Design of Steel I-Section (BS5950)

Project Irish Cement Ltd. - Kiln 3 Cement Mill-Silos Structure Subject Design of Steel Beam at EL 56.40

E. B. NALDOZA Engineering Design Registered in Dublin, Ireland (with Branch Office in Cavite, Philippines) Eddie Naldoza, BSc MIES MPICE

Ref. No. PB9729

Beam No. 192-195, 47 Calculated Checked Date E. Naldoza E. Naldoza 26/09/07

E: [email protected] W: www.eddienaldoza.com

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3

DESIGN OF SIMPLY SUPPORTED ROLLED STEEL I or H SECTION (Spreadsheet Calculation Beam No. : 215-216

Gridlines :

6

(Developed by: Eddie Naldoza, BSc MIES MPICE

/

C-E

Code / 1. Loadings : (Ultimate Loads) Clause Maximum Moment, Mx = 2 kN-m BS5950 Maximum Shear at Support,Fv = 3 kN Part 1 : Total Span, L = 700 mm 2000 / Load factor for Imposed Loads, γf i = 1.5 2. Member Checks : 2.1 Trial Section

:

Limited to Plastic Section & Major Axis Bending only)

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Distance of point load from nearer support, ae

Point Load, P

= =

2 350

kN mm

Maximum Shear at Point Load or at Point of Maximum Moment, Fvc

=

2

kN

127x76x13

UB

Steel Grade = S 275

From Member Capacity Tables (Blue Book) Moment Capacity, Mc = 0.00 kN-m 2.2 Section Properties : D = 127.0 mm b B = 76.00 mm d t = 4.00 mm Sx T = 7.60 mm Zx

= 38 mm = 96.60 mm 3 = 84 cm 3 = 75 cm

r = 7.60 mm Local Buckling Ratios : 2 E = 2.05E+05 N/mm b/T = 5.00 cm4 Ix = 473 d/t = 24.10 ry = 18.4 mm x = 16.3

2.3 Section Classification :

ε √

√

3.1.1 T < 16 mm 275 = 275 = 2 Table 9 Therefore; py = 275 N/mm py 275 3.5.2 For Rolled Section; Outstand Element of Compression Flange Table 11

4.2.3

4.2.3

= 1.00

ֶ ֶ

(3.5.2-Table 11 b )

Flange is Class 1 Plastic For Flange : b/T = 9 ε = 9 x 1.0 = 9.0 ; Actual b/T = 5.00 < 9.0 Web of an I-, H- or Box Section; Neutral Axis at Mid-depth For Web : d/t = 80 ε = 80 x 1.0 = 80.0 ; Actual d/t = 24.10 < 80.0 Web is Class 1 Plastic Therefore : Section is Class 1 Plastic 2.4 Check For Shear Buckling if : d/t > 70 ε 70 ε = 70 x 1.00 = 70.0 ; Actual d/t = 24.10 < 70.0

ֶ

No need to check for Shear Buckling

2.5 Check For Shear Capacity : Fv or Fvc ≤ Pv 2 mm2 py = 275 N/mm ; Av = t D = 4.0 x 127.0 = 508 2 Pv = 0.6 py Av ; Pv = 0.6 x 275 N/mm x 508 mm2 = 83820

N ≈

4.2.5 2.6 Check For Moment Capacity : Mx ≤ Mcx 4.2.5.2/ Check for High/Low Shear (at point of Maximum Moment, Fvm) 4.2.5.3 Shear at point of Maximum Moment ; Fvc = 2 kN

Limiting Shear, Fve = 0.60 x Pv = 0.60 x

84

kN = 50.3 kN >

For Low Shear (for Class 1 Plastic or Class 2 Compact cross-sections) Mcx = py Sx = 275 N/mm² x 84 cm³ = 23.2 kN-m >

4.2.5.1

2

2.0

kN

84 kN > 3.0 kN ; Adequate for Shear

ֶ

Low Shear

kN-m ; Adequate for Flexure

2.7 Check Limit To Avoid Irreversible Deformation Under Serviceability Loads : For simply supported beam : Mcx ≤ 1.2 py Zx 2 3 Mcx = 1.2 x 275 N/mm x 75 cm 23.2 < 24.618 kN-m ; Adequate for Flexure

E. B. NALDOZA Engineering Design Registered in Dublin, Ireland (with Branch Office in Cavite, Philippines) Eddie Naldoza, BSc MIES MPICE

Project Irish Cement Ltd. - Kiln 3 Cement Mill-Silos Structure Subject Design of Steel Beam at EL 56.40

Ref. No. PB9729

Beam No. 192-195, 47 Calculated Checked Date E. Naldoza E. Naldoza 26/09/07

E: [email protected] W: www.eddienaldoza.com

Page of

3

DESIGN OF SIMPLY SUPPORTED ROLLED STEEL I or H SECTION (Spreadsheet Calculation Beam No. : 215-216

Gridlines :

6

(Developed by: Eddie Naldoza, BSc MIES MPICE

/

C-E

Limited to Plastic Section & Major Axis Bending only)

www.eddienaldoza.com)

Part 1 : 2.8 Web Bearing and Buckling Under the Point Load :

2000 / 4.5.2.1

2.8.1 Bearing capacity of the unstiffened web : P ≤ Pb w ; Pb w = (b 1 + n k ) t py w

; py w

= 275

N/mm2

= 275

N/mm2

b1 = t + 1.6r + 2T

= 31.4 mm ; n = 5 (not at the end of the member) ; k = T+r = 15.2 mm Pbw = [ 31.4 + ( 5 x 15.2 ) ] 4.0 x 275 = 118.1 kN > 2 kN ; Therefore Bearing Capacity of the unstiffened web under the point load is adequate

4.5.3.1

2.8.2 Buckling resistance of the unstiffened web : P ≤ P x ; Check if ; a e ≥ 0.7d ; a e is the distance from the load to the nearer end of the member = 350 mm 350 ≥ 0.7 x 96.60 ; 350 > 67.6 mm ; Therefore 25 ε T 25 x 1.0 x 7.6 x 118.1 Px= Pbw ; P x= √ (b1 + nk ) d √ ( 31 + 5 x 15.2 ) x 96.60 P x = 220.33 kN > 2 kN ; Therefore Buckling Resistance of the unstiffened web under the point load is adequate

4.5.2.1

2.9 Web Bearing and Buckling at the Support : 2.9.1 Bearing capacity of the unstiffened web : Fv ≤ Pb w ; Pb w = (b 1 + n k ) t py w

; py w

b1 = t + T + 0.8r - g

= 4.0 + 7.6 + 0.8 x 7.6 - 10 = 7.68 mm (properties of the supporting element ) n = 2 + 0.6b e /k ≤ 5 ; b e is the distance to the nearer end of the member from the end of the stiff bearing = 0 mm n = 2 + 0.6 x 0 / 15.2 = 2.0 < 5 ; Therefore use n = 2.0 ; k = T+r = 54.1 + 30.0 = 84.1 mm Pbw = [ 7.7 + ( 2.0 x 15.2 ) ] 4.0 x 275 = 41.9 kN > 3.0 kN ; Therefore Bearing Capacity of the unstiffened web at the support is adequate 4.5.3.1

2.9.2 Buckling resistance of the unstiffened web : Fv ≤ P x ; Check if ; a e ≥ 0.7d ; a e is the distance from the reaction to the nearer end of the member = 25 mm 25 ≥ 0.7 x 96.60 ; 25 < 67.6 mm ; Therefore ae + 0.7d 25 ε T 25 x 1.0 x 7.6 x 41.9 25 + 67.6 x Px= Pbw ; P x= x 1.40 x 96.60 √ (b1 + nk ) d √ ( 7.7 + 2 x 84.1 ) x 96.60 1.4 d P x = 61.059 kN x > 3.0 kN

0.68

=

41.8

kN

; Therefore Buckling Resistance of the unstiffened web at the support is adequate 2.5.2

2.10 Serviceabilty Limit States : Deflection Check The serviceability Loads are taken as unfactored imposed loads. In this case, as only ultimate loads are the available data, the unfactored imposed loads will be taken as Mc or P / γf i conservatively, for simplicity; γf i to be the load factor = 1.5

