Steel Design to Eurocode 3 University of Sheffield Structural Engineering Masters
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Transcript of Steel Design to Eurocode 3 University of Sheffield Structural Engineering Masters
1
Introduction
For this coursework, two steel beams; one secondary, one primary and a column has been designed according to Eurocode 3.
The figure shows the first floor of the structure, which in total consists of three. The secondary beam designed is the one in the very centre of the structure, carrying load acting on the slab from three meters from each side, transferring it to the primary beam it is connected to as a point load, which transfers the load to the column, which finally transfers it with the moment acting on it due to eccentricity of the structure to the foundation of the structure.
While designing the column, attention paid to the size of the primary and secondary beams, which are connected to the column. Thickness of web and flange and the height and width of the sections were taken in account, so the connections and design was possible and logical.
2
Design of Secondary Beam A-B to Eurocode 3
Ultimate Load
Ultimate Load is given by; 1.35 q + 1.50 q
gk = (2.5 + 0.2 + 0.8 + 1) = 4.5 kN/m2
qk = 3.5 kN/m2
1.35 gk + 1.50 qk = 11.325 kN/m2
Assume S355 Steel is used
Loading on the Beam
There are secondary beams at each 3 meters, so each single beam will get 3 meters of the Ultimate Load from the slab,
Loading on the Beam = 11.325 × 3 = 33.975 kN/m
Maximum Moment is given by q × l2 / 8 for a simply supported beam with a uniform loading,
MEd = 33.975 × 92 / 8 = 344 kN m
VEd = 33.975 × 9 / 2 = 152.8875 kN
Assume the chosen beam to be Class 1
MC,Rd = Mpl,RD = Wpl × fy / γMO
Wpl = Mpl,Rd × γMO / fy
Wpl > MEd × γMO / fy
Plastic Modulus should be greater than MEd / fy × γMO
Wpl > MEd × γMO / fy
Wpl > 344 × 106 × 1 / 355 = 969.01 cm3
Choose Section : 457 × 152 × 74UB
Section Properties
h = 462 mm b = 154.4 mm tw = 9.6 mm tf = 17 mm r = 10.2 mm Wpl = 1627 cm3
I = 32670 cm4 A = 94.5 cm2
3
Class Determination of the Section
For Flanges
c = ( b – tw – 2 × r ) / 2
c = ( 154.4 – 9.6 – 2 × 10.2 ) / 2 = 62.20 mm
c / tf = 3.6588
For Web
c = ( h – 2 × tf – 2 × r ) / 2
c = ( 462 – 2 × 17 – 2 × 10.2 ) = 407.6 mm
c / tw = 42.4583
Flange thickness tf = 17 mm < 40 mm
hence from table 4.2 fy = 355 N/mm2
ε = ( 235 / fy ) 0.5 = 0.8136
From table 4.4
c / tf = 3.6588 < 9 × ε = 7.3224
c / tw = 42.4583 < 72 × ε = 58.57
The section is Class 1 As Assumed
Resisting Of Cross Section
Bending Resistance
MEd = 344 kN m
Mpl,Rd = Wpl × fy / γMO [ Plastic Moment Of Resistance ]
Mpl,Rd = 1627 × 103 × 355 / 1 = 577.58 kN m
Mpl,Rd > MEd
Shear Resistance
VEd = 152.8875 kN
Vpl,Rd = Av × ( fy × 30.5 ) / γMO [ Design Plastic Shear Resistance ]
Av = A – 2 × b × tf + ( tw + 2 × r ) × tf
Av = 9450 – 2 × 154.4 × 17 + ( 9.6 + 2 × 10.2 ) × 17 = 4710.40 mm2
Av > n × hw × tw = 1 × = ( 462 – 2 × 17 ) × 9.6 = 4108.80 mm2
Av = 4710.40 mm2
Vpl,Rd = 4710.40 × ( 355 / 30.5 ) / 1
Vpl,Rd = 965.4405 kN > 152.8875 kN
4
Bending and Shear
Check : VED < 0.5 × Vpl,Rd
