DESIGN AID J.1-1
Areas of Reinforcing Bars
Total Areas of Bars (in.2)
Bar
Size
1
0.11
0.20
0.31
0.44
0.60
0.79
1.00
1.27
1.56
No. 3
No. 4
No. 5
No. 6
No. 7
No. 8
No. 9
No. 10
No. 11
2
0.22
0.40
0.62
0.88
1.20
1.58
2.00
2.54
3.12
3
0.33
0.60
0.93
1.32
1.80
2.37
3.00
3.81
4.68
4
0.44
0.80
1.24
1.76
2.40
3.16
4.00
5.08
6.24
Number of Bars
5
6
7
8
9
10
0.55 0.66 0.77 0.88 0.99 1.10
1.00 1.20 1.40 1.60 1.80 2.00
1.55 1.86 2.17 2.48 2.79 3.10
2.20 2.64 3.08 3.52 3.96 4.40
3.00 3.60 4.20 4.80 5.40 6.00
3.95 4.74 5.53 6.32 7.11 7.90
5.00 6.00 7.00 8.00 9.00 10.00
6.35 7.62 8.89 10.16 11.43 12.70
7.80 9.36 10.92 12.48 14.04 15.60
Areas of Bars per Foot Width of Slab (in.2/ft)
Bar
Size
Bar Spacing (in.)
6
7
8
9
10
11
12
13
14
15
16
17
18
No. 3
0.22
0.19
0.17
0.15
0.13
0.12
0.11
0.10
0.09
0.09
0.08
0.08
0.07
No. 4
0.40
0.34
0.30
0.27
0.24
0.22
0.20
0.18
0.17
0.16
0.15
0.14
0.13
No. 5
0.62
0.53
0.46
0.41
0.37
0.34
0.31
0.29
0.27
0.25
0.23
0.22
0.21
No. 6
0.88
0.75
0.66
0.59
0.53
0.48
0.44
0.41
0.38
0.35
0.33
0.31
0.29
No. 7
1.20
1.03
0.90
0.80
0.72
0.65
0.60
0.55
0.51
0.48
0.45
0.42
0.40
No. 8
1.58
1.35
1.18
1.05
0.95
0.86
0.79
0.73
0.68
0.63
0.59
0.56
0.53
No. 9
2.00
1.71
1.50
1.33
1.20
1.09
1.00
0.92
0.86
0.80
0.75
0.71
0.67
No. 10
2.54
2.18
1.91
1.69
1.52
1.39
1.27
1.17
1.09
1.02
0.95
0.90
0.85
No. 11
3.12
2.67
2.34
2.08
1.87
1.70
1.56
1.44
1.34
1.25
1.17
1.10
1.04
DESIGN AID J.1-2
$SSUR[LPDWH%HQGLQJ0RPHQWVDQG6KHDU)RUFHVIRU&RQWLQXRXV%HDPV
DQG2QHZD\6ODEV
Uniformly distributed load wu (L/D d3)
Two or more spans
Prismatic members
,QWHJUDOZLWK
6XSSRUW
"n"nd"n
"n
"n
wu" n
wu" n
wu" n
wu" navg
wu" n
Spandrel
Support
6LPSOH
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wu" n
wu" n
wu" navg
Positive
Moment
wu" n
Column
Support
wu " navg
wu" n
Note A
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w " u n wu " navg
wu" n
wu " n
wu" n
Negative
Moment
wu" n
wu" n
wu" n
wu" n
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wu " n
wu" n
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DESIGN AID J.1-3
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DESIGN AID J.1-4
Simplified Calculation of As Assuming Tension-Controlled Section and
Grade 60 Reinforcement
f c′ (psi)
As (in.2)
3,000
Mu
3.84d
4,000
Mu
4.00d
5,000
Mu
4.10d
M u is in ft-kips and d is in inches
In all cases, As =
Mu
can be used.
4d
Notes:
Mu
0.5 ρf y 

