Week 5 Challenge

Solution.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 (`f_y`) = 250 MPa
• Factored Shear Force = 1000 kN.

Procedure:

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

Solution.2)

Aim: To check whetherintermediate 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 (`f_y`) = 250MPa

Procedure:

Result: Since, the plate girder is not adequate to carry a load of 2000 kN, a intermediate stiffener is needed with center to center spacing of

Solution.3)

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

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

Procedure:

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

Solution.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 cary the maximum moment developed.

Given:

• Span of Beam = 15 m
• UDL = 100 kN/m
• Flange is classified as Plastic Section.

Procedure:

Result: We conclude that, 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.