Design of Concrete

source: www.CE-Ref.com www.budhicivileng.blogspot.com 1

General - RC design Topics:

• Section properties of reinforcing bars

• Type of cements

• Elastic modulus of concrete

• Modulus of rapture

• Design load combinations from ASCE 7-98

• Design load combinatons from ASCE 7-05

• Strength reduction factor, φ, ACI 318-05

Section properties of reinforcing bars

Bar Nominal diameter Cross Section Perimeter Normal weight

No. in mm in mm in mm in mm

3 0.375 9.52 0.11 71 1.178 29.9 0.376 0.56

4 0.5 12.7 0.2 129 1.571 39.9 0.668 0.994

5 0.625 15.88 0.31 200 1.963 49.9 1.043 1.552

6 0.75 19.05 0.44 284 2.356 59.8 1.502 2.235

7 0.875 22.22 0.6 387 2.749 69.8 2.044 3.042

8 1.000 25.4 0.79 510 3.142 79.8 2.67 3.973

9 1.128 28.65 1.00 645 3.544 90.0 3.4 5.06

10 1.27 32.26 1.27 819 3.990 101.4 4.303 6.404

11 1.541 35.81 1.56 1006 4.43 112.5 5.313 7.907

14 1.693 43.00 2.25 1452 5.32 135.1 7.65 11.38

18 2.257 57.33 4.00 2581 7.09 180.1 13.6 20.24

Type of cements

Type Name Description

I Normal General purpose

II Moderate Moderate sulfate and heat resistance

III High early strength Fast strength development

IV Low-heat Generate low-heat during hydration

V Sulfate-resisting High sulfate resistance

Elastic modulus of concrete

The elastic modulus of concrete, Ec (lb/in2) is (ACI 318-05 Sec 8.5)

Ec = w

c

1.5 (33) √f

c’ lb/in

2


Ec = wc

1.5 (0.043) √f

c’ MPa

where wc is unit weight of concrete between 90 to 155 lb/ft3. fc’ is compression strength of concrete in

lb/in2.

For normal weight concrete Ec = 57000 √f

c’

Modulus of rapture

fr = 7.5 √ f

c’ lb/in

2.

fr = 0.62 √ f

c’ MPa

Concrete strength Elastic modulus Modulus or rapture

psi MPa ksi MPa psi MPa

3000 20.7 3156 21759 411 2.8

3500 24.1 3409 23503 444 3.1

4000 27.6 3644 25126 474 3.3

4500 31.0 3865 26650 503 3.5

5000 34.5 4074 28091 530 3.7

5500 37.9 4273 29462 556 3.8

6000 41.4 4463 30772 581 4.0

6500 44.8 4645 32029 605 4.2

7000 48.3 4821 33238 627 4.3

7500 51.7 4990 34405 650 4.5

8000 55.2 5154 35533 671 4.6

Design load combinations from ASCE 7-98

Load combination or strength design (ASCE 7-98, sec, 2.3.2)

1. 1.4(D+H)

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

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

4. 1.2D+1.6W+0.5L+0.5(Lr or S or R)

5. 1.2D+1.0E+0.5L+0.5(Lr or S or R)

6. 0.9D+1.6W+1.6H

7. 0.9D+1.0E+1.6H

Load combination for allowable stress design (ASCE 7-98, Sec. 2.4.1)

1. D

2. D+L+F+H+T+(Lr or S or R)

3. D+(W or 0.7E)+L+(Lr or S or R)

4. 0.6D+W+H

5. 0.6D+0.7E+H


Where D is dead load, L is live load, Lr is roof live load, W is wind load, E is earthquake load, F is fluid

pressure, R is rain load, S is snow load, T is temperature force.

Note: A 75% load reduction is allowed on two or more loads plus dead load for allowable stress design on

load combination 2 to 5. For example, load combination can be D+0.75(W+L).

Design load combinations from ACI 318-05

Load combination or strength design (Sec 9-2)

1. 1.4(D+F) (9-1)

2. 1.2(D+F+T)+1.6(L+H)+0.5(Lr or S or R) (9-2)

3. 1.2D+1.6(Lr or S or R)+(1.0L* or 0.8W) (9-3)

4. 1.2D+1.6W**+1.0L*+0.5(Lr or S or R) (9-4)

5. 1.2D+1.0E***+1.0L*+0.2S (9-5)

6. 0.9D+1.6W**+1.6H (9-6)

7. 0.9D+1.0E***+1.6H (9-7)

*1.0L can be reduced to 0.5L except garages, public assembly and area that has 100 lb/ft2 of live load.

**1.6W can be reduced to 1.3W when wind load W is not reduced by directional factor (See ASCE 7-02

wind calculation)

*** Where seismic load, E is calculated based on service load, 1.4E shall be used instead of 1.0E.

Strength reduction factor, φ, ACI 318-05

1. Tension-controlled sections, 0.9 (Sec. 9.3.2.1)

2. Compression-controlled sections (Sec. 9.3.2.2)

(a) Column with spiral reinforcement, 0.7

(b) Tie columns, 0.65.

3. φ is permitted to vary from 0.65 to 0.9 when the net tensile strain in extreme tension steel at nominal

strength varies from compressive control strain to 0.005.

4. Shear and torsion, 0.75 (Sec. 9.3.2.3)

5. Bearing on concrete, 0.65 (Sec. 9.3.2.4)


Reinforced concrete beam design Beam stresses under loads

Moment and shear diagram of a beam under dead and live loads are shown below.

Failure modes and reinforcements

1. Concrete is assumed to resist compression only, tension shall be resisted by reinforcements.

Reinforcements shall be placed at the side of the beam that has tension. For a simply supported

beam, tension is at the bottom of beam. For a cantilever end, tension is at the top of the beam.

2. Shear is at its maximum at edge of supports. Diagonal shear cracks is normally developed close to

the support. Stirrup for shear reinforcement is normally placed vertically to intercept the crack. They

are normally closer spaced near the support and gradually spread out toward center of the beam.


Ultimate Strength design of flexural reinforcements

Reinforcement calculation

Design assumption:

1. Strain distribute linearly across the section.

2. Concrete resists only compressive stresses.

Therefore, the stress distribution across the section of the beam is as shown below.