From Maximum Moment, Mc; wi = Unfactored Distributed Load wi Le2 8x 2 kN-m 8 Mc Mc = 1.5 ; wi = = = 2 2 1.5 L 8 1.5 x 0.7 m e From Point Load P; W i = Unfactored Point Load P 2.1 P = 1.5 W i ; Wi = = = 1.4 kN 1.5 1.5 The total deflection is given by: Wi Le³ 5 wi Le4 δ = + 384 E I 48 EI 5 x 21.8 x 0.7 4 1.4 x 0.7 3 = + 384 x 2.E+05 x 473 48 x 2.E+05 x 473 ; Adequate for deflection = 0.081 mm < 3.89 mm

21.8

kN/m

Limits for calculated deflection (2.5.2 Table 8) for : Cantilever Allowable deflection, δa Le 700 δa = = = 180 180

3.89

Therefore : 127x76x13 is Adequate, Adopt Section

mm

UB

Project Irish Cement Ltd. - Kiln 3 Cement Mill-Silos Structure Subject Design of Steel Beam at EL 56.40

E. B. NALDOZA Engineering Design Registered in Dublin, Ireland (with Branch Office in Cavite, Philippines) Eddie Naldoza, BSc MIES MPICE

Ref. No. PB9729

Beam No. 192-195, 47 Calculated Checked Date E. Naldoza E. Naldoza 26/09/07

E: [email protected] W: www.eddienaldoza.com

Page of

3

DESIGN OF SIMPLY SUPPORTED ROLLED STEEL I or H SECTION (Spreadsheet Calculation Beam No. : 215-216

Gridlines :

6

/

C-E

(Developed by: Eddie Naldoza, BSc MIES MPICE

4.3.0

2.11 Lateral Torsional Buckling : Mx ≤ Mb/m LT ; Effective Length of the segment, L E

Maximum Major axis Moment in the segment, Mx = 4.3.6.7 4.3.6.8

Limited to Plastic Section & Major Axis Bending only)

www.eddienaldoza.com)

λ=LE/ry = 700 / 18.4 = u = 0.9

2 38

; λ/x = 38.0 / 16.3 = 2.3 ; η = 0.5 (Equal flange)

(for Rolled I of Equal flange-4.3.6.8 )

; λLT = uvλ√βw

Equivalent Slenderness

β W = 1.0

; ;

= 7.0 x ( 32 -

0.5 ; λLO = 0.4(π ²E /py) 0.5 = 0.4 x ( π ² x 2.05E+05 / 275 ) = 34.31

λLO

; η LT = a LT(λLT-λLO)/1000 ; a LT = 7.0 34.31 )/ 1000 = -0

p E = π ² E/ λLT2 = π ² x 2.05E+05 / 32 Bending Strength,

4.3.6.4

(for Class 1 Plastic or Class 2 Compact Sections-4.3.6.9 )

Perry factor & Robertson constant

η LT B.2.1

; v = 0.93 (Table 19)

Limiting equivalent slenderness

λLT = 0.9 x 0.93 x 38.0 x √ 1.0 = 32 B.2.2

= 700 mm

kN-m

pb =

2

= 1980 pE py

; Φ LT =

Φ LT +(Φ LT 2 -p E p y )

Buckling Resistance Moment;

Mb = p b Sx

0.5

p y +( η LT +1)p E 2 =

=

275 + (

-0

+ 1 ) 1980 2

1980 x 275 1111 + ( 1111

= 280 x

84200

2

- 1980 x 275 )

= 1111

= 280

0.5

= 23.606 kN-m

Uniform Moment Factor for Lateral Torsional Buckling (General case); m LT

Table 18

4.3.6.2

Moments at quarter points

Moments at mid point

M2 = M4 =

M3 =

2 2

m LT = 0.2 +

Mb m LT

=

kN-m kN-m

2

kN-m

0.15M 2+0.5M 3+0.15M 4 0.15 x = 0.2 + Mx

23.606 = 1.00

24

kN-m >

2

kN-m

2

+ 0.50 x 2

2

+ 0.15 x

; Therefore Buckling Resistance Moment is Adequate

2

= 1.00

N/mm2

PB9729

3

r Axis Bending only)

kN quate for Shear

PB9729

3

r Axis Bending only)

N/mm2

N/mm2

mm

PB9729

3

r Axis Bending only)

(Table 19)

N/mm2