Week 3 Challenge
Objective of this challenge is to:
a) Understand the concept of designing the compression members.
b) Know to determine the axial strength of the column.
c) Know how to design the circular strut member for a given load.
d)Design the laced column and battened column, knowledge of IS 800-2007.
1. Effective length factor determination problem:
Solution:
Here the column is effectively restrained in the mid height in one direction but free to move in other direction.
Effective length = 2L=2x10mtr=20mtr/2=10mtr(since from the mid height)
Bottom of the column is fixed and top is free.
Effective length= 2L=2x10=20mtr.
Fig a : Braced Frame and shear Wall.
Hear the shear wall will acts as a fixed at bottom end as well as mid end.
Effective length =2L
Fig b: Unbraced frame with pinned base.
Effective Length=0.8L
Fig c:Unbraced frame with fixed base
Effective Length=1.2L
2. Design Axial strength determination of a column with its end conditions
Given Parameters:
Column=ISMB300
Flange cover plate=200x12mm
Length of beam=3m
Fy=250 N/sqmm
As per table 46 IS 800-2007
Radius of Gyration r=28.4mm
Effective Length KL=2L=2x3=6m
fy=250n/sqmm
KL/r=6000/28.4=211.4
As per table 9c fcd=23.3Mpa
Therefore, Ag=L/r2=6000/(28.4x28.4)=7.43sqmm
fcd=Pcr/Ag
Pcr=FcdxAg=23.3x7.43=173.12N-mmsq
3. Design circular tube strut member for certain load
Given Parameters:
Length=2mtr
Factored Axial Loads=400Kn
fy=210Mpa
Provide Solution
4. Design of laced column
Given Parameters:
Factored Axial load= 1000Kn
L=10 mtr
Effective length=2L=20 mtr
For Single Lacing t>Le/40
Vt=2.5% of axial Load=2.5/100x1000=25KN
Axial Load on single lacing=Vt/2Sin45=17.6Kn
Check for the compressive strength
Assume fy=250mpa and Channel Section 150
As per table 9c
radium of gyration =2.21
KL/r=20/2.21=9
Thus fcd=227mpa
Pd=(bxt)xfcd
=3x60x227=40860>1000KN Hence Ok
Thus thickness of lacing is 3mm and width of lacing is 60 mm to be used at an degree of 45.
5. Design of battened column
Provide solution