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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 shall be placed at where tensile stress exist.

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

U = 1.2D

U = 1.2D+1.6L

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

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.

Interior span: M_{u} =
W_{u }*l*_{n}^{2}/16

End span (discontinuous end unrestrained): M_{u} =
W_{u }*l*_{n}^{2}/11

End span (discontinuous integral with support): M_{u} =
W_{u }*l*_{n}^{2}/14

Negative moments at exterior face of first interior support: M_{u} =
W_{u }*l*_{n}^{2}/10

Negative moments at other face of first interior support: M_{u} =
W_{u }*l*_{n}^{2}/11

Negative moments at interior face of exterior support by spandrel
beam: M_{u} =
W_{u }*l*_{n}^{2}/24

Negative moments at interior face of exterior support by column:
M_{u} = W_{u }*l*_{n}^{2}/16

End span (discontinuous end unrestrained): M_{u} =
W_{u }*l*_{n}^{2}/11

End span (discontinuous integral with support): M_{u} =
W_{u }*l*_{n}^{2}/14

Negative moments at exterior face of interior support: M_{u} =
W_{u }*l*_{n}^{2}/9

Negative moments at interior face of exterior support by spandrel
beam: M_{u} =
W_{u }*l*_{n}^{2}/24

Negative moments at interior face of exterior support by column:
M_{u} = W_{u }*l*_{n}^{2}/16

Negative moments at face of support: M_{u} =
W_{u }*l*_{n}^{2}/12

Shear in end members at face of first interior support: V_{u} =
1.15 W_{u }*l*_{n}/2

Shear at all face of all other support: V_{u} =
W_{u }*l*_{n}/2

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 f_{c}’. 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 = 0.85f_{c}’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 = A_{s}f_{y} =
C = 0.85f’_{c} ab

where f_{y} is
the yield strength of reinforcing steel and As is the area of steel. Therefore,

The depth of stress block,

a = A_{s}f_{y}/(0.85f’_{c} b),
or a = A_{s}f_{y}d/(0.85f’_{c} bd),

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

a = rf_{y}d/0.85f’_{c}

Let m = f_{y}/0.85f’_{c} ,
then, a = r d
m The nominal moment strength of the section,

M_{n} = C (d-a/2) =
0.85f’_{c} ab(d-a/2)

Then, The nominal moment strength of the section,

M_{n} = A_{s}f_{y} (d-a/2)
= A_{s}f_{y} (d-rdm/2)
= A_{s}f_{y} d-
A_{s}f_{y} drm/2

ACI code requires that the factored moment,

M_{u} £ f M_{n}

Where, f =
0.9, is the strength reduction factor for beam design. Let M_{u} = f M_{n} ,
We have M_{u} = f (A_{s}f_{y} d-
A_{s}f_{y} drm/2)

Divide both side by bd^{2}, we have M_{u}/fbd
= (A_{s}/bd)f_{y} -(A_{s}/bd)
f_{y} rm/2)
= rf_{y} -
f_{y} r^{2}m/2)

Let R_{n} = Mu/fbd^{2},
and we cab rewrite the equation as

r^{2}(m/2) - r -
R_{n}/f_{y} =
0

Solving the equation, the reinforcement ratio,

r =
(1/m)(1-2mR_{n}/f_{y})^{1/2}

The area of reinforcement is A_{s} = r b
d

There are two situations when a reinforced concrete beam fails due
to bending. One
is when the reinforcing steel reaches its yield stress, f_{y}. 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

r_{b} =
(0.85f’_{c}/f_{y}) b_{1} [87000/(87000+f_{y})] [f’_{c} and
f_{y} are in psi
(lb/in^{2})]

r_{b} =
(0.85f’_{c}/f_{y}) b_{1} [600/(600+f_{y})] [f’_{c} and
f_{y} are in MPa
(MN/m^{2})]

Where b_{1} =
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 0.75r_{b}. 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 r_{min} =
200/f_{y} (psi).

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.

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, f
_{y}, exceeds 60,000 psi measured at a yield strain of 0.35%: 0.0018x60,000/f_{y}.

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.

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 |

The elastic modulus of concrete, E_{c}
(lb/in^{2}) is (ACI 318-05 Sec 8.5)

E_{c} =
w_{c}^{1.5} (33) Öf_{c}’ lb/in^{2}

E_{c} = w_{c}^{1.5} (0.043) Öf_{c}’ MPa

where w_{c} is
unit weight of concrete between 90 to 155 lb/ft^{3}.
f_{c}’ is compression strength of concrete in
lb/in2.

For normal weight concrete
E_{c} =
57000 Öf_{c}’

The deflection shall be calculated using effective moment of
intertia, I_{e} instead
of gross moment of inertia, I_{g}.

I_{e} = (M_{cr}/M_{a})^{3} I_{g} +
[1-(M_{cr}/M_{a})^{3}] I_{cr}

where Icr is moment of inertia of crack section, M_{a} is
apply moment, and the cracking moment (in-lbs)

M_{cr} = f_{r} I_{g} /
y_{t}

and the modulus of rapture, f_{r} =
7.5 Ö f_{c}’ (lb/in^{2}),
y_{t} is the
distance from neutral axis to top of slab

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

For prismatic members, I_{e} can
be taken at mid-span for simple and continuous spans, and at support
for contilevers.

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

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