Design of Singly Reinforced Sections.doc

7/28/2019 Design of Singly Reinforced Sections.doc

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DESIGN FOR FLEXURE -

SINGLY REINFORCED RECTANGULAR SECTIONS

The design of singly reinforced rectangular cross-sections for flexure may follow one of

several methods. All of the methods described below are based on satisfying the same set

of fundamental equations of equilibrium. Note that it is assumed that the cross-section isunderreinforced, such that the tension reinforcement has yielded before failure.

b

dh

As sy

.003

c

NA

a= 1c

.85f'c

C = .85f'c ab

T = As fy

Force Equilibrium ("C=T"):

b'f85.

fAa

c

ys= (1)

Moment Equilibrium (assuming flexural strength provided = required strength):

== 2a

dfAMM ysnu (2)

Design for Cross-Sections of Known Dimensions

In this case, b and d (or h) are assumed to be known prior to design. These dimensions

may have been established by architectural considerations, or for other reasons (such as

repeating beam dimensions across a span for formwork optimization). This is also

typically the case for one-way slab systems, in which the width is assumed to be a 12-inch wide strip and the depth is established by shear or deflection criteria.

All three methods involve simultaneously satisfying both of the above equations. For thecase of fixed dimensions, only the parameters a and As are unknown in these equations.

Method 1: Trial and error (iterative) approachMethod 2: Simultaneous algebraic solution

Method 3: Solution by design aids

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Design Problem

The dimensions of the beam shown below are fixed as 10" x 20" for architectural reasons.The beam supports an indoor walkway on a 20-foot simply-supported span. The

superimposed dead load is 1000 lb/ft and the superimposed live load is 1000 lb/ft. Select

flexural reinforcement for the beam if the material properties to be used are f'

c = 4000 psiand fy = 60,000 psi. (Ignore the presence of any top bars used to facilitate stirrup

placement.)

h=20"

b=10"

LL = 1.00 k/ft

DL = 1.00 k/ft

20 ft.

Beam self-weight (assume concrete weighs 145 pcf):

( )( ) ft/k20.0ft

k145.

in12

ft1.in20.in10DL

3

2

sw =

=

Total factored load:

( ) ( ) ft/k04.3ft/k00.16.1ft/k20.000.12.1w6.1w2.1w LDu =++=+=

Total factored moment (design for maximum moment along span):

( ) ( )kips.in1824kips.ft152

8

ft20ft/k04.3

8

wM

22u

u ====

Assume 1.5 in. clear cover, 0.5 in. stirrup, and 1.0" diameter bars d = 17.5"

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Method 1: Trial and Error (Iterative) Approach

For this method, one of the variables a or As must be assumed, and then Equations 1 and2 are solved iteratively until convergence.

Typically, a is assumed and As is calculated by Equation 1. Then a is calculated byEquation 2 using the value for As calculated from Equation 1. The computed value of a is

checked against the original assumption and the process is repeated until convergence. A

typical starting assumption is that a = d/4 for beams a = d/6 for slabs. In most cases,these will yield slightly conservative values for a, and thus slightly conservative values

for As.

An alternative starting assumption is to assume the moment arm jd, where jd = (d - a/2).In this case, typical starting assumptions are jd = .875d for beams and jd = .925d for

slabs. Using this process, As is first calculated from Equation 2, and then the assumption

is checked using Equation 1.

Note: When using this method, it is not necessarily required that the designer iterate

until convergence is achieved. If the initial assumption is slightly conservative, then theresult will be a slightly conservative design. The designer may check the capacity of the

section based on the assumption, and if adequate and not overly conservative, complete

the design at this stage. This philosophy is used in the solution below.

