Week 5 Challenge

Aim: To understand the analysis and design of plate girder. The behavior of its components such as web, flanges and stiffeners under different loading conditions.

Introduction: A plate girder is a built up I-beam section, used to carry heavy loads which cannot be carried economically by rolled I-sections. It is made by riveting or welding the steel plates in I-beam shape. Additionally horizontal and vertical stiffeners are added to increase the load carrying capacity.

Design of Plate Girder

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.

Procedure:

Web thickness tw = 10mm

d = 1200-2x25 = 1150mm

d/tw = 1150/10 = 115

for no stiffeners, c/d = α Table

Clause 8.4.2.2

Vn = Vcr

Vn = Av x

Av = d*tw =1150x10 = 1150mm²

dw = SQRT(fyw/sqrt(3).) = SQRT(250/1.732.)

c/d = α

kv = 5.35

k_v t_w/d)²

= ((5.35 xx 2 x 10^5)/12(1-0.3^2)) x (10/1150)^2 = 73

= √(fyw/√3.)

= √(250/1.732x73) = 1.406

Here >= 1.2

Therefore, = fyw / √3. = 250/1.732 x 1.406^2 = 73 MPa

Vcr = x Av = 73 x 11500 = 839.5kN < 1000kN

Results:

1. Factored Shear force is greater than shear capacity
2. Therefore Web is inadequate.

• Higher Web thickness is required.

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.

Procedure:

Factored shear force = 2000 kN

fy = 250 Mpa

d = 1500 -2x30 = 1440mm

d/t = 1440/12 = 120

Assume no stiffener required

Therefore c/d = α

Kv = 5.35

k_v t_w/d)²

= 5.35 x x 2x10^5 / 12(1-0.3^2) x (12/1440)^2 = 67.16

= √(fyw/√3.) = √(250/√3x67.16) = 1.47

Here >= 1.2

Therefore, = fyw / √3.

= 250/3 x 1.47² = 66.8 MPa

Vn = Vcr = x Av = 66.8 x (1140 x12) = 913.85 kN

Vcr = 913.85 < 2000kN

Therefore,

Stiffeners is required

Now, Reverse calculation

= 2000x10^3 / 1440 x 2 = 115.74 N/mm^2

Case 1:

115.74 = 250 / sqrt(3) x dw

dw= 1.247

= ?

1.247 = sqrt(250/1.732x)

= 92.82

92.82 = kv x Pi^2 x 2 x 10^5 / 157248

kv = 7.394

kv = 5.35 + 4/ (c/d)^2

7.394 = 5.35 + 4/(c/d)^2

c/d = 1.399

c = 1.399 x 1440 = 2.014m

Provide 12 thick stiffener spaced @ 2014m

If thickness of stiffener is reduced spacing is d/tw is calculated accordingly in calculation

Case 2:

Assume 0.8 < <1.2

1.574 = (1-0.8(dw-0.8))*(250/√3)

dw= 1.047

= ?

1.047 = sqrt(250/√3 x )

= 131.67

131.67 = kv x Pi^2 x 2x10^5 / 12(1-0.3^2)(1440/12)^2

kv = 10.487

10.487 = 5.35 + 4/(c/d)^2

(c/d) = 0.88

c = 0.88 x 1440 = 1270mm = 1.27m

Results:

Out of both cases 1.27m is critical

Therefore stiffener is spaced at 1.27m

3. Check whether intermediate stiffener required for given load

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

Procedure:

To avoid stiffener

Web thickness enhanced

Trial and error procedure

Assume t =16 mm

d/t = 1440/16 = 90

c/d = α

kv = 5.35

= 5.35x7^2 x 2x10^5 / 12(1 -0.3^2)*(1440/16)^2

= 119.4

= sqrt(250/√3 x119.4)

= 1.099

=(1- 0.8 (1.099 – 0.8) (250/√3) = 109.82

Vcr = Av = 109.82 x 1440 x 16 x 10^-3 =2530 kN > 2000 kN

Results:

The web thickness required to avoid the stiffeners is 16mm

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.

Procedure:

If stiffeners are to be avoided take the ratio

k = d/tw < 67ɛ

Optimum depth of plate girder

d = (Mk/f_yf)^0.33

Max BM = wl²/8 = 1.5x100x15^2 / 8 = 4218.75 kN.m

d= (4218.75 x 10^6 x 67 / 250)^0.33 =971.8mm

Assume d = 1000mm

Optimum thickness of web

t_w = (M/fyk²)^0.33 = (4218.75x10^6 / 250 x67^2)^0.33 = 15.12mm

take tw = 16mm

Two condition of design requirement for no intermediate stiffener

1. d/tw < 200ɛ (Serviceability)
2. d/tw < = 345 ɛ² (To avoid flange buckling)

if d/t < 67 ɛ, these 2 will be met deliberately

Minimum Web thickness

tw>d/67 ɛ

tw > 1000 / 67 (tw > 1492)

Provide tw = 16mm

For plastic section

b/tf < 84 ɛ

bf = 0.3 x dw = 0.3 x 1000 = 300mm

tf> bf/2 x 8.4

tf > 300/2 x 84

tf > 17.86mm

take tf = 20mm

Check Adecuacy of the moment

From IS800-2007:Clause 8.4

Vd = Av fyw / √3 x Ymo

Vd = 16000 x 250 x 10^-3 / √3 x 1.1 = 2099.5 kN

V = wl/4 = 1.5x100x15 / 4 = 5625 kN

Vd > Vn

Safe in shear

Moment capacity clause 8.2.1.2

Md = Zp fy /

Zp = tw.d²/4 + 2((bf x tf)x((D-tf)/2)) = (16x1000²/4) + 2(300x20x(1400-20)/2) = 12.28x10⁶

Md = (250 x 12.28 x10⁶)/1.1 = 2.79x10⁹ N.mm = 2709.9kN.m < 4218.75kN

Hence safe to carry applied moment.

Results:

The minimum depth & web thickness requirement of a beam are 1000mm and 16mm are respectively.