1. Steel Beam Design

C

ETab

STRUCTURASTEEL BEAMNOV13 – M

Clause

EN1990 le A1.2(B)

AL STEELWORK &M DESIGN MAC14

Example o A steel beGk = 8 kN/on 100mmbeam)

1. DESIGN Combinat = ( = 1 Design Ben (FEd x L2) / 8 Design She (FEd x L) / 2

& TIMBER DESIGN

of Design fo

am section/m and Qk

m bearings a

LOADING /

tion of Actio1.35 x 8kN/9.8 kN/m

nding Mom

8 = (19.8 k = 158.4 k

ear Force, V

2 = (19.8 k = 79.2 kN

N - ECS328

1

or Laterally R

n is loaded w= 6 kN/m. B

at each en

/ ACTION

on, FEd = γG

m) + (1.5 x

ment, MEd

kN/m x 8m xkNm

VEd

kN/m x 8m) N

Restrained

with uniformBeam was fd. (Ignore s

G Gk + γQ Qk

6kN/m)

x 8m) / 8

/ 2

Beam

mly distributfully restrainself-weight

k (eq 6.10)

ted loadingned and se of steel

MAT

R

g at

MEd = VEd =

Remarks

= 158.4 kNm

= 79.2 kN

m

STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14

2 MAT

6.1 (1)

Table 3.1

2. SELECTION OF SECTION Required Plastic Modulus , Wply = MEd / (fy / γMO ) Partial factor for particular resistance, γM Resistance of cross-sections whatever the class, γMO = 1.0 Assume the nominal thk. of flange and web is less than 40mm and advance UKB S275 to be used, .: yield strength, fy = 275N/mm2 .: Required Plastic Modulus , Wply = MEd / (fy / γMO ) =158.4 x 106 / (275 / 1.0) =576cm3 From Table of Properties, select 356 x 171 x 51kg/m UKB provided 896cm3 of Plastic Modulus Section Properties for 356 x 171 x 51kg/m UKB Depth of Section, h = 355 mm Width of Section, b = 171.5 mm Thk. Flanges, tf = 11.5 mm Thk. of Web, tw = 7.4 mm Root Radius, r = 10.2 mm Depth between Fillet, d = 311.6 mm Ratio Local Buckling (flange), cf/tf = 6.25 mm Ratio of Local Buckling (web), cw/tw = 42.1mm Second Moment Area, Iyy = 14,100 cm4 Second Moment Area, Izz = 968cm cm4 Radius of Gyration, iy = 14.8 cm Radius of Gyration, iz = 3.86 cm Elastic Modulus, Wely = 796 cm3 Elastic Modulus, Welz = 113 cm3 Plastic Modulus, Wply = 896 cm3 Plastic Modulus, Wplz =174 cm3 Area of Section, A =64.9 cm2

356 x 171 x 51kg/m UKB


3 MAT

Table 3.1

Table 5.2(1)

Table 5.2(2)

6.2.6

6.2.6(1)

6.2.6(2)

6.2.6(3a)

3. CLASIFICATION OF SECTION Take steel grade as S275 and tf = 11.5 mm < 40 mm .: fy = 275 N/mm2 Web ε = 0.92 and cw/tw = 42.1 < 72 ε .: Web section was Class 1 Flange ε = 0.92 and cf/tf = 6.25 < 9 ε .: Flange section was Class 1 .: Both section was Class 1 4.RESISTANCE OF CROSS- SECTION SUBJECT TO BENDING AND SHEAR Shear The design value of the shear force VEd at each cross section shall satisfy: VEd /Vc,Rd ≤ 1 (eq. 6.17) Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) Shear area, Av for rolled I section Av = A – 2btf + (tw + 2r) tf but not less than nhwtw Av = 6400 – 2( 171.5)(11.5) + (7.5 + 2(10.2))11.5 = 2776.35 mm2 hw = h – 2tf hw = 355 – 2(11.5) = 332 mm n may be conservatively taken equal to 1.0. .: nhwtw = 1.0 x 332 x 7.5 = 2490 mm2 Since Av > nhwtw .: take Av = 2776.35 mm2

fy = 275 N/mm2 Both section was Class 1 Av = 2776.35 mm2


4 MAT

6.2.8(1)(2)(3)

