1. Guide to Simplified
Design for Reinforced
Concrete Buildings
(For Buildings of Limited Size and Height,
based on ACI 318-14 and ACI IPS-1, “Essential
Requirements for Reinforced Concrete
Buildings”)
Reported by ACI Committee 314
ACI
314R-16
2. First Printing
June 2016
ISBN: 978-1-942727-93-4
Guide to Simplified Design for Reinforced Concrete Buildings
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135.
Members of ACI Committee 314, Simplified Design of Concrete Buildings, created and reviewed
a series of helpful design aids for reinforced concrete.
Design Aid 1‐1 Areas of Reinforcing Bars
Design Aid 1‐2 Approximate Bending Moments and Shear Forces for Continuous Beams and One‐
way Slabs
Design Aid 1‐3 Variation of φ with Net Tensile Strain in Extreme Tension Steel εt and c/dt – Grade
60 Reinforcement and Prestressing Steel
Design Aid 1‐4 Simplified Calculation of As Assuming Tension‐Controlled Section and Grade 60
Reinforcement
Design Aid 1‐5 Minimum Number of Reinforcing Bars Required in a Single Layer
Design Aid 1‐6 Maximum Number of Reinforcing Bars Permitted in a Single Layer
Design Aid 1‐7 Minimum Thickness h for Beams and One‐Way Slabs Unless Deflections are
Calculated
Design Aid 1‐8 Reinforcement Ratio ρt for Tension‐Controlled Sections Assuming Grade 60
Reinforcement
Design Aid 1‐9 Simplified Calculation of bw Assuming Grade 60 Reinforcement and ρ = 0.5ρmax
Design Aid 1‐10 T‐beam Construction
Design Aid 1‐11 Values of φVs = Vu ‐ φVc (kips) as a Function of the Spacing, s
Design Aid 1‐12 Minimum Shear Reinforcement Av,min /s
Design Aid 1‐13 Torsional Section Properties
Design Aid 1‐14 Moment of Inertia of Cracked Section Transformed to Concrete, Icr
Design Aid 1‐15 Approximate Equation to Determine Immediate Deflection, Δi, for Members
Subjected to Uniformly Distributed Loads
Two Way Slabs – Direct Design Method, includes Design Aid 2.1 through D.8.
Design Aid 2‐1 Conditions for Analysis by the Direct Design Method
Design Aid 2‐2 Definitions of Column Strip and Middle Strip
Design Aid 2‐3 Definition of Clear Span, ℓn
Design Aid 2‐4 Design Moment Coefficients used with the Direct Design Method
Design Aid 2‐5 Effective Beam and Slab Sections for Computation of Stiffness Ratio, αf
Design Aid 2‐6 Computation of Torsional Stiffness Factor, βt, for T‐ and L‐Sections
Design Aid 2‐7 Moment Distribution Constants for Slab‐Beam Members without Drop Panels
Design Aid 2‐8 Stiffness and Carry‐Over Factors for Columns
1
137. Positive
Moment
Negative
Moment
Shear
n,2 n,1 ≤ 1.2n,2 n,2
Prismatic members
n,2
Integral with
Support
Simple
Support
14
2
1
,
n
u
w
16
2
2
,
n
u
w
11
2
2
,
n
u
w
24
2
1
,
n
u
w
10
*
2
,avg
n
u
w
11
2
,avg
n
u
w
11
2
2
,
n
u
w
10
*
2
2
,
n
u
w
0
Spandrel
Support
Column
Support
16
2
1
,
n
u
w
2
1
,
n
u
w
2
2
,
n
u
w
2
2
,
n
u
w
2
2
,
n
u
w
2
15
.
1 1
,
n
u
w
2
15
.