 × d
φf y 1 −
 0.85 f ' c 
•
As =
•
•
For all values of ρ < 0.0125, the simplified As equation is slightly conservative.
It is recommended to avoid ρ > 0.0125 when using the simplified As equation.
DESIGN AID J.1-51
0LQLPXP1XPEHURI5HLQIRUFLQJ%DUV5HTXLUHGLQD6LQJOH/D\HU
Assumptions:
x *UDGHUHLQIRUFHPHQW f y SVL x &OHDUFRYHUWRWKHWHQVLRQUHLQIRUFHPHQW cc LQ
x &DOFXODWHGVWUHVV f s LQUHLQIRUFHPHQWFORVHVWWRWKHWHQVLRQIDFHDW
VHUYLFHORDG SVL
Beam Width (in.)
Bar
Size
1R
1R
1R
1R
1R
1R
1R
1R
Minimum number of bars, nmim:
bw 2(cc 0.5db )
1
s
nmin
GV
where
§ 40,000 ·
¸¸ 2.5cc
s 15¨¨
© fs ¹
§ 40,000 ·
¸¸
d 12¨¨
f
s ¹
©
1
db
FF
sFOHDU
U ôsIRU1RVWLUUXSV
sIRU1RVWLUUXSV
s
&OHDUVSDFHt
FV
FF
EZ
GE
PD[DJJUHJDWHVL]H
Alsamsam, I.M. and Kamara, M. E. (2004). Simplified Design Reinforced Concrete Buildings of
Moderate Size and Heights, EB104, Portland Cement Association, Skokie, IL.
DESIGN AID J.1-61
0D[LPXP1XPEHURI5HLQIRUFLQJ%DUV3HUPLWWHGLQD6LQJOH/D\HU
Assumptions:
x *UDGHUHLQIRUFHPHQW f y NVL x &OHDUFRYHUWRWKHVWLUUXSV cs LQ
x ôLQDJJUHJDWH
x 1RVWLUUXSVDUHXVHGIRU1RDQG1RORQJLWXGLQDOEDUVDQG1R
VWLUUXSVDUHXVHGIRU1RDQGODUJHUEDUV
Beam Width (in.)
Bar
Size
1R
1R
1R
1R
1R
1R
1R
1R
Maximum number of bars, nmax:
nmax
bw 2(cs d s r )
1
(Clear space) db
GV
db
FF
sFOHDU
ôsIRU1RVWLUUXSV
sIRU1RVWLUUXSV
s
&OHDUVSDFHt
FV
FF
1
U EZ
GE
PD[DJJUHJDWHVL]H
Alsamsam, I.M. and Kamara, M. E. (2004). Simplified Design Reinforced Concrete Buildings of
Moderate Size and Heights, EB104, Portland Cement Association, Skokie, IL.
DESIGN AID J.1-7
0LQLPXP7KLFNQHVVhIRU%HDPVDQG2QH:D\6ODEV8QOHVV'HIOHFWLRQVDUH
&DOFXODWHG
Beams or
Ribbed
One-way
Slabs
h t " 1 / 18.5
h t " 2 / 21
2QHHQG
FRQWLQXRXV
%RWKHQGV
FRQWLQXRXV
"1
Solid
One-way
Slabs
&DQWLOHYHU
"2
h t " 1 / 24
h t " 2 / 28
2QHHQG
FRQWLQXRXV
%RWKHQGV
FRQWLQXRXV
"1
h t "3 /8
"3
h t " 3 / 10
&DQWLOHYHU
"2
"3
x Applicable to one-way construction not supporting or attached to partitions or other construction likely to be
damaged by large deflections.
x Values shown are applicable to members with normal weight concrete ( wc
reinforcement. For other conditions, modify the values as follows:
3
145 lbs/ft ) and Grade 60
ƒ For structural lightweight having wc in the range 90-120 lbs/ft3, multiply the values by
1.65 0.005 wc t 1.09.
ƒ For f y other than 60,000 psi, multiply the values by 0.4 f y / 100,000 .
x For simply-supported members, minimum h
­" / 20 for solid one - way slabs
®
¯" / 16 for beams or ribbed one - way slabs
DESIGN AID J.1-8
Reinforcement Ratio ρ t for Tension-Controlled Sections Assuming Grade 60
Reinforcement
f c′ (psi)
ρ t when εt = 0.005
ρ t when εt = 0.004
3,000
0.01355
0.01548
4,000
0.01806
0.02064
5,000
0.02125
0.02429
Notes:
1. C = 0.85 f ' c (β1c )b
T = As f y
C = T ⇒ 0.85 f ' c (β1c )b = As f y
a. When εt = 0.005, c/dt = 3/8.
0.85 f ' c β1 3 d t b = As f y
8
(
ρt =
)
0.85β1 f c′( 3 )
As
8
=
bd t
fy
b. When εt = 0.004, c/dt = 3/7.
0.85 f ' c β1 3 d t b = As f y
7
(
)
0.85β1 f c′( 3 )
As
7
=
ρt =
bd t
fy
2. β1 is determined according to 10.2.7.3.
DESIGN AID J.1-9
Simplified Calculation of bw Assuming Grade 60 Reinforcement and
ρ = 0.5ρ max
f c′ (psi)
3,000
4,000
5,000
bw (in.)*
31.6 M u
d2
23.7 M u
d2
20.0 M u
d2
* M u is in ft-kips and d is in inches
In general:
bw =
36,600 M u
ρ β1 f c′ (1 − 0.2143ρ β1 )d 2
where ρ = ρ / ρ max , f c′ is in psi, d is in inches and M u is in ft-kips
and
ρ max =
0.85β1 f c′
0.003
(10.3.5)
fy
0.004 + 0.003
DESIGN AID J.1-10
T-beam Construction
8.12
be 2
be1
h = hf
bw1
bw3
bw2
s1
Span length

