Week 3 Challenge

Aim: To design the compression members of different sections, using the theory learnt in module.

Introduction: Compression members are commonly used as columns in building structures, chords or webs in trusses, bridge piers or braces in framed structures. The maximum strength of a steel compression member depends, to a large extent, on the member length and end support conditions.

Design of Compression Member

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

Results:

a) Effective length factor in Z direction = 0.8L = 0.8X10 = 8m

Effective length factor in Y direction = 2L = 2X10 = 20m

b) Effective length factor in Z direction = 1L = 1X10 = 10m

Effective length factor in Y direction = 2L = 2X10 = 20m

c)Effective length factor in Z direction = 0.8L = 0.8X10 = 8m

Effective length factor in Y direction = 2L = 2X10 = 20m

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/mm²

Procedure:

For ISMB300

Area = 5626 mm²; Izz =8603.6 x10⁴ mm⁴; Iyy= 453.9 x10⁴ mm⁴

• Determining the radii of gyration for the compound section

Area = 5626 + (2x200x12) = 10426 mm²

Izz for plates = 2[I + Ay₁²]

= 2 x [200x12³ /12 + 200 x 12 x 156² ]

= 11687.04 x 10⁴ mm³

Total Izz = 8603.6 x10⁴ + 11687.04 x 10⁴ = 20290.64 x 10⁴ mm³

Hence, r_z = sqrt(20290.64 x 10⁴/10426) = 139.5 mm

Iyy for ISMB 300 = 453.9 x10⁴ mm⁴

Iyy of plates = 2 x 12 x 200³/12 = 1600 x 10⁴ mm⁴

Total Iyy = 453.9 x10⁴ + 1600 x 10⁴ = 2053.9 x 10⁴ mm⁴

Hence, r_y = sqrt(2053.9 x 10⁴/10426) = 44.38 mm

• Buckling curve classification

From Table 10 of the code for buckling about any axis use curve c

• Since r_y is small, the buckling will be about the y-y axis

Effective length = 2 x 3000 = 6000 mm

𝜆 = KL/r = 6000/44.38 = 135.2 < 180

From table 9c of the code, for

Fy = 250MPa and 𝜆 = 135.2

f_cd = 70.25 N/mm²

^Result:

Hence design strength = 10426 x 70.25 / 1000 = 732.42 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 sections.

For Circular Tube:

Procedure:

fy= 210 MPa

ɛ = sqrt(250/210) = 1.091

As per IS 1161 (1998), Assume medium class pipe of outside diameter 165.1 mm with thickness 5.4mm, weight = 21.3 kg/m , r =52 mm and area = 2500 mm²

• Checking for limiting width to thickness ratio

D/t = 165.1/5.4 = 30.57 < 42ɛ² (42x1.091²=50)

Hence the tube is a plastic section

• Buckling curve classification

From Table 10 of the code, for hot-rolled tubes, use buckling curve ‘a’

• Resistance to flexural buckling

KL/r = 1 x 2000/52 = 38.46

From table 9a of the code, for

Fy = 210MPa and 𝜆 = 38.46

f_cd = 182 N/mm²

Hence design strength = 2500 x 182 / 1000 = 455 kN >400 kN

^Result:

Hence, assumed section is safe.

For Square Tube:

Procedure:

As per IS 4923 (1997), Assume SHS of size 150x150x5 mm with thickness 5mm, weight = 22.26 kg/m , r =58.9 mm and area = 2836 mm²

• Checking for limiting width to thickness ratio

D/t = 150/5 = 30 < 42ɛ² (42x1.091²=50)

Hence the section is a plastic section

• Buckling curve classification

From Table 10 of the code, for hot-rolled tubes, use buckling curve ‘a’

• Resistance to flexural buckling

KL/r = 1 x 2000/58.9 = 33.96

From table 9a of the code, for

Fy = 210MPa and 𝜆 = 33.96

f_cd = 184 N/mm²

Hence design strength = 2836 x 184 / 1000 = 522 kN >400 kN

^Result:

Hence, assumed section is safe.

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.

Procedure:

Given: P = 1000 kN

L = 10m

KL = 1 x 10 = 10m

Assuming design stress =125 N/mm²

Area required = 1000x10³ / 125 = 8000 mm²

Select two ISMC 300.

Area provided =2×4630=9260mm²

r_zz= 118.0 mm

Distance will be maintained so as to get r_yy>r_zz.

L/r_zz = 10x10³ /118 = 84.74

The L/r of built up column should be taken as 1.05 x (L/r_zz) = 1.05x84.74 = 88.98

For L/r_zz = 88.98 and fy = 250Mpa

Using table 9c of IS800:2007, fcd=122.53 MPa

Load carrying capacity=A_e.f_cd = 9260×122.53/1000

=1135KN>1000KN

Hence the column is safe.