1 2
,
n
u
w
9
2
n
u
w
2
2
,
1
,
,
n
n
avg
n
+
=
(2 spans)
(8.3.3)
Note A
12
2
1
,
n
u
w
12
2
,avg
n
u
w
12
2
,avg
n
u
w
12
2
2
,
n
u
w
12
*
*
2
2
,
n
u
w
0
Note A: Applicable to slabs with spans ≤ 10 ft
and beams where the ratio of the sum of
column stiffness to beam stiffness 8 at
each end of the span.
Two or more spans
Uniformly distributed load wu (L/D ≤ 3)
10
2
2
,
n
u
w
(for beams)
*
**
DESIGN AID 1-2
Approximate Bending Moments and Shear Forces for Continuous Beams
and One-way Slabs
8.3.3
3
138. DESIGN AID 1-3
Variation of φ with Net Tensile Strain in Extreme Tension Steel t
ε and t
d
c / –
Grade 60 Reinforcement and Prestressing Steel
9.3.2
0.65
0.75
Spiral
Other
Compression-controlled
0.002
=
εt
φ
Transition Tension-controlled
0.002)(50)
(
0.75 −
ε
+
=
φ t
0.90
Spiral: (5/3)]
)
/
0.15[(1/
0.75 −
+
=
φ t
d
c
Other: (5/3)]
)
/
0.25[(1/
0.65 −
+
=
φ t
d
c
/3)
0.002)(250
(
0.65 −
ε
+
=
φ t
c = 0.6dt
εt = 0.002
0.003
dt
0.005
=
εt
c = 0.375dt
εt = 0.005
0.003
dt
4
139. DESIGN AID 1-4
Simplified Calculation of s
A Assuming Tension-Controlled Section and
Grade 60 Reinforcement
c
f ′ (psi) s
A (in.2
)
3,000
d
Mu
84
.
3
4,000
d
Mu
00
.
4
5,000
d
Mu
10
.
4
u
M is in ft-kips and d is in inches
In all cases,
d
M
A u
s
4
= can be used.
Notes:
•
d
f
f
f
M
A
c
y
y
u
s
×
−
=
'
85
.
0
5
.
0
1
ρ
φ
• 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.
5
140. DESIGN AID 1-51
Minimum Number of Reinforcing Bars Required in a Single Layer
10.6.4
Assumptions:
• Grade 60 reinforcement ( y
f = 60,000 psi)
• Clear cover to the tension reinforcement c
c = 2 in.
• Calculated stress s
f in reinforcement closest to the tension face at
service load = 40,000 psi
Bar
Size
Beam Width (in.)
12 14 16 18 20 22 24 26 28 30 36 42 48
No. 4 2 2 3 3 3 3 3 4 4 4 5 5 6
No. 5 2 2 3 3 3 3 3 4 4 4 5 5 6
No. 6 2 2 3 3 3 3 3 4 4 4 5 5 6
No. 7 2 2 3 3 3 3 3 4 4 4 5 5 6
No. 8 2 2 3 3 3 3 3 4 4 4 5 5 6
No. 9 2 2 3 3 3 3 3 4 4 4 5 5 6
No. 10 2 2 3 3 3 3 3 4 4 4 5 5 6
No. 11 2 2 3 3 3 3 3 4 4 4 5 5 6
Minimum number of bars, nmim:
1
)
5
.
0
(
2
+
+
−
=
s
d
c
b
n b
c
w
min
where
≤
−
=
s
c
s
f
c
f
s
000
,
40
12
5
.
2
000
,
40
15
1
Alsamsam, I.M. and Kamara, M. E. (2004). Simplified Design Reinforced Concrete Buildings of
Moderate Size and Heights, EB104, Portland Cement Association, Skokie, IL.
c
cc
c b
bw
w
1
1-
-1
1/
/2
2″
″ c
cl
le
ea
ar
r
c
cc
c
db
C
Cl
le
ea
ar
r s
sp
pa
ac
ce
e ≥
≥
d
ds
s
c
cs
s
r
r =
=
¾
¾″
″ f
fo
or
r N
No
o.
