ARCH 631 Note Set 10.

1 F2015abn

Reinforced Concrete Design

Notation:
a = depth of the effective compression Es = modulus of elasticity of steel
block in a concrete beam f = symbol for stress
A = name for area fc = compressive stress
Ag = gross area, equal to the total area fc′ = concrete design compressive stress
ignoring any reinforcement fs = stress in the steel reinforcement for
As = area of steel reinforcement in concrete design
concrete beam design fs′ = compressive stress in the
As′ = area of steel compression compression reinforcement for
reinforcement in concrete beam concrete beam design
design fy = yield stress or strength
Ast = area of steel reinforcement in fyt = yield stress or strength of transverse
concrete column design reinforcement
Av = area of concrete shear stirrup F = shorthand for fluid load
reinforcement G = relative stiffness of columns to
ACI = American Concrete Institute beams in a rigid connection, as is Ψ
b = width, often cross-sectional h = cross-section depth
bE = effective width of the flange of a H = shorthand for lateral pressure load
concrete T beam cross section hf = depth of a flange in a T section
bf = width of the flange Itransformed = moment of inertia of a multi-
bw = width of the stem (web) of a material section transformed to one
concrete T beam cross section material
c = distance from the top to the neutral k = effective length factor for columns
axis of a concrete beam (see x)
b = length of beam in rigid joint
cc = shorthand for clear cover
C = name for centroid
c = length of column in rigid joint
= name for a compression force ld = development length for reinforcing
Cc = compressive force in the steel
compression steel in a doubly l dh = development length for hooks
reinforced concrete beam ln = clear span from face of support to
Cs = compressive force in the concrete face of support in concrete design
of a doubly reinforced concrete L = name for length or span length, as is
beam l
d = effective depth from the top of a = shorthand for live load
reinforced concrete beam to the Lr = shorthand for live roof load
centroid of the tensile steel LL = shorthand for live load
d´ = effective depth from the top of a Mn = nominal flexure strength with the
reinforced concrete beam to the steel reinforcement at the yield
centroid of the compression steel stress and concrete at the concrete
db = bar diameter of a reinforcing bar design strength for reinforced
D = shorthand for dead load concrete beam design
DL = shorthand for dead load Mu = maximum moment from factored
E = modulus of elasticity or Young’s loads for LRFD beam design
modulus n = modulus of elasticity
= shorthand for earthquake load transformation coefficient for steel
Ec = modulus of elasticity of concrete to concrete
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n.a. = shorthand for neutral axis (N.A.) wLL = load per unit length on a beam from
pH = chemical alkalinity live load
P = name for load or axial force vector wself wt = name for distributed load from self
Po = maximum axial force with no weight of member
concurrent bending moment in a wu = load per unit length on a beam from
reinforced concrete column load factors
Pn = nominal column load capacity in W = shorthand for wind load
concrete design x = horizontal distance
Pu = factored column load calculated = distance from the top to the neutral
from load factors in concrete design axis of a concrete beam (see c)
R = shorthand for rain or ice load y = vertical distance
= radius of curvature in beam β1 = coefficient for determining stress
deflection relationships (see ρ) block height, a, based on concrete
Rn = concrete beam design ratio = strength, fc′
Mu/bd2 ∆ = elastic beam deflection
s = spacing of stirrups in reinforced ε = strain
concrete beams
ε t = strain in the steel
S = shorthand for snow load
ε y = strain at the yield stress
t = name for thickness
λ = modification factor for lightweight
T = name for a tension force
concrete
= shorthand for thermal load
φ = resistance factor
U = factored design value
Vc = shear force capacity in concrete φc = resistance factor for compression
Vs = shear force capacity in steel shear γ = density or unit weight
stirrups ρ = radius of curvature in beam
Vu = shear at a distance of d away from
deflection relationships (see R)
the face of support for reinforced
= reinforcement ratio in concrete
concrete beam design
beam design = As/bd
wc = unit weight of concrete
wDL = load per unit length on a beam from ρ balanced = balanced reinforcement ratio in
dead load concrete beam design
υ c = shear strength in concrete design

