Moment of Inertia of Cracked Section

560.325 Lecture Notes © T. Igusa 10

III. FLEXURAL ANALYSIS AND DESIGN OF BEAMSCHAPTER 9, Sections 1–2

1. IntroductionThe analysis of beams will proceed as follows:

Chapter III: Dimensioning of the concrete cross section.Selection and placement of reinforcing steel.

Chapter IV: Shear reinforcement.Chapter V: Bond and anchorage of rebars.Chapter VI: Serviceability, deflections, and cracking.

2. Bending of Homogeneous BeamsFor elastic analysis, the elementary equations from mechanics of materials is used:

f = M y / I

f = stressM = moment applied to the beamI = moment of inertia of the cross sectiony = distance from the neutral axis

Example 3.1Consider a plain concrete beam that is 10 inches wide and 30 inches deep. Thestrength of the concrete is given by f'c = 4000 psi and fr = 475 psi.

What is the maximum (nominal) moment that can be applied to the beam?

maximum compression strength = .85 f'c =maximum tension strength = fr = 475 psi

moment of inertia = I = b h3 / 12 =concrete stress = f = M y / I =

maximum (nominal) moment = Mn =


3. Reinforced Concrete Beam BehaviorPlain concrete beams are not practical because as soon as a crack is formed, the entirebeam splits into two parts causing immediate and sudden structural failure. Thebasic mechanics problem is that concrete is very weak under tension stresses.

To increase both the strength and the safety of concrete beams, reinforcing steel isadded where the stresses are in tension. The figures below illustrate how thestresses in the beam change as the moment load is increased.

Four-point bending test

Failure configurations


Stages of stress-strain behaviorStresses elastic and section uncracked:

h d

b

no tension cracks if: fct < fr

Stresses elastic and section cracked:

d

no tension stress (fct = 0) because of cracks

compression stress is linear elastic if: fc < f'c/2

steel stress is linear elastic if: εs < εyStresses inelastic:

d

steel is yielding

compression stress is inelastic

concrete is still intact


Review of section propertiesBefore beginning the analysis of reinforced concrete beams, some of the properties ofthe cross section are reviewed.

A , I1 1

A , I2 2

The centeroid is given by:

y = 1Area

y dA

y = Ai yi∑

Ai∑

The moment of inertia is given by:

I = (y – y)2 dA

I = Ii∑ + Ai∑ (y – yi)2


A rectangular stress distribution can be replaced by a single force as follows:

b

a

f

a/2

C = a b f

A triangular stress distribution can be replaced by a single force as follows:

b

a

f

a/3

C = a b f /2

a. Stresses elastic and section uncrackedTo find the stresses in an uncracked beam, it is useful to replace the reinforcedconcrete cross section with an equivalent, purely concrete, transformed section.Basically, the steel is replaced by an equivalent amount of concrete. Since themodulus of elasticity of steel, Es, is much larger than that of concrete, Ec, the area ofthe steel must be replaced by a much larger area of concrete.

The idea of a transformed section is best explained using diagrams. First, somedefinitions are introduced:


As = area of reinforcing steel (under tension)b = width of a rectangular beamh = total depth of a rectangular beamd = distance from the top of the beam to the centeroid of the steelIut = moment of inertia of the uncracked, transformed section

fc = compression stress in the concretefct = tension stress in the concretefs = tension stress in the steel

n = Es / Ec = modular ratio

In the transformed section, the area of the steel, As, is replaced by a larger area, nAs,of concrete. Some of the concrete "fills" the area, As, originally occupied by the steel.The remaining concrete, which has area (n–1)As, is placed outside of the rectangulardimensions of the beam at the centeroid of the steel.

h d

b(n–1)A s

To prove that the transformed concrete section is equivalent section, it isnecessary only to examine the stresses at the steel:

stress at steel rebars = fs = Es εs =

total tension forcefrom steel rebars = stress × area =

total tension forcefrom equivalentconcrete area =

One important relation from the preceeding proof is:

fs = n fct (at the location of the steel, As)


Example 3.2Consider the same concrete beam examined in Example 3.1 (10" wide, 30" deep,f'c = 4000 psi, and fr = 475 psi). Two inches from the bottom of the beam, twoNumber 9 bars are inserted. The strength of the steel is fy = 60,000 psi.

