Project 2

• Classify the sections according to Class and calculate effective widths for the beams

To classify the sections according to the Class (Class 1, Class 2, or Class 3) and calculate effective widths for the beams, we'll use the given cross-sectional details for each section: BD, DE, and EG. Classifications are based on IS 800 criteria for compactness parameters. Let's start with each section:

Section BD (Outer Sections):

Given cross-sectional details for BD:

• Width of Top Flange (b_f): 3100 mm
• Haunch: 25 mm
• Thickness (d): 250 mm
• Height of I Girder (h): 1225 mm
• Web dimensions (b_w x t_w): 1160 mm x 25 mm
• Top Flange dimensions: 400 mm x 25 mm
• Bottom Flange dimensions: 400 mm x 40 mm

Step 1: Section Classification for BD

Calculate the compactness parameters:

Flange width-to-thickness ratio (b_f / t_f):

• b_f = 3100 mm
• t_f = 25 mm
• b_f / t_f = 3100 mm / 25 mm = 124

Web slenderness ratio (h / t_w):

• h = 1225 mm
• t_w = 25 mm
• h / t_w = 1225 mm / 25 mm = 49

Based on IS 800 criteria:

• Flange is slender (Class 4)
• Web is compact (Class 1)

Step 2: Calculate Effective Width for Bending Resistance (b_eff) for BD

For Class 4 flanges, b_eff = b_f:

• b_eff = 3100 mm

For Class 1 webs, b_eff = 0.95 * b_w (IS 800):

• b_eff = 0.95 * 1160 mm = 1102 mm

Section DE (Inner Section):

Given cross-sectional details for DE:

• Width of Top Flange (b_f): 3100 mm
• Haunch: 25 mm
• Thickness (d): 250 mm
• Height of I Girder (h): 1225 mm
• Top Flange dimensions: 400 mm x 20 mm
• Web dimensions (b_w x t_w): 1175 mm x 12.5 mm
• Bottom Flange dimensions: 400 mm x 30 mm

Step 1: Section Classification for DE

Calculate the compactness parameters:

Flange width-to-thickness ratio (b_f / t_f):

• b_f = 3100 mm
• t_f = 20 mm
• b_f / t_f = 3100 mm / 20 mm = 155

Web slenderness ratio (h / t_w):

• h = 1225 mm
• t_w = 12.5 mm
• h / t_w = 1225 mm / 12.5 mm = 98

Based on IS 800 criteria:

• Flange is compact (Class 1)
• Web is compact (Class 1)

Step 2: Calculate Effective Width for Bending Resistance (b_eff) for DE

For Class 1 flanges, b_eff = b_f:

• b_eff = 3100 mm

For Class 1 webs, b_eff = 0.95 * b_w (IS 800):

• b_eff = 0.95 * 1175 mm = 1116.25 mm

Section EG (Outer Sections - Similar to BD):

We can follow the same steps as for section BD because the cross-sectional details for EG are identical to BD.

In summary, the classifications and effective widths for the beams in the sections BD, DE, and EG are as follows:

• BD:
  • Flange: Class 4
  • Web: Class 1
  • Effective Width (b_eff): 3100 mm (Flange), 1102 mm (Web)
• DE:
  • Flange: Class 1
  • Web: Class 1
  • Effective Width (b_eff): 3100 mm (Flange), 1116.25 mm (Web)
• EG (Similar to BD):
  • Flange: Class 4
  • Web: Class 1
  • Effective Width (b_eff): 3100 mm (Flange), 1102 mm (Web)

Apply loadings on the bridge considering the carriageway width as 9.3m, footpath on both sides 2m each, 500mm height and create load combinations (Surfacing depth of 100m g = 20 kN/m3)

• Carriageway width: 9.3 meters
• Footpath width on both sides: 2 meters each
• Bridge deck height: 500 mm
• Surfacing depth: 100 mm
• Density of surfacing material (g): 20 kN/m⊃3; (typical for asphalt or concrete)

1. Dead Load (DL):

a. Self-weight of Bridge Deck:

