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The concrete beam design flow charts address the following subjects:


For a rectangular beam with given dimensions: Analyzi ng the beam
section to determine its moment strength and thus defini ng the beam
section to be at one of the following cases:


Case 1: Rectangular beam with tension rei nforcement only. This
case exists if the moment strength is larger that the ultimate
(factored) moment.


Case 2: Rectangular Beam with tension and compression
reinforcement. This case may exist if the moment strength is l ess
than the ultimate (factored) moment.


For T-section concrete beam: Analyzing the beam T-section to determi ne
its moment strength and thus defi ning the beam section to be one of the
following cases:


Case 1: The depth of the compression block is withi n the flanged
portion of the beam, i.e, the neutral axis N.A. depth is less than the
slab thi nness, measured from the top of the slab. This case exists if
moment strength is larger than ultimate moment.


Case 2: The depth of the compression block is deeper t han the
flange thickness, i.e. the neutral axis is located below the bottom of
the slab. This case exists if the moment strength of T-section beam
is less that the ultimate (factored) moment.


Beam Section Shear Strength: two separate charts outline i n det ails Shear
check. One is a basic shear check, and two is detailed shear check, i n
order to handle repetitive beam shear reinforcement selection. See shear
check i ntroduction page for further details.

In any of the cases mentioned above, detailed procedure s and equations are
shown withi n the charts cover all design aspects of the element under
investigations, with ACI respecti ve provisions.

Introduction to Concrete Beam Design Flow Charts
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Notations for Concrete Beam Design Flow Charts
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a
= depth of equi valent rectangular stress block, in.
a
b

= depth of equi valent rectangular stress block at balanced condition, i n.
a
max

= depth of equi valent rectangular stress block at maximum ratio of
tension-rei nforcement, in.
A
s

= area of tension reinforcement, i n
2
.
A’
s

= area of reinforcement at compression side, in
2
.
b
= width of beam i n rectangular beam section, in.
b
e
= effecti ve width of a flange i n T-section beam, i n.
b
w
= width of web for T-section beam, in.
c
= distance from extreme compression fiber to neutral axis, i n.
c
b

= distance from extreme compression fiber to neutral axis at balanced
condition, in.
C
c
= compression force i n equivalent concrete block.
C
s
= compression force i n compression reinforcement.
d
= distance from extreme compression fiber to centroid of tension -side
reinforcement.
d’
= distance from extreme compression fiber to centroid of compression -
side reinforcement.
E
s

= modulus of elasticity of reinforcement, psi.
f’
c

= specified compressive strength of concre te.
f
y

= specified tensile strength of reinforcement.
M
n

= nomi nal bendi ng moment.
M
n bal

= moment strength at balanced condition.
M
u

= factored (ultimate) bendi ng moment.
R
u

= coefficient of resistance.
t
= slab thickness in T-section beam, i n.
β
1

= factor as defi ned by ACI 10.2.7.3.
ε
c
= concrete strain at extreme compression fibers, set at 0.003.
ε
'
s

= strain in compression-side reinforcement.
ε
y

= yield strai n of reinforcement.
ρ = ratio of tension reinforcement.
ρ
b
= ratio of tension reinforcement at balanced condition.
ρ
f

= ratio of reinforcement equi valent to compression force i n slab of T -
section beam.
ρ
max
= maximum ratio of tension reinforcement permitted by ACI 10.3.3.
ρ
min

= minimum ratio of tension rei nforcement permitted by ACI10.5.1.
ρ
req’d
= required ratio of tension reinforcement.
φ = strength reduction factor.

ACI 9.3.2.1
ACI 8.4.3
ACI 10.3.3
YES
NO
try rectangular
beam with tension
and compression
steel
use rectangular
beam with tension
steel only
Rectangular
Beam
finding balanced
moment strength
Given:
b, d,f
c
'
,f
y
, M
u
,V
u
NO
YES
ACI 10.2.7.3
ACI 10.3.3
ACI 10.2.7.3
ACI 10.3.3
Strunet.com: Concrete Beam Design V1.01 - Page 3
0 003
29 000 000
c
y
y
s
s
.
f
E
E,,psi
ε
ε
=
=
=
finding
max
ρ
4000
c
f psi


c
b
c y
c d
ε
ε ε
 
=
 
 
+
 
1
4000
0 85 0 05 0 65
1000
c
f
...β


 
= − ≥
 
 
1
0 85.β =
1
β
1
0 85 87 000
87 000
c
b
y y
.f,
f,f
ρ β
 

=
 
 
+
 
0 75
max b
.ρ ρ=
1b b
a cβ=
0 75
max b
a.a=
0 9.φ=
bal
0 85
2
max
n c max
a
M.f ba dφ φ
 
