Week 3 Challenge

Design of Compression member:

Introduction: A member which carries an axial compression is known as a compression member. It is also called a strut. A compression member may be called short or long depending on the slenderness ratio (KL/r), where KL is the effective length and r is the least radius of gyration of the cross section.

PROBLEMS:

1) Effective length factor determination problem:

Given: L = 10m

a) Braced frame and shear wall:

Solution:

L_eff = 0.65 x L

= 0.65 x 10000

L_eff = 6500 mm

b) Unbraced frame with a pinned base:

Solution:

L_eff = 1 x L

= 1 x 10000

L_eff = 10000 mm

c) Unbraced frame with a fixed base:

Solution:

L_eff = (0.8 x L) + (1 x L )

= (0.8 x 3333) + ( 1 x 6673)

L_eff = 9334 mm

2) Design axial strength determination of a column with its end conditions:

Given:

ISMB 300

Cover plate 200x12mm

Length of beam = 3m

Fy = 250N/mm²

End Condition: Rotation fixed and translation free

Solution:

For ISMB 300 (300X140X13.1)

MOI = 8990 X 10⁴mm⁴

A = 58.6 x 10²

Wt = 46kg/m

For composite beam:

I = 8990 x 10⁴+ 2(Ax̅²)

= 8990 x 10⁴+ 2(200 x 12 x 156²)

I = 205.81 x 10⁶mm⁴

Total area = 4800 mm²

r = √205.81x10⁶ / 4800

r = 207.068

Slenderness ratio:

λ = leff / r

= (1.2 x 3000) / 207.068

λ = 17.39

Using buckling class ‘c’,

λ = 17.39; fy = 250N/mm²

By interpolation,

fcd = 227 – [(17.39-10)/(20-10) x (227-224)]

fcd = 224.78N/mm²

Design axial strength:

Pd = Ae x fcd

= 4800 x 224.78

Pd = 1078.944 kN

3) Design circular tube struct member for certain loads:

Given : L = 2m (Simply supported)

P = 400kN; fy = 210 N/mm2²

To find: Circular and square hollow sections

1. Circular tube struct member:

Solution:

Assuming fcd = 0.7 x 210

fcd = 147N/mm²

Required area, A_reqd = 400x10³/147

A_reqd = 2721.09mm²

Assuming light tubular section,

A = 32.3x10²mm²

r = 75.8

h = 2m

Effective length, Leff = KL

= 1 x 2000

Leff = 2000mm

Slenderness ratio, λ = leff / r

= 2000/ 75.8

λ = 26.39

Buckling class ‘a’ as per code for circular section

Calculation of fcd:

λ = 26.39; fy = 210N/mm²

By interpolation,

fcd = 187.8N/mm²

Design compressive strength:

Pd = Ae x fcd

= 32.3x10²x187.8

Pd = 606.6kN > 400kN

HENCE SAFE.

2. Hollow square section:

Assuming 150x150x5mm section,

A = 28.36x10²mm²

r = 5.89

h = 2m

Effective length, Leff = KL

= 1 x 2000

Leff = 2000mm

Slenderness ratio, λ = leff / r

= 2000/ 58.9

λ = 33.96

Buckling class ‘a’ as per code for circular section

Calculation of fcd:

λ = 33.96; fy = 210N/mm²

By interpolation,

fcd = 184.02N/mm²

Design compressive strength:

Pd = Ae x fcd

= 28.36x10²x184.02

Pd = 521.89kN > 400kN

HENCE SAFE.

4) Design of laced column:

Given: Single lacing system

L = 10m

P = 1000N

Solution:

Assuming fcd = 120N/mm²

A_reqd = 1000x10³/120 = 8333.33mm²

Try ISMC 300, A_provd = 2 x 4564

= 9128 mm² > A_reqd

r = 118.1

Actual Slenderness ratio,

λ = leff / r

= 10000/ 118.1

λ = 84.67

Since it is laced column,

λ_e = 1.05 x 84.67

λ_e = 88.907

From table 9c, by interpolation

fcd = 122.64N/mm²

Load carrying capacity:

P_d = A_e x fcd

= 9128 x 122.64

P_d = 1119.092kN>1000kN

Spacing between channels,

Let clear distance be ‘d’

Ixx = 2x6362.6x10⁴

Iyy = 2[3108x10⁴+4564(d/2+23.6)²]

Equating Iyy to Ixx we get,

d = 151.7 mm

d = 150mm

Lacing design:

Let the lacing be provided at 45 degree.

Horizontal spacing of lacing = 150+60+60

= 270mm

Vertical spacing = 270 tan 45^o = 875mm

r = 26.1

Slenderness ratio,

λ = 875/26.1 = 33.52 < 50

Transverse shear to be resisted by lacing,

Vt = 0.025 x 1000 x 10³

= 25000N

Shear to be resisted by each lacing system

= 25000/2 = 12500N

Length of lacing,

cos 45^o= 270/L

L = 490.9mm

Minimum thickness of lacing = ¼ x 490.9

t = 12.27mm = 14mm

Minimum width of lacing if 20mm dia bolt are used

= 3 x 20 = 60mm

Use 60 ISF 14,

Area of lacing = 60 x 14 = 840mm²

r_min = √I/A = √60x14/12x60x14

r_min = 4.014

Slenderness ratio,

λ = 490.9/24.014 = 122.29 < 145

Check strength of lacing,

Flat belongs to buckling class ‘c’

By interpolation, λ = 122.29; fy = 250N/mm²

fcd = 81.55 N/mm²

Load carrying capacity of lacing,

P_d = A_e x fcd

= 60 x 14 x 81.55

P_d = 68502N

= 12500/Sin45^o

= 14705N < 68802N

HENCE SAFE

Provide ISMC 300 @ 150mm spacing with flat 60x14mm.

5) Design of batten column:

Given: L = 5m

P = 2000kN

Solution:

Assume channel ISMC 350

Spacing d = 220mm

Slenderness ratio,

λ = 1x5000/136.6 = 36.6

Distance between center of channel,

S = 20+60+60 = 340mm

Batten design:

Let ‘c’ be spacingof battens

C/r_yy < 50

C = 1415

C/r_yy < 0.7 λ

C = 725.05

Let us select C = 725.05

Transverse shear,

Vt = 0.025 x 2000 x 10³

= 50000N

Vb = VtC/NS = 50000x725.05/2x340

= 53302.5N

M = VtC/2N = 50000x725.05/2x2

= 9063125N

Size of battens,

For end battens,

a) Eff depth d’ = S’+2Cyy

= 220+(2x24.4)

d’ = 268.8mm

Provide 270mm depth for end batten

b) Overall depth d = 270+(2x35)

= 340mm

For Intermediate batten,

a) Eff. depth = ¾ x 270 = 202.5mm

d₁’ = 210mm

b) Overall depth d₁ = 210+(2x35)

= 280mm

Thickness of battens,

t = 1/50 x 340 = 6.8mm

Use 8mm thick plate.

Check for stresses in batten plate,

Shear stress = 53312.5/280x8

= 23.8N/mm² < (fy/√3)x(1/1.1)

Bending stress = 6M/td²

= 6x9063125/(8x280²)

= 86 < fy/1.1x1.2