Self-weight = ρ_concrete * d_concrete * L * (width of deck + width of footpaths) Self-weight = 24 kN/m⊃3; * 0.1 m * 13,300 mm * (9.3 m + 2 m + 2 m) = 59,587.2 kN

b. Self-weight of Surfacing:

Surfacing weight = g * d_surfacing * L * (width of deck + width of footpaths) Surfacing weight = 20 kN/m⊃3; * 0.1 m * 13,300 mm * (9.3 m + 2 m + 2 m) = 29,793.6 kN

c. Total Dead Load (DL):

Total DL = Self-weight of Deck + Self-weight of Surfacing Total DL = 59,587.2 kN + 29,793.6 kN = 89,380.8 kN

2. Live Load (LL):

Total LL = Live Load Intensity * Area Total LL = 20 kN/m⊃2; * (9.3 m + 2 m + 2 m) * 13,300 mm = 265.2 kN

3. Footpath Load (FL):

Total FL = Footpath Load Intensity * Area of One Footpath * 2 (for both sides) Total FL = 5 kN/m⊃2; * 2 m * 13,300 mm * 2 = 266 kN

4. Load Combinations:

Load Combination 1 (LC1): 1.2DL + 1.5LL LC1 = 1.2 * Total DL + 1.5 * Total LL

LC1 = 1.2 * 89,380.8 kN + 1.5 * 265.2 kN = 108,361.6 kN

Given Data for Section BD:

• Width of Top Flange (bf​): 3100 mm
• Thickness (d): 250 mm
• Height of I Girder (ℎh): 1225 mm
• Web dimensions (bw​ x tw​): 1160 mm x 25 mm
• Top Flange dimensions: 400 mm x 25 mm
• Bottom Flange dimensions: 400 mm x 40 mm
• Assumed steel yield strength (σy​): 250 MPa (E350 B0 grade)
• Assumed shear stud properties (σstud_yield​): 385 MPa (HT material)

Step 1: Calculate Section Properties

a. Elastic Section Modulus (S):

Substitute the values:

Calculate S:

b. Plastic Section Modulus (Z):

To calculate Z, we need to determine the effective width (beff​) of the web, which is typically taken as 0.95 times the actual width of the web (bw​) for Class 1 sections:

Now, calculate Z:

Step 2: Elastic Bending Resistance (Melastic​)

Calculate elastic Melastic​ using the elastic section modulus (S) and the yield strength of the steel (σy​):

Step 3: Plastic Bending Resistance (Mplastic​)

Calculate Mplastic​ using the plastic section modulus (Z) and the yield strength of the steel (σy​):

Step 4: Shear Capacity (Vshear​)

To calculate the shear capacity (Vshear​), we first need to determine the number of shear studs required along the effective width (beff​).

Assuming a shear stud spacing (s) of 200 mm:

Since we cannot have a fraction of a shear stud, let's round this to 6 shear studs.

Next, calculate the shear capacity per stud using the shear stud properties (σstud_yield​) and the assumed shear stud diameter (Dstud​) of 16 mm:

Finally, calculate Vshear​:

Step 5: Check Against Applied Forces

To check the section's capacity against applied forces, you would compare the calculated elastic bending resistance (Melastic​), plastic bending resistance (Mplastic​), and shear capacity (Vshear​) with the actual forces and moments applied to the section from the load combinations.

• Check for buckling capacities of the members

To check for the buckling capacity of the members in the composite bridge section BD, we need to calculate the critical buckling load for both the top flange and the web of the I-girder. We'll use the Euler's formula for buckling analysis. 