What is the largest moment that can be applied to the beam before it cracks?

Solution

First, find the modular ratio, n:

Ec = 57,000 √f'c (see page 7) =

n = Es / Ec =

Next, the basic geometric properties of the cross section are found:

No. 9 bar has dia. = 9/8 in.The area of one bar is π (dia.)2 / 4 =

As =

A1 = b h =

A2 = (n–1) As =

I1 = b h3 / 12 =

I2 ≈ 0 (usual approximation)

(n–1)A s

The neutral axis and uncracked moment of inertia are:

y = Ai yi∑

Ai∑ =

A1y1 + A2y2A1 + A2

=

Iut = Ii∑ + Ai∑ (y – yi)2 =


Next, the stresses are computed using the usual formula, f = M y / I :

(n–1)A s

Make sure the concrete tension stress is less than the rupture strength, fr:

fct = M y / Iut =

Make sure the steel tension stress is less than the yield strength, fy:

fs = n fct (at As) = n M y / Iut =

Make sure the concrete compression stress is less than the linear elastic limit, f'c/2:

fc = M y / Iut =


b. Stresses elastic and section crackedTo find the stresses in a cracked beam, the equivalent, purely concrete, transformedsection is still used. The difference between the analysis of cracked and uncrackedsections is that no tensile stresses are allowed in the cracked case.

The following definitions are useful:

kd = distance from the top of the beam to the neutral axisjd = moment arm for the equivalent compression and tension forcesIcr = moment of inertia of the cracked, transformed section

C = equivalent compression force in the concreteT = equivalent tension force in the steelM = moment applied to beam

ρ = As / bd = ratio between steel and concrete areas

Using basic principles of mechanics, it is possible to find the location of the neutralaxis, kd, and the stresses in the concrete and steel. The final results are:

k = (ρn )2 + 2ρn – ρn (9-3)

j = 1 – k/3

fs = MAsj d

(9-5)

fc = 2Mkj bd2

(9-4)

Icr = b k2 j d3 / 2

The derivations are based on the diagram of the concrete stresses on the next page.


d

nA sb

kd

The properties of the two areas, A1 and A2 are:

A1 = b k d y1 = k d / 2 y = k d

A2 = n As = n ρ b d y2 = d

The equations for the forces and moments are:

C = area × average stress =

T = area × average stress =

M = T j d =

= C j d =

Equation (9-3) for the neutral axis is derived from the following:

y = Ai yi∑Ai∑

= A1 y1 + A2 y2

A1 + A2 or y ( A1 + A2 ) = A1 y1 + A2 y2

It is necessary to substitute the terms for A1, A2, y1, y2, and y:

kd ( bkd + nρbd ) = bkd ( kd/2 ) + nρbd ( d )


Example 3.3Consider the same concrete beam examined in Example 3.2 (10" wide, 30" deep,f'c = 4000 psi, fr = 475 psi, fy = 60,000 psi, two No. 9 bars 2" from bottom).

Determine the maximum moment that can be carried without stressing the concretebeyond f'c/2 or the steel beyond fy/2.

Solution

First, the neutral axis is determined:

ρ = As / bd =

ρn =

k = (ρn )2 + 2ρn – ρn =

kd =

j = 1 – k/3 =

Then, the maximum stresses in the steel and concrete are found:

fs = MAsj d

=

fc = 2Mkj bd2

=

Just for checking, the stresses are recalculated using the cracked moment of inertia,Icr, and the usual stress formula, f = My/I:

Icr = b k2 j d3 / 2 =

fc = M y / Icr =

fs = n fct (at As) = n M y / Icr =


4. Design of Tension-Reinforced Rectangular BeamsCHAPTER 4, Sections 1–2, 3 (up to "Upper Limit ...", Example 4-2, skip SI units), 4.

In this section, the strength of rectangular beams is analyzed and a design method isdeveloped. The strength capacity is the maximum moment a beam can handlebefore it becomes essentially destroyed.

A fundamental assumption is that the steel is yielding and the concrete strains arenot high enough to cause crushing. To make sure this type of failure occurs, thesteel ratio must be less than the so-called balanced steel ratio, i.e., ρ ≤ ρb.