bw1 +
12

be1 ≤ bw1 + 6h
 3b
b
s
 w1 − w2 + 1
4
2
 4
s2
 Span length

4

be 2 ≤ bw2 + 16h
b
b +b
s +s
 w2 − w1 w3 + 1 2
4
2
 2
be ≤ 4bw
h = hf ≥
bw
Isolated T-beam
bw
2
DESIGN AID J.1-11
Values of φVs = Vu − φVc (kips) as a Function of the Spacing, s*
s
d/2
No. 3 U-stirrups
19.8
No. 4 U-stirrups
36.0
No. 5 U-stirrups
55.8
d/3
29.7
54.0
83.7
d/4
39.6
72.0
111.6
* Valid for Grade 60 ( f yt = 60 ksi) stirrups with 2 legs (double the tabulated values for
4 legs, etc.).
In general:
φVs =
φAv f yt d
s
(11.4.7.2)
where f yt used in design is limited to 60,000 psi, except for welded deformed wire
reinforcement, which is limited to 80,000 psi (11.4.2).
DESIGN AID J.1-12
Minimum Shear Reinforcement Av,min / s *
f c′ (psi)
Av,min  in.2 


 in. 
s


≤ 4,500
0.00083bw
5,000
0.00088bw
* Valid for Grade 60 ( f yt = 60 ksi) shear reinforcement.
In general:
Av,min
s
= 0.75 f c′
bw 50bw
≥
f yt
f yt
Eq. (11-13)
where f yt used in design is limited to 60,000 psi, except for welded deformed wire
reinforcement, which is limited to 80,000 psi (11.4.2).
DESIGN AID J.1-13
Torsional Section Properties
Section*
Edge
bw
Acp
pcp
Aoh
ph
bwh + behf
2(h + bw + be)
x1y1
2(x1 + y1)
bw(h - hf) + behf
2(h + be)
x1y1
2(x1 + y1)
b1h1 + b2h2
2(h1 + h2 + b2)
x1y1 + x2y2
2(x1 + x2 + y1)
b1h1 + b2h2
2(h1 + h2 + b2) x1y1 + 2x2y2 2(x1 + 2x2 + y1)
be = h - hf ≤ 4hf
h
hff
yyo
hh
1
x1 = bw - 2c - ds
y1 = h - 2c - ds
x1
Interior
be = bw + 2(h - hf) ≤ bw + 8hf
xxo1
hhf
f
yyo
h
1
x1 = bw - 2c - ds
y1 = h - 2c - ds
bw
L-shaped
bb11
x1
x2
h1
yy11
y2 h2
x1 = b1 - 2c - ds
y1 = h1 + h2 - 2c - ds
x2 = b2 - b1
y2 = h2 - 2c - ds
b2
Inverted tee
b1
x1
x2
h1
y1
y2
h2
x1 = b1 - 2c - ds
y1 = h1 + h2 - 2c - ds
x2 = (b2 - b1)/2
y2 = h2 - 2c - ds
b2
* Notation: xi, yi = center-to-center dimension of closed rectangular stirrup
c = clear cover to closed rectangular stirrup(s)
ds = diameter of closed rectangular stirrup(s)
2
Note: Neglect overhanging flanges in cases where Acp
/ pcp calculated for a beam with
flanges is less than that computed for the same beam ignoring the flanges (11.5.1.1).
DESIGN AID J.1-14
Moment of Inertia of Cracked Section Transformed to Concrete, I cr
Gross Section
Cracked Transformed
Section
b
Cracked Moment of Inertia, I cr
b
I cr =
kd
As
nAs
b
b
where
d
n.a.
h
kd =
I cr =
d′
kd
A′s
n.a.
h
As
(n – 1)A′s
b(kd )3
+ nAs (d − kd ) 2
3
2dB + 1 − 1
B
b(kd ) 3
+ nAs (d − kd ) 2
3
+ (n − 1) As′ (kd − d ′) 2
d
where
nAs
kd =
 rd ′ 
2
2dB + 1 +
 + (1 + r ) − (1 + r )
d


B
---continued next page--I g = bh 3 / 12
n = E s / Ec
B = b /(nAs )
r = (n − 1) As′ /(nAs )
DESIGN AID J.1-14
Moment of Inertia of Cracked Section Transformed to Concrete, I cr
(continued)
Cracked
Transformed
Section
Gross Section
b
Cracked Moment of Inertia, I cr
b
hf
I cr =
kd
h
nAs
As
12
b (kd ) 3
+ w
3
hf

