design of concrete
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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
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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
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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)
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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.
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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.
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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
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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).
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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.
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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.
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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.
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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
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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.
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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.
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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.
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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
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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
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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
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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
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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).
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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.
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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.
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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
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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
Factored moment: Mu2 = 6.6 ft-kip
Factor: Rn = 322.7 psi m = 17.7
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Reinforcement ratio, ρ = 0.0057
Area of reinforcement, As = ρ b d = 0.323 in2.
Use #5 at 11" O.C. area of reinforcement = 0.327 in2.
3. Negative moment reinforcement Design negative reinforcement for first interior support,
Factored moment: Mu4 = 9.5 ft-kip
Factor: Rn = 469.3 psi m = 17.7
Reinforcement ratio, ρ = 0.00845
Area of reinforcement, As = ρ b d = 0.482 in2.
Use #6 at 10" O.C. area of reinforcement = 0.528 in2.
Design negative reinforcement for Exterior support
Factored moment: Mu4 = 3.0 ft-kip
Factor: Rn = 149.4 psi m = 17.7
Reinforcement ratio, ρ = 0.00255
Use minimum reinforcement, area of reinforcement, As = 1.33 ρ b d = 0.194 in2.
Use #4 at 12" O.C. area of reinforcement = 0.2 in2.
Temperature reinforcement: As = 0.0018*6*12 = 0.13 in2.
Use #4 at 18" O.C., area of reinforcement 0.133 in2.
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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
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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
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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
Eccentricity: eu4 = Mu4/Pu4 = 0.297 in < 8 in /6 = 1.33 in (O.K.)
Case 5:
Pu5 = 0.9 PD + 1.6 PW = 3.24 kip
Mu5 = 0.9 MD + 1.6 MW = 0.346 ft-kip
Eccentricity: eu4 = Mu4/Pu4 = 1.28 in < 8 in /6 = 1.33 in (O.K.)
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.
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Design of Reinforced Concrete Shear wall
Design code: ACI 318-05
Design of shear reinforcement
Shear strength of concrete
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.
Shear strength of concrete
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
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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
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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 code: ACI 318-05
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
Width of wall: h = 10 in
Concrete strength: fc' = 4000 psi
Yield strength of steel: fy = 60 kis
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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:
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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.
Reinforcement ratio, ρ = 0.0094
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
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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 code: ACI 318-05
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
Width of wall: h = 12 in
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
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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
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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.
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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.
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