The analysis is more difficult than in Section 3 because the concrete stresses are nolonger linear. This means that the linear relation between stress and strain is nolonger valid, i.e., fc ≠ Ec εc. However, researchers have found a moderately simpleapproximation for beam analysis which closely agrees with experiments. Thisapproximation is presented in the following.

a. Equivalent rectangular stress distributionThe concrete stresses just before the beam fails has a nonlinear distribution. Tomake the analysis relatively simple, the nonlinear stresses are replaced with anequivalent rectangular (constant) stress distribution. The magnitude and location ofthe equivalent stress distribution is defined in the diagram below.

ACTUAL STRESS DISTRIBUTION

d c

EQUIVALENT STRESS DISTRIBUTION

a/2

dca

d – a/2


The relationship between the distance to the neutral axis, c, and the depth of thestress distribution, a, is given by:

Equivalent stress parameters

a = β1 c

β1 =

0.85,

if f'c ≤ 4000 psi;

0.85 – 0.05(f'c – 4000)/1000,

if 4000 psi ≤ f'c ≤ 8000 psi;

0.65, if 8000 psi ≤ f'c

(4-8)

With the preceding information, it is now possible to derive the following:

Location of the neutral axis

a = As fy

0.85f'c b =

ρ fy d

0.85 f'c(4-11, 4-14a)

c = a / β1

Maximum nominal moment

Mn = As fy (d – a/2) = R bd2 (4-12a)

R = ρ fy 1 – 0.59ρ fyf'c

design strength = φ Mn = 0.9 Mn

Minimum and maximum values for the steel ratio, ρ

ρmin ≤ ρ ≤ ρmax

ρmin = 200 / fy (f'c < 4440psi) or ρmin = 3 f' c / fy (f'c > 4440psi) (4-31)

ρmax = 0.75 ρb

ρb = 0.85 β1f'cfy

87,000

87,000 + fy(4-25)


To determine the location of the neutral axis, it is necessary only to enforceequilibrium. The equivalent forces, C and T, are:

C = compression stress × area =

T = tensile stress × area =

Solving for a:

a =

The maximum nominal moment is obtained from statics:

Mn = force × moment arm =

b. Balanced steel ratio

The balanced steel ratio, ρb, is the steel ratio where the concrete crushes and thesteel yields at the same time. For other steel ratios, we have:

ρ < ρb under-reinforced, εs > εy steel yields before concrete crushesρ > ρb over-reinforced, εs < εy concrete crushes before steel yields

The analytical formula for ρb is derived by looking at the strains instead of thestresses. The strains we are interested are:

εu = 0.003 compressive strain where concrete crushesεy = fy / Ey = fy / 29,000,000 tensile strain where steel yields

Using similar triangles, we find the following ratios where the maximum concretestrain is εc = εu and the steel strain is εs = εy:


c. Under-reinforced beams

If the steel ratio, ρ, is less than the balanced steel ratio, ρb, then the beam isunder-reinforced, and the steel will yield before the concrete crushes. However, ifthe steel ratio is larger than ρb, then the beam is over-reinforced, and the concretewill crush before the steel yields. As explained in Section 3 (page 11), yielding steelmakes the beam fail slowly with large deflections, while crushing concrete makesthe beam fail explosively with immediate, catostrophic collapse.

To allow for possible variations in the concrete and steel properties, it is safer tomake ρ significantly less than ρb. The ACI Code gives the following design limit:

ρ ≤ ρmax = 0.75 ρb

Thus, the steel ratio must be at least 25% less than the balanced steel ratio.

d. Minimum steel ratioIf the moment capacity of an uncracked beam (Section 3a) is greater than themoment capacity of a cracked beam (Section 4), then the beam will fail as soon as acrack is formed. This type of failure is not as catostrophic as concrete crushing,because the steel will still yield. However, it is sudden, because it will occurimmediately after the concrete cracks.

To eliminate this possibility, the ACI Code gives a lower limit for the steel ratio:

ρ ≥ ρmin where


e. Examples of rectangular beam review and designThere are two type of problems to consider:

Review is where the beam dimensions and reinforcing steel sizes are given, andthe beam strength is to be computed. Review is very straightforward since it isnecessary only to "plug in" the dimensions into a set of formulas.

Design is where the required beam strength (or load) is given, and thedimensions of the beam and reinforcing steel are determined. Design is morechallenging than review for two reasons: First, it may be necessary to iterate tofind the optimal beam dimensions. Second, there is, in general, no uniquesolution to the problem.

In the following are examples of review and design problems.


Example 3.4. Review of a tension-reinforced beam.Consider the same concrete beam examined in Example 3.2 (10" wide, 30" deep,f'c = 4000 psi, fr = 475 psi, fy = 60,000 psi, two No. 9 bars 2" from bottom).

Determine the maximum nominal moment, Mn, and the design strength, φMn.

Solution

First, the steel ratio must be checked against the limits, ρmin and ρmax:

ρ = As/bd = 2.00 / ( 10 × 28 ) = 0.00714

ρb = 0.85 β1f' cfy

87,000

87,000 + fy(4-25)

=

ρmax = 0.75 ρb =


Then, the nominal moment and design strength are computed:

a = As fy

0.85f'c b = (4-11)

Mn = As fy (d – a/ 2) = (4-12a)

design strength = φ Mn = 0.9 Mn =

Summary of Examples 3.1 to 3.4.

Example neutral axis moment stresses (ksi)(inches) (kip-inch) concrete (fc) steel (fs)

3.1 No steel h /2 = 15.0 713 0.3

3.2 Uncracked y = 15.6 817 0.5 3.3

3.3 Cracked, elastic kd = 8.0 1,520 1.5 30.0

3.4 Inelastic a = 3.5 3,150 3.4 60.0

Design 2,830


Basic design stepsThere are two sets of well-defined steps which can be used in design. The objectiveis to satisfy the strength equation,

Mu ≤ φ Mn = φ R b d2 (4-16)

The design is based on the strengths of the concrete and steel, as defined by f'c and fy,and the factored load, Mu.

DESIGN WITH SPECIFIC CONCRETE DIMENSIONS

1. Choose the beam width, b, and effective depth, d, using the following rules:

a. The total depth, h, and width, b, are in even inches.

b. The difference between d and h is usually:

h – d = 2.5 in. for a single layer of steel bars;h – d = 4.0 in. for a double layer of steel bars.

c. The effective depth, d, is usually 2 to 3 times the width, b.

Find the required value for R corresponding to b and d:

R = Mu / φ b d2

2a. Substitute the preceding value for the equation for R on page 22. The result is aquadratic equation with unknown, ρ. The smallest of the two solutions for thesteel ratio, ρ, is the correct solution.

2b. Check the steel ratio with the minimum and maximum values:

ρmin ≤ ρ ≤ ρmaxUse the equations on the bottom of page 22.

3. Calculate the required area of steel, As = ρbd. Use Table A-8 (pg. 874) to findsuitable rebars. Preference is for fewer and same-sized bars. The rebars shouldbe symmetrically placed (in both size and location). For close cases, redo step 2b.

4. Use Table A-6 (pg. 870) to check for placement and concrete cover.

DESIGN WITH AN INITIAL STEEL RATIO

1'. Find the balanced steel ratio, ρb, using equation (4-25). Then choose an initialvalue for the steel ratio ρ. An economical and practical choice is ρ = 0.5ρb.

2'. Find the corresponding value for R using the equation on page 22.

3'. Solve the strength equation in terms of bd2:

b d2 ≥ Mu / φ R

Find b and d which satisfies the above. Then follow the preceding steps for"design with specific concrete dimensions".


Example 3.5. Design of a tension-reinforced beam.A rectangular, tension-reinforced beam is to be designed for dead load of 1400 lb/ftplus self weight and service live load of 800 lb/ft, with a 20 ft simple span. Materialstrengths will be fy = 60 ksi and f'c = 5 ksi for steel and concrete respectively. Thetotal beam depth (h) must not exceed 20 in. Calculate the required beam width andtensile steel requirement, using a steel ratio of 0.5 ρb. The effective depth (d) may beassumed at 2.5 in. less than the total depth (h).

For the first iteration, assume the self weight is 100 lb/ft.

Solution

Before proceeding with the design steps, it is necessary to find the load.

wd = 1400 + 100 = 1500 lb/ft,

wl = 800 lb/ft

wu = 1.4 wd + 1.7 wl = 1.4 (1500) + 1.7 (800) = 3460 lb/ft = 3.46 k/ft

Mu = wu L2 / 8 = 3.46 (20)2 / 8 = 173 k-ft

Also, it is useful to find the parameter using equation (4-8):

β1 = 0.85 – 0.05 (f'c – 4000)/1000 =

The design then proceeds as follows:

1'. Find the balanced steel ratio, ρb, using equation (4-25).