6.2.5

6.2.5(1)

6.2.5(2)

.: Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) = [ 2776.35 (275/√3)] / 1.0 = 440.8 kN VEd /Vc,Rd ≤ 1 (eq. 6.17) 79.2 / 440.8 = 0.18 Since VEd /Vc,Rd ≤ 1 .: Section was satisfactory for shear resistance Bending & Shear If VEd > 0.5 Vc,Rd .: moment resistance, Mc,Rd should be taken as the design resistance of the cross-section, calculated using a reduced yield strength, (1 – ρ) fy where ρ = ((2VEd/Vpl,Rd) - 1)2 In this case, VEd < 0.5 Vc,Rd .: no reduction of moment resistance, Mc,Rd Bending The design value of the bending moment MEd at each cross-section shall satisfy: MEd /Mc,Rd ≤ 1 (eq. 6.12) For Class 1 section, Mc,Rd = Mpl,Rd = (Wpl x fy)/ γMO (eq. 6.13) .: Mc,Rd = (896x103 x 275)/ 1.0 = 246.4kNm MEd /Mc,Rd =0.64 Since MEd /Mc,Rd < 1 .: section was satisfactory for moment resistance

Vc,Rd =440.8 kN Mc,Rd = 246.4kNm


5 MAT

6.2.6(6)

BS EN 1993-1-5:2006

8(1)

5. RESISTANCE TO SHEAR BUCKLING In addition the shear buckling resistance for webs without intermediate stiffeners should be according to section 5 of EN 1993-1-5, if: hw/tw > 72ε/n (eq. 6.22) hw/tw = 332 / 7.5 = 44.3 72ε/n = 72(0.92)/1.0 = 66.24 Since hw/tw < 72ε/n .: no further check on resistance for shear buckling 6. RESISTANCE TO FLANGE INDUCED BUCKLING To prevent the compression flange buckling in the plane of the web, the following criterion should be met: hw/tw ≤ k (E/fy)( √(Aw/Afc)) (eq.8.1) Aw = hw x tw

= 332 x 7.5 = 2490 mm2 Af = b x tf

= 171.5 x 11.5 = 1972.25 mm2 For Class 1, plastic rotation utilize, hence k = 0.3 k (E/fy)( √(Aw/Afc) = 0.3 (210x103/275) (√(2490/1972.25)) = 257.41 Since hw/tw ≤ k (E/fy)( √(Aw/Afc)) .: section is prevent the compression flange buckling in the plane of the web 7. RESISTANCE OF WEB TO TRANSVERSE FORCE (WEB BUCKLING)


6 MAT

6.2(1)

6.4(1)

Figure 6.1 (c)

6.5(3)

6.5(1)

For un-stiffened or stiffened webs the design resistance to local buckling under transverse forces should be taken as: FRD = (fyw x Leff x tw)/γm (eq. 6.1) Effective length for resistance to transverse forces, Leff Leff = XF ly (eq. 6.2) Reduction factor, XF XF = 0.5 / λF ≤ 1.0 (eq. 6.3) λF = √ ( ly tw fyw / Fcr ) (eq. 6.4) Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) For webs without longitudinal stiffeners kF should be obtained from Figure 6.1. For webs with longitudinal stiffeners kF may calculated from the equation 6.6 kF = 2 + 6 [(ss + c)/ hw] For type (c) in Figure 6.1, effective loaded length, ly should be taken as the smallest value obtained from the equation 6.11 or equation 6.12 : ly = le + tf √ [(m1/2) + (le/tf)2 + m2] (eq. 6.11) ly = le + tf √m1 + m2 (eq. 6.12) le = kF E tw2/2fyw hw ≤ ss + c (eq.6.13) m1 = fyf bf / fyw tw (eq. 6.8) m2 = 0.02 ( hw/tf)2 if λF > 0.5 (eq. 6.9) m2 = 0 if λF ≤ 0.5 (eq. 6.9) kF = 2 + 6 [(ss + c)/ hw] (Figure 6.1 (c)) = 2 + 6 [(100 + 0)/ 332]


