Week 5 Challenge

Design of Plate Girder

Question No: 1. Shear resistance of web problem

Determine PG Web thickness for given load without stiffener

For a PG of size 1200 x 300 x 25x 10, Check if the web is adequate to carry a factored shear force of 1000 kN. Yield strength of plate fy is 250 Mpa.

Aim:

To be determine the Plate Girder web thickness for given load without stiffener

Given data:

Plate Girder size = 1200 x 300 x 25 x 10 mm

Factored Shear force = 1000 kN

Yield Strength (f_y) = 250 Mpa.

Procedure:

Minimum Web thickness requirement:

d/t_w = (1200 – 2 x 25) / 10

= 1150 / 10 = 115 > 67ε [ε = √(250 / fy) = 1]

Hence it is said to be Plate Girder.

d/t_w = 115 < 200ε

Shear Capacity check:

As per IS 800 : 2007 Clause 8.4.1

V_d = (A_v x f_yw) / (√3 x γ_mo)

= 1150 x 10 x 250 / (√3 x 1.1)

= 1508983 N

= 1508.983 kN

0.6 V_d = 0.6 x 1508.983 = 905.39 kN < 1000 kN

Section is not Safe in Shear.

Shear Buckling check:

As per IS 800 : 2007 Clause 8.4.2.2

V_cr = A_v x 𝜏_b

Shear stress corresponding to web buckling

𝜏_b = f_yw / (√3 x λ²_w)

λ_w = √(f_yw / (√3 x t_cr) )

t_cr = K_v x ᴨ² x E/[12 (1 - µ²) (d/t_w)²]

K_v = 5.35 when transverse stiffners are provided only at support

E = 2 x 10⁵ N/mm²

µ (Poison’s Ratio) = 0.3

t_cr = 5.35 x ᴨ ² x 2 x 10⁵ / [12 (1 – 0.3²) (1150/10)²]

= 73.12 N/mm²

λ_w = √(250 / (√3 x 73.12) )

= 1.974

𝜏_b = 250 / (√3 x 1.974²)

= 37.042

V_cr = 1150 x 10 x 37.042

= 425983 N = 425.983 kN < 1000 kN

Result:

The web shear force corresponding to the web buckling is less than maximum Factored shear force. Hence, this web section is not safe.

Question No: 2. Check whether intermediate stiffener required for given load

For a PG of size 1500 x 500 x 30 x 12,check if the intermediate web stiffener is required to carry a factored shear force of 2000 kN, if required, find the spacing of stiffener. Yield strength of plate fy is 250 Mpa.

Aim:

To be check whether intermediate stiffener required for the given load

Given data:

Plate Girder size = 1500 x 500 x 30 x 12 mm

Factored shear force = 2000 kN

Yield Strength f_y = 250 Mpa

Procedure:

Minimum Web thickness requirement:

d/t_w = (1500 – 2 x 30) / 12

= 1440 / 12 = 120 > 67ε [ε = √(250 / fy) = 1]

Hence it is said to be Plate Girder.

d/t_w = 120 < 200ε

Shear Capacity check:

As per IS 800 : 2007 Clause 8.4.1

V_d = (A_v x f_yw) / (√3 x γ_mo)

= 1440 x 12 x 250 / (√3 x 1.1)

= 2267478.47 N

= 2267.47 kN

0.6 V_d = 0.6 x 2267.47

= 1360.48 kN < 2000 kN

Section is not Safe in Shear.

Shear Buckling check:

As per IS 800 : 2007 Clause 8.4.2.2

V_cr = A_v x 𝜏_b

Shear stress corresponding to web buckling

𝜏_b = f_yw / (√3 x λ²_w)

λ_w = √(f_yw / (√3 x t_cr) )

t_cr = K_v x ᴨ² x E/[12 (1 - µ²) (d/t_w)²]

K_v = 5.35 when transverse stiffners are provided only at support

E = 2 x 10⁵ N/mm²

µ (Poison’s Ratio) = 0.3

t_cr = 5.35 x ᴨ² x 2 x 10⁵ /[12 (1 – 0.3²) (1440/12)²]

= 67.09 N/mm²

λ_w = √(250 / (√3 x 67.09) )

= 1.466

When λ_w ≥ 1.2

𝜏_b = 250 / (√3 x 1.466²)

= 67.16

V_cr = 1440 x 12 x 67.16

= 1160524.8 N

= 1160.52 kN < 2000 kN

Check for Stiffener requirement:

If d/t_w > 85ε then Vertical stiffener is required

If d/t_w > 200ε then Longitudinal stiffener is required

If d/t_w > 250ε then additional Vertical stiffener is required

λ = d/t_w = (1500 – 2 x 30) / 12 = 120 > 85ε

Vertical stiffener only required.

Check for bearing capacity:

As per IS 800:2007 Clause 8.7.4

F_w = (b₁ + n₂)t_w x f_yw / γ_mo

b₁ = 2 (2.5 x t_f)

= 2 (2.5 x 30)

= 150 mm

F_w = (150 + 0.0) x 12 x 250 / 1.1 = 409090.9 N = 409.09 kN < 2000 kN

Web is not capable to carry the factored shear force. So stiffener to be provided.