Assumptions:

Given section BD:
  - Width of Top Flange (\(b_f\)): 3100 mm
  - Thickness (\(d\)): 250 mm
  - Height of I Girder (\(h\)): 1225 mm
  - Web dimensions (\(b_w\) x \(t_w\)): 1160 mm x 25 mm
  - Top Flange dimensions: 400 mm x 25 mm
  - Bottom Flange dimensions: 400 mm x 40 mm
  - Assumed steel yield strength (\(σ_y\)): 250 MPa (E350 B0 grade)
  - Assumed modulus of elasticity (\(E\)) for steel: 200 GPa (200,000 MPa)
  - Assumed length of member (\(L\)): 13300 mm (13.3 meters)

Step 1: Calculate Effective Length for Buckling (\(L_e\)):

The effective length for buckling depends on the boundary conditions. Assuming the ends are pinned (both ends), the effective length can be taken as the actual length \(L\).

Step 2: Calculate Critical Buckling Load for the Top Flange:

The critical buckling load for the top flange can be calculated using Euler's formula:

\[
P_{\text{cr, top}} = \frac{\pi^2 \cdot E \cdot I_{\text{top}}}{(K_{\text{top}} \cdot L_e)^2}
\]

Where:
- \(P_{\text{cr, top}}\) = Critical buckling load for the top flange
- \(E\) = Modulus of elasticity for steel (200,000 MPa)
- \(I_{\text{top}}\) = Moment of inertia for the top flange
- \(K_{\text{top}}\) = Effective length factor for the top flange (for pinned-pinned ends, \(K_{\text{top}}\) is 1.0)
- \(L_e\) = Effective length (equal to \(L\) for pinned ends)

To calculate \(I_{\text{top}}\), we can use the formula for the moment of inertia of a rectangle:

\[
I_{\text{top}} = \frac{b_f \cdot d^3}{12}
\]

Substitute the values:

\[
I_{\text{top}} = \frac{3100 \, \text{mm} \cdot (250 \, \text{mm})^3}{12}
\]

Step 3: Calculate Critical Buckling Load for the Web:

The critical buckling load for the web can also be calculated using Euler's formula:

\[
P_{\text{cr, web}} = \frac{\pi^2 \cdot E \cdot I_{\text{web}}}{(K_{\text{web}} \cdot L_e)^2}
\]

Where:
- \(P_{\text{cr, web}}\) = Critical buckling load for the web
- \(I_{\text{web}}\) = Moment of inertia for the web
- \(K_{\text{web}}\) = Effective length factor for the web (for pinned-pinned ends, \(K_{\text{web}}\) is 1.0)
- \(L_e\) = Effective length (equal to \(L\) for pinned ends)

To calculate \(I_{\text{web}}\), we can use the formula for the moment of inertia of a rectangle:

\[
I_{\text{web}} = \frac{b_w \cdot t_w^3}{12}
\]

Substitute the values:

\[
I_{\text{web}} = \frac{1160 \, \text{mm} \cdot (25 \, \text{mm})^3}{12}
\]

Step 4: Calculate Critical Buckling Loads:

Now, we can calculate the critical buckling loads for both the top flange and the web.

For the top flange:

\[
P_{\text{cr, top}} = \frac{\pi^2 \cdot 200,000 \, \text{MPa} \cdot \frac{3100 \, \text{mm} \cdot (250 \, \text{mm})^3}{12}}{(1.0 \cdot 13300 \, \text{mm})^2}
\]

For the web:

\[
P_{\text{cr, web}} = \frac{\pi^2 \cdot 200,000 \, \text{MPa} \cdot \frac{1160 \, \text{mm} \cdot (25 \, \text{mm})^3}{12}}{(1.0 \cdot 13300 \, \text{mm})^2}
\]

Now, calculate these values:

\[
P_{\text{cr, top}} \approx 170.58 \, \text{kN}
\]

\[
P_{\text{cr, web}} \approx 1.36 \, \text{kN}
\]

Step 5: Compare with Applied Forces:

To check the buckling capacity, compare the critical buckling loads (\(P_{\text{cr, top}}\) and \(P_{\text{cr, web}}\)) with the applied axial loads and shear forces in the member. If the applied loads are less than the critical buckling loads, the member is stable against buckling.

Please note that the calculated critical buckling loads are theoretical values. In practice, factors such as lateral-torsional buckling, imperfections, and boundary conditions can affect the actual behavior of the member. Consulting with a structural engineer and considering local design codes is essential for a comprehensive structural analysis and design.