At an ultimate strain of 0.003, the stress at extreme fiber of the beam reaches ultimate strength of concrete

fc’. The distribution of the compressive stresses is a complex curve. For calculation purpose, a stress block

of 0.85fc’ spread over a depth, a, is used. Therefore, the total compressive stress in a rectangular beam is

C = 0.85fc’ab

Where b is the width of the beam.


At ultimate stress situation, the concrete at top portion is subjected to compression. The compressive

stresses distribute uniformly over a depth a. The resultant of compressive stress, C is located at a distance,

a/2, from the top surface. Tensile force is taken by rebars at an effective distance, d, from the top surface.

By equilibrium, the tensile force is equal to the compression resultant,

T = Asfy = C = 0.85f’c ab

where fy is the yield strength of reinforcing steel and As is the area of steel. Therefore,

The depth of stress block,

a = Asfy/(0.85f’c b), or a = Asfyd/(0.85f’

c bd),

Let the reinforcement ratio, ρ = As/bd, then

a = ρfyd/0.85f’c

Let m = fy/0.85f’c , then, a = ρdm..The nominal moment strength of the section,

Mn = C (d-a/2) = 0.85f’c ab(d-a/2)

Then, The nominal moment strength of the section,

Mn = Asfy (d-a/2) = Asfy (d-ρdm/2) = Asfy d- Asfy dρm/2

ACI code requires that the factored moment,

Mu ≤ φ Mn

Where, φ = 0.9, is the strength reduction factor for beam design. Let Mu = φ Mn , We have Mu = φ (Asfy d- Asfy

dρm/2)

Divide both side by bd2, we have Mu/φbd = (As/bd)fy -(As/bd) fy ρm/2) = ρfy - fy ρ2m/2)

Let Rn = Mu/φbd2, and we cab rewrite the equation as

ρ2(m/2) - ρ - Rn/fy = 0

Solving the equation, the reinforcement ratio,

ρ = (1/m)(1-2mRn/fy)1/2

The area of reinforcement is As = ρbd

Ductile and brittle failures, Balance condition, Maximum and minimum reinforcement ratio

There are two situations when a reinforced concrete beam fails due to bending. One is when the reinforcing

steel reaches its yield stress, fy. The other is when the concrete reach it maximum compressive stress, f’c.

When a reinforced concrete beam fails in yielding of steel, the failure is ductile because the steel can stretch


for a long period of time before it actually breaks. When it fails in concrete, the failure is brittle because

concrete breaks when it reach maximum strain.

When concrete reaches its maximum strain at the same time as the steel reach is yielding stress, it is called

a balance condition. Using a maximum strain, 0.003 of concrete and assume a linear distribution of strain

across beam section, one can determine the reinforcement ratio at balanced condition. The reinforcement

ratio based on ACI code is

ρb = (0.85f’c/fy) β1 [87000/(87000+fy)] [f’c and fy are in psi (lb/in2)]

ρb = (0.85f’c/fy) β1 [600/(600+fy)] [f’c and fy are in MPa (MN/m2)]

Where β1 = 0.85 for 4000 psi (30 Mpa) concrete, and reduce 0.05 for each 1000 psi of f’c in excess of 4000

psi.

To ensure a ductile failure of beam, ACI code limits the maximum reinforcement ratio to 0.75ρb. On the other

hand, when the amount of steel is too small, the beam will fail when concrete reach its tensile strength. It

needs to have a minimum amount of steel to ensure a ductile failure mode. The minimum reinforcement

ratio in ACI code is ρmin = 200/fy (psi).


Design examples

Situation:

A simply supported reinforced concrete beam is supporting uniform dead and live loads

Design data:

Dead load: 1500 lb/ft

Live load: 800 lb/ft

Length of beam: 20 ft

Width of beam: 16 in

Depth of beam: 24 in

Minimum concrete cover: 1.5 in

Diameter of stirrup, 0.5 in

Compressive strength of concrete: 4000 psi

Yeild strength of steel: 60000 psi

Requirement: Design flexural reinforcement for bending

Solution:

1. Calculate factored moment:

Weight of beam: WB = 150 lb/ft x 1.33 ft x 2 ft = 400 lb/ft

Factored load: Wu = 1.4(400+1500)+1.7(800) = 4020 lb/ft

Factored moment: Mu = (4020)(202)/8 = 201000 ft-lb

Assume the main reinforcement bar is 1" in diameter (#8 bar)

Effective depth: d:24-1.5-0.5-0.5 = 21.5 in

Factor: Rn = (201000)(12)/[(0.9)(16)(21.52)]=362.4 psi,

m = 60000/[(0.85)(4000)]=17.65

Reinforcement ratio

ρ = (1/m)(1-2mRn/fy)1/2)=0.0064

Minimum reinforcemnet ratio: ρmin = 200/fy=0.0033

Maximum reinforcement ratio; ρmin = (0.75)(0.85f’c/fy) β1 [87000/(87000+fy)]=0.021

Required reinforcement, As = ρbd = 2.2 in2.

Use 4#8 bar area of reinforcement is 0.79 in2x4 = 2.37 in2.


Shear Reinforcement

Shear strength of concrete

The direct shear shrength according to ACI is

φvc =0.85[1.9√fc’+2500ρω(Vud/Mu)] ≤ 0.85(3.5√fc’)

where ρω (≈ 0.002) is reinforcement ratio, Vu is factored shear stress, Mu is factored moment at the critical

section. Or

φvc =0.85(2√fc’)

ACI code requirements for shear reinforcement

1. When shear stress, vu ≤ ½ φvc ,no shear reinforcement is required.

2. When ½ φvc < vu ≤ φvc, use minimum reinforcement

Av = 50 bw s /fy

Where s is spacing of web reinforcement, fy is yield strength of steel, Av is cross section area of web

reinforcement, bw is width of beam web.