7 MAT

= 3.81 le = kF E tw2/2 fyw hw ≤ ss + c (eq.6.13) le = kF E tw2/2 fyw hw = (3.81x 210 x103 x 7.52)/(2 x 275 x 332) = 246.47mm ss + c = 100 + 0 = 100mm Since le = kF E tw2/2 fyw hw ≥ ss + c .: take le = 100mm m1 = fyf bf / fyw tw (eq. 6.8) =(275 x 171.5) / (275 x 7.5) = 22.87 m2 = 0.02 ( hw/tf)2 = 0.02 ( 332/11.5) 2 = 16.70 ly = le + tf √ [(m1/2) + (le/tf)2 + m2] (eq. 6.11) = 100 + 11.5 √ [(22.87/2) + (100/11.5)2 + 16.7] = 217.13mm ly = le + tf √m1 + m2 (eq. 6.12) = 100 + 11.5 √22.87 + 16.7 = 172.34mm Choose the lesser value from eq. 6.11 and eq. 6.12 .: ly = 172.34 Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) = 0.9 (3.81) (210x103) ( 7.53 / 332) = 915, 024 N

kF = 3.81 le = 100mm m1 = 22.87 m2 = 16.70 ly = 172.34mm


8 MAT

λF = √ ( ly tw fyw / Fcr ) (eq. 6.4) = √ (( 172.34 x 7.5 x 275) / 915,024) = 0.62 Since λF = 0.62 > 0.5 .: value for m2 acceptable. XF = 0.5 / λF ≤ 1.0 (eq. 6.3) = 0.5 / 0.62 = 0.81 ≤ 1.0 .: take XF = 0.81 Leff = XF ly (eq. 6.2) = 0.81 x 172.34 = 139.6mm FRD = (fyw x Leff x tw)/γm (eq. 6.1) = (275 x 139.6 x 7.5)/1.0 = 287.93 kN Since FRD > VEd .: section is adequate for resistance of web to transverse force 8. DEFLECTION dactual = 5wL4 / 384 EI = (5x6x80004)/(384x210x103x141x106) = 10.8mm dallow = span/360 = 8000/360 = 22.22mm Since dallow > dactual .: section is satisfactory for deflection

Fcr = 915, 024 N λF = 0.62 XF = 0.81 Leff = 139.6mm FRD = 287.93 kN

Clause Remarks

Example of Design for Laterally Restrain Beam Support for a conveyor. Part of the support for a conveyor consists



of a pair ois connecton across Lateral resconnectedlimit state.

1. DESIGN

2. SELECTIO Required P

& TIMBER DESIGN

of identical ted to a sta beam at Dstraint is prd to rigid s

LOADING /

ON OF SECT

Plastic Mod

N - ECS328

9

beams as anchion at D by boltinrovided bysupports. Th

/ ACTION

TION

dulus , Wply =

9

shown in fi end A by

ng through y transversehe loads sh

= MEd / (fy /

igure belowa cleat an the connee beams ahown are a

γMO )

