Week 3 Challenge
Aim:
To calculate effective length factors, compression member, laced column & battended column for the given problem statement.
1. Determination of effective length of column factor
(a) Data:
Column height = 10 meter
Restraints: Restrained perpendicular to frame (Z-direction) + Free in Horizontal direction
End conditions = Fixed bottom + Free top
As per Table-11 from IS-800, Type-1 will apply for this frame. K=2.0
(b)
Bottom condition = Fixed
Top condition = Translation restrained + Rotation restrained
This will fall under Type-6 as per Table-11 of IS-800 (Both end fixed), K=0.65
(c)
Bottom condition = Pinned
Top condition = Restrained against Rotation + Free in translation
This will fall under Type-2 as per Table-11 of IS-800, K=2.0
(d)
Bottom condition = Fixed
Top condition = Restrained against Rotation + Free in translation
This will fall under Type-4 as per Table-11 of IS-800, K=1.2
2. Design axial strength determination
Data:
ISMB300 with flange cover plate 200x12
Section height 3m
End condition = Bottom fixed + Top roller
Yield strength of section Fy=250 MPa
Area of ISMB300 = 5860 mm2
Area of plate = 2x2400 = 4800 mm2
Total area = 10660 mm2
Inertia of ISMB300 = 8.9x10^7 mm4
Inertia of Plate = 2x(200x(12^3)/12) + (2400x126^2) = 7.6x10^7 mm4
Total inertia (I) = 16.5x10^7 mm4
Radius of gyration (r) = sqrt(I/A) = sqrt (16.5x10^7 / 10660)
r = 124.4 mm
effective length factor K = 1.2 (As per Table-11)
kL/r = 1.2x3000/124.4 = 28.9
From clause 7.1.2,
As per Table-10 of IS800, Any builtup section will come under Buckling class-C
Refering to Table-9c, The design compressive stress (Fcd) = 211 MPa
Pd = 211 x 10660 = 2249 KN
3. Design of circular tube strut
Data:
Length (l) = 2m
Factored axial load = 400 kN
End condition = Pinned both ends
Fy = 210 MPa
From Table-2 of IS800, The limiting width to thickness ratio for hollow tube sections under axial compression,
D/t = 88e^2 where e=sqrt(250/Fy)
D/t = 88x250/210 = 104
Selecting a IS-1161-1998 Structural Steel tube of thick 5.4mm and outer diameter 152.4mm (Page-3)
D/t=152.4/5.4 = 28.2 < 104 hence okay
kL/r = 1x2000/52 = 38.5
As per Table-10 of IS800, Any hotrolled tubular section will come under Buckling class-a
Refering to Table-9a, The design compressive stress (Fcd) = 181 MPa
Design compressive strength Pd = Ae x Fcd = 2500 x 181 = 452.5 KN > 400 KN hence okay
4. Design of laced column
Data:
Single lacing system
2 channel back to back
Height = 10m
Factored axial load Pd = 1000 KN
bolted lacing
Transverse shear (V)=2.5/100 x 1000 = 25 KN
Longitudinal force component F = V/2 sin45 = (25/2) sin45 = 8.8 KN
Assume yield strength of lacing material (Fy) = 250 MPa
All rectangular bars will come under buckling class-c
Assume ISMC300 to be involved in the column design. Width (B) of section=90 mm
Making the back to back section as square, the center to center distance between channels is = 300-45-45 = 210mm
The length of lacing = 210 / cos 45 = 296 ~ 300 mm
Effective length of lacing (Le) = 300 mm (For single lacing K=1.0)
Thickness of lacing = 300/40 = 7.5 ~ 8mm thick bar
Assume 20mm dia bolts and Width of lacing plate = 3 x 20 = 60 mm
sqrt(I/A) = sqrt(8x60^3/12 / 480) = 17.3 mm
kL/r = 300/17.3 = 17.3 which is less than maximum 145. Hence okay
Refering to Table-9c, The design compressive stress (Fcd) = 224 MPa
Pd = 8x60x224 = 107.5 KN > 8.8 KN Hence okay.
5. Design of battened column
Data:
Battened column made up of 2 channels back to back carrying 2000 KN axial load with 5 m height.
Welded battens
Refering to the channel sections from previous problem, ISMC300
Thickness of batten (t) = l/50 = 210/50 = 4.2 ~ 6 mm
Effective depth of batten (d) = 3/4th of centroid of whole column = 3/4 x 150 = 112.5 ~ 120mm
Taking the effective depth as Overall depth due to welded connection. D=120mm and t=6mm
Effective length (Le) = 0.7 x 210 = 147mm
Radius of gyration (r) = sqrt(I/A) = sqrt(864000/720) = 34.6 mm
kL/r = 147/34.6 = 4.25 which is less than limit 145
Refering to Table-9c, The design compressive stress (Fcd) = 227 MPa
Spacing of batten (C) = 50xr = 50x34.6 = 1730 mm
number of battens in column height = 5000/1730 = 2.9 ~ 3 numbers
Shear = VC/NS = 0.025x2000x1000x1730 / (3x210)
Acting shear (Vt) = 137.3 KN
Pd = bxtx Fcd = 120x6x227 = 163.4 KN > Vt Hence okay