3. When φvc < vu , use vu ≤ φ(vc + vs), where vs is shear strength provided by shear reinforcement.

Stirrup reinforcements

The shear force that is resisted by shear reinforcements is Vs = (Vu - φVc). Normally, stirrup is spaced

vertically at a spacing, s, for shear reinforcement. Within an effective depth d, the shear strength provided by

Avfyd/s, where Av is area of stirrup, fy is yield strength of reinforcing steel. The shear strength multiply by a

reduction factor, φ, needs to be larger than Vs. Therefore, Vs = φ (Avfyd/s). The spacing of stirrup is

calculated as

s = (φAvfyd)/Vs

ACI code requirements for placing stirrup are as follows.

1. When ½ φvc < vu ≤ φvc, max s = d/2 ≤ 24 in.

2. When φvc < vs ≤ 4√fc, max s = d/2 ≤ 24 in.

3. When φvc < vs ≤ 8√fc, max s = d/4 ≤ 12 in.


Design example

Situation:

Design shear reinforcement for the beam in the previous example

Support column size: 12”x12”

Solution:

1. Calculate factored shear:

Clear distance between support, Ln = 19 ft

Factor shear Vu = WuLn/2 = 38.2 kips

Shear strength of concrete:

φVc = 0.85(2√4000) d b = 37 kips

1/2φVc = 18.5 kips

The length that required no shear reinforcement is

L1 = (Ln /2)(18.5/38.2) = 4.6 ft

Distance from center of beam that required minimum reinforcment is

L2 = (Ln/2)( φVc /Vu) = 9.2 ft close to Ln/2 = 9.5 ft

Use #3 stirrup the area of stirrup, area of steel: Av = 2(0.11 in2) = 0.22 in2.

Maximum spacing, s = (0.22 in2)(60000 psi) /[(50 psi)(16 in)] = 16.5 in

Maximum spacing d/2 = 10.75 in (Govern)

Use 6 stirrups at 10.75 inch spacing, with first stirrup at 5". Total length cover by stirrups is Ls =

(5)(10.75 in)+5 in = 4.9 ft O.K.


Design of reinforced concrete columns

Type of columns

Failure of reinforced concrete columns

Short column

Column fails in concrete crushed and bursting.

Outward pressure break horizontal ties and bend

vertical reinforcements

Long column

Column fails in lateral buckling.

See test picture from web-site below

part-3.html

See picture from web-site below

struct-walls.htm


Short column or Long column?

ACI definition

For frame braced against side sway: For Frame not braced against side sway:

Long column if klu/r > 34-12(M1/M2) or 40 Long column if klu/r > 22

Where k is slenderness factor, lu is unsupported length, and r is radius of gyration. M1 and M2 are the

smaller and larger end moments. The value, (M1/M2) is positive if the member is bent in single curve,

negative if the member is bent in double curve.

Determine the slenderness factor, k

The slender factor, k should be determined graphically from the Jackson and Moreland Alignment charts.

(Charts will be added later)

where ψ Ε( ∑ =cΙc/lc) of column /∑ Ε(bΙb/lb) of beam, is the ratio of effective length factors.

Ec and Ec are younger modulus of column and beams.

lc and lc are unbraced length of column and beams.

The cracked moment of inertia, Ιc is taken as 0.7 times gross moment of column and Ιb is taken as 0.35

times gross moment of inertia of beam.

Alternatively, k can be calculated as follows:

1. For braced frame with no sway,

k can be taken as the smaller value of the two equations below.

k = 0.7 + 0.05 (ψA+ψB) ≤ ,1

k = 0.8 + 0.05 (ψmin) ≤ 1

ψA and ψB are the ψ at both ends, ψmin is the smaller of the two ψ values.

2. For unbraced frame with restrains at both ends,

For ψm < 2

k = [(20- ψm)/20] √(1+ψm)

For ψm ≥ 2

k = 0.9 √(1+ψmin)

ψm is the average of the two ψ values.

2. For unbraced frame with restrain at one end, hinge at the other.

k = 2.0 + 0.3 ψ

ψ is the effective length factor at the restrained end.


Example:

Beam information:

Beam size: b = 18 in, h = 24 in

Beam unsupported length: lb = 30 ft

Concrete strength: 4000 psi

Young's modulus, Eb = 57 0004√ 5063 = ksi

Moment of inertia of beam: Ib = 0.35bh3/12 = 7258 in4.

Column information:

Square Column: D = 18 in, unsupported length lc =10 ft

Concrete strength: 5000 psi

Young's modulus: Ec = 57 0005√ 0304 = ksi

moment of inertia of column: Ic = 0.7D4/12 = 6124 in4.

Column top condition:

There are beams at both sides of column at top of column, no column stop above the beams

The effective length factor: ψΑ Ε( =cΙc/lc) /[2 Ε(bΙb/lb)] = 1.4

Column bottom condition:

There are beams at both sides of column at bottom of column and a column at bottom level

The effective length factor: ψΑ Ε( 2[ =cΙc/lc)] / [2 Ε(bΙb/lb)] = 2.8

From chart:

If the column is braced: k ≈ 0.84

If the column is unbraced: k ≈ 1.61

From equation

If the column is braced:

k = 0.7 + 0.05 (ψA+ψB) = 19.0

k = 0.8 + 0.05 (ψmin) = 29.0

If the column is unbraced: ψm = (ψA+ψB)/2 = 2.12

k = 0.9 √(1+ψmin) = 1.6

Design of reinforced concrete columns

Short column Long column:

1. Column shall be designed to resist factored axial

compressive load and factored moments.

2. Column strength shall be determined based on

strain compatibility analysis.

1. Column shall be designed to resist factored axial

compressive load. Factored moment shall be

magnified with magnification factors.

2. Column strength shall be determined based on

strain compatibility analysis.


Column ties and spiral

ACI code requirements for column ties

1. No. 3 ties for longitudinal reinforcement no. 10 bars or less, no. 4 ties for no. 11 bars or larger and

bundled bars.

2. Tie spacing shall not exceed 16 diameter of longitudinal bars, 48 diameters of tie bars, nor the least

dimension of column.

3. Every corner bar and alternate bars shall have lateral tie provide the angle shall not exceed 135

degree.