w. Each bed is suppor

ecting flangat A, B anat the ultim

MAT

eam rted ges. d E ate

MEd = VEd =

= 825 kNm

= 375 kN


10 MAT

6.1 (1) Table 3.1 Table 3.1

Partial factor for particular resistance, γM Resistance of cross-sections whatever the class, γMO = 1.0 Assume the nominal thk. of flange and web is less than 40mm and advance UKB S355 to be used, .: yield strength, fy = 355N/mm2 .: Required Plastic Modulus , Wply = MEd / (fy / γMO ) =825 x 106 / (355 / 1.0) =2324cm3 From Table of Properties, select 533 x 210 x 101 kg/m UKB provided 2610cm3 of Plastic Modulus Section Properties for 533 x 210 x 101 kg/m UKB Depth of Section, h = 536.7 mm Width of Section, b = 210 mm Thk. Flanges, tf = 17.4 mm Thk. of Web, tw = 10.8 mm Root Radius, r = 12.7 mm Depth between Fillet, d = 476.5 mm Ratio Local Buckling (flange), cf/tf = 4.99 mm Ratio of Local Buckling (web), cw/tw = 44.1 mm Second Moment Area, Iyy = 61,500 cm4 Second Moment Area, Izz = 2690 cm4 Radius of Gyration, iy = 21.9 cm Radius of Gyration, iz = 4.57 cm Elastic Modulus, Wely = 2290 cm3 Elastic Modulus, Welz = 256 cm3 Plastic Modulus, Wply = 2610 cm3 Plastic Modulus, Wplz =399 cm3 Area of Section, A =129 cm2

3. CLASIFICATION OF SECTION Take steel grade as S355 and tf = 17.4 mm < 40 mm .: fy = 355 N/mm2

533 x 210 x 101 kg/m UKB fy = 355 N/mm2


11 MAT

Table 5.2(1) Table 5.2(2) 6.2.6 6.2.6(1) 6.2.6(2) 6.2.6(3a)

Web ε = 0.81 and cw/tw = 44.1 < 72 ε .: Web section was Class 1 Flange ε = 0.81 and cf/tf = 4.99 < 9 ε .: Flange section was Class 1 .: Both section was Class 1 4.RESISTANCE OF CROSS- SECTION SUBJECT TO BENDING AND SHEAR Shear The design value of the shear force VEd at each cross section shall satisfy: VEd /Vc,Rd ≤ 1 (eq. 6.17) Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) Shear area, Av for rolled I section Av = A – 2btf + (tw + 2r) tf but not less than nhwtw Av = 12,900 – 2( 210)(17.4) + (10.8 + 2(12.7))17.4 = 6221.8 mm2 hw = h – 2tf hw = 536.7 – 2(17.4) = 501.9 mm n may be conservatively taken equal to 1.0. .: nhwtw = 1.0 x 501.9 x 10.8 = 5420.52 mm2 Since Av > nhwtw .: take Av = 6221.8 mm2 .: Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) = [6221.8 (355/√3)] / 1.0 = 1275.22 kN

Both section was Class 1 Av = 6221.8 mm2 Vc,Rd =1275.22 kN


12 MAT

6.2.8(1)(2)(3) 6.2.5 6.2.5(1) 6.2.5(2)

VEd /Vc,Rd ≤ 1 (eq. 6.17) 375 / 1275.22 = 0.29 Since VEd /Vc,Rd ≤ 1 .: Section was satisfactory for shear resistance Bending & Shear If VEd > 0.5 Vc,Rd .: moment resistance, Mc,Rd should be taken as the design resistance of the cross-section, calculated using a reduced yield strength, (1 – ρ) fy where ρ = ((2VEd/Vpl,Rd) - 1)2 VEd = 375kN and 0.5 Vc,Rd = 0.5 x 1275.22 = 637.61kN In this case, VEd < 0.5 Vc,Rd .: no reduction of yield strength, fy for moment resistance, Mc,Rd Bending The design value of the bending moment MEd at each cross-section shall satisfy: MEd /Mc,Rd ≤ 1 (eq. 6.12) For Class 1 section, Mc,Rd = Mpl,Rd = (Wpl x fy)/ γMO (eq. 6.13) .: Mc,Rd = (2.61 x 106 x 355) / 1.0 = 926.6kNm MEd /Mc,Rd =0.89 Since MEd /Mc,Rd < 1 .: section was satisfactory for moment resistance 5. RESISTANCE TO SHEAR BUCKLING In addition the shear buckling resistance for webs without intermediate stiffeners should be according to section 5 of EN

