Design Methods and Requirements

ACI Building Code

The ACI Building Code Requirements for Reinforced Concrete is based partly on empirical and mostly rational data.

Strength Design and Working Stress Design Methods (old code)

The working stress method focuses on conditions at service load and the strength design method focuses on loads when failure may be imminent.

Working Stress Method

The working stress method or alternate design method has set limits on the stresses allowed under service loads (working loads).

Alternate Design Method ACI Appendix A of old code.

Some of the obstacles to the working stress method are as follows:

1. No account for different types of loads.2. Creep and shrinkage are not easily accounted for in elastic stresses.3. Stress is not proportional to strain at concrete crushing therefore

the inherent factor of safety is unknown.

Strength Design Method

The strength design method or ultimate strength method has the service loads increased by sufficient factors to obtain the load at which failure is considered to be imminent. The strength provided must be greater than the required strength to carry these factored loads.

Unified Design Procedure

This is a major part of the ACI 318-02 code revision.

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Comments on Design Methods

Although strength design is currently the philosophy employed most widely, serviceability must be maintained. Working stress is stilled required to calculate deflections and cracking of structure in service load conditions.

General

The two primary factors used to provide safety in the ultimate strength method of design are U the overload factors (load factors) in ACI 9.2 (new factors) or ACI C.2 (old factors) and the understrength factors (strength reduction factors) in ACI 9.3 (new factors) or ACI C.3 (old factors).

Design Strength<Required Strength

Traditional ACI Code - Appendix C

The structure must be designed for the most severe of any load combination. The load factors for some basic combinations are as follows per ACI C.2 (new code and old factors):

1. U = 1.4D + 1.7L Dead – D & Live - L2. U = 0.75(1.4D + 1.7L +1.7W) Wind – W

U = 0.75(1.4D + 1.7L) +1.6W 1.3W w/o direction factor 3. U = 0.9D + 1.3W

U = 0.9D + 1.6W 1.3W w/o direction factor4. U = 0.75(1.4D + 1.7L +1.7(1.1E)) Seismic – E

U = 0.75(1.4D + 1.7L) +1.0E 1.4E for service load5. U = 0.9D + 1.3(1.1E)

U = 0.9D + 1.0E 1.4E for service load6. U = 1.4D + 1.7L + 1.7H Earth pressure - H7. U = 0.9D + 1.7H8. U = 1.4D + 1.7L + 1.4F Fluid pressure - F 9. U = 0.9D + 1.4F10. U = 0.75(1.4D + 1.4T + 1.7L) Settle, creep, shrink, temp - T

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11. U = 1.4D + 1.4T

No account for D+L+(2 others), but old codes use 2/3 of basic factors.

In addition, for intermediate moment frames shear must be check with seismic load doubled per ACI 21.12(b). Applies to members resisting seismic loading (frame).

The strength reduction factors accounting for adverse variations in material strength, workmanship, dimensions, control & degree of supervision, and importance of the member to the building structure are as follows per ACI C.3 (new code and old factors):

1. = 0.90 Flexure, with or without axial tension2. = 0.90 Axial tension

Tension controlled sections (t > 0.005)3. = 0.85 Shear and torsion4. = 0.75 Compression members, spirally reinforced

Compression controlled sections (t < 0.002 i.e. balance)5. = 0.70 Compression members, others

Compression controlled sections (t < 0.002 i.e. balance)6. = 0.70 Bearing on concrete7. = 0.65 Bending in plain concrete

NEW ACI Code – Chapter 9

The structure must be designed for the most severe of any load combination. These are the factors in the ACI 318-02 code. The load factors for some basic combinations are as follows per ACI 9.2 (ASCE 7-88):

1. U = 1.4DU = 1.4(D + F)

2. U = 1.2D + 1.6L + 0.5(Lr or S or R)U = 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3. U = 1.2D + 1.6(Lr or S or R) + (0.5L or 0.8W)U = 1.2D + 1.6(Lr or S or R) + (1.0L or 0.8W)

4. U = 1.2D + 1.3W + 0.5L + 0.5(Lr or S or R)

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U = 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R)5. U = 1.2D + 1.5E + (0.5L or 0.2S)

U = 1.2D + 1.0E + 1.0L + 0.2S6. U = 0.9D - (1.3W or 1.5E) (1.0E – steel) Typo in text

U = 0.9D + 1.6W + 1.6HU = 0.9D + 1.0E + 1.6H

All the same notes on W and E as in C.2 (direction & service)

The strength reduction factors accounting for adverse variations in material strength, workmanship, dimensions, control & degree of supervision, and importance of the member to the building structure are as follows per ACI 9.3 (ASCE 7-88):

1. = 0.80 Flexure, with or without axial tension = 0.90 Tension controlled sections

2. = 0.80 Axial tension = 0.90 Tension controlled sections

3. = 0.75 Shear and torsion4. = 0.70 Compression members, spirally reinforced

Compression controlled sections5. = 0.65 Compression members, others

Compression controlled sections6. = 0.65 Bearing on concrete7. = 0.55 Bending in plain concrete

Basis of Nominal Flexural Strength

The modern analytical approach to reinforced concrete beam design should include a linear strain distribution and a nonlinear stress distribution as shown below. The following equations can be derived statically assuming a ductile failure of steel at yield stress:

k1 - fraction of stress area to rectangular blockk2 - fraction of x to centroidk3 - fraction of f’c at maximum stress

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C k k f xbc 1 3 '

T A fs y

F T Cx 0

xA fk k f b

s y

c

1 3 '

M M T d k xc n ( )2

M A f dkk k

A ff bn s ys y

c

2

1 3 '

It should be noted that only the ratio of k2/(k1k3) need be know and not the individual constants. Tests indicate that the variation is between 0.55 and 0.63.

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Whitney Rectangular Stress Distribution

The computation of flexural strength based on the approximately parabolic stress may be done, but the designer needs a simpler method based on static equilibrium. The Whitney rectangular stress block as shown below is the basis for standard design. The average stress of 0.85fc' is used over a rectangular depth a=1x (ACI 10.2.7.1). The value for 1 should be taken as 0.85 for fc' < 4000 psi, and 0.05 less for each 1000 psi of fc' in excess of 4000 psi (10.2.7.3). The value of 1 need not be less than 0.65 (See ACI SP-17(97) FLEXURE 1 for this and more). This can be written as follows:

1 085 0 054000

10000 65

. .

'.

f c

The static derivation is as follows:

C f bac0 85. '

T A fs y

F T Cx 0

6

aA ff b

s y

c

0 85. '

M M A f d ac n s y ( )

2

It should be noted that the ratio of k2/(k1k3) turn out to be 0.59 which is right in the middle between 0.55 and 0.63.

Nominal Moment Strength Mn - Rectangular Sections Having Tension Reinforcement Only

The above definitions are for sections that are singly reinforced. Minimum concrete cover around the reinforcing per ACI 7.7 provides a minimum, but protection from fire may require more. Typical values for cast-in-place concrete include the following:

3” Cast against and permanently exposed to earth (All)2” Exposed to weather or earth (#6 & Larger)1-1/2” Beam & columns not exposed to weather or earth (All)3/4” Slab walls & joists not exposed to weather or earth (#11

& smaller)

It should be obvious that all the concrete in tension was ignored and the effective depth d (distance from tensile steel to compressive face of concrete) was used in computation.

The ACI Strength Design Method includes the provision for basic load factors and a strength reduction factor of 0.9 (tension controls) down to 0.65 (tied members). The basic equations for the ultimate moment caused by the load Mu and the nominal moment strength of the beam Mn is as follows:

M Mn u

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The following assumptions apply (ACI 10.2)

1. The strength of members shall satisfy equilibrium and compatibility.2. Strain in the steel and concrete shall be directly proportional to the

distance from the neutral axis.3. The maximum usable strain cu at the extreme compression fiber shall

be 0.003.4. The tensile strength of the concrete is neglected5. The modulus of elasticity of steel is 29000 ksi.6. Whitney stress block applies as previously defined.

Problem 3.3 case 2 - Compute the nominal flexural strength Mn and the service moment capacity (MD + ML) assuming the total service moment is 60% live load. Use d=19.5 in, bw=12 in, 3-#10 bars, f’c=3500psi and fy=60,000psi.

Since f’c=3500psi < 4000 psi

Else OR FLEXURE 1

Find what value to use from t

Since a = 6.403 = 1x , x = a/1x = 6.403/0.85 = 7.533

Find t from the strain diagram by geometry as follows:

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Since t > 0.004, meets code, but t < 0.005, transition zone

t = 0.005 = 0.9 tension controlst = 0.004766 = 0.8805 Our caset = 0.002 = 0.65 compression controls (tied)

Since ML=0.6Ms, MD=0.4Ms and Mu=1.2MD+1.6ML then:

Problem 3.3 case 5 - Compute the nominal flexural strength Mn and the service moment capacity (MD + ML) assuming the total service moment is 60% live load. Use d=36.25 in, bw=18 in, 8-#11 bars, f’c=4000psi and fy=60,000psi.