Let us provide two channels back to back and connect them by lacing and denote S as the spacing between two channels

Spacing between the channels:

2Izz = 2[Iyy + A(S/2 + Cyy)²]

2x6420x10⁴ = 2[313 x 10⁴ + 4630(S/2 + 23.5)²]

S = 182.70mm

Provide S=183 mm

Lacings:

Let the lacings be provided at 45⁰ to horizontal.

Spacing of lacing bar, L₀=2(183+50+50) cot45⁰= 2x283x1= 566 mm

[note: g=50 is gauge distance]

L_o/r_yy should be < 0.7 x L/r of whole column

21.77 < 0.7x88.98 = 62.3

Hence safe

Traverse shear to be resisted by lacing systems= (2.5/100) x 1000 x 10³ = 25000 N

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

Compressive force in the lacing bar = (V/N)cosec45⁰ =12500 x 1.414 = 19445 N

Assuming 16mm diameter bolts,

Minimum width of lacing flat = 3 x 16, say 50mm (Clause 7.6.2 of the code)

Minimum thickness = (1/40)(183 + 50 + 50) cosec 45⁰ = 10.01 mm

Provide 12 mm thick plate with a width of 50mm

Minimum r = t/sqrt(12) = 12 / sqrt(12) = 3.464 mm

L/r of the lacing bar = 283 x cosec45⁰ /3.464 = 115.5 < 145

Hence safe.

For L/r = 115.5 and fy = 250 Mpa, using Table 9c of the code

Fcd = 88.6 MPa

Load carrying capacity = 88.6 x 50 x 12 =53163N > 19445N

Hence lacing is safe

Tensile strength of lacing flat = 0.9(B-d)tf_u/Y_ml or f_yA_g/Y_mo

= 0.9(50 – 18) x 12 x 410/1.25 or 250 x 50 x12/1.1

= 113356 N or 136363N

Hence, Tensile strength of lacing flat = 113356 > 19445N

^Result:

Hence the lacing flat is safe.

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.

Procedure:

Given: P = 2000 kN

L = 5m

Required area of cross section = 2000 x 10³ / (0.6 x 250) = 13333.33 mm²

Take ISMC 400 @501 N/m

Area provided = 2 x 6380 = 12760 mm²

𝜆 = KL/r = 1.1 x 5000/154 = 35.71

From table 9c of the code, for

Fy = 250MPa and 𝜆 = 35.71

f_cd = 205 N/mm²

Capacity of Built up column = 12760 x 205 / 1000 = 2615.8 kN > 2000kN

Hence column is Safe.

Let us provide two channels back to back and connect them by lacing and denote S as the spacing between two channels

Spacing between the channels:

2Izz = 2[Iyy + A(S/2 + Cyy)²]

2x15200x10⁴ = 2[508 x 10⁴ + 6380 (S/2 + 24.2)²]

S = 255.1mm

Provide S=256 mm

Spacing of battens

L_o/r_yy should be < 0.7 x slenderness ratio of the built up column. Hence,

L_o/r_yy < 0.7(L/r)

L_o< 0.7(L/r) r_yy

< 0.7(35.71) x 28.2

< 704.9

Also L_o/r_yy should be less than 50

L_o/r_yy < 50

L_o < 50 x 28.2 = 1210

Provide the batten at a spacing of 840mm

Size of end battens

Overall depth = 256 x 2 x c_yy

= 256 + 2 x 24.2 = 304.4 mm

Provide a 6c mm overlap of batten on channel flange for welding

Length of batten = 256+2x62 = 380mm

Thickness of batten = 1/50 x 380 = 7.60 mm

Provide 380x310x8 mm end batten plate.

Design forces:

Traverse shear to be resisted by lacing systems= (2.5/100) x 2000 x 10³ = 50000 N

Longitudinal shear V_b = V_tL_o/ns = 50000x840/[2(256 + 2 x 62/2)] = 66038 N

Moment M = V_tL_o/2n = 50000x840/[2x2] = 10.5 x 10⁶ N.mm

Check:

• For end battens

Shear stress = 66038/(310*8) = 26.63 < (250/sqrt(3) x 1.1) = 131.2 Mpa

Shear stress = 10.5 x 10⁶ x5/(310²*8) = 68.29 < (250/x 1.1) = 227 Mpa

• For intermediate battens

Shear stress = 66038/(230*8) = 35.89 < (250/sqrt(3) x 1.1) = 131.2 Mpa

Shear stress = 10.5 x 10⁶ x5/(230²*8) = 124 < (250/x 1.1) = 227 Mpa

^Result:

Hence the battens are safe