Stiffener Design:

Assume Stiffener thickness (t_q) as 12 mm.

As per IS 800: 2007 Clause 8.7

Outstanding Length L₀ = 14t_q ε = 14 x 12 x 1 = 168 mm.

Outstand from the face of the

web on either side of stiffener = 20t_q ε

= 2 x 20 x 12 x 1

= 480 mm

Core Area = 2 x 168 x 12 + 480 x 12 = 9792 mm²

I_{c or e} = 480 x 12³/12 + 2 (12 x 168³/12 + 168 x 12 x 90²)

= 42.211 x 10⁶ mm⁴

r_min = √(I/A) = √(42.211 x 10⁶ / 9792) = 65.66 mm

Slenderness Ratio kL/r = 0.7 x 1440 / 65.66 = 15.35

From IS 800 : 2007 Table 9c

f_cd = 225.395 N/mm²

F_c = A_{c or e} x f_cd

= 9792 x 225.395

= 2207067.84 N = 2207.067 kN > V_{u max} = 2000 kN

Hence Safe.

Spacing of Intermediate stiffener (c):

3d ≥ c ≤ d

Let c / d = √2 = 1.4

c = 1.4 x d = 1.4 x 1440 = 2016 Say 2000 mm c/c

3 x 1440 ≥ 2000 ≤ 1440

Hence OK.

c / d = 2000 / 1440 = 1.39 < √2

I_s = 1.5 d³ x t³_w / c²

= 1.5 x 1440³ x 12³ / 2000²

= 19.35 x 10⁵ mm⁴

I_provided = 12 x 1440³ / 12 = 29.85 x 10⁸ mm⁴

Result:

• Thus it is concluded that intermediate stiffeners are required for the given Plate Girder to resist the shear force of 2000kN. Hence 12mm thick stiffener provided at 2000mm c/c.

Question No: 3. Check whether intermediate stiffener required for given load

For the above problem, determine the web thickness required to avoid the stiffeners.

Aim:

To be check whether intermediate stiffener required for the given load and determine the web thickness required to avoid the stiffeners.

Given data:

Plate Girder size = 1500 x 500 x 30 x 12

Factored shear force = 2000 kN

Yield Strength f_y = 250 Nmm²

Procedure:

Minimum Web thickness requirement:

d/t_w = (1500 – 2 x 30) / 12

= 1440 / 12

= 120 > 67ε [ε = √(250 / fy) = 1]

Hence it is said to be Plate Girder.

d/t_w = 120 < 200ε

Shear Capacity check:

As per IS 800 : 2007 Clause 8.4.1

V_d = (A_v x f_yw) / (√3 x γ_mo)

= 1440 x 12 x 250 / (√3 x 1.1)

= 2267478.47 N

= 2267.47 kN

0.6 V_d = 0.6 x 2267.47 = 1360.48 kN < 2000 kN

Section is not Safe in Shear.

Shear Buckling check:

As per IS 800 : 2007 Clause 8.4.2.2

V_cr = A_v x 𝜏_b

Shear stress corresponding to web buckling

𝜏_b = f_yw / (√3 x λ²_w)

λ_w = √(f_yw / (√3 x t_cr) )

t_cr = K_v x ᴨ² x E/[12 (1 - µ²) (d/t_w)²]

K_v = 5.35 when transverse stiffeners are provided only at support

E = 2 x 10⁵ N/mm²

µ (Poison’s Ratio) = 0.3

t_cr = 5.35 x ᴨ² x 2 x 10⁵ /[12 (1 – 0.3²) (1440/12)²]

= 67.09

λ_w = √(250 / (√3 x 67.09) )

= 1.466

When λ_w ≥ 1.2

𝜏_b = 250 / (√3 x 1.466²)

= 67.16

V_cr = 1440 x 12 x 67.16

= 1160524.8 N = 1160.52 kN < 2000 kN

Check for Bearing capacity:

As per IS 800 : 2007 Clause 8.7.4

F_w = (b₁ + n₂)t_w x f_yw / γ_mo

b₁ = 2 (2.5 x t_f) = 2 (2.5 x 30) = 150 mm

F_w = (150 + 0.0) x 12 x 250 / 1.1 = 409090.9 N = 409.09 kN < 2000 kN

Web is not capable to carry the factored shear force. So stiffener to be provided.

Thickness of Web to be increased to resist the Factored shear force without stiffener.

Let’s assume the Thickness of web is 16 mm.