Reinforced Concrete Design

Structural design standards for reinforced concrete are established

by the Building Code and Commentary (ACI 318-14) published by
the American Concrete Institute International, and uses strength
design (also known as limit state design).

f’c = concrete compressive design strength at 28 days (units of psi

when used in equations)

Materials

Deformed reinforcing bars come in grades 40, 60 & 75 (for 40 ksi,

60 ksi and 75 ksi yield strengths). Sizes are given as # of 1/8” up

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to #8 bars. For #9 and larger, the number is a nominal size (while the actual size is larger).
Reinforced concrete is a composite material from a mixture of cement, coarse aggregate, fine
aggregate and water, and the average density is considered to be 150 lb/ft3. It has the properties
that it will creep (deformation with long term load) and shrink (a result of hydration) that must be
considered.

Plane sections of composite materials can still be

assumed to be plane (strain is linear), but the stress
distribution is not the same in both materials because
the modulus of elasticity is different. (f=E⋅ε)

E1 y E2 y
f1 = E1ε = − f 2 = E2 ε = −
R R
where R (or ρ) is the radius of curvature

In order to determine the stress, we can define n as the

ratio of the elastic moduli: E 2
n=
E1

n is used to transform the width of the second material such that it sees the equivalent element stress.

Transformed Section y and I

In order to determine stresses in all types of

material in the beam, we transform the materials
into a single material, and calculate the location of
the neutral axis and modulus of inertia for that
material.

ex: When material 1 above is concrete and material 2 is steel:

E2 E steel
to transform steel into concrete n= =
E1 E concrete
to find the neutral axis of the equivalent concrete member we transform the width of the
steel by multiplying by n

to find the moment of inertia of the equivalent concrete member, Itransformed, use the new
geometry resulting from transforming the width of the steel
My
concrete stress: f concrete = −
I transforme d
Myn
steel stress: f steel = −
I transforme d

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Reinforced Concrete Beam Members

Strength Design for Beams

Strength design method is similar to LRFD. There is a nominal strength that is reduced by a
factor φ which must exceed the factored design stress. For beams, the concrete only works in
compression over a rectangular “stress” block above the n.a. from elastic calculation, and the
steel is exposed and reaches the yield stress, fy

For stress analysis in reinforced concrete beams

• the steel is transformed to concrete
• any concrete in tension is assumed to be cracked and
to have no strength
• the steel can be in tension, and is placed in the bottom
of a beam that has positive bending moment

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The neutral axis is where there is no stress and no strain. The concrete above the n.a. is in
compression. The concrete below the n.a. (shown as x, but also sometimes named c) is
considered ineffective. The steel below the n.a. is in tension. (Shown as x, but also sometimes
named c.)

Because the n.a. is defined by the moment areas, we can solve for x knowing that d is the
distance from the top of the concrete section to the centroid of the steel: x
bx ⋅ − nA s ( d − x ) = 0
2
x can be solved for when the equation is rearranged into the generic format with a, b & c in the
− b ± b 2 − 4 ac
binomial equation: ax 2 + bx + c = 0 by x=
2a

T-sections
f f

hf hf
If the n.a. is above the bottom of a flange in a T
section, x is found as for a rectangular section.
bw
If the n.a. is below the bottom of a flange in a T
section, x is found by including the flange and the
stem of the web (bw) in the moment area calculation: bw

 h 
b f h f  x − f  + ( x − h f ) bw
( x − h f ) − nA (d − x) = 0
 2  2
s

Load Combinations - (Alternative values allowed)

