Design Aid J.1-1 Areas of Reinforcing Bars

Design Aid J.1-2 Approximate Bending Moments and Shear Forces for Continuous Beams and One-way Slabs

Design Aid J.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 J.1-4 Simplified Calculation of As Assuming Tension-Controlled Section and Grade 60 Reinforcement

Design Aid J.1-5 Minimum Number of Reinforcing Bars Required in a Single Layer

Design Aid J.1-6 Maximum Number of Reinforcing Bars Permitted in a Single Layer

Design Aid J.1-7 Minimum Thickness h for Beams and One-Way Slabs Unless Deflections are

Calculated

Design Aid J.1-8 Reinforcement Ratio ρt for Tension-Controlled Sections Assuming Grade 60 Reinforcement

Design Aid J.1-9 Simplified Calculation of bw Assuming Grade 60 Reinforcement and ρ = 0.5 ρmax

Design Aid J.1-10 T-beam Construction

Design Aid J.1-11 Values of φVs = Vu - φVc (kips) as a Function of the Spacing, s

Design Aid J.1-12 Minimum Shear Reinforcement Av, min / s

Design Aid J.1-13 Torsional Section Properties

Design Aid J.1-14 Moment of Inertia of Cracked Section Transformed to Concrete, Icr

Design Aid J.1-15 Approximate Equation to Determine Immediate Deflection, Δi, for Members Subjected to Uniformly Distributed Loads

Design Aids J.2 Two-Way Slabs – Direct Design method, includes the following:

• Conditions for Analysis by the Direct Design Method

• Definitions of Column Strip and Middle Strip

• Definition of Clear Span,

• Design Moment Coefficients used with the Direct Design Method

• Effective Beam and Slab Sections for Computation of Stiffness Ratio, αf

• Computation of Torsional Stiffness Factor, βt, for T- and L-Sections

• Moment Distribution Constants for Slab-Beam Members without Drop Panels

• Stiffness and Carry-Over Factors for Columns

DESIGN AID J.1-1 Areas of Reinforcing Bars

Total Areas of Bars (in.2)

Bar Size

Number of Bars 1 2 3 4 5 6 7 8 9 10

No. 3 0.11 0.22 0.33 0.44 0.55 0.66 0.77 0.88 0.99 1.10 No. 4 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00 No. 5 0.31 0.62 0.93 1.24 1.55 1.86 2.17 2.48 2.79 3.10 No. 6 0.44 0.88 1.32 1.76 2.20 2.64 3.08 3.52 3.96 4.40 No. 7 0.60 1.20 1.80 2.40 3.00 3.60 4.20 4.80 5.40 6.00 No. 8 0.79 1.58 2.37 3.16 3.95 4.74 5.53 6.32 7.11 7.90 No. 9 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 No. 10 1.27 2.54 3.81 5.08 6.35 7.62 8.89 10.16 11.43 12.70 No. 11 1.56 3.12 4.68 6.24 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

PositiveMoment

NegativeMoment

Shear

n n n n

Prismatic members

n

nuw nuw nuw

nuw avgnuw avgnuw nuw nuwSpandrelSupport

ColumnSupport

nuw

nuw nuwnuw nuwnuw nuw

nuw

nnavgn

Note A nuw avgnuw avgnuw nuw nuw

Two or more spans

Uniformly distributed load wu (L/D 3)

nuw

DESIGN AID J.1-2

DESIGN AID J.1-3

DESIGN AID J.1-4

Simplified Calculation of sA Assuming Tension-Controlled Section and Grade 60 Reinforcement

cf ′ (psi) sA (in.2)

3,000 d

M u

84.3

4,000 d

M u

00.4

5,000 d

M u

10.4

uM is in ft-kips and d is in inches

In all cases, d

MA us 4

= can be used.

Notes:

• d

ff

f

MA

c

yy

us

×

−

=

'85.05.0

1ρ

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.

DESIGN AID J.1-51

Assumptions:

yf

cc

sf

Bar Size

Beam Width (in.)

Minimum number of bars, nmim:

1)5.0(2

sdcb

n bcwmin

where

s

cs

f

cf

s

000,4012

5.2000,4015

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.

db

DESIGN AID J.1-61

Assumptions:

yf

sc

Bar Size

Beam Width (in.)