4. No longitudinal bar shall be spacing more than 6 inches without a lateral tie.

ACI code requirements for spiral

1. Sprial shall be evenly space continuous bar or wire, no. 3 or larger.

2. Sprial spacing shall not exceeds 3 in, nor be less than 1 in.

3. Anchorage of spiral shall be provided by 1-1/2 extra turn.


One way slab design

Aspect Ratio: B > 2 L

Placement of Dead and Loads: Live load shall be placed at such that it produces maximum positive and negative moments in slabs:

Load case 1: Dead load and live load

Load case 2: Dead load + skip live load 1.

Loca case 3: Dead load + skip live load 2.

Reinforcement layout: Reinforcement shall be placed at where tensile stress exist.

Load factors and combination

One way slab is normally designed for gravity load only. The design load combinations are as follows

ACI 318-99

U = 1.4D

U = 1.4D+1.7L

ACI 318 02 & 05

U = 1.2D

U = 1.2D+1.6L

where U is factored loads, D is dead load, and L is live load


ACI design factors for calculating shear and moment in slab

Limitation

1. There are two or more spans.

2. Spans are approximately equal. The two adjacent spans shall not be more than 20 percent difference in

length.

3. Load distributes uniformly.

4. Live load shall not exceed three times of dead load.

5. Members are prismatic.

Factored Moments:

Three or more spans

Positive moment: Interior span: Mu = Wu ln2/16

End span (discontinuous end unrestrained): Mu = Wu ln2/11

End span (discontinuous integral with support): Mu = Wu ln2/14

Negative moments: Negative moments at exterior face of first interior support: Mu = Wu ln2/10

Negative moments at other face of first interior support: Mu = Wu ln2/11

Negative moments at interior face of exterior support by spandrel beam: Mu = Wu ln2/24

Negative moments at interior face of exterior support by column: Mu = Wu ln2/16


Two spans Positive moment: End span (discontinuous end unrestrained): Mu = Wu ln2/11

End span (discontinuous integral with support): Mu = Wu ln2/14

Negative moments: Negative moments at exterior face of interior support: Mu = Wu ln2/9

Negative moments at interior face of exterior support by spandrel beam: Mu = Wu ln2/24

Negative moments at interior face of exterior support by column: Mu = Wu ln2/16

Slabs with span not exceeding 10 ft Negative moments at face of support: Mu = Wu ln2/12

Factored Shear: Shear in end members at face of first interior support: Vu = 1.15 Wu ln/2

Shear at all face of all other support: Vu = Wu ln/2


Slab reinforcement design Strength Design of Flexural reinforcement Design assumption: Strain distribute linearly across the section.

Concrete resists only compressive stresses.

Therefore, the stress distribution across the section of the beam is as shown below.

At an ultimate strain of 0.003, the stress at extreme fiber of the beam reaches ultimate strength of concrete

fc’. The distribution of the compressive stresses is a complex curve. For calculation purpose, a stress block

of 0.85fc’ spread over a depth, a, is used. Therefore, the total compressive stress in a rectangular beam is

C = 0.85fc’ab

Where b is the width of the beam.

At ultimate stress situation, the concrete at top portion is subjected to compression. The compressive

stresses distribute uniformly over a depth a. The resultant of compressive stress, C is located at a distance,

a/2, from the top surface. Tensile force is taken by rebars at an effective distance, d, from the top surface.

By equilibrium, the tensile force is equal to the compression resultant,

T = Asfy = C = 0.85f’c ab

where fy is the yield strength of reinforcing steel and As is the area of steel. Therefore,

The depth of stress block,

a = Asfy/(0.85f’c b), or a = Asfyd/(0.85f’

c bd),

Let the reinforcement ratio, ρ = As/bd, then


a = ρfyd/0.85f’c

Let m = fy/0.85f’c , then, a = ρ d m The nominal moment strength of the section,

Mn = C (d-a/2) = 0.85f’c ab(d-a/2)

Then, The nominal moment strength of the section,

Mn = Asfy (d-a/2) = Asfy (d-ρdm/2) = Asfy d- Asfy dρm/2

ACI code requires that the factored moment,

Mu ≤ φ Mn

Where, φ = 0.9, is the strength reduction factor for beam design. Let Mu = φ Mn , We have Mu = φ (Asfy d- Asfy

dρm/2)

Divide both side by bd2, we have Mu/φbd = (As/bd)fy -(As/bd) fy ρm/2) = ρfy - fy ρ2m/2)

Let Rn = Mu/φbd2, and we cab rewrite the equation as

ρ2(m/2) - ρ - Rn/fy = 0

Solving the equation, the reinforcement ratio,

ρ = (1/m)(1-2mRn/fy)1/2

The area of reinforcement is As = ρ b d

Ductile and brittle failures, Balance condition, Maximum and minimum reinforcement ratio

There are two situations when a reinforced concrete beam fails due to bending. One is when the reinforcing

steel reaches its yield stress, fy. The other is when the concrete reach it maximum compressive stress, f’c.

When a reinforced concrete beam fails in yielding of steel, the failure is ductile because the steel can stretch

for a long period of time before it actually breaks. When it fails in concrete, the failure is brittle because

concrete breaks when it reach maximum strain.

When concrete reaches its maximum strain at the same time as the steel reach is yielding stress, it is called

a balance condition. Using a maximum strain, 0.003 of concrete and assume a linear distribution of strain

across beam section, one can determine the reinforcement ratio at balanced condition. The reinforcement

ratio based on ACI code is

ρb = (0.85f’c/fy) β1 [87000/(87000+fy)] [f’c and fy are in psi (lb/in2)]

ρb = (0.85f’c/fy) β1 [600/(600+fy)] [f’c and fy are in MPa (MN/m2)]

Where β1 = 0.85 for 4000 psi (30 Mpa) concrete, and reduce 0.05 for each 1000 psi of f’c in excess of 4000

psi.

To ensure a ductile failure of beam, ACI code limits the maximum reinforcement ratio to 0.75ρb. On the other

hand, when the amount of steel is too small, the beam will fail when concrete reach its tensile strength. It

needs to have a minimum amount of steel to ensure a ductile failure mode. The minimum reinforcement

ratio in ACI code is ρmin = 200/fy (psi).


