Project 2

Project 2:

Step 1: Loads on the Abutment

For the loads on the abutment, We would need to consider dead loads, live loads, superimposed dead loads, and any other relevant loads. Since We've mentioned using IRC 6, follow the load combinations and factors provided in the code to calculate the design loads.

Step 2: Live Load Analysis and Effective Width Calculation

Perform live load analysis using appropriate load distribution factors as per IRC 6 for class A and the given wheel load (70 R). Calculate the reactions on the abutment due to these loads. Effective width calculation involves determining the width of the slab that contributes to the load distribution. This can be based on the characteristics of the bridge and the code recommendations.

Step 3: Load Combinations

As per IRC 6, create load combinations for Ultimate Limit State (ULS), Serviceability Limit State (SLS), rare load cases, and quasi-permanent loads. Apply appropriate load factors as per the code.

Step 4: Checks for ULS and SLS

Perform checks for ULS and SLS using the calculated loads and load combinations. Check factors such as bending, shear, deflection, and bearing capacity. Design the piles for vertical and horizontal capacities considering the soil parameters, pile spacing, and arrangement.

Step 5: Anchorage Length and Bar Minimization

Calculate the anchorage length required for the reinforcement bars in the pile cap. This involves ensuring that the bars can transfer the forces safely from the piles to the cap. Minimize the bars wherever possible to optimize the design and economy while satisfying the strength and detailing requirements.

Step 6: Pile and Pile Cap Design

Design the pile cap for the calculated loads, considering the shear, bending, and torsional forces. Determine the dimensions of the pile cap and the arrangement of piles under it. Design the reinforcement layout for the pile cap considering the calculated loads and moments.

Step 7: Rebar Detailing

Detail the reinforcement bars in the pile cap as per the design calculations. Provide information about the size, spacing, cover, and arrangement of the bars. Ensure that the detailing meets the code requirements for seismic and non-seismic zones.

Step 8: Drawings and Quantity Calculations

Draw the pile cap and pile reinforcement layout to scale, indicating all the relevant details such as bar sizes, spacing, and clear cover. Calculate the quantity of concrete and steel reinforcement required per unit volume (kg/cumec) for the pile cap and piles.

Given and Assume Data:

• Span Length (L): 15 m
• Abutment Height (h): 5 m
• Bridge Width (b): 12 m
• Pile Diameter (d): 0.6 m (600 mm)
• Concrete Cover (c): 75 mm
• Soil Bearing Capacity (q): 200 kN/m²
• Density of Soil (γ): 20 kN/m³
• Grade of Concrete (f_ck): M35 (35 MPa)
• Grade of Steel (f_y): Fe500 (500 MPa)
• Scour Depth at Abutment: 4 m
• Live Load: IRC Class A loading
• Load Factors: As per IRC 6

Step 1: Loads on the Abutment

• Dead Load (DL): 25 kN/m²
• Live Load (LL): 70 kN/m (Assuming a simplified point load)
• Superimposed Dead Load (SDL): 5 kN/m²
• Impact Factor (IF): 1.5 (IRC 6)

Total Factored Load (ULS):

• Factored DL = 1.5 * DL = 1.5 * 25 = 37.5 kN/m²
• Factored LL = IF * LL = 1.5 * 70 = 105 kN/m

Step 2: Live Load Analysis and Effective Width Calculation

Assuming a 1000 mm slab for self-weight reactions, the effective width for live load distribution can be estimated conservatively as half the span length: beff = L/2 = 15/2 = 7.5 m.

Step 3: Load Combinations

As per IRC 6:

• ULS: 1.5 * DL + 1.5 * LL = 1.5 * 37.5 + 1.5 * 105 = 255 kN/m²
• SLS: DL + LL = 25 + 70 = 95 kN/m²

Step 4: Checks for ULS and SLS

Assuming a simplified pile arrangement (e.g., 4 piles under the pile cap) and simplified analysis, calculate moments and shears using basic structural analysis equations.

Using the formulas for simply supported beams:

• Maximum Bending Moment at the face of the abutment due to ULS: Mu_ULS = (w * L^2) / 8 = (255 * 15^2) / 8 = 716.25 kNm

• Maximum Shear Force at the face of the abutment due to ULS: Vu_ULS = w * L / 2 = 255 * 15 / 2 = 1912.5 kN

• For SLS, the moments and shears will be lower.

Step 5: Anchorage Length and Bar Minimization

Let's calculate the anchorage length required for the reinforcement bars in the pile cap. Using simplified code provisions, we'll use 30 times the diameter of the largest bar as the anchorage length.

Anchorage Length (Ld) = 30 * Bar Diameter = 30 * 20 mm = 600 mm

Step 6: Pile and Pile Cap Design

Assume pile cap dimensions: Width (b) = 5 m, Length (L) = 6 m. Calculate the effective depth of the pile cap (d_eff) based on bending moment and shear calculations.

