Week 3 Challenge

Q.1

Answer:-

Given data:-

Height of column = 10m

Column is effectively restrained in mid height in one direction (in Z direction) but free to move in other direction (Y direction).

Therefore the actual length (L) is 5m.

a. Braced frame and shear wall:

As per IS 800 : 2007, Table 11.

The effective length factor for Bottom of column is fixed and Top of column is free = 2.0L

L_e = 2.0 x 5m = 10m

b. Unbraced frame with a pinned base:

As per IS 800 : 2007, Table 11.

The effective length factor for Bottom of column is pinned and Top of column is free = 2.0L

L_e = 2.0 x 5m = 10m

c. Unbraced frame with a fixed base:

As per IS 800 : 2007, Table 11.

The effective length factor for Bottom of column is fixed and Top of column is free = 1.2L

L_e = 1.2 x 5m = 6m

Q.2

Given data:-

• Section - Hot rolled ISMB 300 with flange cover plate of 200 x 12 mm on each flange.
• Unsupported Length of column is 3 m.
• Bottom of Column is fixed and Top as Roller (Rotation fixed, translation free)
• F_y is 250 N/mm²

Answer:-

Step 1:-

Properties of built-up section:

The built-up column consist of one ISMB 300 and Two cover plates of 200 x 12 mm as shown above.

From steel section tables (SP 6 - Part 1),

Cross section area of ISMB 300 = 5626 mm²

I_xx of ISMB 300 = 8603.6 x 10⁴ mm⁴

I_yy of ISMB 300 = 453.9 x 10⁴ mm⁴

Cross section area of Plate = 2400 mm²

Moment of inertia of built-up section about xx-axis

I_xx = [8603.6 x 10⁴ + 2 x 200 x 12³ x (1/12) + 2 x Ah²]

= [8603.6 x 10⁴ + 5.76 x 10⁴ + 2 x 2400 x (150+12/2)]

= [8603.6 x 10⁴ + 5.76 x 10⁴ + 74.88 x 10⁴]

I_xx = 8684.24 x 10⁴ mm⁴

Moment of inertia of built-up section about yy-axis

I_yy = [453.9 x 10⁴ + 2 x 12 x 200³ x (1/12)]

= [453.9 x 10⁴ + 1600 x 10⁴ ]

I_yy = 2053.9 x 10⁴ mm⁴

Step 2:-

Slenderness ratio:

I_yy is less than I_xx. Therefore r_yy is Minimum.

Minimum radius of gyration r_min = (I_yy / A)^1/2

= (2053.9 x 10⁴ / 10426)^1/2

r_min = 44.38 mm

Unsupported Length (L) = 3 m = 3000 mm.

Effective Length (L_e) = 1.2 x L = 3600mm

Slenderness ratio of the column = (L/r_min ) = 3600/44.38 = 81.12

Step 3:-

Safe Load:

From IS 800 : 2007

For L/r = 81.12 and the steel having yeild stress, F_y is 250 N/mm²

Allowable working stress in compression

f_cd = 138 N/mm²

Safe load carrying capacity of builtu-up section P = f_cd x A = 138 x 10426 = 1438788 N

P = 1438.788 kN

Result:-

Design Strength = 1438.788 kN

Q.3

Given data:-

Length of strut member = 2 m

Factored axial load = 400 kN

F_y = 210 N/mm²

Assume the ends of member are simply supported

Hence L_e = 1.0 x L = 2 m = 2000 mm.

Answer:-

Required area of cross section = P / F_y

= 400 x 10³ / 210

=1904.76 mm²

From IS 4923 : 1997, Table 1

For a circular hollow section of 125 outer bore dia = 139.7 thickness = 4.5, Area of circular section = 1910 mm²

For a square hollow section of 100x100x6 has a Area of square section = 2163 mm²

From IS 800 : 2007, Table 2

Limiting Width to Thickness ratio of circular section should be lesser than 88 e²

e = (250/F_y)^1/2 = 1.09

D/t = 139.7 / 4.5 = 31.04

88 e² = 95.92

Width to Thickness ratio of circular section is lesser than 88 e²

Hence safe.

From IS 800 : 2007, Table 10

Square tube Sections comes under class a

For circular section Slenderness ratio λ=L/r_min = 2000/47.8 = 41.84 < 180

For square section Slenderness ratio λ= L/r_min =2000/37.9

L/r_min = 52.77 < 180

Hence safe.