1 0 85 0 054000

10000 85 0 05

4000 40001000

0 85 0 65

. .

'. . . .

f c FLEXURE 1

C f ba a ac 0 85 0 85 4000 18 61200. ' .

T A ff y 8 1 56 60000 748800.

F T C C T ax 0 61200 648800

aA ff bs y

c

0 85

8 1 56 600000 85 4000 18

12 24. '

..

. OR a 74880061200

12 24.

M M A f da

ib in k ftc n s y

( ) . .

.2

8 1 56 60000 36 2512 24

222560000 1880

OR

M M T da

C da

ib in k ftc n

( ) ( ) .

.2 2

648800 36 2512 24

222560000 1880

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OR

M A f dA ff b

in lb k ftn s ys y

c

0 59 8 1 56 60000 36 25 0 59

8 1 56 600004000 18

22560000 1880.'

. . ..

M k ftn 1880

Find what value to use from t

Since a = 12.24 = 1x , x = a/1x = 12.24/0.85 = 14.40

Find t from the strain diagram by geometry as follows:

Since t > 0.004, meets code, but t < 0.005, transition zone

t = 0.005 = 0.9 tension controlst = 0.004552 = 0.8618 Our caset = 0.002 = 0.65 compression controls (tied)

Since ML=0.6Ms, MD=0.4Ms and Mu=1.2MD+1.6ML then:

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Balanced Strain Condition – Is it still needed?

At the balanced strain condition the maximum strain cu at the extreme concrete compression fiber just reaches 0.003 simultaneously with the tension steel reaching y = fy/Es (0.002069 for fy = 60 ksi) as shown below. The amount of steel corresponding to the balanced condition is Asb.

If the actual steel provided was more than Asb then a sudden failure would occur at ultimate loading since the concrete would fail first in a brittle manner (BAD).

If the actual steel provide was less than Asb then a gradual failure would occur at ultimate loading since the steel would fail first in a ductile manner (GOOD).

Derive the balanced strain condition for a rectangular beam with tension reinforcement only.

Abds (3.5.1)

x db

cu cu y

11

xdb cu

cu y

xd fb

y

0 0030 003 29000000

.. /

xd fb

y

8700087000 (3.5.2)C f b xb c b0 85 1. '

T A f bdfb sb y b y

F T Cx 0

0 85 1. 'f b x bdfc b b y

bc

y

bff

xd

0 851

. '(3.5.3)

bc

y y

ff f

0 85 87000870001

. ' (3.5.4)

Maximum Reinforcement Ratio

In order to assure a ductile failure the ACI limits (ACI 10.3.3) the amount of tension steel to not more than 75 % of the amount in the balanced strain condition (max = 0.75b). Table 3.6.1 pp. 57 gives maximum reinforcement ratios, also FLEXURE 1 SP-17(97). This is all from the 1995 or 1999 ACI code. The 2002 code states in 10.3.5 that if t>0.004 to ensure ductility (about 72.4% for grade 60).

A more direct way of controlling ductility is prescribe a maximum value for the neutral axis distance x at the failure imminent condition (max x = 0.75 xb).

For the 2002 code for flexural tension reinforcement at t = 0.004 the strength reduction factor turns out to be = 0.81667 using t = 0.002 for grade 60 steel. Other important strain-phi relationships are as follows:

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t = 0.005 = 0.9 tension controlst = 0.002 = 0.65 compression controls (tied)t = 0.002 = 0.70 compression controls (spiral)t = 0.004 = 0.81667 maximum for singly reinforced (tied)t = 0.003667 = 0.78889 balanced condition (tied)

Minimum Reinforcement Ratio

This is all a very nice derivation of the amount of reinforcement required to provide the same moment capacity as the cracking moment of the concrete. This led to ACI 10.5.1 that states:

Aff

b dsc

yw,min

'

3

but not less than

OR

For statically determinate T-section with the flange in tension ACI 10.5.2 that states:

Aff

b dsc

yw,min

'

6 (old) (new)

but not less than

Aff

b dsc

yE,min

'

3 (old) none (new)

Looks like the same requirements to me. Is it?

The amount of reinforcing may be less than minimum per ACI 10.5.3 provided the area provided is one-third more than is required. For Slab or footings the minimum area is per ACI 7.12 with spacing not to exceed 3t or 18”.