+ (b − bw )h f  kd −
2

d
n.a.
(b − bw )h 3f




2
+ nAs (d − kd ) 2
bw
where
kd =
b
b
hf
A′s
yt
d′
I cr =
kd
h
As
(n – 1)A′s
nAs
bw
n.a.
d
C (2d + h f f ) + (1 + f ) 2 − (1 + f )
C
(b − bw )h 3f
12
b (kd ) 3
+ w
3
hf

+ (b − bw )h f  kd −
2





2
+ nAs (d − kd ) 2 + (n − 1) As′ (kd − d ′) 2
where
kd =
C (2d + h f f + 2rd ′) + (1 + r + f ) 2 − (1 + r + f )
C
yt = h − {0.5[(b − bw )h 2f + bw h 2 ] /[(b − bw )h f + bw h]}
I g = (b − bw )h 3f / 12 + bw h 3 / 12 + (b − bw )h f (h − 0.5h f − yt ) 2 + bw h( yt − 0.5h) 2
n = E s / Ec
C = bw /(nAs )
f = h f (b − bw ) /(nAs )
r = (n − 1) As′ /(nAs )
DESIGN AID J.1-15
Approximate Equation to Determine Immediate Deflection, ∆ i , for Members
Subjected to Uniformly Distributed Loads
∆i =
5 KM a  2
48 Ec I e
where
M a = net midspan moment or cantilever moment
 = span length (8.9)
Ec = modulus of elasticity of concrete (8.5.1)
= w1c.5 33 f c′ for values of wc between 90 and 155 pcf
wc = unit weight of concrete
I e = effective moment of inertia (see Flowchart A.1-5.1)
K = constant that depends on the span condition
Span Condition
Cantilever*
2.0
Simple
1.0
Continuous
*
K
1.2 − 0.2( M o / M a )**
Deflection due to rotation at supports not included
** M o = w 2 / 8 (simple span moment at midspan)
DESIGN AID J.2-1
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DESIGN AID J.2-2
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Minimum of "1/4 or ("2)B/4
(""2)A
½-Middle strip
Column strip
½-Middle strip
"1
Minimum of "1/4 or ("2)A/4
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DESIGN AID J.2-3
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K
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K
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K
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Page 3 of 11
DESIGN AID J.2-4
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Flat Plate or Flat Slab
Flat Plate or Flat Slab with Spandrel Beams
6HH 'HVLJQ $LG - IRU GHWHUPLQDWLRQ RI E t
Page 4 of 11
DESIGN AID J.2-4
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FRQWLQXHG Flat Plate or Flat Slab with End Span Integral with Wall
Flat Plate or Flat Slab with End Span Simply Supported on Wall
Page 5 of 11
DESIGN AID J.2-4
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a
C
C
"2
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CL
Slab, Is
Slab, Is
h
h
a
Beam, Ib
Beam, Ib
b
b
beff = b + 2(a – h) d b + 8h
beff = b + (a – h) d b + 4h
Ecb I b
(T Ecs I s
Df
Ec
Edge Beam
PRGXOXVRIHODVWLFLW\RIFRQFUHWH wc f cc IRUYDOXHVRI wc EHWZHHQDQGSFI
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Ib
ah·
h
·
§
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b a h b a h ¨ yb ¸ beff h beff h¨ a yb ¸ ¹
¹
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Interior Beam
C
C
"2
h
a
b
beff = b + 2(a – h) d b + 8h
Case A
y2
y2
x2C A
x1
§
x ·x y
¨¨ ¸¸ y ¹ ©
§
x · x y ¨¨ ¸¸
y ¹ ©
y1
Case B
y2
x2C B
x1
§
x · x y
¨¨ ¸¸
y ¹ ©
§
x · x y ¸
¨
¨ ¸
y ¹ ©
y1
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FRQWLQXHG CL
"2
Edge Beam
h
a
b
beff = b + (a – h) d b + 4h
Case A
y2
x2
x1
CA
§
x · x y
¨¨ ¸¸ y ¹ ©
§
x · x y
¨¨ ¸¸ y ¹ ©
y1
Case B
y2
x2
x1
CB
§
x ·x y
¨¨ ¸¸ y ¹ ©
§
x · x y ¨¨ ¸¸
y ¹ ©
y1
C
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k NF Ecs I sb " )L[HGHQGPRPHQW FEM m FN qu " " $SSOLFDEOHZKHUH c N c F DQG c N c F DQG DXQLIRUPO\GLVWULEXWHGORDG qu DFWVRYHUWKHHQWLUHVSDQOHQJWK6HHPCA Notes on ACI 318-11IRURWKHUFDVHVLQFOXGLQJ
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Page 10 of 11
DESIGN AID J.2-8
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6HHPCA Notes on ACI 318-11IRURWKHUFDVHVLQFOXGLQJIDFWRUVIRUPHPEHUVZLWKGURS
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Page 11 of 11