ρb = 0.85 β1f' cfy

87,000

87,000 + fy =

Then choose an initial value for the steel ratio ρ.

ρ = 0.5 ρb =

2'. Find the corresponding value for R using the equation on page 22.

R = ρ fy 1 – 0.59ρ fyf'c

=

3'. Solve the strength equation in terms of bd2:

b d2 ≥ Mu / φ R =

Find b and d which satisfies the above. Since h ≤ 20" and h = d + 2.5", we knowthat d ≤ 17.5". Use d = 17.5" and solve for b:


Then check the weight and repeat any necessary calculations.

weight per foot = area (ft2) × 145 pcf = bh (ft2) × 145 pcf =

wd =

wu = 1.4 wd + 1.7 wl =

Mu = wu L2 / 8 =

b d2 ≥ Mu / φ R =

Next follow the steps for "design with specific concrete dimensions".

1. Find the required value for R corresponding to b and d:

R = Mu / φ b d2 =

2a. Substitute the preceding value for R into the equation on page 22. The result isa quadratic equation with unknown, ρ. The smallest of the two solutions forthe steel ratio, ρ, is the correct solution.

R = ρ fy 1 – 0.59ρ fyf'c

=

2b. Check the steel ratio with the maximum and minimum values:

ρmax = 0.75 ρb = 0.75 × 0.0335 = 0.0251

ρmin = 3 f' c / fy = 3 × 5000 / 60,000 = 0.00354 (4-31)

3. Calculate the required area of steel, As = ρbd. Use Table A-8 to find suitablerebars. Preference is to fewer and same-sized bars. In any case, the rebars shouldbe symmetrically placed (in both size and location). For close cases, redo step 2b.

As = ρbd =


4. Use Table A-6 to check for placement and concrete cover. A well-drawndiagram is essential, particularly in professional practice.

5. Design AidsThere are many tables which can be used to simplify some of the calculations.However, some of the tables require interpolation, which requires as much work assolving the original analytical expressions. The following tables can be used:

TABLE A-8 Areas of groups of standard bars, in2

This table is used to select the number and sizes of steel rebars which will fulfillthe required steel area, As. To explain the columns A, B, C, the second row ofentries for the areas is examined. Under columns A, 0.40 is for two No. 4, 1.40 isfor two+five No. 4, 0.51 is for two No. 4 + one No. 3, 0.62 is for two No. 4 + twoNo. 3. Column C is similar to Column B.

TABLE A-6 Maximum number of bars as a single layer in beam stems

This table gives an estimate of how many steel rebars will fit in a single layer in abeam with width bw. For most cases (unless stated otherwise) the maximumaggregate size will be 3/4 in.. Columns A, B, C are interpreted as in Table A-8.


6. Practical Considerations in the Design of Beamsa. Concrete protection for reinforcementThe outermost steel must be protected against fire and corrosion. Fortunately,concrete provides excellent insulation for fire, and is reasonably good in keeping outwater and oxygen to prevent corrosion.

The minimum distances between the outermost steel and the surface of the concreteare as follows:

CAST IN PLACE

3/4 in. slabs & walls

1-1/2 in. beams & columns

CAST IN PLACE,EXPOSED TO WEATHER OR GROUND

1-1/2 in. No. 5 & smaller

2 in. No. 6 & larger

CAST IN GROUND WITOUT FORMS

3 in. All cases

Some commonly used dimensions are given below for slabs and beams:

SLABS

No. 3 ~ 4 bars

Depth is:. . . 4.5", 5", 5.5", 6", 7", 8", . . .

Required cover is 3/4 in.


BEAM WITH STIRRUPS

No. 3 stirrupsNo. 4 ~ 10 bars

b & h are in even inches

Required cover is 1-1/2 in.

b. Concrete proportionsThe usual effective depth, d, of a beam is about 2 to 3 times the width b (or bw for T-beams). However, architectural or construction considerations may place limits oneither d or b or both.

c. Selections of bars and bar spacingThe steel rebars should always be symmetric with respect to the vertical centerline ofa beam or column. Different sizes of bars can be used in a single layer, but thedifference between the size numbers should be 2 or less. For example, No. 10 andNo. 8 bars can be used in a single layer, but No. 10 and No. 7 bars cannot.

