Week 5 Challenge

1)

Aim: To check the adequacy of Plate Girder so that it can carry a factored shear force of 1000 kN.

Given:

Size of Plate Girder 1200 x 300 x 25 x 10

Yield Strength of Plate (fy) = 250 MPa

Factored Shear Force = 1000 kN.

Procedure:

Result: Since, the Web Shear Force carrying capacity corresponding to Web Buckling is less than Maximum Factored Shear Force, So the given Plate Girder is not safe ( or adequate) to carry a load of 1000 kN.

2)

Aim: To check whether intermediate stiffener is required or not to carry a factored load of 2000 kN.

Given:

Size of Plate Girder is 1500 x 500 x 30 x 12

Factored Shear Force = 2000 kN

Yield Strength of Plate (fy) = 250MPa

Procedure:

Result: The plate girder is not adequate to carry a load of 2000 kN, a intermediate stiffener is needed with center-to-center spacing of 2000 mm.

3)

Aim: To determine the thickness of the web of Plate Girder to avoid the use of Stiffener for carrying a factored shear force of 2000 kN.

Given: Size of Plate Girder is 1500 x 500 x 30 x 12

Procedure:

Result: The use of intermediate stiffeners can be avoided by increasing the web thickness of the Plate Girder from 12 mm to 16mm. Hence, the final Web Thickness will be 16mm.

4)

Aim: To Design the Plate Girder, determine the dimension of Web and Flanges without using any stiffeners and check whether the Plate Girder is safe to carry the maximum moment developed.

Given:

The span of Beam = 15 m

UDL = 100 kN/m

The flange is classified as Plastic Section.

Procedure:

Result: Web of 1500 x 16 mm and Flange of 500 x 30 mm should be used to built up Plate Girder to carry the load of 100 kN/m and the Beam is safe to resist the maximum moment developed.