Mc,Rd = 926.6kNm


13 MAT

6.2.6(6) BS EN 1993-1-5:2006 8(1) 6.2(1)

1993-1-5, if: hw/tw > 72ε/n (eq. 6.22) hw/tw = 501.9 / 10.8 = 46.47 72ε/n = 72(0.81)/1.0 = 58.32 Since hw/tw < 72ε/n .: no further check on resistance for shear buckling 6. RESISTANCE TO FLANGE INDUCED BUCKLING To prevent the compression flange buckling in the plane of the web, the following criterion should be met: hw/tw ≤ k (E/fy)( √(Aw/Afc)) (eq.8.1) Aw = hw x tw

= 501.9 x 10.8 = 5,420.52 mm2 Af = b x tf

= 210 x 17.4 = 3,654 mm2 For Class 1, plastic rotation utilize, hence k = 0.3 k (E/fy)( √(Aw/Afc) = 0.3 (210x103/355) (√(5,420.52/3,654)) = 216.15 Since hw/tw ≤ k (E/fy)( √(Aw/Afc)) .: section is prevent the compression flange buckling in the plane of the web 7. RESISTANCE OF WEB TO TRANSVERSE FORCE (WEB BUCKLING) For un-stiffened or stiffened webs the design resistance to local buckling under transverse forces should be taken as:


14 MAT

6.4(1) Figure 6.1 (c) 6.5(2) 6.5(1)

FRD = (fyw x Leff x tw)/γm (eq. 6.1) Effective length for resistance to transverse forces, Leff Leff = XF ly (eq. 6.2) Reduction factor, XF XF = 0.5 / λF ≤ 1.0 (eq. 6.3) λF = √ ( ly tw fyw / Fcr ) (eq. 6.4) Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) For webs without longitudinal stiffeners kF should be obtained from Figure 6.1. For webs with longitudinal stiffeners kF may calculated from the equation 6.6 kF = 2 + 6 [(ss + c)/ hw] For type (a) and (b) in Figure 6.1, effective loaded length, ly should be taken as : ly = ss + 2tf ( 1 + √ m1+m2 ) (eq. 6.11) but ly ≤ distance between adjacent transverse stiffeners. m1 = fyf bf / fyw tw (eq. 6.8) m2 = 0.02 ( hw/tf)2 if λF > 0.5 (eq. 6.9) m2 = 0 if λF ≤ 0.5 (eq. 6.9) At D, two UB intersect which 533×210×101kg/m upper load carrying beam ABCDE and 610×229×140kg/m lower support beam at D. From Table of Properties, UB section for 610×229×140kg/m


15 MAT

Depth of Section, h 617.2 mm Width of Section, b 230.2 mm Thk. Flanges, tf 22.1 mm Thk. of Web, tw 13.1 mm Root Radius, r 12.7 mm

hw = (h−2tf) = 617.2 – (2×22.1) = 573 mm ss = 2tf + tw + (2 − √2)rb = (2 × 22.1) + 13.1 + (2 − √2)12.7 = 64.74 < hw = 573 mm kF = 6 + 2 (hw/α)2 (Figure 6.1 (a)). Assume α is large value = 6 m1 = fyf bf / fyw tw (eq. 6.8) =(355 x 210) / (355 x 10.8) = 19.4 m2 = 0.02 ( hw/tf)2 = 0.02 (501.9/17.4) 2 = 16.64 ly = ss + 2tf ( 1 + √ m1+m2 ) = 64.74 + 2(17.4) (1+√19.4 + 17.4) = 310.65 mm Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) = 0.9 (6) (210x103) ( 10.83 / 501.9) = 2,846.2 x 103 N

kF = 6 m1 = 22.87 m2 = 16.64 ly = 310.65 mm Fcr = 2,846.2x103 N


16 MAT

λF = √ ( ly tw fyw / Fcr ) (eq. 6.4) = √ ((310.65 x 10.8 x 355) / 2,846.2 x 103) = 0.65 XF = 0.5 / λF ≤ 1.0 (eq. 6.3) = 0.5 / 0.65 = 0.77 ≤ 1.0 .: take XF = 0.77 Leff = XF ly (eq. 6.2) = 0.77 x 310.65 = 239.2 mm FRD = (fyw x Leff x tw)/γm (eq. 6.1) = (355 x 239.2 x 10.8)/1.0 = 917.1 kN Since FRD > VEd at D .: section is adequate for resistance of web to transverse force Additional notes on resistance of web to transverse force (web buckling) BS EN 1993-1-5distinguishes between two types of forces applied through a flange to the web:

λF = 0.65 XF = 0.77 Leff = 239.2 mm FRD = 917.1 kN



(a) Forces (b) Forces for loading

Figure 6.1:

For loading(i) crushingyielding ofas web cru(ii) Localizeflange, thecrippling.

For loading(i) web cru(ii) Buckling

& TIMBER DESIGN

resisted by transferredg type (b)

Buckling Co

g types (a) g of the wef the flangeushing ed bucklinge combined

g type (b) tushing g of the we

N - ECS328

17

y shear in thd through th

oefficient forEN 1

and (c) theb close to t

e, the comb

g and crushd effect som

the web is l

eb over mo

7

he web for lhe web dire

r different typ1993-1-5)

e web is likethe flange bined effec

hing of the wmetimes re

ikely to fail

st of the de

oading typectly to the

pes of load a

ely to fail asaccompan

ct sometime

web beneaferred to as

as a result

epth of the

pes (a) ande other flang

application

s a result ofnied by es referred t

ath the s web

of

member

MAT

(c) ge

(BS

f

to



Clause 6.3be taken adistributedshould not

Simply sup The beamlength an

& TIMBER DESIGN

3 (1) mentioas the dista

d at slope ot be taken a

pported late

shown in fnd has bea

N - ECS328

18

oned that leance over wof 1:1 (BS ENas larger th

erally restra

figure belowaring lengt

8

ength of stifwhich applieN 1993-1-5, F

an hw

ained beam

w is fully latths of 50

ffness bearied load is eFigure 6.2).

m

terally restramm at the

ing ss shouldeffectively However ss

ained alonge un-stiffen

MAT

d

s

g its ned



supports resistance loading of= 30kN/m

1. DESIGN Combinat Combinat

DESIGN BE (FEd1 x L2) / (FEd2 x L) / 4 ∑MEd = 33 = 53 DESIGN SH (FEd1 x L) / 2 FEd2 / 2 = = ∑FEd = 2672. SELECTIO Section Pro

& TIMBER DESIGN

and 75 m of 533 x f uniformly dand conce

LOADING /

tion of Actio

tion of Actio

ENDING MO

8 = (62.5 kN = 330.1 k

4 = (129 kN = 209.6 k

30.1 kNm + 239.7 kNm

HEAR FORC

2 = (62.5 k = 203.13

129 kN / 2 64.5 kN

.63 kN ON OF SECT

operties for

N - ECS328

19

mm under 210 x 92 Udistributed entrated loa

/ ACTION

on UDL, FEd

on PL, FEd

OMENT, MEd

N/m x 6.5mNm

N x 6.5m) / kNm

209.6 kNm

E, VEd

kN/m x 6.5m3 kN

TION

r 533 x 210 x

9

the concUKB using Sload (UDL) ad Gk = 40k

= γG Gk + = (1.35 x 15 = 62.5 kN/m

= γG Gk + = (1.35 x 40 = 129 kN

m x 6.5m) / 8

4

m) / 2

x 92kg/m U

centrated S275 steel was gk = 1

kN and Qk =

γQ Qk (eq 5kN/m) + (1m

γQ Qk (eq 0kN) + (1.5 x

8

KB

load. Chegrade for 5kN/m and

= 50kN

6.10)