Solution by Method 1

Assume jd:

( ) .in31.15.in5.17875.d875.2adjd === =

Calculate As from Equation 2:

( )( )( )2

y

us

ysnu

.in21.2.in31.15ksi6090.

kips.in1824

2

adf

MA

2

adfAMM

==

=

==

At this point, iterate further or assume result is acceptable and analyze section:

*We will not iterate, but if we had, we would get a = 3.83 in. and As = 2.17 in.2

Try 3 #8 bars (As = 2.37 in.2)

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Analyze section:

Actual d: d = 20" - 1.5" - 0.5" - 1/2(1.0") = 17.5"

Check reinforcement limits (b from Text Table A-5):

( ( )( )( )( )

)OK(319.0)85(.375.0375.0d

a239.0

.in5.17

.in18.4

d

a

.in18.4.in10ksi485.

ksi60.in37.2

b'f85.

fAa

1

TCLtt

2

c

ys

===


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or, in alternative form:

=

'c

2u

y

'c

f85.

bd

M2

11f

f85.(5)

Either Equation 4 or Equation 5 may be used to solve directly for the required amount of

reinforcement.


Solve for required As using Equation 4:

( )( )( )

( )( )( ) ( )

( )( )

( ) ( )2

s

22

u

.in17.2psi400085.

psi662211

psi60000

psi400085..in5.17.in10A

psi662ksi662..in5.17.in1090.

kips.in1824bdM

=

=

===

Try 3 #8 bars (As = 2.37 in.2)

Check reinforcement limits, analyze section, check cover & spacing as before

Method 3: Solution by Design Aids

This method is identical to the method outlined above (Method 2), except design aids are

used for the solution of the closed form equation.

Repeating Equation 5 from above:

=

'

c

2

u

y

'

c

f85.

bd

M2

11f

f85.(5)

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Let us define the resistance coefficient Rn:

2

un

bd

MR

=

Several design aids have been published which tabulate Rn as a function of for variouscombinations of f'c and fy. In using these design aids, Rn is typically stated in units of psi.

An example of such a design aid is shown below:

From: McCormac, Design of Reinforced Concrete

Note: MacGregor Text has a similar set of design aid tables (Table A-3), except for one

important difference. MacGregor defines the resistance coefficient as kn, which is

identical to Rn. However, values are tabulated for kn, not kn. Therefore, one must

essentially enter the tables after calculating Mu/bd2 (in units of psi) rather than Mu/bd

2.

The MacGregor design aid is shown on the next page:

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From: MacGregor, Reinforced Concrete - Mechanics and Design

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To simplify the process, a similar design aid has been published by several authors. This

alternate design aid is a nondimensionalized design aid based on a modified form of

Equation 5. Only one table is needed, regardless of the choice of f'c and fy.

Define the reinforcement index :

'c

y

ff=

Now rewrite Equation 5 in terms of:

( )

=

=

=

+

=

=

'

c

2

u2

'

c

2

u

2

'

c

2

u

2

'

c

2

u

'

c

2

u

fbd

M

7.1

85.

fbd

M2

85.2

85.

85.

fbd

M2

185.85.

21

85.

fbd

M2

185.

1

85.

fbd

M2

1185.

( )=

59.01fbd

M'

c

2

u (6)

A single table provides values for the term on the left side of the equation as a function of

the reinforcement index w. This table is found in several publications, including PCA's

Notes of 318-99, and is shown below:

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From: PCA, Notes on ACI 318-99

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Solve for required using design aids for Equation 5:

( )( )( )

( )( )psi596ksi596.

.in5.17.in10

kips.in1824

bd

Mk

or

psi662ksi662.

.in5.17.in1090.

kips.in1824

bd

MR

22

un

22

un

==

==

==

==

Entering table in MacGregor text (Table A-3) with kn = 596 psi, f'c = 4 ksi, fy = 60 ksi,

and interpolating linearly:

( )( )( ) 2requiredrequired,s

required

.in17.2.in5.17.in100124.bdA

0124.

