Week 3 Challenge

. i,

Effective length factor determination problem

• 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 column is Free.
• Calculate effective length factor of columns in following cases

The height of the column is 10m and is restrained in mid height.  Since they are restrained in both y-y and z-z direction we take the actul length as 5m.

 a. braced frame and shear wall = Here top is free and bottom is fixed ; Eff Length = 2L = 2x5=10m

 b. Unbraced frame with pinned base =  Eff length = 2L =  2x5 =10m

 c. Unbraced frame with a  fixed base =  Eff. length = 1.2L =1.2 x 5=6m

ii. 

Design Axial strength determination of a column with its end conditions

Calculate the Axial compressive capacity of Column made up of Hot rolled ISMB 300 with flange cover plate of 200x12 mm on each flange having length of beam as 3 m. Assume the bottom of the column is Fixed and Top as Roller (Rotation Fixed, translation free).  Yield strength of material fy= 250 N/mm²

Tip Steps:

• Determine new radius of gyration of compound section
• Using Buckling curve classification & slenderness ratio using table 9c in code, calculate compressive strength
• Using compressive strength, calculate design strength multiplying the cross sectional area.

ans: 

Fy =250N/mm2

Fu = 410N/mm2

effective length = 1.2 x L =1.2x 3= 3.6m

height of ISMB 300 = 300mm , flange cover= 200 , Tf cover=12mm, tf =12.4mm . breadth of flange bf= 140mm

h/Bf =300/200 = 1.5 -> which should be greater than or equal to 1.2 acc. to IS800 tabl 10.

Hence the class is c, because of built up member.

rmin =ryy =25.4 from SP6-1 steel

From tble 9c , under fy=250

KL/r = 3600/25.4 = 141.7

for 140 => 66.2 & 150 => 59.2

by extrapolating we get 141.7 => 65.01 as design compressive strength.

To calculate the design strngth = 

imperfection factor- alpha = 0.49

Pd =Ac Fcd

fcc =122.84N/mm

lamba = root of (250/122.84) = 1.426

alpha =0.49

 `O/`= 3.38

Fcd = 92.19N/mm2

Pd = 5186.6KN

iii, 

Design circular tube strut member for certain load

Design 2 m long tubular strut member to carry factored axial load of 400 kN. Assume the ends of the member are simply supported and the yield strength fy=210 Mpa. Also find Equivalent Square hollow sections.

Tip Steps:`lambda =`

• Calculate the area required from design compressive stress and Factored load on strut
• Choose a section satisfying the area calculated.
• Check limiting width to thickness ratio
• Find buckling class & slenderness ratio
• Using Buckling curve classification & slenderness ratio using table 9c in code, calculate compressive strength
• Using compressive strength, calculate design strength multiplying the cross sectional area.

Ans: 

L= 2m ,P= 400kN, fy = 210Mpa, K=1, KL =2000mm, let fcd = 80N/mm2

Area = 400x 10^3 / 80 = 5000mm2

`(pi*d^2)/4 = 5000`

d= 80mm

For circular hollow section (semi compact)

D/t <=88 `E^2`

e=`root 2 (250 / (fy) )`= 1.09

let t =8 mm

80/8 = 88 x1.09^2

`lambda =` (KL)/r

r = 80/2 = 40mm

`lambda`= 2000/40 = 50

Circular sections ae under bukling class a, table 9a

Kl/r = 50 at   fcd @fy =210

Pd = Ac x fcd 

= 5000x 1.76

=880000 KN

iv, 

Design of laced column

Design a single laced column made up of two channels back to back of 10 m height to carry factored axial load of 1000kN. The lacings are bolted. Design of bolted connection not required.

Tip Steps:

• Calculate required Cross section area
• Assume a channel section and calculate the compressive capacity
• Find the appropriate spacing of channel
• Find spacing of lacings
• Calculate design forces –Transverse, compressive force in lacing bars
• Check if the assumed lacings are safe for calculated design forces.
• Check if effective slenderness ratio is within limit

ans :

 Assuming fcd = 135N/mm2

Area reuired = 1000x1000/ 135 = 7407.40mm2

Assuming 2 ISMC  250 @ 298.2N/mm2

Area provided = 2 x 3867 = 7734 mm2

rzz=99.4m

Distance will be maintained so as to get ryy<rzz

Actual slenderness ratio = KL/r = 1x10000/99 = 100.60

For laced column effective slenderness ratio = 1.05 x actual slenderness rtio = 105.63

From table 9c of "IS 800 = buckling class c, table 10

fcd = 107 -(105.63 -100)/(110-100) x (107.94.6)

fcd = 100.018 N/mm2

Load carrying capacity = 7734 x 100.015

7735.39KN > 100KN hence ok.

Spacing between channels 

Let it be a clear distance "d"

Now Ixx= 2 3816.8 x10^4= 7633.6 x 10^4mm4

Iyy =2(219.1x10^4 + 3867(d/2 +23)^2)

d= 86.18mm

Provide 90mm spacing

Horizontal spacing = 90+23+23 =136mm\

Vertical spacing = 136sin45 =192.304mm

ryy =23.8

kl/r of channel between lacing = 192.3/23.8=8.08<50;

Traverse shear to be resisted by lacing system = 2.5/100 * 1000 *10^3 =12500N

Shear to be resisted by each lacing system = 25000/2 = 12500N

Length of lacing: 90+23+23 * 1/cos45 = 192.30

Min thickness of lacing = 1/40 * 192.30 = 480mm

Min width of lacing if 20mm bolts are used = 3*20 =60mm

v,

5. Design of battened column

Design a battened column made up of two channels back to back of 5 m height to carry factored axial load of 2000kN. The Battens are welded. Design of weld not required. Use Hot rolled channel sections.

Tip Steps:

• Calculate required Cross section area
• Assume a channel section and calculate the compressive capacity
• Find the appropriate spacing of channel
• Find spacing of battens & size of end/intermediate battens
• Calculate design forces –Transverse, longitudinal shear, Moment
• Check if the assumed battens are safe for calculated design forces.

  Ans:

P = 2000KN,  fcd = 135 assumed,

L= 5m

Effective length = 1.1x 5= 5.5m

C/s area required = (2000x10^3)/135 = 14814.8mm2

Provide 2ISMC 350 @ 413N/mm2

A =5365mm2 , rxx =136.6mm , ryy =28.3mm

Ixx =10008x10^4mm4, Iyy = 430.6x10^4mm4

Area provided =2*5365

=10730mm2

l/r =x10^3 /136.6 =36.60

for l/r =36.6 & fy = 250 Mpa

Since it is a battened column KL/r = 1.1x5x10^3 /136.6 =40.26

From table 9c IS 800,

fcd = 211- (40.26-30)/(40-30)x (211-198)

=197.60N/mm2

Load carrying capacity =10.730 x 197.66 =2120.89 KN >2000KN , hence ok.

Spacing of battens,

C/28.3 <50, C<1415

It should also satisfy the condition 

c/28.3< 0.7*40.26 ie C<797.55

Let us select C= 1200mm

vt =2.5/100x2000x1000 =50000N

Vb =vtc/ns =50000*120 / 2*250 =12000N

M=Vtc/2N = 50000*1200/ 2*2 =150000n-m

Thickness of battens t>Lb/50

Overall depth D = de+2xend distance

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