1.4D
1.2D + 1.6L + 0.5(Lr or S or R)
1.2D + 1.6(Lr or S or R) + (1.0L or 0.5W)
1.2D + 1.0W + 1.0L + 0.5(Lr or S or R)
1.2D + 1.0E + 1.0L + 0.2S
0.9D + 1.0W
00.9D + 1.0E

b 0.85f’c
Internal Equilibrium a/2 C
C
c or x a=
C = compression in concrete = h d β1x
stress x area = 0.85 f´cba n.a.
As T T
T = tension in steel =
stress x area = Asfy
actual stress Whitney stress block

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C = T and Mn = T(d-a/2)

where f’c = concrete compression strength

 f c′ − 4000 
a = height of stress block β1 = 0.85 −   (0.05) ≥ 0.65
β1 = factor based on f’c  1000 
x or c = location to the neutral axis
b = width of stress block
fy = steel yield strength
As = area of steel reinforcement
d = effective depth of section
(depth to n.a. of reinforcement)
As f y
With C=T, Asfy = 0.85 f´cba so a can be determined with a = = β1c
0.85 f c′b

Criteria for Beam Design

For flexure design:

Mu ≤ φMn φ = 0.9 for flexure (when the section is tension controlled)

so, Mu can be set =φMn =φT(d-a/2) = φ Asfy (d-a/2)

Reinforcement Ratio
The amount of steel reinforcement is limited. Too much reinforcement, or over-reinforced will
not allow the steel to yield before the concrete crushes and there is a sudden failure. A beam
with the proper amount of steel to allow it to yield at failure is said to be under reinforced.
A
The reinforcement ratio is just a fraction: ρ = s (or p) and must be less than a value
bd
determined with a concrete strain of 0.003 and tensile strain of 0.004 (minimum).

When the strain in the reinforcement is 0.005 or greater, the section is tension controlled. (For
smaller strains the resistance factor reduces to 0.65 because the stress is less than the yield stress
in the steel.) Previous codes limited the amount to 0.75ρbalanced where ρbalanced was determined
from the amount of steel that would make the concrete start to crush at the exact same time that
the steel would yield based on strain (εy) of 0.002.

d −c fy
The strain in tension can be determined from ε t = (0.003) . At yield, ε y = .
c Es

The resistance factor expressions for transition and compression controlled sections are:

0.15
φ = 0.75 + (ε t − ε y ) for spiral members (not less than 0.75)
(0.005 − ε y )
0.25
φ = 0.65 + (ε t − ε y ) for other members (not less than 0.65)
(0.005 − ε y )

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Flexure Design of Reinforcement

One method is to “wisely” estimate a height of the stress block, a, and solve for As, and calculate
a new value for a using Mu.

1. guess a (less than n.a.)

0 .85 f c′ba
2. As =
fy
3. solve for a from
Mu = φAsfy (d-a/2) :
 M u 
a = 2 d −
 φ As f y 

4. repeat from 2. until a found from step 3 matches a used in step 2.
from Reinforced Concrete, 7th,
Wang, Salmon, Pincheira, Wiley & Sons, 2007
Design Chart Method:
Mn
1. calculate R n =
bd 2
2. find curve for f’c and fy to get ρ
3. calculate As and a
As f y
A s = ρ bd and a =
0.85 f c′b

Any method can simplify the size of d

using h = 1.1d

Maximum Reinforcement
Based on the limiting strain of
0.005 in the steel, x(or c) = 0.375d so
a = β1 ( 0.375d ) to find As-max
(β1 is shown in the table above)

Minimum Reinforcement
Minimum reinforcement is provided
even if the concrete can resist the
tension. This is a means to control
cracking.
3 f c′
Minimum required: As = ( bw d )
fy
(tensile strain of 0.004)
but not less than: As = 200 ( bw d )
fy
3 f c′ 200
where f c′ is in psi. This can be translated to ρ min = but not less than
fy fy
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Compression Reinforcement

If a section is doubly reinforced, it means there is steel

in the beam seeing compression. The force in the
compression steel at yield is equal to stress x area, Cs =
Ac’Fy. The total compression that balances the tension
is now: T = Cc + Cs. And the moment taken about the
centroid of the compression stress is
Mn = T(d-a/2)+Cs(a-d’)
where As‘ is the area of compression reinforcement, and d’ is the effective depth to the
centroid of the compression reinforcement

T-sections (pan joists)

T beams have an effective width, bE, that sees

compression stress in a wide flange beam or
joist in a slab system.