===

=

Try 3 #8 bars (As = 2.37 in.2)


Alternatively, using the design aid based on Equation 6:

( )( )( ) ( )1654.

ksi4.in5.17.in1090.

kips.in1824

fbd

M2'

c2

u =

=

Entering PCA Notes table with value of 0.1654, and interpolating linearly:

( )( )

( )


y

'c

requiredrequired

required

.in17.2.in5.17.in100124.bdA

0124.ksi60

ksi41858.

f

f

1858.

===

===

=

Try 3 #8 bars (As = 2.37 in.2)


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Design for Cross-Sections of Unknown Dimensions

In this case, b and d (or h) are not known prior to design. Therefore, we have threeparameters that the designer must choose - b, h, and As (assuming f

'c and fy are

established). These three parameters must only satisfy two equilibrium equations, so

there are mathematically an infinite number of possible design combinations. Typically,an assumption is made regarding one of the parameters, usually A s (). Examples of

common design assumptions are:

= 0.5b

= 0.375b (this corresponds to 0.50max for previous editions of ACI 318)

= .01 = 1%

The design at this stage is also complicated by the fact that the self-weight of the beam isunknown.

Once a design assumption is made, the process of beam design generally relies on one ofthe design aids presented in the previous section. It is again emphasized that the designaids are identical to the equations derived in the previous section (Equations 3, 4, 5, & 6),

such that solution by these equations is essentially identical to solution using design aids.

Design Problem (same problem as before with b and d unknown)

The beam below supports an indoor walkway on a 20-foot simply-supported span. The

superimposed dead load is 1000 lb/ft and the superimposed live load is 1000 lb/ft. Select

flexural reinforcement for the beam if the material properties to be used are f 'c = 4000 psiand fy = 60,000 psi. (Ignore the presence of any top bars used to facilitate stirrup

placement.)

h

b

LL = 1.00 k/ft

DL = 1.00 k/ft

20 ft.

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Beam self-weight (assume concrete weighs 145 pcf; assume 12"x24" cross-section):

( )( ) ft/k29.0ft

k145.

in12

ft1.in24.in12DL

3

2

sw =

=




( )( )kips.in1890kips.ft5.157

8

ft20ft/k15.3

8

wM

22u

u ====

Assume 1.5 in. clear cover, 0.5 in. stirrup, and 1.0" diameter bars d = h - 2.5"

Design assumption:

( )

( )( )( )

1605.ksi4

ksi600107.

f

f

0107.0285.375.0375.0

'c

y

b

===

===

Enter design aid ( design aid used here):

1453.fbd

M'c

2

u =

Solve for required bd2:

( )( ) ( )( )( )

3

'c

urequired

2 .in36131453.ksi490.

kips.in1890

1453.f

Mbd =

==

Possible combinations of b and d:

b = 10", d = 19" bd2 = 3610 in.3

b = 12", d = 17.5" bd2 = 3675 in.3

b = 14", d = 16" bd2 = 3584 in.3

Note: This calculation gets us "in the ballpark". Selection of a b-d combination thatgives us less than the calculated requirement of 3625 in.2 will simply mean that we need a

slightly higher reinforcement ratio. Any of these three combinations will therefore work.

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Try b = 12", h = 20" (d = 17.5")

Now, proceed as for a design with b and d known.

Beam self-weight (assume concrete weighs 145 pcf):

( )( ) ft/k242.0ft

k145.

in12

ft1.in20.in12DL

3

2

sw =

=




( ) ( )kips.in1854kips.ft5.1548

ft20ft/k09.3

8

wM

22

uu ====

Assume 1.5 in. clear cover, 0.5 in. stirrup, and 1.0" bars d = 17.5"

Using the design aid based on Equation 6:

( )( )( ) ( )1401.

ksi4.in5.17.in1290.

kips.in1854

fbd

M2'

c2

u =

=

Entering PCA Notes table with value of 0.1401, and interpolating linearly:

( )( )

( )


y

'c

requiredrequired

required

.in16.2.in5.17.in120103.bdA

0103.ksi60

ksi41541.

f

f

1541.