Effective Depth (d_eff) = h - c - Bar Diameter - Clear Cover Assuming clear cover of 50 mm, and using h_eff = d_eff + Bar Diameter + Clear Cover:

Effective Depth (d_eff) = 500 - 75 - 20 - 50 = 355 mm

Step 7: Rebar Detailing

• Moment due to ULS (Mu_ULS) = 716.25 kNm
• Shear due to ULS (Vu_ULS) = 1912.5 kN
• f_ck = 35 MPa
• f_y = 500 MPa

Assuming simplified distribution of reinforcement, let's calculate the required area of tensile reinforcement (A_s) using simplified methods. For instance, use a balanced design approach:

Assume a bar diameter of 20 mm and provide 2 layers of reinforcement:

Tensile Reinforcement:

• Using simplified formula for balanced design: Mu_ULS = 0.87 * f_y * A_s * d * (1 - (0.425 * Mu_ULS) / (f_ck * b * d^2))

  Solve for A_s: A_s = (Mu_ULS * 1000) / (0.87 * f_y * d * (1 - (0.425 * Mu_ULS) / (f_ck * b * d^2)))

  Plugging in values: A_s = (716.25 * 1000) / (0.87 * 500 * 0.355 * (1 - (0.425 * 716.25) / (35 * 500 * 500 * 0.355^2))) = 2723.54 mm²

  Provide 20 mm bars in two layers: A_s_per_layer = 2723.54 / 2 = 1361.77 mm²

  Assuming a bar spacing of 150 mm, calculate the number of bars in one layer: Number_of_bars_per_layer = A_s_per_layer / (bar_spacing * bar_diameter) Number_of_bars_per_layer = 1361.77 / (150 * 20) = 4.537 (Round up to 5 bars)

  Number of bars in two layers = 5 * 2 = 10 bars

Step 8: Drawings and Quantity Calculations

Create scaled drawings of the pile cap and pile reinforcement layout, indicating bar sizes, spacing, and clear cover. Calculate the quantities of concrete and steel required per unit volume (kg/cumec) for the pile cap and piles.

Reinforcement Calculation:

Recall from the previous steps that we calculated the required area of tensile reinforcement (A_s) as 2723.54 mm² for a 600 mm deep pile cap, assuming 20 mm bars in two layers.

Number of bars in two layers = 10 bars

Concrete Calculation:

Calculate the volume of the pile cap:

Volume = Width * Length * Depth Volume = 5 m * 6 m * 0.6 m = 18 m³

Quantities:

Assuming the weight of 1 m³ of concrete to be 2400 kg (considering M35 concrete with 35 MPa compressive strength):

• Concrete Quantity = Volume * Unit Weight
• Concrete Quantity = 18 m³ * 2400 kg/m³ = 43200 kg

Steel Reinforcement Quantities:

Assuming a density of 7850 kg/m³ for steel:

• Total Weight of Reinforcement = Number of bars * Length of one bar * Area of one bar * Density of Steel
• Total Weight of Reinforcement = 10 bars * 6 m * (π * 0.02^2 / 4) m² * 7850 kg/m³ = 9.42 kg

Summary of Quantities:

• Concrete Quantity: 43200 kg (43.2 tons)
• Steel Reinforcement Quantity: 9.42 kg (0.00942 tons)

These quantities represent the total amount of concrete and steel reinforcement needed for the pile cap of the bridge foundation.

2)Live load analysis and effective width calculation for 70 r and class A to be done to calculate the reactions on the abutment:

a) 1000 mm slab can be assumed for 15 m span (For self weight reactions)

The live load analysis for IRC Class A loading using the 70 R wheel load and calculate the reactions on the abutment. Since we mentioned that a 1000 mm slab can be assumed for self-weight reactions, we'll consider this for the calculations.

Assumptions:

• Span Length (L): 15 m
• Live Load: IRC Class A loading
• Wheel Load (70 R): 70 kN

Live Load Analysis:

IRC Class A loading consists of two loads of 70 kN each placed at 1.5 m from the edge and 7.5 m apart. We will consider only one load acting at a time for calculating reactions. Let's calculate the reactions due to the live load:

Reaction due to a Single Wheel Load:

1. Calculate the reaction at the support closest to the load:

Load = 70 kN Distance from the edge = 1.5 m

Reaction at support A = (Load * Distance from the edge) / Span Reaction at support A = (70 * 1.5) / 15 = 7 kN

2. Calculate the reaction at the support farthest from the load:

Reaction at support B = Load - Reaction at support A Reaction at support B = 70 - 7 = 63 kN

Total Reaction due to Live Load:

For one wheel load: Total reaction at support A = Reaction due to wheel load = 7 kN Total reaction at support B = Reaction due to wheel load = 63 kN

Effective Width Calculation:

Given that a 1000 mm slab can be assumed for self-weight reactions, and considering the spacing of the two wheel loads, we'll calculate the effective width for the live load distribution factor.

Effective Width (beff) = Clear width of slab + Distance between loads Effective Width (beff) = 1000 mm + 7.5 m = 7.5 m

3) Load combinations as per IRC 6 for ULS,SLS rare and quasi: