Design of slab beam column

Design of Slab

* If ≥ 2, design as one way slab

< 2, design as two way slab.

* In case of simply supported slab if ≥ 25, there is no need to check for deflections.

* Spacing of bars may be found using the relation S = × 1000.

* Spacing should not exceed 3 × depth of slab or 300 mm whichever is smaller.

* A minimum of 0.15% of total cross section distribution steel is provided if mild steel is used and it

is 0.12%, if Fe 415 steel is used.

In case of continuous slab, assume a depth of th span and design moment at support next to end

support is

Mmax =

* In case of cantilever assume depth of slab as . In this case Mu = and Vu = wuL and

reinforcements are required at top. Sufficient anchorage depth should be provided at the support.

* In case of simply supported slabs with two adjacent edges at corner, if there is discontinuity,

torsional reinforcements are required. Torsional reinforcement consists of a mesh of

reinforcements at top and bottom of slab, the area of reinforcement in each layer shall be 3/4th of

the area required at mid-span of slab. The length shall be of shorter span.

Design of Columns

* Table 28 of IS 456–2000 gives the effective lengths for columns with various end conditions.

* Slenderness ratio limits prescribed for columns are:

(i) Unsupported length, l shall not be more than 60 times the least lateral dimension.

(ii) If in any given plane, one end of the column is unrestricted l 100 .

* Minimum eccentricity to be considered is

emin =

* The assumptions made are

1. The maximum compressive strain in axial compressive is 0.002.

2. In case of column subjected to axial load and bending, maximum strain permitted is

emax = 0.0035 – 0.75 e2

where e2 = strain at the least compressed extreme fibre.

* Design:

Pu = 0.4 fck Ac + 0.67 fy Asc

.

The above value may be multiplied by 1.05 if helical reinforcements are provided, if the volume of

helical reinforcement to the volume of core is greater than or equal to 0.36 .

* Reinforcement: Maximum 0.8%

Minimum 6%.

It is preferable to use from 0.8% to 4%

* Minimum number of bars: 4 for rectangular sections and 6 for circular sections

* Minimum diameter of bars to be used is 12 mm.

* Spacing of longitudinal bars measured along the periphery should not exceed 300 mm.

Diameter of transverse reinforcement should not be less than

1. th of larger main reinforcement

2. 5 mm

* Pitch shall not be more than

1. Least lateral dimension

2. 16 × dia. of smallest longitudinal bar

3. 300 mm.

* Pitch of helical spring should not be

1. more than 75 mm

2. more than th core diameter

3. less than 25 mm

4. less than 3 × diameter of helix.

* Design of columns subjected to combined axial load and uniaxial moment: Use interaction chart

given in SP–16 in which interaction diagrams are available for vs. for different values of

, where p is the percentage reinforcement.

Design of Isolated column footing

* For masonry walls, footing is one way reinforced type and for columns, it is two-way reinforced

footing.

* Column footing may be isolated, combined or raft

* Minimum depth of footing is given by

h =

where p = SBC, w = Unit weight of soil and f = Angle of friction of soil.

* Minimum cover to reinforcement shall be 50 mm. If plain concrete bed is not provided it shall be

75 mm.

Minimum thickness shall be

– 150 mm for footing on soil

– 300 mm for footing on piles.

* To find area of footing, add 10% of vertical load to account for self-weight. Then

A =

* Soil reaction for the factored load

qu = .

* Determine the minimum depth required from the consideration of bending moment, single shear and

double shear.

* Bending moment is maximum at the face of column.

* Critical section for one way shear is at a distance d from the face of column.

* Double shear is critical at distance d/2 from the face of column. Thus, if b × d is size of column,

area of concrete resisting punching shear is

A = 4(b + d)d.

Punching shear S = B × D – h × d.

where B × D = area of footing.

* Footing may be of uniform thickness or sloping.

* Bending moments and shear forces may be found by the direct design method or by equivalent frame

method

* The direct design method is applicable, only if the following conditions are satisfied:

1. There are at least three continuous spans in each direction.

2. £ 2.

3. The successive span length in each direction shall not differ by more than × longer span.

4. The design live load 3 × design dead load

* Total design moment

Mo =

where W = Design load on the area L2 × Ln

Ln = Clear span but not less than 0.65 L1

Note: Circular supports shall be treated as square supports having the same area for finding clear

span. If transverse span on either side varies it shall be taken as the average.

In an interior span,

–ve design moment = 0.65 Mo

In an end span,

Interior –ve design moment =

Interior +ve design moment =

Exterior –ve design moment =

where ac =

kc = Flexural stiffness of columns meeting at joint

ks = Flexural stiffness of the slab

* In the column strip the distribution of moment is

–ve moment at exterior support = 100%

–ve moment at interior support = 75%

+ve moment = 60%

* The critical section for shear shall be at a distance d/2 from the periphery of the column/capital

drop.