Week 3 Challenge
AIM AND INTRODUCTION : DESIGN OF COMPRESSION MEMBER
PROCEDURE : TO FIND THE EFFECTIVE LENGTH OF COLUMN
CASE A the frame is braced bottom is pinned and top is also pinned
bottom top
translation rotation translation rotation
restrained free restrained free hence effective lenght factor =1.0L
CASE B unbraced frame with pinned base
free restrained free restrained hence effective lenght factor =2.0L
CASE C unbraced frame with FIXED base
restrained restrained free restrained hence effective lenght factor =1.2L
AIM AND INTRODUCTION : DESIGN OF AXIAL STRENGTH DETERMINATION OF COLUMN with end condition.
procedure:
DATA GIVEN : ISMB 300
length of beam = 3m
yield strength of material fy= 250 mpa
ISMB 300 A=58.60cm2 Ixx =8990cm4 Iyy=486cm4 bf=140mm D=300 tf=13.1mm tw=7.7
rxx=12.37cm ryy=2.84cm
Cover plates - 200x12(TOP & BOTTOM)
Determination of radii of gyration for the compound section
AREA OF COMPOUND SECTION = 5860mm2+2x(200x12)mm2
=(5860+4800)mm2=10660mm2
Izz for plates = 2x(200x12^3/12 +200x12x156^2)=11687x10^4
total Izz of beam and plates= (11687x10^4+8990x10^4)=20677x10^4mm4
Hence Rzz =`sqrt(I/A)`=(20677x10^4/10660)`=139.24`
Iyy of ISMB 300 = 486x 10^4
Iyy of two plates= 2x (12x200^2/12)=1600x10^4
total Iyy of beam and plates=(486+1600)x10^4 =2086x10^4mm2
ryy= `sqrt(2084 x 10^4/10660) = 44.24mm
Buckling curve classification from table 10 using curve b tw<40mm
Since ryy is small the buckling will be about the YY axis
effective length=1.2L(support is fixed and other end is roller)
Slenderness ratio `lamda`= (1.2x 3 x 10^3)/44.24`=81.4mm<180
From table 9c of the code for fy=250 lamda=81.4
fcd = 134.2
Hence design strength =134.2x 10660/1000= 1430.5 KN
RESULT: The axial strength of he column is 1430.5KN
AIM AND INTRODUCTION : DESIGN OF CIRCULAR TUBE STRUT MEMBER FOR CERTAIN LOAD
procedure: Data given Length of strut = 2 m
Factored axial load = 400 KN
Yield strength of strut fy= 210n/mm2
Assume the ends of member simply supported
Area required= 400x10^3/210=1904.76mm2
for the given area a square hollow section of 100x100x6 Area of square section =2163mm2
a circular hollow section of 125 outer bore dia=139.7 thk=4.5 Area of square section =1910mm2
From table 10 both sections class a
Effective length =1.0L =2.0m(both sections)
For square section Slenderness ratio lamda=2x10^3/r min=2x10^3/37.9=52.77<180
For circular section Slenderness ratio lamda=2x10^3/r min=2x10^3/47.8=41.84<180
From table 9c of the code for fy=210
For square section fcd=174.1N/mm2
HENCE the design strength= 174.1x21.63=376.58KN
For circular section fcd=180 N/mm2
HENCE the design strength= 180x 19.10=343.8KN
RESULT:
For square section fcd=174.1N/mm2
HENCE the design strength= 174.1x21.63=376.58KN
For circular section fcd=180 N/mm2
HENCE the design strength= 180x 19.10=343.8KN
AIM AND INTRODUCTION : DESIGN OF LACED COLUMN
PROCEDURE: DATA GIVEN Height = 10M
AXIAL LOAD= 1000 KN
DESIGN OF COLUMN P =1000x10^3
L =10m
Assume design stress of 125 N/mm2(0.5 fy)
Required area =1000x10^3/125=8000mm2
SELECT two ISMC 300 channel back to back prop A=91.28cm2 rxx=118.1mm ryy=38.7mm
tf=13.6 mm cyy=23.6 mm tw=7.6 mm Ixx=12725.2 cm4
when the radius of gyration of yy axis is increased in such a way that it is equel to radius of gyration
about zz axis .this can be achieved by spcing the section in such away that rzz becomes rmin.
so first checking the safety of the section and then workout the required spacing bw the two channels.