. 3
3 s
st
ti
ir
rr
ru
up
ps
s
1
1″
″ f
fo
or
r N
No
o.
. 4
4 s
st
ti
ir
rr
ru
up
ps
s
1
1″
″
d
db
b
1
1.
.3
33
3(
(m
ma
ax
x.
. a
ag
gg
gr
re
eg
ga
at
te
e s
si
iz
ze
e)
)
6
141. DESIGN AID 1-61
Maximum Number of Reinforcing Bars Permitted in a Single Layer
7.6
Assumptions:
• Grade 60 reinforcement ( y
f = 60 ksi)
• Clear cover to the stirrups s
c = 1.5 in.
• ¾-in. aggregate
• No. 3 stirrups are used for No. 5 and No. 6 longitudinal bars, and No. 4
stirrups are used for No. 7 and larger bars
Bar
Size
Beam Width (in.)
12 14 16 18 20 22 24 26 28 30 36 42 48
No. 4 5 6 8 9 10 12 13 14 16 17 21 25 29
No. 5 5 6 7 8 10 11 12 13 15 16 19 23 27
No. 6 4 6 7 8 9 10 11 12 14 15 18 22 25
No. 7 4 5 6 7 8 9 10 11 12 13 17 20 23
No. 8 4 5 6 7 8 9 10 11 12 13 16 19 22
No. 9 3 4 5 6 7 8 8 9 10 11 14 17 19
No. 10 3 4 4 5 6 7 8 8 9 10 12 15 17
No. 11 3 3 4 5 5 6 7 8 8 9 11 13 15
Maximum number of bars, nmax:
1
space)
(Clear
)
(
2
+
+
+
+
−
=
b
s
s
w
d
r
d
c
b
nmax
1
Alsamsam, I.M. and Kamara, M. E. (2004). Simplified Design Reinforced Concrete Buildings of
Moderate Size and Heights, EB104, Portland Cement Association, Skokie, IL.
c
cc
c b
bw
w
1
1-
-1
1/
/2
2″
″ c
cl
le
ea
ar
r
c
cc
c
db
C
Cl
le
ea
ar
r s
sp
pa
ac
ce
e ≥
≥
d
ds
s
c
cs
s
r
r =
=
¾
¾″
″ f
fo
or
r N
No
o.
. 3
3 s
st
ti
ir
rr
ru
up
ps
s
1
1″
″ f
fo
or
r N
No
o.
. 4
4 s
st
ti
ir
rr
ru
up
ps
s
1
1″
″
d
db
b
1
1.
.3
33
3(
(m
ma
ax
x.
. a
ag
gg
gr
re
eg
ga
at
te
e s
si
iz
ze
e)
)
7
142. 1 2 3
One end
continuous
5
.
18
/
1
≥
h 21
/
2
≥
h
Both ends
continuous
8
/
3
≥
h
Cantilever
1 2 3
One end
continuous
24
/
1
≥
h 28
/
2
≥
h
Both ends
continuous
10
/
3
≥
h
Cantilever
Solid
One-way
Slabs
• Applicable to one-way construction not supporting or attached to partitions or other construction likely to be
damaged by large deflections.
• Values shown are applicable to members with normal weight concrete ( 145
=
c
w lbs/ft3
) and Grade 60
reinforcement. For other conditions, modify the values as follows:
For structural lightweight having c
w in the range 90-120 lbs/ft3
, multiply the values by
( ) .
09
.
1
005
.
0
65
.
1 ≥
− c
w
For y
f other than 60,000 psi, multiply the values by ( ).
000
,
100
/
4
.