 

= −
 
 
 
 
bal

u n
M Mφ<
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Moment Strength of Rectangular Concrete Beam
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NO
ACI 10.5.2
YES
NO
'maxreq d
ρ ρ>
YES
NO
ACI 10.5.1
STOP. go to rectangular
beam with tension and
compression steel
select
reinforcement, A
s
proceed to
shear design
d
T
C
cb
a
0.85f
c
'
b
A
s
rectangular beam
with tension steel
only
YES
2
u
u
M
R
bdφ
=
0 85 2
1 1
0 85
c u
req'd
y c
.f R
f.f
ρ
 

= − −
 
 

 
3
max of
200
c
y
min
y
f
f
f
ρ

=
req'd min
ρ ρ≥
1 33
req'd
.ρ ρ=
min
ρ ρ<
min
ρ ρ=
1 33
req'd
.ρ ρ=
req'd
useρ ρ=
s
A bdρ=
0 85
s y
c
A f
a
.f b
=

0 85
2
n c
a
M.f ba dφ φ
 
 

= −
 
 
 
 
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Rectangular Concrete Beam with
Tension Reinforcement
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d
'
=2.5"
steel at
tension side
steel at
comp. side
select
reinforcement
A
s
& A'
s
find the new d
proceed to check
compression steel
yields
T
C
c
c
a
0.85f
c
'
b
A
s
d
A'
s
rectangular beam
with tension and
compression steel
1
C
s
c
ε
s
ε

d'
u u n
bal
M M Mφ

= −
( )
( )
0 85
u
y c
M
f.f d d bd
ρ
φ


=
′ ′
− −
max
ρ ρ ρ

= +
s
A bdρ=
s
A bdρ
′ ′
=
( )
0 85
s s y
s
c s
A A f
A
a
.f b A



= +

1
a
c
β
=
s c
c d
c
ε ε


 

=
 
 
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Rectangular Beam with Tension & Compression Reinforcement
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compression
steel yields
compression
steel does NOT
yield
compression steel may be
neglected, and thus moment
strength is calculated based on
the tension steel only.
Alternatively:
proceed to
shear design
YES
NO
s y
ε ε

>
alternatively
alternatively
1
( ) ( )
( )( )
( )
2
1
1
87 87 4 0 85 87
2 0 85
s y s s y s c s
c
Af A Af A.f b Ad
c
.f b
β
β
′ ′ ′ ′ ′
− ± − +
=

0 85
c c
C.f ab

=
s s c y
c d
f E f
c
ε


 
= <
 
 
( )
0 85
2
n c s s
a
M.f ba d Af d dφ φ
 
 
′ ′ ′
= − + − 
 
 
 
 
 
( )
0 85
s s y c
C A f.f
′ ′
= −
( )
2
n c s u
a
M C d C d d Mφ φ
 
 

= − + − ≥
 
 
 
 
( )
n s y
M Af d dφ
′ ′ ′
= −
n n n u
bal
M M M Mφ φ φ

= + ≥
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Rectangular Beam with Tension & Compression Reinforcement (cont.)
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b
w
A
s
b
e
t
T-Section
Beam
finding balanced
moment strength
@a=t
Given:
b
w
,b
e
,d ,f'
c
,f
y
, M
u
V
u
ACI 9.3.2.1
NO
YES
ACI 10.2.7.3
ACI 10.3.3
ACI 10.2.7.3
YES
NO
let a=t
ACI 8.4.3
use T-Section
case 2
use T-Section
case 1
d
finding
max
ρ
4000
c
f psi′ ≤
1
0 85.β=
1
4000
0 85 0 05 0 65
1000
c
f
...β
′ −
 