For interior T-sections, bE is the smallest of

L/4, bw + 16t, or center to center of beams

For exterior T-sections, bE is the smallest of

bw + L/12, bw + 6t, or bw + ½(clear distance to next beam)

When the web is in tension the minimum reinforcement required is the same as for rectangular
sections with the web width (bw) in place of b. Mn =Cw(d-a/2)+Cf(d-hf/2) (hf is height of flange or t)

When the flange is in tension (negative bending), the

minimum reinforcement required is the greater value of or

where f c′ is in psi, bw is the beam width, and bf is the effective flange width

Lightweight Concrete

Lightweight concrete has strength properties that are different from normalweight concretes, and
a modification factor, λ, must be multiplied to the strength value of f c′ . for concrete for some
specifications (ex. shear). Depending on the aggregate and the lightweight concrete, the value of
λ ranges from 075 to 0.85, 0.85, or 0.85 to 1.0. λ is 1.0 for normalweight concrete.

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Cover for Reinforcement

Cover of concrete over/under the reinforcement must be provided to protect the steel from
corrosion. For indoor exposure, 3/4 inch is required for slabs, 1.5 inch is typical for beams, and
for concrete cast against soil, 3 inches is typical.

Bar Spacing

Minimum bar spacings are specified to allow

proper consolidation of concrete around the
reinforcement.

Slabs

One way slabs can be designed as “one unit”-wide

beams. Because they are thin, control of
deflections is important, and minimum depths are
specified, as is minimum reinforcement for
shrinkage and crack control when not in flexure.
Reinforcement is commonly small diameter bars
and welded wire fabric. Maximum spacing
between bars is also specified for shrinkage and
crack control as five times the slab thickness not
exceeding 18”. For required flexure reinforcement
spacing the limit is three times the slab thickness
not exceeding 18”.

Shrinkage and temperature reinforcement (and minimum for flexure reinforcement):

A
Minimum for slabs with grade 40 or 50 bars: ρ = s = 0.002 or As-min = 0.002bt
bt
As
Minimum for slabs with grade 60 bars: ρ= = 0.0018 or As-min = 0.0018bt
bt

Shear Behavior

Horizontal shear stresses occur

along with bending stresses to cause
tensile stresses where the concrete
cracks. Vertical reinforcement is
required to bridge the cracks which
are called shear stirrups.

The maximum shear for design, Vu is the value at a distance of d from the face of the support.

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Nominal Shear Strength

The shear force that can be resisted is the shear stress × cross section area: Vc = ν c × bw d
The shear stress for beams (one way) υc = 2λ f c′ so φVc = φ 2λ f c′ bw d
where bw = the beam width or the minimum width of the stem.
φ = 0.75 for shear
λ = modification factor for lightweight concrete
One-way joists are allowed an increase of 10% Vc if the joists are closely spaced.

Av f yt d
Stirrups are necessary for strength (as well as crack control): Vs = ≤ 8 f c′ bw d (max)
s
where Av = area of all vertical legs of stirrup
s = spacing of stirrups
d = effective depth

For shear design:

Vu ≤ φVc + φVs φ = 0.75 for shear

Spacing Requirements
φVc
Stirrups are required when Vu is greater than . A minimum is required because shear failure
2
of a beam without stirrups is sudden and brittle and because the loads can vary with respect to
the design values.

greater of and

smaller of and

NOTE: section numbers are pre ACI 318-14

Economical spacing of stirrups is considered to be greater than d/4. Common φ Av f yt d

spacings of d/4, d/3 and d/2 are used to determine the values of φVs at which φVs =
s
the spacings can be increased.

