 to 4.0 for beam spans > 25 ft.  `b` is taken as an even number  Larger the d/b, the more efficient is the section due to less deflection CLEAR COVER  Not less than 1.5 in. when there is no exposure to weather or contact with the ground  For exposure to aggressive weather 2 in.  Clear distance between parallel bars in a layer must not be less than the bar diameter or 1 in. 24 Jahidur Rahman
  • 25. BAR SIZE  #n = n/8 in. diameter for n 8. Ex. #1 = 1/8 in. …. #8 = 8/8 i.e., I in. Weight, Area and Perimeter of individual bars inch mm 3 0.376 0.375 9 0.11 1.178 4 0.668 0.500 13 0.20 1.571 5 1.043 0.625 16 0.31 1.963 6 1.502 0.750 19 0.44 2.356 7 2.044 0.875 22 0.60 2.749 8 2.670 1.000 25 0.79 3.142 9 3.400 1.128 28 1.00 3.544 10 4.303 1.270 31 1.27 3.990 11 5.313 1.410 33 1.56 4.430 14 7.650 1.693 43 2.25 5.319 18 13.600 2.257 56 4.00 7.091 Perimeter (in.) Stamdard Nominal Dimensions Bar No Wt.per Foot (lb) Diameter db C/S Area, Ab (in2 ) 25 Jahidur Rahman

Hinweis der Redaktion

  1. Emphasizes= গুরুত্ব আরোপ করা।
  2. Adopted= গৃহীত