= − ≥
 
 
1
β
1
0 85 87 000
87 000
c
b
y y
.f,
f,f
ρ β
 

=
 
 
+
 
( )
0 85
c e w
f
y w
.f b b t
f b d
ρ


=
( )
0 75
w
max b f
e
b
.
b
ρ ρ ρ= +
2
n c
t
M C dφ φ
 
= −
 
 
u n
M Mφ<
0 9.φ=
0 85
c c e
C.f b t′=
Moment Strength of T-Section Beam
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NO
ACI 10.5.2
YES
ACI 10.5.1
proceed to
shear design
verify depth of
compression
block
STOP. go to
T-Section case 2
STOP. use
compression steel
at T-Section
NO
YES
Alternatively:
select
reinforcement,A
s
check moment
strength
continuation of
previous sheet
YES
YES
NO
NO
T-Section
case 1
1
1
2
u
e
Mu
R
b dφ
=
0 85 2
1 1
0 85
c u
req'd
y c
.f R
f.f
ρ
 

= − −
 
 

 
req'd max
ρ ρ<
2
1 1
0 85
u
c
R
a d
.f
 
= − −
 
 

 
3
max of
200
c
y
min
y
f
f
f
ρ

=
a t>
0 85
s y
c
A f
a
.f b
=

0 85
2
n c
a
M.f ba dφ φ
 
 

= −
 
 
 
 
0 85
c e
s
y
.f ab
A
f

=
s e
A b dρ=
req'd
useρ ρ=
1 33
req'd
.ρ ρ=
min
ρ ρ=
min
ρ ρ<
1 33
req'd
.ρ ρ=
req'd min
ρ ρ≥
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T-Section Beam Case - 1
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T-Section
case 2
YES
NO
b
w
A
s
b
e
a
d
select
reinforcement,
A
s
STOP. revise to
include
compression steel
proceed to
shear design
t
( )( )
2
2 0 5
2
0 85
e w
u
c w w
t b b d.t
M
a d d
.f b bφ
− − 
= − − −
 

 
[ ]
0 85
c
s w e w
req'd
y
.f
A ab t(b b )
f

= + −
s max e
max
A b dρ=
s s
req'd max
A A≥
( )
0 85
s y
e w
w w
Af
t
a b b
.f b b
= − −

1
0 85
c c w
C.f b a

=
( )
2
0 85
c c e w
C.f t b b

= −
1 2
2 2
n c c u
a t
M C d C d Mφ
   
= − + − ≥
   
   
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T-Section Beam Case - 2
Introduction to Concrete Beam Shear Design
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Concrete Beam Shear Desi
g
n
Introduction and discussion:

The approach of the beam shear check chart is to define the nominal shear strength of the
concrete, then compare it with the ultimat e shear force at the critical section, and subsequent
sections. Shear reinforcement calculation is performed, where applicable.
The shear charts are presented into two parts. One is the Shear Basic Chart, which is outlining
the main procedures of the shear design in accordance with ACI applicabl e code provisions. The
second, Shear Detailed Chart, is outlining the steps required for repetiti ve shear check. The
detailed charts provi de as much variabl es and or scenarios as needed to facilitate the creation of
automat ed shear check applications.

The concept in selecting stirrups is based on an input of the bar diameter ( d
b
) of the stirrups to be
used, usually #3, 4, or 5, as well as the number of legs and thus finding the spacing (s) requi red.
The shear chart intentionally did not include the foll owi ng ACI provisions due to p ractical and
economical justifications:


Detail ed method of ACI §11.3.2.1 for calculating nominal shear strength of concrete, v
c

.
The reason is the value V
ud
/M
u
is not constant along the beam span. Although the stirrups
spacing resulting from the detail ed met hod may be 1.5 larger than that of the direct
method using ACI 11.3.1.1 at the critical section only, the use of the det ailed method is
not practically justified beyond this critical section, i.e. beyond distance d from the face of
support.


Shear rei nforcement as inclined stirrups per §11.5.6.3, and bent up bars per §11.5.6.4
and §11.5.6.5. Only vertical stirrups per §11.5.6.2 are used, since other types of shear
reinforcement are not economically justified.

b
w
= Width of beam (web)
d =

flexural depth of the beam, in.

f’
c
=

concrete compressi ve strength

f
ct
=

average splitting tensile strength of lightwei ght concrete

f
y
=

reinforcement yield strengt h

L =

beam clear span, from support face to other support face.