Minimum flexural reinforcement and maximum spacing: Minimum flexural reinforcement in slab shall be the same as minimum temperature reinforcement: (ACI 318-

05 Sec. 10.5.4)

Maximum spacing shall not exceed 3 time of slab thickness or 18 inches.

Temperature reinforcement: Minimum reinforcement ratio (reinforcment area/ gross area):

Slab with grade 40 or 50 deformed bars: 0.002

Slab with grade 60 deformbars or welded wire reinforcement: 0.0018

Slab with reinforcement with yield stress, fy, exceeds 60,000 psi measured at a yield strain of 0.35%:

0.0018x60,000/fy.

Corner reinforcement: Although the major reinforcement in one way slab span is in the short direction, yet at the corner, the

moment is actual supported by beams at both directions. If the slab is supported in the short direction only,

the slab may crack in the long direction until the reinforcement in the short direction pick up the strength.

The corner reinforcement can be arrange in two way as shown below.


Slab thickness and deflection

1. Minimum thickness for solid one-way slabs unless deflections are calculated for slab not supporting or

attached to partitions or other construction likely to be deamaged by large deflection. (ACI 318-05 Table

9.5a)

Simple supported One end continuous Both ends continuous Cantilever

L/20 L/24 L/28 L/10

2. Deflection limitation (ACI 318-05 Table 9.5b)

Flat roof Not supporting non-structural elements likely to

be damaged by large deflection Live load deflection L/180

Floor Not supporting non-structural elements likely to

be damaged by large deflection Live load deflection L/360

Roof or floor Supporting non-structural elements likely to be

damaged by large deflection)

Long term deflection + live load

deflection L/480

Roof or floor Supporting non-structural elements not likely to

be damaged by large deflection)

Long term deflection + live load

deflection L/240

3. Modulus of Elasticity and moment of inertia

The elastic modulus of concrete, Ec (lb/in2) is (ACI 318-05 Sec 8.5)

Ec = w

c

1.5 (33) √f

c’ lb/in

2

Ec = wc

1.5 (0.043) √f

c’ MPa

where wc is unit weight of concrete between 90 to 155 lb/ft3. fc’ is compression strength of concrete in

lb/in2.

For normal weight concrete Ec = 57000 √f

c’

4. Moment of inertia

The deflection shall be calculated using effectivemoment of intertia, Ieinstead of gross moment of

inertia,Ig.

Ie=(Mcr/Ma)3Ig+[1-(Mcr/Ma)3] Icr

where Icr is moment of inertia of crack section,Mais apply moment, and the cracking moment (in-lbs)

Mcr= frIg/yt

and the modulus of rapture, fr = 7.5 √ f

c’ (lb/in

2), yt is the distance from neutral axis to top of slab

For continuous member, Iecan be taken asthe average value of calculated from critical positive and

negative momentsections.

For prismatic members, Iecan be taken atmid-span for simple and continuous spans, and at support

forcontilevers.


For light weight concrete, if slitting tensilestrength, fctis specified, use fct/6.7 to substitute√fc’ but not

greater than√fc’ . If fct is not specified, fcr shall be multiplied by0.75 for all light weight concrete, 0.85

for sand light weightconcrete.

Design check list

1. Aspect ratio: B/L ≥ 2

2. Location and length of reinforcement

3. Flexural and Temperature reinforcement

4. Slab deflection ratio within the limit


One way slab design example Example 1: Design of interior three span slab as shown below

Design code: ACI 318-05

Design data:

Number of span: 3

Length of spans: L1 = 20 ft, L2 = 24 ft, L3 = 20 ft

Thickness of slab: 6 in

Live load: WL = 50 psf

Superimposed dead load: WSD = 10 psf

Concrete compressive strength: fc' = 4000 psi

Yield strength of steel: fy = 60 ksi

Requirement: design slab reinforcement use ACI factors and check deflection

1. Check if two adjacent spans differ more than 20%. L2/L1 = 120%, L2/L3 = 120 % O.K.

Weight of slab: Ws = 75 psf

Total dead load: WD = 75 + 10 = 85 psf

Live load to dead load ratio: WL/WD = 0.6 < 3 O.K.

Calculate slab moments using ACI factors:

Assume 1 ft width of slab for calculation, b = 1 ft

Factored load: Wu = (1.2*85+1.6*50) * 1 = 182 lb/ft (ACI 318-02, 05)

Positive moment:

Exterior span: Mu1 = 182*202 / 14 /1000 = 5.2 ft-kip

Interior span: Mu2 = 182*242 / 16 /1000 = 6.6 ft-kip

Negative moment:

Exterior face of first interior support: Mu3 = 182*202 / 10 /1000 = 7.3 ft-kip

Interior face of first interior support: Mu4 = 182*242 / 11 /1000 = 9.5 ft-kip

Interior face of Exterior support: Mu1 = 182*202 / 24 /1000 = 3.0 ft-kip

2. Design Positive moment reinforcement Exterior span

Factored moment: Mu1 = 5.2 ft-kip

Concrete cover: 0.75" for interior slab

Effective depth: d = 6 - 0.75-0.5 = 4.75 in

Factor: Rn = Mu1/(0.9 b d2 ) = 256.1 psi m = fy/(0.85 fc') = 17.7

Reinforcement ratio, ρ = 1/m [ 1 - ( 1 - 2 m Rn / fy )] = 0.00444

Area of reinforcement, As = ρ b d = 0.253 in2.

Use #5 at 12" O.C. area of reinforcement = 0.3 in2.

Interior span


Factor: Rn = 322.7 psi m = 17.7


Reinforcement ratio, ρ = 0.0057



3. Negative moment reinforcement Design negative reinforcement for first interior support,


Factor: Rn = 469.3 psi m = 17.7




Design negative reinforcement for Exterior support


Factor: Rn = 149.4 psi m = 17.7


Use minimum reinforcement, area of reinforcement, As = 1.33 ρ b d = 0.194 in2.


Temperature reinforcement: As = 0.0018*6*12 = 0.13 in2.

Use #4 at 18" O.C., area of reinforcement 0.133 in2.