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Development Length in Compression

db f y
ld = ≤ 0.0003 f y db or 8 in. minimum
50 λ f c′
Hook Bends and Extensions
db f y
The minimum hook length is ldh = but not less than the larger of 8db and 6 in
50 λ f c′

Modulus of Elasticity & Deflection

Ec for deflection calculations can be used with the transformed section modulus in the elastic
range. After that, the cracked section modulus is calculated and Ec is adjusted.

Code values:
Ec = 57,000 f c′ (normal weight) Ec = wc1.5 33 f c′ , wc = 90 lb/ft3 - 160 lb/ft3

Deflections of beams and one-way slabs need not be computed if the overall member thickness
meets the minimum specified by the code, and are shown in 7.3.1.1 (see Slabs). The span
lengths for continuous beams or slabs is taken as the clear span, ln.

Criteria for Flat Slab & Plate System Design

Systems with slabs and supporting beams, joists or columns typically have multiple bays. The
horizontal elements can act as one-way or two-way systems. Most often the flexure resisting
elements are continuous, having positive and negative bending moments. These moment and
shear values can be found using beam tables, or from code specified approximate design factors.
Flat slab two-way systems have drop panels (for shear), while flat plates do not.

Two way shear at columns is resisted by the thickness of the slab at a perimeter of d/2 away from
the face of the support by the shear stress × cross section area: V c = ν c × bo d

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The shear stress (two way) υc = 4 λ f c′ so φVc = φ 4 λ f c′ bo d

where bo = perimeter length.
φ = 0.75 for shear
λ = modification factor for lightweight concrete

Criteria for Column Design

(American Concrete Institute) ACI 318-14 Code and Commentary:

Pu ≤ φcPn where
Pu is a factored load
φ is a resistance factor
Pn is the nominal load capacity (strength)

Load combinations, ex: 1.4D (D is dead load)

1.2D + 1.6L (L is live load)
1.2D + 1.6Lr + 0.5W
(W is wind load)
0.90D + 1.0W

For compression, φc = 0.75 and Pn = 0.85Po for spirally reinforced,

φc = 0.65 Pn = 0.8Po for tied columns where Po = 0.85 f c′( Ag − Ast ) + f y Ast and Po
is the name of the maximum axial force with no concurrent bending moment.
Ast
Columns which have reinforcement ratios, ρ g = , in the range of 1% to 2% will usually
Ag
be the most economical, with 1% as a minimum and 8% as a maximum by code.
Bars are symmetrically placed, typically.

Columns with Bending (Beam-Columns)

Concrete columns rarely see only axial force

and must be designed for the combined
effects of axial load and bending moment.
The interaction diagram shows the reduction
in axial load a column can carry with a
bending moment.

Design aids commonly present the

interaction diagrams in the form of
load vs. equivalent eccentricity for
standard column sizes and bars used.

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Eccentric Design

The strength interaction diagram is dependent upon the strain developed in the steel
reinforcement.

If the strain in the steel is less than the yield stress, the section is said to be compression
controlled.

Below the transition zone, where the steel starts to yield, and when the net tensile strain in the
reinforcement exceeds 0.005 the section is said to be tension controlled. This is a ductile
condition and is preferred.

Rigid Frames

Monolithically cast frames with beams

and column elements will have members
with shear, bending and axial loads.
Because the joints can rotate, the effective
length must be determined from methods
like that presented in the handout on Rigid
Frames. The charts for evaluating k for
non-sway and sway frames can be found
in the ACI code.

Frame Columns

Because joints can rotate in frames, the

effective length of the column in a frame
is harder to determine. The stiffness
(EI/L) of each member in a joint
determines how rigid or flexible it is. To find k, the relative stiffness, G or Ψ, must be found for
both ends, plotted on the alignment charts, and connected by a line for braced and unbraced
fames.

Σ EI l
G =Ψ = c

Σ EI l
b

where
E = modulus of elasticity for a member
I = moment of inertia of for a member
lc = length of the column from center to center
lb = length of the beam from center to center

• For pinned connections we typically use a value of 10 for Ψ.