N =

number of stirrups required wit hin a gi ven segment of the beam

N
l
=

number of legs for each stirrup

V
c
=

concrete nominal shear strengt h

V
s
=

nominal shear strength provi de by the shear reinforcement

V
sb
=

nominal shear strength provi ded by shear reinforcement at the section where Vs is the
max permitted by ACI 11.12.1.1

. locating of this section is needed to define which
maximum s provisions applies, i.e. §11.5.4.1 or §11.5.4.3

V
s req’ d
=

required nominal shear strength provided by shear reinforcement.

V
u
=

factored shear force at the face of the beam support

V
u d
=

factored shear force at distance d from the face of the support in

accordance with §11.3.1.1 this is the critical shear force provi ded that:


support is subjected to compressi ve force.


no concentrated load on the beam within the distance d.

V
u req’d
=

factored shear force at the mid-span of the beam, will not be zero if the beam is
partially loaded with superimposed loads (i.e. li ve load on hal f the beam span)

φ
V
n max

=

reduced shear strength of the beam section locat ed al ong the beam span where
minimum shear reinforcement is requi red i n accordance with §15.5.5.1

s
1
=

spacing of stirrups within the critical section.

s
k
=

spacing of stirrups within any section subsequent t o the critical section.

s
max
=

maximum stirrups spacing permitted by §11.5.4.1 or §11.5.4.3

s
req’ d
=

required stirrups spacing at the section considered

x
b
=

the distance along the beam at which V
sb
occurs. for any beam section within the
distance x
b
, V
sb
is based on §11.5.4.3, otherwise is based on §11.5.4.1

x
min
=

distance from the face of the support along the beam span after which minimum shear
reinforcement in accordance to §11.5.5.1 is no longer required.

x
max
=

distance from the face of the support along the beam span after which stirrups shall be
placed with the maximum spacing per §11.5.4.1, and §11.5.4.3

s =

incremental in stirrups spacing between the subsequent sections, suggested to be 1,
2, and or 3 inches


Notations for Concrete Beam Shear Design
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Normal or Light
Wt Concrete
LIGHT
NO
ACI 11.3
ACI 11.3.1.1
ACI 11.2.1
ACI 11.2.1.2
YES
ACI 11.2.1.1
ACI 9.3.2.3
NORMAL
Finding V
c
STOP. increase
f'
c
or d or b
w
s
max
is the min. of:
d/2
24"
h=<10"
h<2.5t
h<0.5b
w
STOP. no min.
shear reinf.req'd
STOP. no min.
shear reinf.req'd
'
100
c
f psi≤
s
max
= min. of:
d/2
24"
s
max
= min. of:
d/4
12"
S
max
s is the min. of:
s
max
s
req'd
loop for other
values of V
u
loop for other
values of V
u
ACI 11.1.1
ACI 11.1.1
ACI 11.12.1.1
ACI 11.12.1.1ACI 11.1.2
ACI 11.5.4.1
ACI 11.2.1.1
ACI 11.5.4.3
ACI 11.5.6.8
ACI 11.5.6.2
ACI 11.1.1
ACI 11.5.6.2
ACI 11.5.5.1
ACI 11.5.5.3
ACI 11.5.5.1
Strunet.com: Concrete Beam Design V1.01 - Page 12

ct
is f given?
2
'
c c w
V f b d=
2
6 7
6 7
ct
c w
'
ct
c
f
V b d
.
f
f
.
 
=
 
 
 

 
 
(
)
(
)
0 75 2
0 85 2
'
c c w
'
c c w
All Light wt:V.f b d
Sand Light Wt:V.f b d
− =
=
0 85.φ=
c

u
V
u c
V Vφ>
s u c
V V Vφ φ= −
2
c
u
V
V
φ
>
req'd
s
s
V
V
φ
φ
=
4
req'd
s c
V V>
n c s
V V Vφ φ φ= +
v y
s
A f d
V
s
=
5000
50
v y
req'd
w c
A f
s
b f
 
=
 

 
50
v y
req'd
w
A f
s
b
=
req'd
v y
s
A f d
s
V
=
n c s
V V Vφ φ φ= +
2
req'd
s c
V V>
ACI 11.5.4.1
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Beam Shear Basic Chart
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Shear Beam
Check
Normal or Light
Wt Concrete
?
ct
is f given
LIGHT
NO
ACI 11.3
ACI 11.3.1.1
ACI 11.2.1
ACI 11.2.1.2
YES
ACI 11.2.1.1
ACI 9.3.2.3
NORMAL
Finding V
c
STOP.x
min
does not exist
1
2
'
c c w
V f b d=
2
6 7
6 7
ct
c w
'
ct
c
f
V b d
.
f
f
.
 
=
 
 
 

 
 
(
)
(
)
0 75 2
0 85 2
'
c c w
'
c c w
All Light wt:V.f b d
Sand Light Wt:V.f b d
− =
=
0 85.φ=
c
V
c

2
min
c
n
V
V
φ
φ =
0 0
mid
u
V.>
mid min
u n
V Vφ>
0 5 1
min
n
min
u
V
x.L
V
φ
 
= −
 
 
0 5 1
min mid
mid
n u
min
u u
V V
x.L
V V
φ 

= −
 
 

 
min
x
v l b
A NA=
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Beam Shear Detailed Chart
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STOP. increase
f'
c
or d or b
s
max
= min. of:
d/2
24"
h=<10"
h<2.5t
h<0.5b
STOP. no min.
shear reinf.req'd
STOP. no min.
shear reinf.req'd
S
max
'
100
c
f psi≤
s = min. of:
s
max
s
req'd
use:N @ s
s
max
= min. of:
d/2
24"
1
2
0 0
mid
u
V.>
( )
0 5
0 5
mid
d mid
u u
u u
V V
V.L d V
.L

 
= − +
 
 
( )
0 5
0 5
d
u
u
V
V.L d
.L
 
= −
 
 
d
u
V
d
u c
V Vφ>
s u c
V V Vφ φ= −
2
c
u
V
V
φ
>
req'd
s
s
V
V
φ
φ
=
4
req'd
s c
V V>
2
req'd
s c
V V>
2
b
s c
let V V=
b
n c s
V V Vφ φ φ= +
0 0
mid
u
V.>
0 5 1
b
n
b
u
V
x.L
V
φ
 
= −
 
 
0 5 1
b mid
mid
n u
b
u u
V V
x.L
V V
φ 

= −
 
 

 
b
x
max
s
50
v y
req'd
A f
s
b
=
5000
50
v y
req'd
c
A f
s
b f
 
=
 

 
v y
s
A f d
V
s
=
n c s
V V Vφ φ φ= +
min
x
N
s
=
STRUNET
CONCRETE DESIGN AIDS
Beam Shear Detailed Chart (cont. 1)
Strunet.com: Concrete Beam Design V1.01 - Page 15
use:N @s
max
STOP. go to
stirrups number
Loop as long:
x
min
>x
k
, and
s
k
<s
max
2
3
4
max
v y
s
max
A f d
V
s
=
max max
n c s
V V Vφ φ φ= +
d max
u s
V Vφ>
0 0
mid
u
V.>
min
x
N
s
=
0 5 1
max
n
max
u
V
x.L
V
φ
 
= −
 
 
0 5 1
max mid
mid
n u
max
u u
V V
x.L
V V
φ 

= −
 
 

 
max
x
1
req'd
v y
s
A f d
s
V
=
1
s
1

k
let s s s= +∆
k max
s s<
k
v y
s
k
A f d
V
s
=
k k
n c s
V V Vφ φ φ= +
STRUNET
CONCRETE DESIGN AIDS
Beam Shear Detailed Chart (cont. 2)
Strunet.com: Concrete Beam Design V1.01 - Page 16
s
max
= min. of:
d/4
12"
STOP. go to
stirrups number
stirrups
number
Loop until:
x
k+1
=x
min
, and
x
k
=x
max
revise S
max
3
4
0 0
mid
u
V.>
0 5 1
k mid
mid
n u
k
u u
V V
x.L
V V
φ 

= −
 
 

 
0 5 1
k
n
k
u
V
x.L
V
φ
 
= −
 
 
k
x
k b
x x>
k min
x x>
1

k k
let s s s
+
= +∆
1k k
s s
+
=
1
1
k
x
N
s
=
1k k
k
k
x x
N
s
+

=
STRUNET
CONCRETE DESIGN AIDS
Beam Shear Detailed Chart (cont. 3)