The two large sized rebars, No. 15 and No. 18, are used mainly in columns.

The minimum distances between the rebars are:

BEAMS

1 in.

bar diameter,or 1 in.

COLUMNS

1.5 bar diameters,or 1.5 in.

These distance requirements limits the number of rebars that can be fit into a singlelayer. Table A-6 can be used as a rough estimate of the maximum number of rebars.


7. Rectangular Beams with Tension & CompressionReinforcementFor added strength, steel can also be added to the compression side of a beam. In thissection, it is shown how this compression reinforcement will affect the momentcapacity of the beam.

Detail derivations for the beam strength formulas are given in the textbook. Thenotes will emphasize physical behavior rather than mathematical analysis.

First, some definitions are needed:

A's = area of reinforcing steel (under compression)d' = distance from the top to the centeroid of the compression steel

ε's = strain in the compression steelf's = stress in the compression steel

ρ ' = A's / bd = ratio between compression steel and concrete areas

As in Section 4, the the equivalent stress distribution is used in the analysis:

s y

ca

d

b

d'

f = f

assume tension steel is yielding

0.85 f'c

s yε ≥ ε

uε = 0.003

s y

d – d' d – a/2d

d'a

0.85 f'c C = 0.85 f' abc

A f


It is clear from the preceding diagram that the total moment of the beam is:

Mn = 0.85 f'c ab ( d – a/2 ) + A's f's ( d – d' ) (5-18)

There are two unknowns in the above: the compression steel stress, f's, and thedepth of the equivalent stress, a. These parameters are given by:

CASE I: Compression steel is yielding, ρ ≥ ρcy

f's = fy

a = ( As – A's )fy

0.85f 'c b =

(ρ – ρ' ) fy d

0.85 f 'c(5-11a)

CASE II: Compression steel is not yielding, ρ < ρcy

f's = εu Es ( c – d' ) / c

a = β1 c

The distance to the neutral axis, c, is determined by the quadratic equation:

0.85 β1 f'c b c2 – As fy c + A's εu Es ( c – d' ) = 0 (5-17)

STEEL RATIOS

For both cases, the critical compression steel ratio, ρcy, is given by

ρcy = 0.85 β1f'cfy

d'd

87,00087,000 – fy

+ ρ' (alternative to 5-4)

Furthermore, it is assumed that the tension steel is always yielding. This is trueonly if the compression steel ratio, ρ, is less than the balanced steel ratio, ρb

ρb = ρb + ( f's / fy ) ρ' (5-15b)

The ACI Code restriction for the compression steel ratio is

ρmin ≤ ρ ≤ ρmax = 0.75 ρb + ( f's / fy ) ρ'

In the above, ρmin and ρb are as in Section 4:


ρb = 0.85 β1f' cfy

87,000

87,000 + fy(4-25)


DERIVATIONS

CASE I: Assume the compression steel is yielding.

If f's = fy, then the only unknown is the parameter a. It is found from equilibriumof the forces for the moment Mn2:

C = 0.85 f'c a b = Asfy – A'sf's =

CASE II: Assume the compression steel is not yielding.

Here, there are two unknowns, and it is necessary to examine the strains. Fromsimilar triangles, the ratio of concrete and compression steel strains is:

ε'sεu

=

Since the compression steel is not yielding, the following linear relation holds:

f's = ε's Es =

This expression for f's is substituted into the equilibrium equation:

C = 0.85 f'c a b = Asfy – A'sf's =

Finally, the preceding equation is multiplied by c and the relation a = β1c is used:

0.85 f'c β1 c2 b =

EFFECT OF TENSION STEEL RATIOIt is important to understand how the tension steel affects the strains and stresses inthe concrete and steel. To satisfy the equilibrium of forces, an increase in thetension steel ratio, ρ, will cause an increase in the compression steel strain, ε's. At asufficiently high steel ratio, ρ ≥ ρcy, the compression steel will yield, ε's ≥ εy.