Design of Reinforced Concrete Wall

ACI 318-05 requirements:

Minimum vertical wall reinforcement: (Section 14.3.2)

1. 0.0012 for deformed bars smaller than #5 and fy > 6000 psi

2. 0.0015 for deformed bars #5 or larger

Minimum horizontal wall reinforcement: (Section 14.3.3)

1. 0.0020 for deformed bars smaller than #5 and fy > 6000 psi

2. 0.0025 for deformed bars #5 or larger

Bar placement: (Section 14.3.4, 14.3.5) 1. Maximum bar spacing: 18 in

2. Walls more than 10" thick shall have two layers of reinforcement.

3. Lateral ties is not required if reinforcement ratio is less than 0.01.

4. Minimum 2#5 bars shall be placed around all window or door openings extend 24" beyond.

Empirical design method:

The empirical design is limited to the following situation:

1. solid rectangular cross section.

2. resultant of all factored loads is located with the middle third of the overall thickness of the wall.

Design of load bearing RC wall with empirical method Limitation: ACI 318-05 Section 14.5

1. solid rectangular cross section.

2. resultant of all factored loads is located with the middle third of the overall thickness of the wall

3. Minimum thickness of wall shall not be less than 1/25 the support height of length or 4 inches.

4. Minimum thickness of exterior basement walls and foundation walls shall not be less than 7.5 inches.

RC wall strength by empirical design method: The empirical equation for calculate axial strength of RC wall is

φPn = 0.55 φ fc' Ag [ 1 - ( k lc / 32 h)2] ACI 318 Eq (14.1)

Where φ = 0.65, lc is height of wall, h is thickness of wall, Ag is gross section area,

fc' is compression strength of concrete, k is effective length factor.

The effective length factor, k shall be

For wall braced top and bottom against lateral translation, and

1. restrained against rotation at one or both ends, k = 0.8.

2. unrestrained against rotation at both ends, k = 1.0

For walls not braced against lateral translation, k = 2.0


Example 1:

Design code: ACI 318-05 Section 14.5

Situation: A RC exterior wall supporting a two story building.

Design data:

Width of wall: h = 8 in

Height of wall: lc = 12 ft.

Compression strength of concrete: fc' = 3000 psi

Yield strength of concrete: fy = 60000 psi

Tributary width: 15 ft

Roof live load: RL = 20 psf

Roof dead load: RD = 80 psf

Floor live load: FL = 100 psf

Floor dead load: FD = 80 psf

Lateral wind pressure: pw = 15 psf

Lateral seismic force: pE = 10 psf

Assumption: wall is continuous and restrained against rotation at top and bottom.

Requirement: Design wall reinforcement using empirical design method

Solution:

Design for 1 ft width of wall: b = 1 ft

Axial dead load: PD = FD *Trib*2*b + RD * Trib. b = 3.6 kips

Axial roof live load: PRL = RL * Trib*b = 0.3 kips

Axial floor live load: PFL = FL*Trib*b = 1.5 kip

Axial wind load: PW = 0

Axial seismic load: PE = 0

Out-of-plan bending dead load moment: MD = 0

Out-of-plan bending Roof ive load moment:: MRL = 0

Out-of-plan bending floor live load moment: MFL = 0

From the continuous beam formula, the maximum negative moment is 0.1time span length lc.

Out-of-plan bending wind load moment: MW = 0.1 * pw*lc2*b = 216 lb-ft

Out-of-plan bending seismic load moment: ME = 0.1 * pE*lc2*b = 144 lb-ft

Load combination:

Case 1: Pu1 = 1.4 PD = 5.04 kip, Mu1 = 0

Eccentricity: eu1 = 0

Case 2:

Pu2 = 1.2 PD + 1.6 PFL + 0.5 PRL = 6.87 kip

Mu2 = 1.2 MD + 1.6 MFL + 0.5 MRL = 0

Eccentricity: eu2 = 0

Case 3:

Pu3 = 1.2 PD + 1.6 PW + 1.0 PFL + 0.5 PRL = 5.97 kip


Mu3 = 1.2 MD + 1.6 MW + + 1.0 MFL + 0.5 MRL = 0.346 ft-kip

Eccentricity: eu3 = Mu3/Pu3 = 0.695 in < 8 in /6 = 1.33 in (O.K.)

Case 4:

Pu4 = 1.2 PD + 1.0 PE + 1.0 PFL = 5.82 kip

Mu4 = 1.2 MD + 1.0 ME + + 1.0 MFL = 0.144 ft-kip


Case 5:

Pu5 = 0.9 PD + 1.6 PW = 3.24 kip

Mu5 = 0.9 MD + 1.6 MW = 0.346 ft-kip


Strength reduction factor: φ 56.0 =

Gross section: Ag = b* h = 96 in2

Effective length factor: k = 0.8

Design axial strength of wall:

φPn = 0.55 φ fc' Ag [ 1 - ( k lc / 32 h)2] = 82.1 kips

Maximum axial load: Pu3 = 5.97 kips O.K.

Check vertical reinforcement for bending moment,

Maximum moment per foot width: Mu = 0.346 ft kip

Use one layer of reinforcement, place at middle of the section:

Effective depth: d = 4 in

Factor: Rn = Mu / (0.9*b*d2) = 24 psi, m = fy / (0.85 fc' ) = 23.5

Reinforcement ratio: ρ = (1/m)*[1-2−1(√ m Rn / fy )] = 0.0004

Less than minimum vertical reinforcement ratio, use ρ 2100.0 =

Vertical reinforcement: As = 0.0012 Ag = 0.115 in2.

Use #4 at 18 in o.c. area of reinforcement, 0.133 in2.

Minimum horizontal reinforcement: ρt = 0.002, area of reinforcement, As = 0.192 in2.

Use #4 at 12" o.c. area of reinforcement, 0.2 in2.


Design of Reinforced Concrete Shear wall


Design of shear reinforcement


Shear reinforcement

Effective depth

Maximum shear strength permitted

Location of Critical Section

Minimum shear reinforcement

Bar placement

Design of Flexural reinforcement

Design of shear reinforcement

1. The shear strength, φVn must equal or be greater than the factored shear, Vu

φVn ≥ Vu ACI eq (11-1)

where φ 57.0 = is strength reduction factor, and

Vn = Vc + Vs ACI eq (11-2)

where Vc is shear strength of concrete, and Vs is strength of shear reinforcement.