Check for Shear Buckling:

As per IS 800 : 2007 Clause 8.4.2.2

V_cr = A_v x 𝜏_b

Shear stress corresponding to web buckling

𝜏_b = f_yw / (√3 x λ²_w)

λ_w = √(f_yw / (√3 x t_cr) )

t_cr = K_v x ᴨ² x E/[12 (1 - µ²) (d/t_w)²]

K_v = 5.35 when transverse stiffeners are provided only at support

E = 2 x 10⁵ N/mm²

µ (Poison’s Ratio) = 0.3

t_cr = 5.35 x ᴨ² x 2 x 10⁵ /[12 (1 – 0.3²) (1440/16)²]

= 119.40 N/mm²

λ_w = √(250 / (√3 x 119.40) )

= 1.10

When 0.8 < λ_w < 1.2

𝜏_b = [1-0.8(λ_w – 0.8)] (f_yw / √3)

= [1 - 0.8(1.1 – 0.8)] (250 / √3)

= 109.696

V_cr = 1440 x 16 x 109.696

= 2527395.84 N = 2527.395 kN > V_max = 2000 kN

Now the Web section is safe in factored shear force.

Result: -

• It is concluded that thickness of the web of Plate Girder is to be increased to 16 mm for resisting the Factored shear load of 2000kN without stiffener.

Question No: 4. Proportioning problem

Calculate the minimum depth & web thickness requirement of a beam of span 15 m to carry a UDL of 100 kN/m. Also calculate the flange depth & thickness considering section to be plastic and check if the assumed size is adequate to carry the moment.

Aim:

To be calculate the minimum depth & web thickness requirement of a beam and also to be calculate the flange depth & thickness considering section to be plastic and check if the assumed size is adequate to carry the moment.

Given data:

Span = 15 m

UDL = 100 kN/m

Section is Plastic

Assume steel grade as Fe 415

γ_mo = 1.1

Procedure:

Load Calculation:

Factored Load (UDL) = 1.5 x 100 = 150 kN/m

Self weight of Plate Girder (w) = w x 15 / 200 = 150 x 15 / 200 = 11.25 kN/m

Total Factored Load = 150 + 11.25 = 161.25 kN/m

Calculation of Bending Moment and Shear Force:

Maximum Bending Moment (M_{u max}) = wL² / 8

= 161.25 x 15 x 15 / 8

M_{u max} = 4535.16 kN m

Maximum Shear Force V_{u max} = wL / 2

= 161.25 x 15 / 2

V_{u max} = 1209.375 kN

Design of Web:

If Stiffeners to be avoided,

K = d / t_w ≤ 67ε [ε = √(250 / fy) = 1]

Optimum depth of web plate d_w = (Mk / f_y)^0.33

= (4535.16 x 10⁶ x 67 / 250)^0.33

= 1067.19mm Say 1100 mm

Thickness of web t_w = [M / (f_y x k²)] ^0.33

= [4535.16 x 10⁶ / (250 x 67²)] ^0.33

= 15.93 mm Say 18 mm

Hence Web size is 1100 mm x 18 mm

Design of Flange:

Area of Flange A_f = (M x γ_mo) / (f_y x d_w)

= (4535.16 x 10⁶ x 1.1) / (250 x 1100)

A_f = 18140.64 mm²

To Keep the Flange in Plastic category

From IS 800 : 2007 Table 2

b / t_f = 8.4ε

b = 8.4 x t_f

A_f = b_f x t_f

= 8.4 x t_f x t_f

18140.64 = 8.4 x t_f x t_f

t_f² = 18140.64 / 8.4

= 2159.6

t_f = 46.47 mm Say 50 mm

Width of Flange = A_f / t_f = 18140.64 / 50 = 362.81 mm Say 400 mm

Hence Flange size is 400 mm x 50 mm

Section Classification:

b / t_f = 400 / 50 = 8 < 8.4

Hence Flange is Plastic.

Shear Capacity check:

Assuming V_{u max} < 0.6 V_d

As per IS 800 : 2007 Clause 8.4.1

V_d = (A_v x f_yw) / (√3 x γ_mo)

= 1100 x 18 x 250 / (√3 x 1.1)

= 2598076.211 N

= 2598.076 kN

0.6 V_d = 0.6 x 2598.076 = 1558.84 kN > 1209.375 kN

Hence Section is Safe in Shear.

Check for Moment capacity:

I_zz­ = 2[1/12 x b_f x t_f³ + b_f x t_f x (D/2)²

D = d_w + 2 x t_f = 1100 + 2 x 50 = 1200 mm

= 2[1/12 x 400 x 50³ + 400 x 50 x (1200/2)²

= 14408.33 x 10⁶ mm⁴

y_max = 600 mm

Z_e = I_zz / y_max = 14408.33 x 10⁶ / 600 = 24.014 x 10⁶ mm³

M_d = (Z_e x f_y)/ γ_mo

= 24.014 x 10⁶ x 250 / 1.1

M_d = 5458.77 kN m > M_{u max} = 4535.16 kN m

Hence the section is safe in Bending moment.

Check for Bearing capacity:

As per IS 800 : 2007 Clause 8.7.4

F_w = (b₁ + n₂)t_w x f_yw / γ_mo

Assuming the minimum stiff bearing provided support (b₁) = 200 mm

n₂ = 2.5 t_f = 2.5 x 50 = 125 mm

F_w = (200 + 125) x 18 x 250 / 1.1

= 1329545.45 N

= 1329.545 kN > 1209.375 kN

Result:

• Hence it is OK. And end stiffener is also not required.

Conclusion:

• Here the given all sums are calculated and checked.