• For fixed connections we typically use a value of 1 for Ψ.

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Braced – non-sway frame Unbraced – sway frame

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Factored Moment Resistance of Concrete Beams, φMn (k-ft) with f’c = 4 ksi, fy = 60 ksia
Approximate Values for a/d
0.1 0.2 0.3
Approximate Values for ρ
b x d (in) 0.0057 0.01133 0.017
10 x 14 2 #6 2 #8 3 #8
53 90 127
10 x 18 3 #5 2 #9 3 #9
72 146 207
10 x 22 2 #7 3 #8 (3 #10)
113 211 321
12 x 16 2 #7 3 #8 4 #8
82 154 193
12 x 20 2 #8 3 #9 4 #9
135 243 306
12 x 24 2 #8 3 #9 (4 #10)
162 292 466
15 x 20 3 #7 4 #8 5 #9
154 256 383
15 x 25 3 #8 4 #9 4 #11
253 405 597
15 x 30 3 #8 5 #9 (5 #11)
304 608 895
18 x 24 3 #8 5 #9 6 #10
243 486 700
18 x 30 3 #9 6 #9 (6 #11)
385 729 1074
18 x 36 3 #10 6 #10 (7 #11)
586 1111 1504
20 x 30 3 # 10 7#9 6 # 11
489 851 1074
20 x 35 4 #9 5 #11 (7 #11)
599 1106 1462
20 x 40 6 #8 6 #11 (9 #11)
811 1516 2148
24 x 32 6 #8 7 #10 (8 #11)
648 1152 1528
24 x 40 6 #9 7 #11 (10 #11)
1026 1769 2387
24 x 48 5 #10 (8 #11) (13 #11)
1303 2426 3723
a
Table yields values of factored moment resistance in kip-ft with reinforcement indicated. Reinforcement choices
shown in parentheses require greater width of beam or use of two stack layers of bars. (Adapted and corrected from
Simplified Engineering for Architects and Builders, 11th ed, Ambrose and Tripeny, 2010.

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Column Interaction Diagrams

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Column Interaction Diagrams

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Beam / One-Way Slab Design Flow Chart

Collect data: L, ω, γ, ∆llimits, hmin; find beam

charts for load cases and ∆actual equations
(self weight = area x density)

Collect data: load factors, fy, f’c

Find V’s & Mu from constructing diagrams

or using beam chart formulas with the
factored loads (Vu-max is at d away
from face of support)

Determine Mn required by
Assume b & d (based Mu/ φ, choose method
on hmin for slabs) Chart (Rn vs ρ)

Find Rn off chart with fy, f’c and

Select ρmin ≤ ρ ≤ ρmax
select ρmin ≤ ρ ≤ ρmax

Choose b & d combination

based on Rn and hmin (slabs),
estimate h with 1” bars (#8)

Calculate As = ρbd

Select bar size and spacing to fit

width or 12 in strip of slab and not Increase h, find d*
exceed limits for crack control

Find new d / adjust h; NO

Is ρmin ≤ ρ ≤ ρmax ?

YES

Increase h, find d
Calculate a, φMn

NO
Is Mu ≤ φMn?

Yes (on to shear reinforcement for beams)

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Beam / One-Way Slab Design Flow Chart - continued

Beam, Adequate for Flexure

Determine shear capacity of plain

concrete based on f’c, b & d, φVc

NO NO
Is Vu (at d for beams) ≤ φVc? Beam?

YES

Determine φVs = (Vu - φVc) Increase h and re-evaluate

NO
Beam? flexure (As and φMn of
previous page)*

YES NO
Is φVs ≤ φ ?
NO (4φVc)
Is Vu < ½ φVc?
YES

YES Determine s & Av

Find where V = φVc

and provide minimum
Av and change s

Find where V = ½ φVc

and provide stirrups
just past that point

Yes (DONE)

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