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Grade 40 or 50 As = 0.0020*Ag

Grade 60 As = 0.0018*Ag

Grade 40 or 50 As = 0.0018*Ag(60/fy)

Design of Rectangular Sections in Bending Having Tension Reinforcement Only Under ACI-10.3 and 10.5

The problem is to determine b, d, and As from the required value of Mn = Mu/ and the material properties fc' and fy. Since there are only two applicable equations of equilibrium, but three unknown, many possible solutions exist. Assume = 0.9, if required, and check at the end.

If the reinforcement ratio is preset then the following equation can be derived from the previously solved equilibrium conditions:

(3.8.4b)

where

mff

y

c

0 85. ' (3.8.4a)

In some situations the values of b and d are preset are the equations would be:

11 1

2m

mRf

n

y(3.8.5)

The procedure to be used in strength design of rectangular sections with tension reinforcement only is as follows (Flexure 2.1 - 2.4 from ACI SP-17(97) typ., cover this NOW):

Design Table for Singly Reinforced Rectangular Beams

ACI 340R-97 SP-17(97)

FLEXURE 2.1 - 2.4 is set up to find As given bd as follows:

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Enter FLEXURE 2.1 - 2.4 with Kn and interpolate to find .

A bds

It should be noted that if Kn is above the line then is must be increased to 1.33As (<min). If 1.33As is less than minimum then use it

(<min). If 1.33As is more than minimum then use minimum (min). Also if Kn is off the bottom of the chart compression steel is required

(>max).

FLEXURE 2.1 - 2.4 is also set up to find bd given as follows:

Enter FLEXURE 2.1 - 2.4 with and interpolate to find Kn.

It should be noted that if Kn is above the line then is must be decreased to 0.75Kn (<min). Also if Kn is off the bottom of the chart

compression steel is required (>max).

bd F2 12000

FLEXURE 2.1 - 2.4 is also set up to find Mn given As and bd as follows:

Abds

15

Enter FLEXURE 2.1 - 2.4 with and interpolate to find Kn.It should be noted that if Kn is above the line then is must be decreased

to 0.75Kn (<min). Also if Kn is off the bottom of the chart compression steel is required (>max).

F bd

2

12000

Problem #3.8 case 2 - Design a rectangular beam with tension reinforcing only such that excessive deflections would not be expected under normal circumstances. Use the following data and include the actual beam weight (d/b=1-5-2.0[1.75], wD=0.8k/ft, wL=1.8k/ft, L=30ft, f’c=3000psi and fy=60,000psi.)

Assume beam weight in lb/ft=20 lb/ft*span in feet 20 lb/ft*30=600 lb/ft

FLEXURE 1 or 2.1

Enter Table 2.1 with to find Kn and check = 0.9

and and Fbd

2

12000

16

For total height add 2.5’’ and round up to h=34 (d=31.5) and b=18

w k ftinin ft

inin ft

k ftbeam

0 15

1812

3412

0 63753. // /

. /

The following revised values include the beam weight

Fbd

2 2

1200018 31 5

120001

..488

Enter Table 2.1 with Kn and interpolate to find and check = 0.9

0.007031

Choose 3-#11 from REINFORCEMENT 14 (bmax = 23.5 int.)

The following is a check

Enter Table 2.1 with to find Kn and check = 0.9

Kn=446.9

Fbd

2 2

1200018 31 5

120001

..488

17

There are many answers to this question, but I will use the following

18x34 w/3-#11

Problem #3.8 case 4 - Design a rectangular beam with tension reinforcing only such that excessive deflections would not be expected under normal circumstances. Use the following data and include the actual beam weight (d/b=1-5-2.0[1.75], wD=1.2k/ft, wL=2.5k/ft, L=30ft, f’c=4000psi and fy=60,000psi.)

FLEXURE 1 or 2.2

Enter Table 2.2 with to find Kn check = 0.9

and and Fbd

2

12000

Increase 5% add 2.5’’ and round up to h=34 (d=31.5) and b=18 inches

w k ftinin ft

inin ft

k ftbeam

0 15

1812

3412

0 63753. // /

. /

The following revised values include the beam weight

18

Enter Table 2.2 with Kn and interpolate to find and check

0.00948

Choose 4-#11 from REINFORCEMENT 14

The following a check

Enter Table 2.2 with to find Kn and check = 0.9

Kn=596.2

There are many answers to this question, but I will use the following

18x34 w/4-#11

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Revised Design Procedure for Singly Reinforced Rectangular Beams

1. Assume that min < < max (I like 0.5b=0.375max)2. Determine the required bd2 (from Rn)3. Choose values for b and d (reasonable proportions d/b=1.5 to 2.0)4. Determine revised (for new b and d)5. Compute As (from )6. Select reinforcement and calculate strength (Mn > Mu)7. Remember to check = 0.9 an any step

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