2. Shear strength of concrete shear wall shall be calculated as

a. Vc = 2 √fc' h d

Or the lesser of

b. Vc = 3.3 √fc' h d + Nu d/ (4 lw) ACI eq (11-29)

c. Vc = { 0.6 √fc' + lw ( 1.25 √fc' + 0.2 [Nu/(lw* h)]) /( Mu/Vu - lw/2)} h d ACI eq (11-30)

(It does not apply when Mu/Vu - lw/2 < 0)

where fc' is compressive strength of concrete,

h is thickness of the wall,

d is effective depth,

lw is overall length of wall.

Nu is axial force, positive for compression, negative for tension,

Mu is moment parallel to the direction of the wall.

Shear reinforcement


When Vu > φ Vc,

shear strength provided by horizontal shear reinforcement shall be calculated aS

Vs = Av fy d /s ACI eq. (11-31)

where Av is area of reinforcement, fy is yield strength of reinforcement, d is effective depth, and s is spacing

of reinforcement.

Maximum shear strength permitted

φVn = 10 √fc' ACI section 11.10.3

Effective depth

1. d = 0.8 lw

2. Calculate based on strain compatibility analysis (same as in a beam section).

Location of Critical Section

1. One-half of the wall height.

2. lw /2

whichever is smaller.

Minimum shear reinforcement

When φ Vc > Vu ≥ 0.5 φ Vc

1. Minimum horizontal shear reinforcement ratio to gross section area, ρt = 0.0025.

2. Minimum vertical shear reinforcement ratio to gross section area,

ρl = 0.0025 + 0.5 (2.5 - hw/ lw )( ρt - 0.0025) ≥ 5200.0 ACI eq. (11-32)

and need not be greater than calculated based on ACI eq. 11-31.

When Vu < 0.5 φ Vc

1. Minimum vertical wall reinforcement: (Section 14.3.2)

a. 0.0012 for deformed bars smaller than #5 and fy > 6000 psi

b. 0.0015 for deformed bars #5 or larger

2. Minimum horizontal wall reinforcement: (Section 14.3.3)

a. 0.0020 for deformed bars smaller than #5 and fy > 6000 psi

b. 0.0025 for deformed bars #5 or larger.

Bar placement

1. Maximum spacing of horizontal reinforcement

a. lw /5. b. 3 h c. 18 inches.

2. Bar shall be placed uniformly across the length and height of walls


Design of Flexural reinforcement

1. Non-load bearing shear wall

Design as a cantilever beam:

φMn ≥ Mu

2. Load bearing shear wall

Design as a column subject to axial load and bending

φPn ≥ Pu

φMn ≥ Mu

Location of flexural reinforcement:

Flexural reinforcement shall be placed at each end of walls

Design examples

Example 1: Design of reinforced concrete non-load bearing shear wall.


Design data: Seismic shear force: (service load)

Roof: Vr = 100 kips

4th floor: V4 = 75 kips, ,

3rd floor: V3 = 50 kips

2nd floor: V2 = 25 kips

Floor height: H = 15 ft

Length of wall: lw = 18 ft


Concrete strength: fc' = 4000 psi

Yield strength of steel: fy = 60 kis


Assumption:

1.out-of-plan moment is neglectable.

2.The wall is an exterior wall.

Requirement:

Design reinforcement for shear wall

Solution:

Maximum shear occurs at load combination: 1.2D+1.4E+1.0L

Calculate maximum vertical and shear force at first floor

Maximum factored shear: Vu = 1.4 (100+75+50+25) = 350 kips

Check maximum shear strength permitted

Assume effective depth, d = 0.8 (18) = 14.4 ft

Strength reduction factor, φ = 0.75

φVn = 10 √fc' h d = 819 kips > 350 kips O.K.

Critical section for shear at smaller of 18 ft/2 = 9 ft , H/2 = 7.5 ft

Calculate factored overturning moment and weight of wall at critical section

Mu = 1.4 [100 (60-7.5)+75(45-7.5)+50(30-7.5)+25 (25-7.5)] = 13130 ft-kips

Nu = (0.15)(10/12)(18)(60-7.5) = 118.1 kips

Calculate shear strength of concrete:

φVc = 0.75 [3.3 √fc' h d + Nu d/ (4 lw)] = 288.2 kips

Mu/Vu - lw/2 = 28.5 ft

φVc = 0.75 { 0.6 √fc' + lw ( 1.25 √fc' + 0.2 [Nu/(lw* h)]) /( Mu/Vu - lw/2)} h d = 163.8 kips

Or φVc = 0.75 (2 √fc' h d) = 163.9 kips Use

Design horizontal shear reinforcement:

Vs = Vu - φVc = 186.1 kips

Use #5 bar in one layer, area of reinforcement, Av = 0.3 in2.

Spacing: S = φAv fy d /Vs = 12.6 in, Use 12" O.C.

Check maximum spacing: (18x12)/5 = 43 in, 3 (10) = 30 in, or 18 in O.K.

Check minimum reinforcement: ρt = 0.3 in2 / (12x10) = 0.0025 O.K.

Design vertical reinforcement:

ρl = 0.0025 + 0.5 (2.5 - hw/ lw )( ρt - 0.0025) = 0.0025

Use ρl = 0.0025

Area of reinforcement: Av = 0.0025 (10)(12) = 0.3 in2/ft

Use #5 bars at 12" O.C

Design flexural reinforcement:

Calculate factored moment at base:


Mu = 1.4 [(100)(60)+(75)(45)+(50)(30)+(25)(15)]=15750 ft-kip

Tension control section, φ = 0.9

Factor: Rn = (15750)(12000)/[0.9(10)(14.4x12)2] = 703 psi, and m = fy/(0.85fc')=17.7

Reinforcement ratio, ρ = (1/m)[1-(1- 2 m Rn/fy)] = 0.013

Area of reinforcement, As = 0.013 h d = 22.9 in2.

Use #10 bars, number of bars, n = 22.9/1.27 = 18

Check effective depth

Concrete cover = 2" for exterior wall. Use 3" spacing between #10 bars in two layers

Effective depth, d = (18)(12) - 2-(3)(8)/2 =202 in

Recalculate reinforcement, Factored Rn = Mu / φ h d2 = 514.7 psi, m = 17.


