Week 3 Challenge

1. 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

ANS

A- HEIGHT OF COLUMN = 10 M

AS BOTTOM IS FIXED AND TOP IS FREE = K = 2

EFFECTIVE LEGTH = 2*L = 20 M

B -

a

b unbracd frame wth pinned base.

condition : one end pinned and other end fixed

effective length = 0.8 l

c unbraced frame with fixed base.

condition : one nd fixed and other end also fixed.

effective length : 0.65 l

2. 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/mm^2.

GIVEN DATA - ISMB 300

A= 56.26 CM2

H = 300 MM

b =140MM

tf= 2.4 mm

tw= 7.5 mm

rxx= 12.37 cm

ryy= 2.84 cm

IXX= 8603.6 CM4 MAJOR AXIS

Iyy =453.9 cm4 MINOR AXIS

k = 1.2 as per condition

effective length = 1.2*3000 =3600 mm

plate = 200x12 mm

L = 3m= 3000 mm

r = `sqrt(I/A)`

I= Iyy of SMB 300 + Iyy of plates*2

= 453.9X10^4 + 12X200^3/12 *2 = 4939X10^3 mm^4

A = 56.26X10^2 + 200*12 X 2 = 10426 mm^2

r = 58.85 mm

FOR ANY BUILT UP SECTION, BUCKLING CLASS IS C

FOR SLENDERNESS RATIO`lambda` = KL /rmin

= 3600/58.85 = 61.17

compressive stress by table 9c = Fcd = 70-60/70-61.17 =152-168/168-fcd

1.13= -16/168-fcd

fcd = 182.15 n/mm2

design strength = Ae*FCD

10426 x 182.15 = 1899 KN

3. 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 sectioN

GIVEN DATA =

LENGTH OF STRUT MEMBER = 2000 MM

FACTORED AXIAL LOAD = 400KN

END CONDITIO = SIMPLY SUPPORTED , FIXED

Fy = 250 N/MM2

EFFECTIVE LENGTH = 0.65 *2000 = 1300mm

ASSUEME FCD= 100 N/MM2

AREA REQUIRED = P/FCD = 400000/100 = 4000 MM2

REQUIRED SECTION = ISMB 500 , WHOSE AREA IS 110.74 MM^2

4. 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.

ASSUME Fcd = 140 N/MM2

AREA REQURED Ae(REQUIRED)= Pd / Fcd

= 1000000 / 140 = 7142.8mm2 (AREA FOR TWO SECTION)

A/2 = 7142.8/2= 3571.4 MM2 (AREA FOR ONE SECTIN) == 35.71CM^2

CHOOSE SECTION ISMC 300 ,

A= 45.64 CM2 , rxx= 11.81 cm =118.1 mm ( in the channel sectin rxx s the minior axis).

Ixx =6362.6 cm4 , Iyy = 310.8 cm4 , Cyy = 2.36 cm .g =50 mm

`lamda` = kl/rmin = 10000/118.1 = 84.67

as er clause 7.6.1.5

lamda effective = 84.67*1.05 = 88.90

fro table 9c find out the fcd ,

X1 = 80 ,X2 =90, X = 88.90

Y1 =136, Y2 =121 , Y =fcd

90-80/90-88.90 = 121-136/121- fcd

fcd = 122.66 n/mm2

Pd = Ae*fcd

= 9128 *122.66 = 1119 kn >P so ok

APPROPRIATE SPACING OF CHANNEL S = d + 2g (for back to back)

where d = distance between te inner side o the channel ,

Iyy= Ixx

2*( Iyy+A(d/2 +Cyy)^2) =2* Ixx

2*( 310.8x10^4 + 4564 (d/2 +23.6)^2 ) = 2*6362..6x10^4

d=103.55

S = 103.55+ 2*50 =203.55 MM

`theta` =45 `

spacing of lacing c = 2xs* tan45 = 2* 203.55 tan45= 407.1 mm

C = 407.1 mm

length of lacing bar = L = S/COS 45 = 287.86 MM

THICKNESS OF LACING BAR T= L/40 =7.19 = 8 MM

WIDTH OF LACING = 3*D WHERE D= DIA OF BOLT

ASSUME DIA OF BOLT = 16 MM

WIDTH OF LACING B= 48 MM = 50 MM

AREA OF LACING A= B XT

= 50*8

= 400 MM2

RADIS OF GYRTION r = `Squrt`I/A

= T/ SQRT 12 = 2.312 MM

SENDERNESS RATIO = 287.86/2.36 =121.97 < 145

FROM TABLE 9C

X1=120, X2= 130 X = 121.97

Y1 =83.7 ,Y2 =74.3 Y= FCD

FROM ITERPLOTATON METHOD

X2-X1/X2-X = Y2-Y1/Y2-Y

130-120/130-121.97 = 74.3-83.7 /74.3 -fCD

Fcd= 81.79 n/mm2

Pd = A X fcd (of lacing bar)

=400x 81.79 = 32.7 kn

Vt = 2.5 % of total load

25 kn < Pd so ok

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.

LENGTH =5M = 5000 MM

FACTORD LOAD = 2000KN

EFFECTIVE LENGTH = KL = 1*2000 =2000 MM

ASSUME Fcd = 130 N/MM2

AREA REQUIRED = P/ fcd = 2000/ 130 = 15384 mm2 for two channel

A/2 = 15384/2 =7692 MM2 = 76.92 CM2 FOR 1 CHANNEL

A = 76.92 CM2

choose suitable section from is table