Maximum number of bars, nmax:

1space)(Clear

)(2

b

sswdrdcb

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.

db

1 2 3

5.18/1h 21/2h 8/3h

1 2 3

24/1h 28/2h 10/3hSolid 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 ( 145cw lbs/ft3) and Grade 60 reinforcement. For other conditions, modify the values as follows:

For structural lightweight having cw in the range 90-120 lbs/ft3, multiply the values by .09.1005.065.1 cw

For yf other than 60,000 psi, multiply the values by .000,100/4.0 yf

For simply-supported members, minimum slabsway -one ribbedor beamsfor 16/

slabsway -one solidfor 20/h

Beams or Ribbed One-way Slabs

DESIGN AID J.1-7h

DESIGN AID J.1-8

Reinforcement Ratio tρ for Tension-Controlled Sections Assuming Grade 60 Reinforcement

cf ′ (psi) tρ when εt = 0.005 tρ when εt = 0.004

3,000 0.01355 0.01548

4,000 0.01806 0.02064

5,000 0.02125 0.02429 Notes:

1. ( )bcfC c 1'85.0 β=

ys fAT = ( ) ysc fAbcfTC =⇒= 1'85.0 β

a. When εt = 0.005, c/dt = 3/8.

( ) ystc fAbdf =83'85.0 1β

y

c

t

st f

f

bdA )8

3(85.0 1 ′==

βρ

b. When εt = 0.004, c/dt = 3/7.

( ) ystc fAbdf =73'85.0 1β

y

c

t

st f

f

bdA )7

3(85.0 1 ′==

βρ

2. β1 is determined according to 10.2.7.3.

DESIGN AID J.1-9

Simplified Calculation of wb Assuming Grade 60 Reinforcement and maxρ=ρ 5.0

cf ′ (psi) wb (in.)*

3,000 26.31

d

M u

4,000 27.23

d

M u

5,000 20.20

d

M u

* uM is in ft-kips and d is in inches

In general:

( ) 211 2143.01

600,36

df

Mb

c

uw

βρ−′βρ=

where maxρρ=ρ / , cf ′ is in psi, d is in inches and uM is in ft-kips and

003.0004.0003.085.0 1+

′β=ρ

y

cmax f

f (10.3.5)

1s

fhh =

+−

+

+

≤

2443

612lengthSpan

121

1

1

1sbb

hb

b

b

ww

w

w

e

1eb

2s 1wb 2wb

2eb

++

+−

+≤

242

164lengthSpan

21312

22ssbbb

hbb

www

we

3wb

2w

fbhh ≥=

wb

we bb 4≤

Isolated T-beam

DESIGN AID J.1-10 T-beam Construction

8.12

DESIGN AID J.1-11

Values of cus VVV 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=ytf ksi) stirrups with 2 legs (double the tabulated values for

4 legs, etc.).

In general:

sdfA

V ytvs

φ=φ (11.4.7.2)

where ytf 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 */, sA minv

cf ′ (psi) s

A minv,

in.in.2

500,4≤ wb00083.0

5,000 wb00088.0

* Valid for Grade 60 ( 60=ytf ksi) shear reinforcement.

In general:

yt

w

yt

wc

minv

fb

fb

fs

A 5075.0, ≥′= Eq. (11-13)

where ytf 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* 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 - dsy1 = h1 + h2 - 2c - dsx2 = b2 - b1y2 = h2 - 2c - ds

Inverted tee

b1h1 + b2h2 2(h1 + h2 + b2) x1y1 + 2x2y2 2(x1 + 2x2 + y1)x1 = b1 - 2c - dsy1 = h1 + h2 - 2c - ds

x2 = (b2 - b1)/2y2 = 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 cpcp pA /2 calculated for a beam with flanges is less than that computed for the same beam ignoring the flanges (11.5.1.1).

