Project 1

AIM: To design the beam marked B1 and the floor slab as per IS 456-2000.

GIVEN:

• Location: Typical school classroom

• Concrete Grade: M25

• Steel Grade: Fe 415

• Clear Cover: 20 mm for slab, 30 mm for beam

• Superimposed Dead Load: 2.5 kN/m² (including floor finish and ceiling)

SOLUTION:

1.Slab Design:

The slab consists of individual panels with a length of 8 m and width of 3 m. The aspect ratio of the slab (8/3 = 2.67) is greater than 2, indicating it behaves as a one-way slab spanning between the horizontal beams.

The slab acts as a 4-span continuous slab, with each span being 3 m. According to clause 24.2 of IS 456-2000, a slab spanning in one direction and continuous over supports should be designed as a continuous beam.

Estimating Slab Thickness:

Based on IS 456-2000's span/depth ratio, we estimate the slab's thickness.

• Span/Effective Depth Ratio for Continuous Slabs (clause 23.2.1):
  The span/effective depth ratio for continuous members is 26. Let the effective depth of the slab be denoted as d.

  $$$$

$$\frac{3000}{d} = 26 \quad \Rightarrow \quad d = \frac{3000}{26} = 115 \, \text{mm}$$

Considering the clear cover (20 mm) and the diameter of the longitudinal bars (8 mm), we opt for a 150 mm thick slab.

Loads on the Slab:

• Superimposed Dead Load: 2.5 kN/m²

• Self-weight of Slab (thickness = 150 mm):
  Self-weight=0.15×25=3.75kN/m²

• Live Load (Educational Building): 3 kN/m² (from IS 875 Part 2, Table 1)

Thus, the total dead load on the slab is $6.25 \, \text{kN/m²}$6.25kN/m² (sum of superimposed dead load and self-weight), and the live load is 3 kN/m².

Since the slab is one-way and behaves as a continuous beam, we use the bending moment coefficients from IS 456-2000 (Table 12) to determine the design moments.

Design Moments for the Slab:

• End Span (Middle):
  Dead Load Moment =+(1/12)×6.25×3²=4.7kNm
  Live Load Moment =$+(1/10) \times 3 \times 3^2 = 2.7 \, \text{kNm}$+(1/10)×3×3²=2.7kNm
  Ultimate Moment (1.5D + 1.5L) =$1.5 \times (4.7 + 2.7) = 11.1 \, \text{kNm}$1.5×(4.7+2.7)=11.1kNm

• Interior Span (Middle):
  Dead Load Moment =$+(1/16) \times 6.25 \times 3^2 = 3.52 \, \text{kNm}$+(1/16)×6.25×3²=3.52kNm
  Live Load Moment =+(1/12)×3×3²=2.25kNm
  Ultimate Moment (1.5D + 1.5L) =1.5×(3.52+2.25)=8.7kNm

• Support Next to End Support (Negative Moment):
  Dead Load Moment =−(1/10)×6.25×3²=−5.62kNm
  Live Load Moment =−(1/9)×3×3²=−3kNm
  Ultimate Moment (1.5D + 1.5L) =1.5×(−5.62−3)=−13kNm

Reinforcement for the Slab:

We now calculate the required reinforcement using the ultimate moments and the quadratic equation for the reinforcement area:

For the middle of the end span:

• Moment $M_u = 11.1 \, \text{kNm}$M_u​=11.1kNm

• Required reinforcement (bottom):
  Area of Steel = 252mm²

Spacing for 8 mm diameter bars: 250 mm.

For the middle of the interior span:

• Moment $M_u = 8.7 \, \text{kNm}$M_u​=8.7kNm

• Required reinforcement (bottom):
  Area of Steel =196.56mm²

Spacing for 8 mm diameter bars: 333 mm.

For the top reinforcement at supports:

• Moment at support = $-13 \, \text{kNm}$−13kNm

• Required reinforcement (top):
  Area of Steel = 297.36mm²

Spacing for 8 mm diameter bars: 200 mm.

2.Beam Design (Beam B1):

The beam has a typical span of 8 m (8000 mm). We first estimate the required depth of the beam.

Beam Depth Estimation:

Using the span/depth ratio for continuous beams (26), we calculate the effective depth required for the beam:

$$d=\frac{8000}{26}=307.7\text{mm}$$

To accommodate the clear cover (30 mm) and other dimensions (8 mm stirrups, 16 mm longitudinal bars), we select a beam depth of 450 mm.

Load on the Beam:

• Dead Load (from slab): 6.25kN/m²×3m=18.75kN/m

• Live Load (from slab): 3kN/m²×3m=9kN/m

• Weight of Beam: $1.87 \, \text{kN/m}$1.87kN/m

Total Dead Load = 18.75 + 1.87 = 20.6 kN/m
Live Load = 9 kN/m

Moments for the Beam:

• End Span Moment (Ultimate):
  Ultimate Moment =$1.5 \times (109.9 + 57.6) = 251.25 \, \text{kNm}$1.5×(109.9+57.6)=251.25kNm

• Interior Span Moment (Ultimate):
  Ultimate Moment =$1.5 \times (82.4 + 48) = 195.6 \, \text{kNm}$1.5×(82.4+48)=195.6kNm

• Support Next to End Support (Negative Moment):
  Ultimate Moment =1.5×(−131.85−64)=−294kNm

Beam Reinforcement:

After calculating the moment demands, we check the adequacy of the beam depth to ensure the beam is under-reinforced:

• For a beam depth of 450 mm and width of 250 mm, the moment capacity is insufficient to resist the maximum moment of 294 kNm.

We adjust the beam's dimensions:

• Widened Beam (350 mm x 550 mm): The beam’s moment capacity is now adequate.

Final Beam Reinforcement:

For the final beam design, using M30 concrete and Fe 415 steel:

• Bottom Reinforcement (End Span):
  Area of Steel = 1675.8mm²
  Bars: 9 x 16 mm diameter

• Bottom Reinforcement (Interior Span):
  Area of Steel = 1252.44mm²
  Bars: 3 x 25 mm diameter

• Top Reinforcement (Support):
  Area of Steel = 2028.6mm²
  Bars: 3 x 32 mm diameter

Thus, the final beam design is achieved using a 350 mm x 550 mm beam with M30 concrete, ensuring it is adequately reinforced for the applied loads.

RESULT:

Slab Design:

• Thickness: 150 mm based on IS 456-2000 span-to-depth ratios.

• Loads: Dead load = 6.25 kN/m², Live load = 3 kN/m².

• Reinforcement:

  • Bottom reinforcement for the end span: 252 mm² (5-8 mm bars, 250 mm spacing).

  • Bottom reinforcement for the interior span: 196.56 mm² (4-8 mm bars, 333.33 mm spacing).

  • Top reinforcement near supports: 297.36 mm² (6-8 mm bars, 200 mm spacing).

  • Temperature and shrinkage reinforcement: 8 mm bars at 300 mm spacing.

Beam Design (B1):

• Initial Depth: 450 mm, but inadequate for the moment demand.

• Final Beam: 350 mm x 550 mm with M30 concrete and Fe 415 steel.

• Reinforcement:

  • Bottom reinforcement at the end span: 1675.8 mm² (4-25 mm bars).

  • Bottom reinforcement at the interior span: 1252.44 mm² (3-25 mm bars).

  • Top reinforcement near support: 2028.6 mm² (3-32 mm bars).

This design ensures the slab and beam are adequately reinforced to handle the applied loads.