L/rzz=10x10^3/118.1=84.67
L/r of the built up section should be taken as 1.05x84.67=88.90
fcd taken from code for given rzz=120N/mm2
= 9128 x 120=1095 kn> 1000kn. HENCE THE SECTION IS SAFE.
SPACING OF CHANNELES
12725.2x104 =2 (IYY + A(S/2 +cyy)2)
=2 x(310.8 + 4564(s/2 +23.6)2
solving the above equation S =183.1mm
LACING SYSTEM of the builtup section
using single lacing system with the inclination of lacing bar gauge length g=50 mm for 90 mm flange
Spacing of lacing bars Lo= 2x(183+50+50) cot 45
=2x 283 x 1=566 mm
Lo/ryy should be <0.7xL/r of whole column
566/26=21.76 < 0.7 x 88.98=62.3 HENCE SAFE
Maximum shear =(2.5/100)x 1000x10^3=25000 KN
TRANSVERSE shear in each panel =V/N=25000/2=12500 N
COMPRESSIVE force in the lacing bar =(V/N) cosec45
= 12500 x1.414
=17675 N
AIM AND INTRODUCTION : DESIGN OF BATTEN COLUMN
PROCEDURE: DATA GIVEN Height = 5 M
AXIAL LOAD= 2000 KN
DESIGN OF COLUMN P =2000x10^3
L =5m
Assume design stress of 150 N/mm2(0.6 fy)
Required area =2000x10^3/150=13333mm2
SELECT two ISMC 350 channel back to back prop A=53.66 x2=10732cm2 rxx=136.6mm ryy=28.3mm
bf=100 tf=13.5 mm cyy=24.40 mm tw=8.1 mm Ixx=12725.2 cm4
when the radius of gyration of yy axis is increased in such a way that it is equel to radius of gyration
about zz axis .this can be achieved by spcing the section in such away that rzz becomes rmin.
so first checking the safety of the section and then workout the required spacing bw the two channels.
L/rzz= 5x10^3/136.6 =36.60 k=1 fy=250n/mm2 using Table 9c of code.
fcd taken from code for given rzz=36.60 Fcd=202.4/mm2
capactity of built up column = 10732 x 202.4 =2172 KN >2000 KN HENCE SAFE.
SPACING OF CHANNELES
2 x 20016 x104 =2 (IYY + A(S/2 +cyy)2)
=2 x(430.6 x10^4 + 5366(s/2 +24.4)2
solving the above equation S =333.4 mm
spacing of battens Lo/ryy should be less than 0.7x L/ryy
Lo /ryy < 0.7 x36.6
Lo < 0.7 x36.6x28.3 =725 mm
ALSO Lo/ryy should be less than 50
ALSO Lo/ryy < 50 Lo < 50x ryy =50x28.3 =1415 mm
hence provide batten at spacing 725 mm
SIZE OF END BATTENS
OVERALL DEPTH OF BATTEN = 334 + 2X CYY
=334 +2 X24.4= 383MM
PROVIDE 60 MM OVERLAP OF BATTEN ON CHANNEL FLANGE FOR WELDING
LENGTH OF BATTEN=334 X 2 X 60=454MM=460MM
THICKNESS OF BATTENS =1/50X 460=9.2 MM
PROVIDE BATTEN SIZE =460 X 390 X 9.2 MM
SIZE OF INTERMEDIATE BATTENS
OVERALL DEPTH = 3/4 X 390 = 293 MM
PROVIDE 460 X300X 9.2 MM.