From table 9a of the code for f_y = 210

For circular section f_cd = 180 N/mm²

For square section f_cd = 174.1 N/mm²

Hence, the design strength for circular section = 180 x 19.10 = 343.8kN.

Q.4

Given data:-

Height = 10m

Factored axial load = 1000kN

Assume design stress of 125 N/mm² (0.5 fy)

Required area =1000x10³ / 125=8000 mm²

Answer:-

Assume the column to be built up of two ISMC 300 channels placed back-to-back

Cross section area of each channel = 4564 mm²

Cross section area of two columns (A) = 2 x 4564 = 9128 mm²

r_xx = 118.1 mm

r_yy = 26.1 mm

t_f = 13.6 mm

t_w = 7.6 mm

c_yy = 23.6 mm

I_xx for two channel = 2 x 6362.8 x 10⁴ mm⁴ = 12725.2 x 10⁴ mm⁴

I_yy for one channel = 310.8 x 10⁴ mm⁴

The distance back to back 'S' between the channels is so adjusted as to make r_xx = r_yy

L/r_xx = 10 x 10³ / 118.1 = 84.67

L/r of the built-up section should be taken as 1.05 x 84.67 = 88.90

f_cd taken from code for given r_xx = 120 N/mm²

Permissible axial load = 9128 x 120 = 1095 kN > 1000 kN.

Hence the section is safe.

Spacing of channels = 12725.2 x 10⁴ = 2 (I_yy + A (S/2 +C_yy)²)

=2 x (310.8 x 10⁴ + 4564 (S/2 +23.6)²)

solving the above equation S = 183.1mm

Design of Lacing:

Using the single lacing bar inclined at 45 degree with the column axis.

Lacing system of the built-up section using single lacing system with the inclination of lacing bar gauge length g = 50 mm for 90 mm flange

Spacing of lacing bars L_o= 2 x (183 + 50 x 2) cot 45

= 2 x 283 x 1 = 566 mm

For a single column element, slenderness ratio L_o / r_yy should be < 0.7 x L/r of whole column

566 / 26 = 21.76 < 0.7 x 88.98=62.3

Hence safe.

Transverse shear force (V) = (2.5 /100) x axial load = (2.5/100) x 1000 x 10³ = 25 kN = 25000 N

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

Q.5

Given data:-

Height = 5m

Axial load= 2000 kN

Answer:-

Assume design stress of 150 N/mm² (0.6 fy)

Required area =2000 x 10³ / 150 = 13333 mm²

Assume the column to be built up of two ISMC 350 channels placed back-to-back

Cross section area of each channel = 5366 mm²

Cross section area of two columns (A) = 2 x 5366 = 10732 mm²

r_xx = 136.6 mm

r_yy = 28.3 mm

t_f = 13.5 mm

t_w = 8.1 mm

c_yy = 24.4 mm

I_xx for two channel = 2 x 10008 x 10⁴ mm⁴ = 20016 x 10⁴ mm⁴

I_yy for one channel = 430.6 x 10⁴ mm⁴

The distance back to back 'S' between the channels is so adjusted as to make r_xx = r_yy

L/r_xx = 5 x 10³ / 136.6 = 36.60

f_cd taken from IS 800 : 2007 Table 9c for given r_xx = 202.4 N/mm²

Permissible axial load = 10732 x 202.4 = 2172 kN > 2000 kN.

Hence the section is safe.

Spacing of channels = 20016 x 10⁴ = 2 (I_yy + A (S/2 +C_yy)²)

=2 x (430.6 x 10⁴ + 5366 (S/2 +24.4)²)

solving the above equation S = 333.4 mm

Spacing of battens L_o/r_yy should be less than 50 or 0.7x L/r_yy

L_o/r_yy < 0.7 x 36.6

Lo < 0.7 x 36.6 x 28.3 = 725 mm

Also, Lo/ryy should be less than 50

Lo/ryy < 50

Lo < 50 x ryy = 50 x 28.3 = 1415 mm

Hence provide batten at spacing 725 mm

Size of end battens

Overall depth of batten = 334 + 2 x C_yy

= 334 + 2 x 24.4= 383 mm

Provide 60 mm overlap of batten on channel flange for welding

Length of batten = 334 x 2 x 60 = 454 mm = 460 mm

Thickness of battens = 1/50 x 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.