• Check for longitudinal shear

To calculate the longitudinal shear in the composite bridge section BD, we'll follow a step-by-step procedure with all the values and assumptions. We'll calculate the shear capacity of shear studs and compare it to the applied shear force.

Given Data for Section BD:
- Width of Top Flange (\(b_f\)): 3100 mm
- Web dimensions (\(b_w\) x \(t_w\)): 1160 mm x 25 mm
- Assumed shear stud properties (\(σ_{\text{stud\_yield}}\)): 385 MPa (HT material)
- Assumed shear stud diameter (\(d_{\text{stud}}\)): 16 mm
- Applied shear force (\(V_{\text{applied}}\)): To be determined

Assumed Shear Stud Spacing (\(s\)):
- Shear stud spacing is a design parameter typically determined based on the applied shear force and other design considerations. For this calculation, we'll assume a spacing of 150 mm (0.15 meters) as an example.

Step 1: Calculate the Shear Capacity of Shear Studs:

The shear capacity of a single shear stud can be calculated using the formula:

\[
\text{Shear Capacity per Stud} = \frac{\pi}{4} \cdot (d_{\text{stud}}^2) \cdot σ_{\text{stud\_yield}}
\]

Substitute the values:

\[
\text{Shear Capacity per Stud} = \frac{\pi}{4} \cdot (16 \, \text{mm})^2 \cdot 385 \, \text{MPa}
\]

Calculate the Shear Capacity per Stud:

\[
\text{Shear Capacity per Stud} = \frac{\pi}{4} \cdot (256 \, \text{mm}^2) \cdot 385 \, \text{MPa} = 98101.5 \, \text{N}
\]

Step 2: Determine the Number of Shear Studs:

To find the number of shear studs required along the web's length, we need to calculate the effective length (\(L_{\text{eff}}\)) for which shear studs are needed. In this case, we'll consider the entire length of the web as the effective length, which is equal to the spacing between bracing locations.

Given data:
- Length between braces (\(L\)): 13300 mm (13.3 meters)

The number of shear studs (\(N_{\text{studs}}\)) required is given by:

\[
N_{\text{studs}} = \frac{L}{s}
\]

Substitute the values:

\[
N_{\text{studs}} = \frac{13300 \, \text{mm}}{150 \, \text{mm}}
\]

Calculate the Number of Shear Studs:

\[
N_{\text{studs}} = 88.67
\]

Since we can't have a fraction of a shear stud, we'll round up to the nearest whole number:

\[
N_{\text{studs}} \approx 89 \text{ studs}
\]

Step 3: Calculate the Total Shear Capacity Provided by Shear Studs:

The total shear capacity (\(V_{\text{shear}}\)) provided by all the shear studs in the section is given by:

\[
V_{\text{shear}} = N_{\text{studs}} \cdot \text{Shear Capacity per Stud}
\]

Substitute the values:

\[
V_{\text{shear}} = 89 \text{ studs} \cdot 98101.5 \, \text{N/stud}
\]

Calculate the Total Shear Capacity:

\[
V_{\text{shear}} = 8734223.5 \, \text{N} = 8.73 \, \text{kN}
\]

Step 4: Compare with Applied Shear Force:

Now, we need to compare the total shear capacity (\(V_{\text{shear}}\)) provided by the shear studs with the applied shear force (\(V_{\text{applied}}\)). If \(V_{\text{applied}}\) is less than or equal to \(V_{\text{shear}}\), the section can resist the applied shear force.

For example, if the applied shear force is known to be 7 kN, then:

\[
V_{\text{applied}} = 7 \, \text{kN}
\]

We compare:

\[
V_{\text{applied}} \leq V_{\text{shear}}
\]

\[
7 \, \text{kN} \leq 8.73 \, \text{kN}
\]

Since 7 kN is less than 8.73 kN, the section can resist the applied shear force, and it's considered safe against longitudinal shear.