Gross Section Cracked Transformed Section

Cracked Moment of Inertia, crI

23

)(3

)( kddnAkdbI scr −+=

where

BdBkd 112 −+

=

2

23

)()1(

)(3

)(

dkdAn

kddnAkdbI

s

scr

′−′−+

−+=

where

( ) ( )

B

rrddrdB

kd+−++

′

++=

1112 2

---continued next page--- 12/3bhI g =

cs EEn /= )/( snAbB =

)/()1( ss nAAnr ′−=

b

As

A′s

b d′

n.a.

nAs

kd

d

b

n.a.

b

As

DESIGN AID J.1-14

Moment of Inertia of Cracked Section Transformed to Concrete, crI

h

h

nAs

kd

d (n – 1)A′s

Gross Section Cracked

Transformed Section

Cracked Moment of Inertia, crI

2

2

33

)(

2)(

3)(

12

)(

kddnA

hkdhbb

kdbhbbI

s

ffw

wfwcr

−+

−−+

+−

=

where

C

fffhdCkd

f )1()1()2( 2 +−+++=

22

2

33

)()1()(

2)(

3)(

12

)(

dkdAnkddnA

hkdhbb

kdbhbbI

ss

ffw

wfwcr

′−′−+−+

−−+

+−

=

where

C

frfrdrfhdCkd

f )1()1()22( 2 ++−+++′++=

]})/[(])[(5.0{ 22 hbhbbhbhbbhy wfwwfwt +−+−−=

2233 )5.0()5.0()(12/12/)( hyhbyhhhbbhbhbbI twtffwwfwg −+−−−++−=

cs EEn /= )/( sw nAbC =

)/()( swf nAbbhf −=

)/()1( ss nAAnr ′−=

nAs

kd

b

n.a.

d′ b

As

hf

bw

A′s

nAs

kd d

b

n.a.

b

As

hf

bw

DESIGN AID J.1-14

Moment of Inertia of Cracked Section Transformed to Concrete, crI (continued)

h

h yt

d (n – 1)A′s

DESIGN AID J.1-15

Approximate Equation to Determine Immediate Deflection, i∆ , for Members Subjected to Uniformly Distributed Loads

ec

ai IE

KM48

5 2=∆

where =aM net midspan moment or cantilever moment

= span length (8.9)

=cE modulus of elasticity of concrete (8.5.1)

= cc fw ′335.1 for values of cw between 90 and 155 pcf

=cw unit weight of concrete

=eI 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 **)/(2.02.1 ao MM−

* Deflection due to rotation at supports not included

** 8/2wM o = (simple span moment at midspan)

DESIGN AID J.2-1

f

f

scs

bcbf IE

IE

cE

cc fw cw

sb II

Page 1 of 11

DESIGN AID J.2-2

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

Page 2 of 11

DESIGN AID J.2-3

n

n

Page 3 of 11

DESIGN AID J.2-4

Flat Plate or Flat Slab

Flat Plate or Flat Slab with Spandrel Beams

t

Page 4 of 11

DESIGN AID J.2-4

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

Two-Way Beam-Supported Slab

f t

Page 6 of 11

DESIGN AID J.2-5

f

CC2

ha

b

beff = b + 2(a – h) b + 8h

Beam, Ib

Slab, Is

2

ha

b

Beam, Ib

Slab, Is

CL

beff = b + (a – h) b + 4h

Interior Beam Edge Beam

scs

bcbf IE

IE

cE

cc fw cw

hIs

beffeffbb yhahbhbhayhabhabI

habhb

habhahby

eff

eff

b

Page 7 of 11

DESIGN AID J.2-6

t

CC2

ha

b

beff = b + 2(a – h) b + 8h

Interior Beam

Case A

yxyx

yxyxCA

Case B

yxyx

yxyxCB

C AC BC

scs

cbt IE

CE

hIs cc fwE cw

x2x1

y1

y2y2

x2

x1

y1

y2

Page 8 of 11

2

ha

b

CL

beff = b + (a – h) b + 4h

DESIGN AID J.2-6

t

Edge Beam

Case A

yxyx

yxyxCA

Case B

yxyx

yxyxCB

C AC BC

scs

cbt IE

CE

hIs cc fwE cw

x2x1

y1

y2

x2

x1

y1

y2

Page 9 of 11

DESIGN AID J.2-7

Nc NcNFk NFC NFm

sbcsNFsb IEkK

uFN qmFEM

FN cc FN cc uqPCA Notes on ACI 318-11

Nc

Nc

Fc

Fc

Page 10 of 11

DESIGN AID J.2-8

cHABk ABC

ccccBABAc

ccccABABc

IEkK

IEkK

PCA Notes on ACI 318-11

H c

Page 11 of 11