STEPS FOR REVIEW OF DOUBLY REINFORCED BEAMS

1. Find the balanced moment, ρb, using equation (4-25).

2. Check if it is necessary to include the effect of the compression steel area, A's.

If ρ ≤ 0.75 ρb: Ignore A's and use equations on page 22.

If ρ > 0.75 ρb: Continue with the steps below.

3. Find the critical ratio, ρcy, using equation (5-4).

4. Find the compression steel stress, f's, and the depth of the stress distribution, a.

If ρ ≥ ρcy: This is Case I, with f's = fy.Find a using equation (5-11a).

If ρ < ρcy: This is Case II, with f's < fy.First solve for c using equation (5-17).Use a = β1 c.Then find f's using the on page 33.

5. Check if ρmin ≤ ρ ≤ ρmax = 0.75 ρb + ( f's / fy ) ρ'.

6. Find the nominal moment, Mn, using equation (5-18).As usual, the design strength is φMn = 0.9Mn.

Example 3.6. Review of a doubly reinforced beam.A rectangular concrete beam measures 14" wideand has an effective depth of 20". Compressionsteel consisting of three No. 9 bars is located 2.5"from the compression face. If f'c = 3000 psi andfy = 60,000 psi, what is the design moment capacityof the beam, according to the ACI Code, fortension steel consisting of six No. 9 bars?

Solution

The basic parameters of the beam are:

d

b

d'

As = 6 in2, ρ = As / bd =

A's = 3 in2, ρ' = A's / bd =


1. Find the balanced moment:

ρb = 0.85 β1f'cfy

87,000

87,000 + fy = 0.85 × 0.85 × 3

60 87

87 + 60 = 0.0214 (4-25)

2. Check if it is necessary to include the effect of the compression steel area, A's:0.75ρb =

3. Find the critical steel ratio:

ρcy = 0.85 β1f'cfy

d'd

87,00087,000 – fy

+ ρ' =

4. Since ρ < ρcy, this is Case II, with f's < fy.

The distance to the neutral axis, c, is determined by the quadratic equation:

0.85 β1 f'c b c2 – As fy c + A's εu Es ( c – d' ) = 0 (5-17)

The stress on the compression steel is:

f's = εu Es ( c – d' ) / c =

The depth of the equivalent stress is:

a = β1 c =

5. Check if ρmin ≤ ρ ≤ ρmax:

ρmin = 200 / fy = 200 / 60,000 = 0.0033

ρmax = 0.75 ρb + ( f's / fy ) ρ' =

6. Finally, the nominal moment is:

Mn = 0.85 f'c ab ( d – a/2 ) + A's f's ( d – d' )

=

As usual, the design strength is φMn = 0.9Mn =Recall that the design strength without compression steel is: φMn = 4840 k-in.See Example 5-6 in the text (pp. 168-170) for Case II: ρ ≥ ρcy.


8. T BeamsMost reinforced concrete floors, roofs, and decks are cast in one step. In particular,beams are integrated into floor slabs as shown below. The basic dimensions are:

hf = floor slab thicknessbw = beam web widthb = effective flange widthL = beam span length

a. Effective flange widthAccording to the ACI Code, the effective flange width, b, is given by:

b = minimum of: L / 416 hf + bwcenter-to-center distance between beam webs

b. Strength analysisIf the depth of the equivalent stress block is less than the thickness of the slab, a ≤ hf,then the T-beam can be replaced by a rectangular beam with width b, as shownbelow. For such cases, the analysis of the preceding sections would be unchanged.


a

s yf = f

If a > hf, then it is necessary to express the nominal moment as a sum of twomoments, as shown in the diagrams below. The following definitions are used:

As = total area of tension steel (in the web)Asf = area of tension steel which balances the flange compression forceρw = As / bwd = steel ratio for the steel in the webρf = Asf / bwd = steel ratio for the steel area, Asfρ = As / bd = steel ratio for the equivalent rectangular beam

ca

d

0.85 f'cc uε = 0.003

s yf = f

0.85 f'c

s yf = f

0.85 f'c

s yf = f


It is clear from the preceding diagram that the total moment of the beam is:

Mn = Mn1 + Mn2 = Asf fy ( d – hf/2 ) + 0.85 f'c bw a ( d – a/2 ) (5-5)

There are two unknowns in the above: the steel area, Asf, and the depth of theequivalent stress block, a. These parameters are given by:

Asf = 0.85 f'cfy

( b – bw ) hf (5-1b)

a = ( As – Asf ) fy

0.85f 'c bw(5-3)

The ACI Code restriction for the compression steel ratio is

ρmin ≤ ρw ≤ ρmax = 0.75 ( ρb + ρf )

In the above, ρmin and ρb are as in Section 4:


ρb = 0.85 β1f' cfy

87,000

87,000 + fy(4-25)

c. Proportions of cross sectionsThe slab thickness, hw, is usually determined by the loads on the slab. The webdimensions are determined by the following three criteria:

1. To keep the steel ratio, ρw, within the ACI Code limits.

2. To keep the shear stresses small (to be examined in Chapter IV).

3. For resisting negative moments at the supports.

d. Examples of review and design of T beamsThe basic steps for review and design are given next. An review example is inExample 3.13 in the text. A design example is given in the following.

STEPS FOR REVIEW OF T BEAMS1. Find the effective flange width, b, using the ACI Code equations.

2. Assume that a ≤ hf and do the analysis of rectangular beams in Section 4.

3. If the assumption, a ≤ hf, is valid, then stop.If a > hf, then use the preceding T beam equations.


STEPS FOR DESIGN OF T BEAMS

1. Find the effective flange width, b, using the ACI Code equations.

2. Assume that a ≤ hf and do the design of rectangular beams in Section 4 (page 26).

3. If the assumption, a ≤ hf, is valid, then stop.If a > hf, then continue the design.

4. Find the area Asf using equations (5-16).

5. Set the design strength equal to the factored moment, φMn = Mu, usingequation (5-5). This is a quadratic equation which must be solved to find theunknown parameter, a.

6. Use equation (5-3) to find the required tension steel area, As.

7. Check if ρmin ≤ ρw ≤ ρmax = 0.75 ( ρb + ρf ).

8. Use Table A-8 to find suitable rebars.

9. Use Table A-6 to check for placement and concrete cover.

Example 3.7. Design of a T beam.A T beam is to be designed for a moment of 6400 k-in. Material strengths will be fy =60 ksi and f'c = 3 ksi for steel and concrete respectively. The beam span length is L =24 ft, the webs are spaced 47 in apart, and the floor slab thickness is hf = 3 in. Thebeam web dimensions are bw = 11 in and d = 20 in.

Calculate the required tensile steel requirement.


Solution

1. Find the effective flange width, b, using the ACI Code equations.

b = minimum of: L / 4 =16 hf + bw =

center-to-center distance between beam webs =

2. Assume that a ≤ hf and do the design of rectangular beams in Section 4:

Find the required value for R corresponding to b and d:

R = Mu / φ b d2 =

Substitute the preceding value for R into the equation on page 22. The resultis a quadratic equation with unknown, ρ. The smallest of the two solutionsfor the steel ratio, ρ, is the correct solution.

R = ρ fy 1 – 0.59ρ fyf'c

=

Solve for the equivalent stress block depth, a:

a = ρ fy d

0.85f'c =

3. If the assumption, a ≤ hf, is valid, then stop.If a > hf, then continue the design.

4. Find the area Asf using equations (5-16).

Asf = 0.85 f'cfy

( b – bw ) hf =

ρf = Asf

bw d =


5. Set the design strength equal to the factored moment, Mu = φMn, usingequation (5-5). This is a quadratic equation which must be solved to find theunknown parameter, a.

Mu / φ = Mn = Asf fy ( d – hf/2 ) + 0.85 f'c bw a ( d – a/2 ) (5-5)

6. Use equation (5-3) to find the required tension steel area, As.

a = ( As – Asf ) fy

0.85f 'c bw =

7. Check if ρmin ≤ ρw ≤ ρmax:

ρmin = 200 / fy = 200 / 60,000 = 0.00333

ρb = 0.0214 (balanced steel ratio, calculated from equation 4-25)

ρmax = 0.75 ( ρb + ρf ) =

8. Use Table A-8 to find suitable rebars.

9. Use Table A-6 to check for placement and concrete cover.

Moment of Inertia of Cracked Section

Documents

concrete stress

transformed concrete

concrete fs

equivalent concrete

stress area

inelastic concrete

larger area of concrete

stress moment