1.5 x 30kN/m

6.10)

x 50N)

MAT

eck the

d qk

m)


20 MAT

Depth of Section, h = 533.1 mm Width of Section, b = 209.3 mm Thk. Flanges, tf = 15.6 mm Thk. of Web, tw = 10.1 mm Root Radius, r = 12.7 mm Depth between Fillet, d = 476.5 mm Ratio Local Buckling (flange), cf/tf = 5.57 mm Ratio of Local Buckling (web), cw/tw = 47.2mm Second Moment Area, Iyy = 55,200 cm4 Second Moment Area, Izz = 2390 cm4 Radius of Gyration, iy = 21.7 cm Radius of Gyration, iz = 4.51 cm Elastic Modulus, Wely = 2070 cm3 Elastic Modulus, Welz = 228 cm3 Plastic Modulus, Wply = 2360 cm3 Plastic Modulus, Wplz =355 cm3 Area of Section, A =117 cm2

3. CLASIFICATION OF SECTION Take steel grade as S275 and tf = 15.6 mm < 40 mm .: fy = 275 N/mm2 Web ε = 0.92 and cw/tw = 47.2 < 72 ε .: Web section was Class 1 Flange ε = 0.92 and cf/tf = 5.57 < 9 ε .: Flange section was Class 1 .: Both section was Class 1 4.RESISTANCE OF CROSS- SECTION SUBJECT TO BENDING AND SHEAR Shear


21 MAT

The design value of the shear force VEd at each cross section shall satisfy: VEd /Vc,Rd ≤ 1 (eq. 6.17) Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) Shear area, Av for rolled I section Av = A – 2btf + (tw + 2r) tf but not less than nhwtw Av = 11,700 – 2(209.3)(15.6) + (10.1 + 2(12.7))15.6 = 5723.64 mm2 hw = h – 2tf hw = 533.1 – 2(15.6) = 501.9 mm n may be conservatively taken equal to 1.0. .: nhwtw = 1.0 x 501.9 x 10.1 = 5069.2 mm2 Since Av > nhwtw .: take Av = 5723.64 mm2 .: Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) = [ 5723.64 (275/√3)] / 1.0 = 908.75 kN VEd /Vc,Rd ≤ 1 (eq. 6.17) 267.63 / 908.75= 0.29 Since VEd /Vc,Rd ≤ 1 .: Section was satisfactory for shear resistance Bending & Shear If VEd > 0.5 Vc,Rd .: moment resistance, Mc,Rd should be taken as the design resistance of the cross-section, calculated using a reduced


22 MAT

yield strength, (1 – ρ) fy where ρ = ((2VEd/Vpl,Rd) - 1)2 In this case, VEd < 0.5 Vc,Rd .: no reduction of moment resistance, Mc,Rd Bending The design value of the bending moment MEd at each cross-section shall satisfy: MEd /Mc,Rd ≤ 1 (eq. 6.12) For Class 1 section, Mc,Rd = Mpl,Rd = (Wpl x fy)/ γMO (eq. 6.13) .: Mc,Rd = (2360x103 x 275)/ 1.0 = 649kNm MEd /Mc,Rd = 539.7 kNm/649 kNm = 0.83 Since MEd /Mc,Rd < 1 .: section was satisfactory for moment resistance 5. RESISTANCE TO SHEAR BUCKLING In addition the shear buckling resistance for webs without intermediate stiffeners should be according to section 5 of EN 1993-1-5, if: hw/tw > 72ε/n (eq. 6.22) hw/tw = 501.9/10.1 = 49.7 72ε/n = 72(0.92)/1.0 = 66.24 Since hw/tw < 72ε/n .: no further check on resistance for shear buckling 6. RESISTANCE TO FLANGE INDUCED BUCKLING To prevent the compression flange buckling in the plane of the web, the following criterion should be met:


23 MAT

hw/tw ≤ k (E/fy)( √(Aw/Afc)) (eq.8.1) Aw = hw x tw

= 501.9x10.1 = 5069.2 mm2 Af = b x tf

= 209.3 x15.6 = 3265.08 mm2 For Class 1, plastic rotation utilize, hence k = 0.3 k (E/fy)( √(Aw/Afc) = 0.3 (210x103/275) (√(5069.2/3265.08)) = 285.45 Since hw/tw ≤ k (E/fy)( √(Aw/Afc)) .: section is prevent the compression flange buckling in the plane of the web 7. RESISTANCE OF WEB TO TRANSVERSE FORCE (WEB BUCKLING) For un-stiffened or stiffened webs the design resistance to local buckling under transverse forces should


24 MAT

be taken as: FRD = (fyw x Leff x tw)/γm (eq. 6.1) Effective length for resistance to transverse forces, Leff Leff = XF ly (eq. 6.2) Reduction factor, XF XF = 0.5 / λF ≤ 1.0 (eq. 6.3) λF = √ ( ly tw fyw / Fcr ) (eq. 6.4) Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) For webs without longitudinal stiffeners kF should be obtained from Figure 6.1. For webs with longitudinal stiffeners kF may calculated from the equation 6.6 kF = 2 + 6 [(ss + c)/ hw] For type (c) in Figure 6.1, effective loaded length, ly should be taken as the smallest value obtained from the equation 6.11 or equation 6.12 : ly = le + tf √ [(m1/2) + (le/tf)2 + m2] (eq. 6.11) ly = le + tf √m1 + m2 (eq. 6.12) le = kF E tw2/2fyw hw ≤ ss + c (eq.6.13) m1 = fyf bf / fyw tw (eq. 6.8) m2 = 0.02 ( hw/tf)2 if λF > 0.5 (eq. 6.9) m2 = 0 if λF ≤ 0.5 (eq. 6.9) kF = 2 + 6 [(ss + c)/ hw] (Figure 6.1 (c)) = 2 + 6 [(50 + 0)/ 501.9] = 2.6 le = kF E tw2/2 fyw hw ≤ ss + c (eq.6.13)


25 MAT

le = kF E tw2/2 fyw hw = (2.6 x 210 x103 x 10.12)/(2 x 275 x 501.9) = 201.77 mm ss + c = 50 + 0 = 50mm Since le = kF E tw2/2 fyw hw ≥ ss + c .: take le = 50mm m1 = fyf bf / fyw tw (eq. 6.8) =(275 x 209.3) / (275 x 10.1) = 20.72 m2 = 0.02 ( hw/tf)2 = 0.02 (501.9/15.6) 2 = 20.7 ly = le + tf √ [(m1/2) + (le/tf)2 + m2] (eq. 6.11) = 50 + 15.6 √ [(20.72/2) + (50/15.6)2 + 20.7] = 150.28 mm ly = le + tf √m1 + m2 (eq. 6.12) = 50 + 15.6 √20.72 + 20.7 = 150.4 mm Choose the lesser value from eq. 6.11 and eq. 6.12 .: ly = 150.29 Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) = 0.9 (2.6) (210x103) (10.13 / 501.9) = 1008.75 x 103 N λF = √ ( ly tw fyw / Fcr ) (eq. 6.4)


26 MAT

= √ ((150.29 x 10.1 x 275) / 1008.75 x 103) = 0.64 XF = 0.5 / λF ≤ 1.0 (eq. 6.3) = 0.5 / 0.64 = 0.78 ≤ 1.0 .: take XF = 0.78 Leff = XF ly (eq. 6.2) = 0.78 x 150.29 = 117.23 mm FRD = (fyw x Leff x tw)/γm (eq. 6.1) = (275 x 117.23 x 10.1)/1.0 = 325.61 kN VEd / FRD = 267.63 / 325.61 = 0.82 Since FRD > VEd .: section is adequate for resistance of web to transverse force

1. Steel Beam Design

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