===

===

=

Try 3 #8 bars (As = 2.37 in.2)


(This section works)

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Design by Simplified Methods

Simplified design methods involve the same basic approach as the methods outlinedpreviously, but with further simplifications (i.e. assumptions) that reduce the effort

involved in the design process. The method outlined in this section is adapted from the

following references:

Fanella, D. and Ghosh, S.K. (editors), Simplified Design - Reinforced Concrete

Buildings of Moderate Size and Height, Portland Cement Association, SecondEdition, 1993.

Fanella, D., "Time-saving design aids for reinforced concrete," Structural

Engineer, August 2001, pp. 38-41.

Consider the simplifying assumption that

b375. = (7A)

A set of design aids (Tables 1A through 5A) have been developed based on this

simplifying assumption.

Note: A second set of design aids (Tables 1B through 5B) have been developed based on

a similar simplifying assumption:

b21 = (7B)

Using the assumption of a given as a multiple ofb, the values of and are tabulated

in Table 1A and 2A (or 1B and 2B) for various combinations of f'c and fy.

A constant C1 can then be calculated where:

kipsftofunitsinisMand,inchesofunitsinaredandbwhere

MCbd

u

u1required2

=

(8)

Equation 8 can be used to help size the cross-section when dimensions have not been

established prior to design.

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The constant C1 can be determined by manipulation of Equation 6:

( )

( )u1'

c

u2

'

c

2

u

MC59.01f

Mbd

59.01fbd

M

=

=

=

( )=

59.01f

1C

'

c

1

(9)

Values of C1 are tabulated in Tables 3A and 3B.

The value of j, where jd is the length of the moment arm between the resultant internaltensile and compressive forces, can also be computed directly as a function of f 'c and fy:

d2

bf85.

fA

1j

d2

a1j

2

adjd

'c

ys

=

=

=

7.11j

= (10)

Values of j are tabulated in Tables 4A and 4B.

Finally, a required quantity of reinforcement can be computed based on the following

expression:

kipsftofunitsinMand,.inofunitsinA,inchesofunitsinisdwhere

dC

MA

u2

s

2

u

required,s

=

(11)

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The constant C2 can be determined by manipulation of Equation 2:

dC

M

jdf

MA

jdfA2

adfAM

2

u

y

us

ysysu

=

=

=

=

jfC y2 = (12)

Values of C2 are tabulated in Tables 5A and 5B.

Design Problem (same problem as before with b and d unknown)

Consider the design example solved previously, and use the design assumption:

b375.0 =

Note that as before, we assume a self weight for the beam such that Mu = 157.5 ft.-kips:

From Table 1A = .0107

From Table 2A = .1604

From Table 3A C1 = 23.0

(bd2)required = 23.0(157.5) = 3623 in.2

Try b = 12", d = 17.5" (h = 20")

Correcting our self-weight assumption to the values for the selected beams size, we get

Mu = 154.5 ft.-kips:

From Table 4A j = 0.906

From Table 5A C2 = 4.08(As)required = (154.5)/[(4.08)(17.5)] = 2.16 in.

2

Try 3 #8 bars (As = 2.37 in.2)

Check reinforcement limits, analyze section, check cover &

spacing as before (This section works)

Technically, the designer only needs to look up values in Table 3A and Table 5A. The

values in tables 1A, 2A, and 4A are only provided for reference.

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40000 50000 60000 75000

3000 0 .0 139 0. 01 03 0. 00 80 0 .0 058

3500 0 .0 162 0. 01 20 0. 00 94 0 .0 068

4000 0 .0 186 0. 01 38 0. 01 07 0 .0 0784500 0 .0 203 0. 01 50 0. 01 17 0 .0 085

5000 0 .0 218 0. 01 62 0. 01 26 0 .0 091

6000 0 .0 246 0. 01 82 0. 01 41 0 .0 103

7000 0 .0 267 0. 01 98 0. 01 54 0 .0 112

8000 0 .0 284 0. 02 11 0. 01 63 0 .0 119

9000 0 .0 319 0. 02 37 0. 01 84 0 .0 134

10000 0 .0 355 0. 02 63 0. 02 04 0 .0 148

11000 0 .0 390 0. 02 89 0. 02 25 0 .0 163

12000 0 .0 426 0. 03 16 0. 02 45 0 .0 178

40000 50000 60000 75000 40000 50000 60000 75000

3000 0 .18 56 0 .1 721 0 .16 04 0. 14 55 3000 26.9 28.7 30.6 33.43500 0 .18 56 0 .1 721 0 .16 04 0. 14 55 3500 23.0 24.6 26.2 28.6

