Week 3 Challenge

1)

Aim: To determine the effective length factor for the given cases.

• Height of column is 10 m and it is effectively restrained in mid height in one direction (in Z direction) but free to move in other direction (Y direction). Bottom of column is Fixed and top of the column is Free.
• Calculate effective length factor of columns in following cases

Procedure:

a.

Given

Height of column = 10 m

At the mid-height, the column is effectively restrained in one direction, but free to move in other direction.

Also the bottom of column is fixed and top of column is free.

In path Ab of the column

One is fixed for translation and rotation while other end is fixed for translation.

Kx = 0.8 (Since one end is fixed and other end is hinged.)

Ky = 2.0 (Since one end is fixed and other end is free.)

b. Since the frame is braced.

There will be no translational motion. (Non-Sway Frame)

K = 0.8 (One end is fixed)

c. Since the frame is unbraced.

There will be translational motion. (Sway Frame)

K = 2.0 (One end is hinged)

d. Since the frame is unbraced.

There will be translational motion. (Sway Frame)

K = 1.2 (One end is fixed)

Result:

Effective length factor for the following cases has been found.:

a) Kx = 0.8 and Ky = 2.0

b) K = 0.8

c) K = 2.0

d) K = 1.2

2)

Aim: To determine the Design Axial Compressive Strength of Column with given conditions.

Given:

Section Used: ISMB 300 with flange cover plate of 200 x 12 mm on each flange length of Beam as 3 m.

Bottom of the column is Fixed and Top as Roller (Rotation Fixed, translation free).

Yield strength of material fy= 250 N/mm²

Procedure:

According to SP 6 (1),

Properties of ISMB 300 are as follow:

Sectional Area = 56.2 cm²

Radius of Gyration, rxx = 12.37 cm

ryy = 2.84 cm

Moment of Inertia, Ixx = 8603.6 cm⁴

Iyy = 453.9 cm⁴

For the compound Section,

Now, new Radius of Gyration of the compound section,

3)

Aim: To design a circular strut member for the given load conditions.

Procedure:

Result: A circular hollow section has been designed for the given load, with an Outer diameter of 100 mm and

the inner diameter of 60 mm.

4)

Aim: Design a single laced column made up of two channels back to back of 10 m height to carry factored axial load of 1000kN.

Procedure:

5)

Aim: To design a battened column made up of two channels back to back of 5 m height to carry factored axial load of 2000kN.

Procedure: