Week 3 Challenge

Design of Compression Member

Question: 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.

Aim:

To be determine the effective length factor of columns for the given problem.

Given Data:

• Height of the Column (h) = 10 m.

Procedure:

Effective Length factor (E_LF) at 1.0

(E_LF) = K x L

L = 10/2 = 5 m

(E_LF) = 1.0 x 5 = 5 m

Effective Length factor (E_LF) at 1.2

(E_LF) = 1.2 x L

(E_LF) = 1.2 x 5 = 6 m

Effective Length factor (E_LF) at 2.0

(E_LF) = 2.0 x L

(E_LF) = 2.0 x 5 = 10 m

Result:

• Effective Length Factor (E_LF) at 1.0 = 5 m
• Effective Length Factor (E_LF) at 1.2 = 6 m
• Effective Length Factor (E_LF) at 2.0 = 10 m

Question: 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

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.

Aim:

To be determine the axial compressive capacity of Column by using above given tip steps.

Given Data:

Grade of used steel for column ISMB 300

Size of Flange cover plate = 200 x 12 mm

Length of beam (L) = 3m.

F_y = 250 N/mm²

Procedure:

Grade of used steel for column ISMB 300. The following mechanical properties values for

A = 58.60 Cm² = 5860 mm², D = 300, I_xx = 8990 cm⁴, I_yy = 486 cm⁴

B_f = 140 mm, T_f = 13.1 mm, T_w = 7.7 mm, R_xx = 12.37 cm⁴, R_yy = 2.84 cm⁴

Determine new radius of gyration of compound section

Area of the compound section (A)

A = 5860 + 2 (200 x 12)

A = 5860 + 4800

A = 10660 mm².

I_z for plates

I_z = 2 x (200 x (12)³/12 + 200 x 12 x (156)²)

I_z = 11687 x 10⁴ mm⁴

Total Izz of beam and plates

I_zz = (11687 x 10⁴ + 8990 x 10⁴)

I_zz = 20677 x 10⁴ mm⁴

R_zz =√ (I_zz /A)

R_zz = √ (20677 x 10⁴ /10660)

R_zz = 139.24 mm

I_xx of ISMB 300 = 486 x 10⁴

I_yy for two plates

I_yy = 2 x (12 x 200² /12)

I_yy = 1600 x 10⁴ mm⁴

Total I_XY for beam and plates

I_XY = (I_xx + I_yy)

I_XY = (486 +1600)

I_XY = 2086 x 10⁴ mm⁴

R_yy = √ (2084 x 10⁴ /10660)

R_yy = 44.24 mm, T_w < 40mm

Using Buckling curve classification & slenderness ratio using table 9 (c) in code

Effective length (E_L) = 1.2 x L

= (1.2 x 3 x 10³)

(E_L) = 3.6 x 10² mm

Slenderness ratio (λ) = (E_L) / R_yy

(λ) = (3.6 x 10²) /44.24

(λ) = 81.4 mm < 180. Its within limit.

From table, taken F_cd = 134.2

The design of axial strength for column (σ_s) = (F_cd x A) / 1000

(σ_s) = (134.2 x 10660) /1000

(σ_s) = 1430.5 KN

Result:

The axial compressive strength of the column (σ_s) = 1430.5 KN

Question: 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.

Tip Steps:

• 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.

Aim:

To be determine the design of circular tube strut member and equivalent Square hollow section using the given tips in this problem.

Given Data:

• Length of tubular strut member (L) = 2 m
• Factored Axial Load (F_L) = 400 KN
• Yield strength Fy =210 Mpa.
• Here all ends of the member is simply supported

Procedure:

Section satisfying Area (A) = F_L / Fy

A = 400x10³ / 210

A = 1904.76 mm²

Here we taken the dimensions from the table 10 and 9 (c) for the square and circular hollow section.

Square Hollow Section:

Using Buckling curve classification & slenderness ratio using table 9c in code.

Here we considered, Size of the square hollow section is 100x100x6mm,

Area of square section (A) = 2163 mm²

Effective length (E_L) = 1.0 x L

(E_L) = 1.0 x 2

(E_L) = 2 m = 2 x 10³ mm.

Slenderness Ratio (λ) = (E_L) / R_min

(λ) = (2 x 10³) / 37.9

(λ) = 52.77 < 180. Its within limit

To find the axial strength of square hollow section

Fy = 210 Mpa ; Compressive strength (F_cd) =174.1 N/mm²

Design of axial strength for square hollow section (σ_s)

(σ_s) = (F_cd x A) / 1000

(σ_s) = 174.1 x 2163 / 1000

(σ_s) = 376.58 KN

Circular Hollow Section:

Using Buckling curve classification & slenderness ratio using table 9c in code.

Circular hollow section of 125 outer bore diameter = 139.7 mm thickness = 4.5 mm, Area of circular section = 1910 mm²

Effective length (E_L) = 1.0 x L

(E_L) = 1.0 x 2

(E_L) = 2 m = 2 x 10³ mm.

Slenderness Ratio (λ) = (E_L) / R_min

(λ) = (2 x 10³) / 47.80

(λ) = 41.84 < 180. Its within limit.

Fy = 210 Mpa ; Compressive strength (F_cd) =180 N/mm2

Design of axial strength for Circular hollow section (σ_s)

(σ_s) = (F_cd x A) / 1000

(σ_s) = 180 x 1910 / 1000

(σ_s) = 343.80 KN

Result:

• Design of axial strength for square hollow section (σ_s) = 376.58 KN
• Design of axial strength for Circular hollow section (σ_s) = 343.80 KN
• The Design circular tube strut member and equivalent square hollow section for certain load is in safe conditions.

Question: 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.

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

Aim:

To be determine the design of a laced column by using the given tip steps

Given Data:

• Height (L) = 10 m
• Axial load (A_L) = 1000 KN
• Lacings are bolted but design of bolted connection not required

Procedure:

Assume,

Design of column P =1000x10³

Design stress (σ) = 125 N/mm²

Required cross section area (A) = P / σ

(A) = 1000x10³ / 125

(A) = 8000 mm²

Here we taken for two ISMC 300 channel mechanical properties are,

A = 91.28 cm²; Rxx =118.1 mm; Ryy =38.7 mm; Tf =13.6 mm; Cyy =23.6 mm;

Tw =7.6 mm; Ixx=12725.2 cm⁴

Slenderness Ratio

Here Rxx = Rzz

L / Rzz =10 x10³ / 118.1

L / Rzz = 84.67

Slenderness Ratio (λ) taken as

(λ) = 1.05 x (L / Rzz)

(λ) = 1.05 x 84.67

(λ) = 88.90 < 180. Its within limit.

F_cd taken from code for given Rzz = 120 N/mm²

= 9128 x 120

=1095 KN > 1000 KN.

Hence the section is safe.

Spacing of channels = 12725.2 x 104 = 2 (Iyy + A ((S/2) + Cyy) 2)

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

S = 183.1 mm

For 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_S) = 2 x (183+50+50) cot 45̊

(L_S) = 2 x 283 x 1

(L_S) = 566 mm

To check its condition L_s / Ryy should be < 0.7 x L / R of whole column

566 / 26 = 21.76 < 0.7 x 88.98 = 62.3

Hence it is in safe condition

Maximum tranverse shear (V_t) = 2.5% of axial force of the column.

(V_t) = (2.5/100) x 1000 x10³

(V_t) = 25000 KN

Transverse shear in each panel (T_s)

(T_s) = (V_t) / n

(T_s) = 25000 / 2

(T_s) =12500 N

Compressive force in the lacing bar (Cf)

(Cf) = (V_t / n) cosec 45̊

(Cf) = 12500 x 1.414

(Cf) = 17675 N

Result:

• Spacing of channels (S) = 183.1 mm
• Spacing of lacing bars (L_S) = 566 mm
• Maximum tranverse shear (V_t) = 25000 KN
• Transverse shear in each panel (T_s) =12500 N
• Compressive force in the lacing bar (Cf) = 17675 N

Question: 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 2000 kN. 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.

Aim:

To be determine the design of battened coloumn made up of two channels by using given tip steps

Given Data:

Height (L) = 5m

Axial load (A_L) = 2000 kN

Procedure:

Assume,

Design of column (P) = 2000 x 10³ N

Design stress (σ) = 150 N/mm²

Required area (A) = (P) / (σ)

• = 2000x10³ / 150

• = 13333 mm²

Here we taken for two ISMC 300 channel mechanical properties are,

A = 53.66 x 2 =10732 cm²; Rxx = 136.6mm; Ryy =28.3mm; Bf =100; Tf =13.5 mm

Cyy = 24.40 mm; Tw =8.1 mm; Ixx =12725.2 cm⁴

Slenderness Ratio

Here Rxx = Rzz

L / Rzz = 5 x 10³ / 136.6

L / Rzz = 36.60

Slenderness Ratio (λ) taken as

(λ) = 1 x (L / Rzz)

(λ) = 1 x 36.60

(λ) = 36.60 < 180. Its within limit.

K =1, Fy =250 N/mm2 fcd taken from code for given Rzz =36.60

F_cd =202.4/mm2

Capacity of built up column (C) = A x F_cd

= 10732 x 202.4

= 2172 KN > 2000 KN

Hence it is in safe condition.

Spacing of channels = 2 x 20016 x104 = 2 (Iyy + A ((S/2) + Cyy) (2))

=2 x (430.6 x10⁴ + 5366 ((S/2) +24.4) (2))

S =333.4 mm

Spacing of battens (Ls) / Ryy < 0.7x L / Ryy

Ls / Ryy < 0.7 x 36.6

(Ls) < 0.7 x 36.6 x 28.3

(Ls) =725 mm

Also, Ls / Ryy < 50

Ls / Ryy < 50

Ls < 50 x Ryy =50 x 28.3 =1415 mm

Hence provide batten at spacing = 725 mm

Maximum tranverse shear (V_t) = 2.5% of axial force of the column.

(V_t) = (2.5/100) x 2000 x10³

(V_t) = 50000 KN

Transverse shear in each panel (T_s)

(T_s) = (V_t) / n

(T_s) = 50000 / 2

(T_s) =25000 N

Compressive force in the lacing bar (Cf)

(Cf) = (V_t / n) cosec 45̊

(Cf) = 25000 x 1.414

(Cf) = 35350 N

Size of end battens

Overall depth of batten (Db) = 334 + 2 x Cyy

(Db) = 334 +2 x 24.4

(Db) = 383 mm

Provide 60 mm overlap of batten on channel flange for welding

Length of batten (Lb) = 334 x 2 x 60

(Lb) = 460 mm

Thickness of battens (Tb) = (1/50) x 460

(Tb) = 9.2 mm

Size of intermediate battens

Overall depth of intermediate battens (Dbi)

(Dbi) = (3/4) x 390

(Dbi) = 300 mm

Result:

Here designed battened column Provide in the size of 460 x 300 x 9.2 mm.

Conclusion:

Here for all 5 questions are solved or designed for the given copression members as per their data and tips in the questions.