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# One way slab design

## Aspect Ratio: B > 2 L

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

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

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.

### 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:

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, r = As/bd, then

a = rfyd/0.85f’c

Let m = fy/0.85f’c , then, a = r 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-rdm/2) = Asfy d- Asfy drm/2

ACI code requires that the factored moment,

Mu £ f Mn

Where, f = 0.9, is the strength reduction factor for beam design. Let Mu = f Mn , We have Mu = f (Asfy d- Asfy drm/2)

Divide both side by bd2, we have Mu/fbd =  (As/bd)fy -(As/bd) fy rm/2) = rfy - fy r2m/2)

Let Rn = Mu/fbd2, and we cab rewrite the equation as

r2(m/2) - r - Rn/fy = 0

Solving the equation, the reinforcement ratio,

r = (1/m)(1-2mRn/fy)1/2

The area of reinforcement is As = r 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

rb = (0.85f’c/fy) b1 [87000/(87000+fy)]            [f’c and fy are in psi (lb/in2)]

rb = (0.85f’c/fy) b1 [600/(600+fy)]                    [f’c and fy are in MPa (MN/m2)]

Where b1 = 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 0.75rb.  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 rmin = 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):

1. Slab with grade 40 or 50 deformed bars: 0.002
2. Slab with grade 60 deformbars or welded wire reinforcement: 0.0018
3. 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 = wc1.5 (33) Öfc                              lb/in2

Ec = wc1.5 (0.043) Öfc                         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 Öfc

### 4. Moment of inertia

The deflection shall be calculated using effective moment of intertia, Ie instead of gross moment of inertia, Ig.

Ie = (Mcr/Ma)3 Ig + [1-(Mcr/Ma)3] Icr

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

Mcr = fr Ig / yt

and the modulus of rapture,  fr = 7.5 Ö fc (lb/in2),  yt is the distance from neutral axis to top of slab

For continuous member, Ie can be taken as the average value of  calculated from critical positive and negative moment sections.

For prismatic members, Ie can be taken at mid-span for simple and continuous spans, and at support for contilevers.

For light weight concrete, if slitting tensile strength, fct is specified, use fct/6.7 to substitute Öfc  but not greater than Öfc .  If fct is not specified, fcr shall be multiplied by 0.75 for all light weight concrete, 0.85 for sand light weight concrete.

## Design check list

1. Aspect ratio: B/L ³ 2

2. Location and length of reinforcement

3. Flexural and Temperature reinforcement

4. Slab deflection ratio within the limit