0 y
f
+
• For simply-supported members, minimum
=
slabs
way
-
one
ribbed
or
beams
for
16
/
slabs
way
-
one
solid
for
20
/
h
Beams or
Ribbed
One-way
Slabs
DESIGN AID 1-7
Minimum Thickness h for Beams and One-Way Slabs Unless Deflections are
Calculated
9.5.2
8
143. DESIGN AID 1-8
Reinforcement Ratio t
ρ for Tension-Controlled Sections Assuming Grade 60
Reinforcement
c
f ′ (psi) t
ρ when εt = 0.005 t
ρ when εt = 0.004
3,000 0.01355 0.01548
4,000 0.01806 0.02064
5,000 0.02125 0.02429
Notes:
1. ( )b
c
f
C c 1
'
85
.
0 β
=
y
s f
A
T =
( ) y
s
c f
A
b
c
f
T
C =
⇒
= 1
'
85
.
0 β
a. When εt = 0.005, c/dt = 3/8.
( ) y
s
t
c f
A
b
d
f =
8
3
'
85
.
0 1
β
y
c
t
s
t
f
f
bd
A )
8
3
(
85
.
0 1
′
=
=
β
ρ
b. When εt = 0.004, c/dt = 3/7.
( ) y
s
t
c f
A
b
d
f =
7
3
'
85
.
0 1
β
y
c
t
s
t
f
f
bd
A )
7
3
(
85
.
0 1
′
=
=
β
ρ
2. β1 is determined according to 10.2.7.3.
9
144. DESIGN AID 1-9
Simplified Calculation of w
b Assuming Grade 60 Reinforcement and
max
ρ
=
ρ 5
.
0
c
f ′ (psi) w
b (in.)*
3,000 2
6
.
31
d
Mu
4,000 2
7
.
23
d
Mu
5,000 2
0
.
20
d
Mu
* u
M is in ft-kips and d is in inches
In general:
( ) 2
1
1 2143
.
0
1
600
,
36
d
f
M
b
c
u
w
β
ρ
−
′
β
ρ
=
where max
ρ
ρ
=
ρ / , c
f ′ is in psi, d is in inches and u
M is in ft-kips
and
003
.
0
004
.
0
003
.
0
85
.
0 1
+
′
β
=
ρ
y
c
max
f
f
(10.3.5)
10
146. DESIGN AID 1-11
Values of c
u
s V
V
V φ
−
=
φ (kips) as a Function of the Spacing, s*
s No. 3 U-stirrups No. 4 U-stirrups No. 5 U-stirrups
d/2 19.8 36.0 55.8
d/3 29.7 54.0 83.7
d/4 39.6 72.0 111.6
* Valid for Grade 60 ( 60
=
yt
f ksi) stirrups with 2 legs (double the tabulated values for
4 legs, etc.).
In general:
s
d
f
A
V
yt
v
s
φ
=
φ (11.4.7.2)
where yt
f 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).
12
147. DESIGN AID 1-12
Minimum Shear Reinforcement *
/
, s
A min
v
c
f ′ (psi)
s
A min
v,
in.
in.2
500
,
4
≤ w
b
00083
.
0
5,000 w
b
00088
.
0
* Valid for Grade 60 ( 60
=
yt
f ksi) shear reinforcement.
In general:
yt
w
yt
w
c
min
v
f
b
f
b
f
s
A 50
75
.
0
,
≥
′
= Eq. (11-13)
where yt
f 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).
13
148. DESIGN AID 1-13
Torsional Section Properties
Section* Acp pcp Aoh ph
Edge
bwh + behf 2(h + bw + be) x1y1 2(x1 + y1)
x1 = bw - 2c - ds
y1 = h - 2c - ds
Interior
bw(h - hf) + behf 2(h + be) x1y1 2(x1 + y1)
x1 = bw - 2c - ds
y1 = h - 2c - ds
L-shaped
b1h1 + b2h2 2(h1 + h2 + b2) x1y1 + x2y2 2(x1 + x2 + y1)
x1 = b1 - 2c - ds
y1 = h1 + h2 - 2c - ds
x2 = b2 - b1
y2 = h2 - 2c - ds
Inverted tee
b1h1 + b2h2 2(h1 + h2 + b2) x1y1 + 2x2y2 2(x1 + 2x2 + y1)
x1 = b1 - 2c - ds
y1 = h1 + h2 - 2c - ds
x2 = (b2 - b1)/2
y2 = h2 - 2c - ds
* 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)
hf
h
hf
yo
yo
xo
h
bw
hf
be = bw + 2(h - hf) ≤ bw + 8hf
h
bw
hf
be = h - hf ≤ 4hf
x1
y1
y1
x1
b1
y1
b1
h1
h2
b2
y1
y2
x1
x2
h1
h2
b2
y2
b1
x1
x2
y1
Note: Neglect overhanging flanges in cases where cp
cp p
A /
2
calculated for a beam with
flanges is less than that computed for the same beam ignoring the flanges (11.5.1.1).
14
149. Gross Section
Cracked Transformed
Section
Cracked Moment of Inertia, cr
I
2
3
)
(
3
)
(
kd
d
nA
kd
b
I s
cr −
+
=
where
B
dB
kd
1
1
2 −
+
=
2
2
3
)
(
)
1
(
)
(
3
)
(
d
kd
A
n
kd
d
nA
kd
b
I
s
s
cr
′
−
′
−
+
−
+
=
where
( ) ( )
B
r
r
d
d
r
dB
kd
+
−
+
+
′
+
+
=
1
1
1
2 2
---continued next page---
12
/
3
bh
Ig =
c
s E
E
n /
=
)
/( s
nA
b
B =
)
/(
)
1
( s
s nA
A
n
r ′
−
=
b
As
A′s
b d′
n.a.
nAs
kd
d
b
n.a.
b
As
DESIGN AID 1-14
Moment of Inertia of Cracked Section Transformed to Concrete, cr
I
h
h
nAs
kd
d
(n – 1)A′s
15
150. Gross Section
Cracked
Transformed
Section
Cracked Moment of Inertia, cr
I
2
2
3
3
)
(
2
)
(
3
)
(
12
)
(
kd
d
nA
h
kd
h
b
b
kd
b
h
b
b
I
s
f
f
w
w
f
w
cr
−
+
−
−
+
+
−
=
where
C
f
f
f
h
d
C
kd
f )
1
(
)
1
(
)
2
( 2
+
−
+
+
+
=
2
2
2
3
3
)
(
)
1
(
)
(
2
)
(
3
)
(
12
)
(
d
kd
A
n
kd
d
nA
h
kd
h
b
b
kd
b
h
b
b
I
s
s
f
f
w
w
f
w
cr
′
−
′
−
+
−
+
−
−
+
+
−
=
where
C
f
r
f
r
d
r
f
h
d
C
kd
f )
1
(
)
1
(
)
2
2
( 2
+
+
−
+
+
+
′
+
+
=
]}
)
/[(
]
)
[(
5
.
0
{ 2
2
h
b
h
b
b
h
b
h
b
b
h
y w
f
w
w
f
w
t +
−
+
−
−
=
2
2
3
3
)
5
.
0
(
)
5
.
0
(
)
(
12
/
12
/
)
( h
y
h
b
y
h
h
h
b
b
h
b
h
b
b
I t
w
t
f
f
w
w
f
w
g −
+
−
−
−
+
+
−
=
c
s E
E
n /
=
)
/( s
w nA
b
C =
)
/(
)
( s
w
f nA
b
b
h
f −
=
)
/(
)
1
( s
s nA
A
n
r ′
−
=
nAs
kd
b
n.a.
d′
b
As
hf
bw
A′s
nAs
kd
d
b
n.a.
b
As
hf
bw
DESIGN AID 1-14
Moment of Inertia of Cracked Section Transformed to Concrete, cr
I
(continued)
h
h
yt
d
(n – 1)A′s
16
151. DESIGN AID 1-15
Approximate Equation to Determine Immediate Deflection, i
∆ , for Members
Subjected to Uniformly Distributed Loads
e
c
a
i
I
E
KM
48
5 2
=
∆
where =
a
M net midspan moment or cantilever moment
=
span length (8.9)
=
c
E modulus of elasticity of concrete (8.5.1)
= c
c f
w ′
33
5
.
1
for values of c
w between 90 and 155 pcf
=
c
w unit weight of concrete
=
e
I effective moment of inertia (see Flowchart A.1-5.1)
=
K constant that depends on the span condition
Span Condition K
Cantilever* 2.0
Simple 1.0
Continuous *
*
)
/
(
2
.
0
2
.
1 a
o M
M
−
* Deflection due to rotation at supports not included
** 8
/
2
w
Mo = (simple span moment at midspan)
17
152. DESIGN AID 2-1
Conditions for Analysis by the Direct Design Method
13.6.1
For a panel with beams between supports on all sides, Eq. (13-2) must also be satisfied:
0
.
5
2
.
0
2
1
2
2
2
1
≤
α
α
≤
f
f
where
s
cs
b
cb
f
I
E
I
E
=
α Eq. (13-3)
=
c
E modulus of elasticity of concrete (8.5.1)
= c
c f
w ′
33
5
.
1
for values of c
w between 90 and 155 pcf
=
s
b I
I , moment of inertia of beam and slab, respectively (see Design Aid J.2-5)
18
153. DESIGN AID 2-2
Definitions of Column Strip and Middle Strip
13.2
½-Middle strip
½-Middle strip
1
Column strip
Minimum of 1/4 or (2)A/4
Minimum of 1/4 or (2)B/4
(2)A
(2)B
19
154. DESIGN AID 2-3
Definition of Clear Span, n
13.6.2.5
1
1
65
.
0
≥
n
0.89h
h
0.93h
h
h
h
h
h
20
155. DESIGN AID 2-4
Design Moment Coefficients used with the Direct Design Method
13.6.3 – 13.6.6
Flat Plate or Flat Slab
Flat Plate or Flat Slab with Spandrel Beams
See Design Aid J 2-6 for determination of t
β
21
156. DESIGN AID 2-4
Design Moment Coefficients used with the Direct Design Method
13.6.3 – 13.6.6
(continued)
Flat Plate or Flat Slab with End Span Integral with Wall
Flat Plate or Flat Slab with End Span Simply Supported on Wall
22
157. DESIGN AID 2-4
Design Moment Coefficients used with the Direct Design Method
13.6.3 – 13.6.6
(continued)
Two-Way Beam-Supported Slab
Notes:
• Mo is defined per 13.6.2
See Design Aids J 2-5 and J 2-6 for determination of f
α and t
β respectively
23
158. DESIGN AID 2-5
Effective Beam and Slab Sections for Computation of Stiffness Ratio f
α
C
C
2
h
a
b
beff = b + 2(a – h) ≤ b + 8h
Beam, Ib
Slab, Is
2
h
a
b
Beam, Ib
Slab, Is
CL
beff = b + (a – h) ≤ b + 4h
Interior Beam Edge Beam
s
cs
b
cb
f
I
E
I
E
=
α Eq. (13-3)
=
c
E modulus of elasticity of concrete (8.5.1)
= c
c f
w ′
33
5
.
1
for values of c
w between 90 and 155 pcf
3
2
12
1
h
Is
=
2
3
2
3
2
12
1
2
)
(
)
(
12
1
−
−
+
+
−
−
−
+
−
= b
eff
eff
b
b y
h
a
h
b
h
b
h
a
y
h
a
b
h
a
b
I
where
( )
)
(
2
2
2
h
a
b
h
b
h
a
b
h
a
h
b
y
eff
eff
b
−
+
−
+
−
=
24
159. DESIGN AID 2-6
Computation of Torsional Stiffness Factor, t
β , for T- and L-Sections
C
C
2
h
a
b
beff = b + 2(a – h) ≤ b + 8h
Interior Beam
Case A
3
63
.
0
1
2
3
63
.
0
1
2
3
2
2
2
1
3
1
1
1
y
x
y
x
y
x
y
x
CA
−
+
−
=
Case B
3
63
.
0
1
3
63
.
0
1
2
3
2
2
2
1
3
1
1
1
y
x
y
x
y
x
y
x
CB
−
+
−
=
=
C maximum of A
C and B
C
s
cs
cb
t
I
E
C
E
2
=
β Eq. (13-5)
where 12
/
3
2h
Is
= and c
c f
w
E ′
= 33
5
.
1
for values of c
w between 90 and 155 pcf (8.5.1)
x2
x1
y1
y2
y2
x2
x1
y1
y2
25
160. 2
h
a
b
CL
beff = b + (a – h) ≤ b + 4h
DESIGN AID 2-6
Computation of Torsional Stiffness Factor, t
β , for T- and L-Sections
(continued)
Edge Beam
Case A
3
63
.
0
1
3
63
.
0
1
2
3
2
2
2
1
3
1
1
1
y
x
y
x
y
x
y
x
CA
−
+
−
=
Case B
3
63
.
0
1
3
63
.
0
1
2
3
2
2
2
1
3
1
1
1
y
x
y
x
y
x
y
x
CB
−
+
−
=
=
C maximum of A
C and B
C
s
cs
cb
t
I
E
C
E
2
=
β Eq. (13-5)
where 12
/
3
2h
Is
= and c
c f
w
E ′
= 33
5
.
1
for values of c
w between 90 and 155 pcf (8.5.1)
x2
x1
y1
y2
x2
x1
y1
y2
26
161. DESIGN AID 2-7
Moment Distribution Constants for Slab-Beam Members without Drop Panels*
1
1 /
N
c 2
2 /
N
c
Stiffness Factor,
NF
k
Carry-over Factor,
NF
C
Fixed-end Moment
Coefficient, NF
m
0.10
0.10 4.18 0.51 0.0847
0.20 4.36 0.52 0.0860
0.30 4.53 0.54 0.0872
0.40 4.70 0.55 0.0882
0.20
0.10 4.35 0.52 0.0857
0.20 4.72 0.54 0.0880
0.30 5.11 0.56 0.0901
0.40 5.51 0.58 0.0921
0.30
0.10 4.49 0.53 0.0863
0.20 5.05 0.56 0.0893
0.30 5.69 0.59 0.0923
0.40 6.41 0.61 0.0951
0.40
0.10 4.61 0.53 0.0866
0.20 5.35 0.56 0.0901
0.30 6.25 0.60 0.0936
0.40 7.37 0.64 0.0971
Slab-beam stiffness 1
/
sb
cs
NF
sb I
E
k
K =
Fixed-end moment 2
1
2
u
FN q
m
FEM =
* Applicable where (1) 1
1 F
N c
c = and 2
2 F
N c
c = and (2) a uniformly distributed load u
q
acts over the entire span length. See PCA Notes on ACI 318-11 for other cases, including
constants for members with drop panels.
1
2
2
N
c
1
N
c
2
F
c
1
F
c
27
162. DESIGN AID 2-8
Stiffness and Carry-Over Factors for Columns*
c
H
/
Stiffness Factor,
AB
k
Carry-over Factor,
AB
C
1.05 4.52 0.54
1.10 5.09 0.57
1.15 5.71 0.60
1.20 6.38 0.62
1.25 7.11 0.65
1.30 7.89 0.67
1.35 8.73 0.69
1.40 9.63 0.71
1.45 10.60 0.73
1.50 11.62 0.74
Column stiffness:
( )
( )
=
=
c
c
cc
BA
BA
c
c
c
cc
AB
AB
c
I
E
k
K
I
E
k
K
/
/
* See PCA Notes on ACI 318-11 for other cases, including factors for members with drop
panels and column capitals.
H
A
B
c
28