4000 0 .18 56 0 .1 721 0 .16 04 0. 14 55 4000 20.2 21.6 23.0 25.1

4500 0 .18 01 0 .1 670 0 .15 56 0. 14 12 4500 18.4 19.7 21.0 22.9

5000 0 .17 47 0 .1 619 0 .15 09 0. 13 69 5000 17.0 18.2 19.4 21.2

6000 0 .16 38 0 .1 518 0 .14 15 0. 12 84 6000 15.0 16.1 17.1 18.7

7000 0 .15 28 0 .1 417 0 .13 21 0. 11 98 7000 13.7 14.7 15.6 17.1

8000 0 .14 19 0 .1 316 0 .12 26 0. 11 13 8000 12.8 13.7 14.7 16.0

9000 0 .14 19 0 .1 316 0 .12 26 0. 11 13 9000 11.4 12.2 13.0 14.3

10000 0 .14 19 0 .1 316 0 .12 26 0. 11 13 10000 10.3 11.0 11.7 12.8

11000 0 .14 19 0 .1 316 0 .12 26 0. 11 13 11000 9.3 10.0 10.7 11.7

12000 0 .14 19 0 .1 316 0 .12 26 0. 11 13 12000 8.5 9.2 9.8 10.7

40000 50000 60000 75000 40000 50000 60000 75000

3000 0.891 0.899 0.906 0.914 3000 2.67 3.37 4.08 5.14

3500 0.891 0.899 0.906 0.914 3500 2.67 3.37 4.08 5.14

4000 0.891 0.899 0.906 0.914 4000 2.67 3.37 4.08 5.14

4500 0.894 0.902 0.908 0.917 4500 2.68 3.38 4.09 5.16

5000 0.897 0.905 0.911 0.919 5000 2.69 3.39 4.10 5.17

6000 0.904 0.911 0.917 0.924 6000 2.71 3.42 4.13 5.20

7000 0.910 0.917 0.922 0.930 7000 2.73 3.44 4.15 5.23

8000 0.917 0.923 0.928 0.935 8000 2.75 3.46 4.18 5.26

9000 0.917 0.923 0.928 0.935 9000 2.75 3.46 4.18 5.26

10000 0.917 0.923 0.928 0.935 10000 2.75 3.46 4.18 5.26

11000 0.917 0.923 0.928 0.935 11000 2.75 3.46 4.18 5.26

12000 0.917 0.923 0.928 0.935 12000 2.75 3.46 4.18 5.26

TABLE 4a TABLE 5a

f'c

(psi)

j

TABLE 3aTABLE 2a

fy (psi)

f'c

(psi)

(corresponding to

0.375b)C1

f'c

(psi)

C2

TABLE 1a

fy (psi)

fy (psi)

fy (psi)

f'c

(psi)

0.375b

f'c

(psi)

fy (psi)

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40000 50000 60000 75000

3000 0. 01 86 0 .013 8 0. 01 07 0 .0 07 8

3500 0. 02 17 0 .016 1 0. 01 25 0 .0 09 1

4000 0. 02 47 0 .018 4 0. 01 43 0 .0 10 34500 0. 02 70 0 .020 0 0. 01 56 0 .0 11 3