Area of reinforcement, As = 18.9 in2.

Use #10 bars, number of bars, n = 18.9 /1.27 = 15, Use 16 # 10

Use #4 closed shape ties to enclose tension reinforcement,

Area of reinforcement for shear As = 0.4 in2.

Check clear spacing between bars, S = 10-(2)(2)-(0.5)(2)-1.27 = 3.73 in O.K.

Reinforcement detail


Example 2: Design of Reinforced Concrete load bearing shear wall Situation: A reinforced concrete load bearing shear wall supporting for a four story building


Design data: Vertical load: (service load)

Dead load at each floor and roof: PD = 40 kips

Live load at each floor and roof: PL = 25 kips

Seismic shear force: (service load)

Roof: Vr = 100 kips

4th floor: V4 = 75 kips, ,

3rd floor: V3 = 50 kips

2nd floor: V2 = 25 kips

Floor height: H = 15 ft

Length of wall: lw = 18 ft


Concrete strength: fc' = 4000 psi

Yield strength of steel: fy = 60 kis

Assumptions:

1. out-of-plan moment is neglectable.

2. The wall is an exterior wall.

Requirement: Design reinforcement for shear wall

Solution

Maximum shear occurs at load combination: 1.2D+1.4E+1.0L

Calculate maximum vertical and shear force at first floor


Maximum factored shear: Vu = 1.4 (100+75+50+25) = 350 kips

Check maximum shear strength permitted

Assume effective depth, d = 0.8 (18) = 14.4 ft

Strength reduction factor, φ = 0.75

φVn = 10 √fc' h d = 819 kips > 350 kips O.K.

Critical section for shear at smaller of 18 ft/2 = 9 ft , H/2 = 7.5 ft

Calculate factored overturning moment and weight of wall at critical section

Mu = 1.4 [100 (60-7.5)+75(45-7.5)+50(30-7.5)+25 (25-7.5)] = 13130 ft-kips

Nu = 1.2 [(0.15)(10/12)(18)(60-7.5)+4 PD ]+1.0 (4 PL ) = 462.1 kips

Calculate shear strength of concrete:

φVc = 0.75 [3.3 √fc' h d + Nu d/ (4 lw)] = 393.9 kips

Mu/Vu - lw/2 = 28.5 ft

φVc = 0.75 { 0.6 √fc' + lw ( 1.25 √fc' + 0.2 [Nu/(lw* h)]) /( Mu/Vu - lw/2)} h d = 228.2 kips (Use)

Or φVc = 0.75 (2 √fc' h d) = 196.7 kips

Design horizontal shear reinforcement:

Vs = Vu - φVc = 112.1 kips

Use #4 bar in two layer, area of reinforcement, Av = 0.4 in2. (Code requires two layers for 12" wall)

Spacing: S = φAv fy d /Vs = 25.7 in

Check maximum spacing: (18x12)/5 = 43 in, 3 (10) = 30 in, or 18 in Use 18"

Check minimum reinforcement: ρt = 0.4 in2 / (18x10) = 0.0019 < 0.0025

Use ρt =0.0025, spacing S = 0.4 in2 / (0.0025)(h) = 13.3 in Use 12 in

Design vertical reinforcemnt

ρl = 0.0025 + 0.5 (2.5 - hw/ lw )( ρt - 0.0025) = 0.0025

Use ρl = 0.0025

Use #4 bars in two layers at 12" O.C

Calculate factored moment and axial load at base:

Mu = 1.4 [(100)(60)+(75)(45)+(50)(30)+(25)(15)]=15750 ft-kip

Nu = 1.2 [(0.15)(10/12)(18)(60)+4 PD ]+1.0 (4 PL ) = 486.4 kips

Design as a column subjected to axial load and bending

Gross area, Ag = (18)(12)(12) = 2592 in2.

Assume tension control section, φ = 0.9

φNu/Ag = 0.141 ksi

φMu/(Ag lw) = 0.253 ksi

From ACI column design chart (See column design section), Area of reinforcement, ρ = 0.011

Area of reinforcement, As = (0.01)(18x12)(12) = 22.8 in2.

Use #10 bars, number of bar, n = 22.8/1.27 = 18


Use 10#10 bars at each end of shear wall, column ties is required since ρ > 0.01. Use #4 ties at 12"

O.C.


Design of concrete slab on grade

Slab Reinforcement and spacing of control joints.

When concrete slab are poured on grade, steel reinforcement and control joints are used to control

shrinkage and expansion. When concrete slab shrink or expansion, it will drag the soil beneath the slab with

it. Friction at bottom of slab is trying to prevent the slab from moving. Steel reinforcement inside the slab is

also trying to prevent the slab from shrink or expansion. It is not practical to place large amount of

reinforcement to completely prevent slab from shrink or expanse. Therefore, Control joints are normal placed

at a distance to allow for the slab to shrink or expanse.

Subgrade drag equation

PCA recommends that the area of steel per linear foot of slab width for slab on grade is

As = F L w / (2 fs) [1]

Where

As = area of steel (in2) per linear foot of slab width.

F = coefficient of subgrade friction (F = 1.2 to 2 for pavement, F = 1.5 for concrete slab on grade)

L = Length of slab between control joint (feet)

w = weight of slab (lb/ft2)

fs = allowable working stress of reinforcement (lb/in2) is usually 0.67fy to 0.75 fy, fy is yield strength of

reinforcing steel)

Example:

Given:

Length between Joint: 30 ft

Thickness of pool slab: 6 in

Slab reinforcement in the long direction:

Weight of slab: w = 150 lb/ft3 x 6 in = 75 lb/ft2

Yield strength of steel: 60 ksi

Allowable stress of reinforcement: fs = 0.75 x 60 ksi = 45 ksi

Coefficient of subgrade fraction: F = 1.5

Reinforcement required, As = F L w / (2 fs) = 0.037 in2/ft

Use Welded Wire fabric, WWW 6x6, W2.0xW2.0, Area = 0.04 in2/ft.

Design of Concrete

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