5000 0. 02 91 0 .021 6 0. 01 68 0 .0 12 2

6000 0. 03 28 0 .024 3 0. 01 89 0 .0 13 7

7000 0. 03 57 0 .026 4 0. 02 05 0 .0 14 9

8000 0. 03 78 0 .028 1 0. 02 18 0 .0 15 8

9000 0. 04 26 0 .031 6 0. 02 45 0 .0 17 8

10000 0. 04 73 0 .035 1 0. 02 72 0 .0 19 8

11000 0. 05 20 0 .038 6 0. 03 00 0 .0 21 8

12000 0. 05 68 0 .042 1 0. 03 27 0 .0 23 7

40000 50000 60000 75000 40000 50000 60000 75000

3000 0. 24 75 0. 229 4 0 .21 38 0 .1 94 0 3000 21.0 22.4 23.8 25.93500 0. 24 75 0. 229 4 0 .21 38 0 .1 94 0 3500 18.0 19.2 20.4 22.2

4000 0. 24 75 0. 229 4 0 .21 38 0 .1 94 0 4000 15.8 16.8 17.8 19.4

4500 0. 24 02 0. 222 7 0 .20 75 0 .1 88 3 4500 14.4 15.3 16.3 17.7

5000 0. 23 29 0. 215 9 0 .20 12 0 .1 82 6 5000 13.3 14.2 15.0 16.4

6000 0. 21 84 0. 202 4 0 .18 86 0 .1 71 2 6000 11.7 12.5 13.3 14.4

7000 0. 20 38 0. 188 9 0 .17 61 0 .1 59 8 7000 10.6 11.3 12.1 13.2

8000 0. 18 92 0. 175 4 0 .16 35 0 .1 48 4 8000 9.9 10.6 11.3 12.3

9000 0. 18 92 0. 175 4 0 .16 35 0 .1 48 4 9000 8.8 9.4 10.0 10.9

10000 0. 18 92 0. 175 4 0 .16 35 0 .1 48 4 10000 7.9 8.5 9.0 9.8

11000 0. 18 92 0. 175 4 0 .16 35 0 .1 48 4 11000 7.2 7.7 8.2 9.0

12000 0. 18 92 0. 175 4 0 .16 35 0 .1 48 4 12000 6.6 7.1 7.5 8.2

40000 50000 60000 75000 40000 50000 60000 75000

3000 0.854 0.865 0.874 0.886 3000 2.56 3.24 3.93 4.98

3500 0.854 0.865 0.874 0.886 3500 2.56 3.24 3.93 4.98

4000 0.854 0.865 0.874 0.886 4000 2.56 3.24 3.93 4.98

4500 0.859 0.869 0.878 0.889 4500 2.58 3.26 3.95 5.00

5000 0.863 0.873 0.882 0.893 5000 2.59 3.27 3.97 5.02

6000 0.872 0.881 0.889 0.899 6000 2.61 3.30 4.00 5.06

7000 0.880 0.889 0.896 0.906 7000 2.64 3.33 4.03 5.10

8000 0.889 0.897 0.904 0.913 8000 2.67 3.36 4.07 5.13

9000 0.889 0.897 0.904 0.913 9000 2.67 3.36 4.07 5.13

10000 0.889 0.897 0.904 0.913 10000 2.67 3.36 4.07 5.13

11000 0.889 0.897 0.904 0.913 11000 2.67 3.36 4.07 5.13

12000 0.889 0.897 0.904 0.913 12000 2.67 3.36 4.07 5.13

TABLE 4b TABLE 5b

f'c

(psi)

j

TABLE 3bTABLE 2b

fy (psi)

f'c

(psi)

(corresponding to

0.5 b)C1

f'c

(psi)

C2

TABLE 1b

fy (psi)

fy (psi)

fy (psi)

f'c

(psi)

0.5 b

f'c

(psi)

fy (psi)

( )

kipsftofunitsinisMand,inchesofunitsinaredandbwhere

MCbd

u

u1required2

=

kipsftofunitsinMand,.inofunitsinA,inchesofunitsinisdwhere

dC

MA

u2

s

2

u

required,s

=

20

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