152.8875 < 0.5 × 965.4405
No Check For Bending & Shear is Required,
Shear Buckling Resistance
hw / tw = 428 / 9.6 = 44.58 < 72 × ε / η = 72 × 0.8136 / 1 = 58.57
No check on shear buckling required,
Deflection Check
Maximum Moment Due To Working Load Mmax = ( gk + qk ) × l2 / 8
Mmax = ( 4.5 × 3 + 3.5 × 3 ) × 92 / 8 = 243 kN m
Elastic Resistance Of The Section is Mc,RD(el) = Wel × fy / γMO
Mc,Rd(el) = 1414 × 103 × 355 / 1= 501.97 kN m
Mc,Rd(el) > Mmax
Deflection Can Be Calculated elastically,
Deflection due to permanent load :
w = 4.5 × 3 + 3.5 × 3 = 24 N / mm
wtot = 5 / 384 × w × l4 / E / I
wtot = 5 / 384 × 24 × ( 9×103 )4 / 210 / 103 / 32670 / 104
wtot = 29.8849 mm
Span / 300 = 30 mm
wtot < span / 300
Deflection Check is OK, The Chosen Section 457 × 152 × 74UB is suitable for the secondary beams,
5
Design of Primary Beam C - D to Eurocode 3
Loading On the Beam
The loading on the beam will be a point load in the middle of the because of the secondary beams connected to it,
Point Load = 305.7750 kN [ Beam Loading ]
VEd = 305.7750 / 2 = 152.8875 kN
MEd = 152.8875 × 6 / 2 = 458.6625 kN m
Assume S355 Steel is used
Assume Section : 533 × 210 × 101
Section Properties
h = 536.7 mm b = 210.00 mm tw = 10.8 mm tf = 17.4 mm r = 12.7 mm Wpl = 2612 cm3
I = 61520 cm4 A = 129 cm2
Class Determination of the Section
For Flanges
c = ( b – tw – 2 × r ) / 2
c = ( 210.00 – 10.8 – 2 × 12.7 ) / 2 = 86.90 mm
c / tf = 4.99
For Web
c = ( h – 2 × tf – 2 × r ) / 2
c = ( 536.7 – 2 × 17.4 – 2 × 12.7 ) = 476.50 mm
c / tw = 44.12
Flange thickness tf = 17.4 mm < 40 mm
hence from table 4.2 fy = 355 N/mm2
ε = ( 235 / fy ) 0.5 = 0.8136
From table 4.4
c / tf = 4.99 < 9 × ε = 7.3224
c / tw = 44.12 < 72 × ε = 58.57
The section is Class 1 As Assumed
6
Calculation of Mcr
L = 6000 mm, C1 = 1.365 for the loading case
Mcr = C1 × π2 × E × Iz / L2cr × [ (Iw / Iz) + ( L2
cr × G × IT / π2 / E / Iz) ]
Mcr = 1.365 × π 2 × 210000 × 2692 × 104 / 60002 × [ 1810 × 109
/ 2692 / 104 + (60002 × 80769 × 1010 × 103 / π2
/ 210000 / 2692 / 104) ]0.5 = 731.8769 kN m
λLT = ( Wy × fy / Mcr )0.5
λLT = ( 2612 × 103 × 355 / 731.8769 × 106 )
0.5 = 1.13
Buckling Factor χLT
Imperfection factor αLT = 0.34
ΦLT = 0.5 × [ 1 + αLT × ( λLT – 0.2 ) + λLT 2 ], χLT = 1 / [ 1.29 + (1.292 – 1.132 ) 0.5 ]
ΦLT = 0.5 × [1 + 0.34 × ( 1.13 – 0.2 ) + 1.132 ]
ΦLT = 1.29
χLT = 1 / * ΦLT + ( ΦLT2 – λLT
2 ) 0.5 ]
χLT = 0.52
Buckling Resistance Of Beam
MB,Rd = χLT × WPl, y × fy / γM1
MB,Rd = 0.52 × 2612 × 103 × 355 / 1
MB,Rd = 482.1752 kN m
MEd = 458.6625 kN m
MB,Rd > MEd
Buckling resistance of the beam is greater than the Design Moment.
Resisting Of Cross Section
Bending Resistance
MEd = 458.6625 kN m
Mpl,Rd = Wpl × fy / γMO [ Plastic Moment Of Resistance ]
Mpl,RD = 2612 × 103 × 355 / 1 = 927.26 kN m
Mpl,Rd > MEd
7
Shear Resistance
VEd = 152.8875 kN
Vpl,Rd = Av × ( fy × 30.5 ) / γMO [ Design Plastic Shear Resistance ]
Av = A – 2 × b × tf + ( tw + 2 × r ) × tf
Av = 12900 – 2 × 210.00 × 17.4 + ( 10.8 + 2 × 12.7 ) × 17.4 = 6221.88 mm2
Av > n × hw × tw = 1 × ( 536.7 – 2 × 17.4 ) × 10.8 = 5420.52 mm2
Av = 6221.88 mm2
Vpl,Rd = 6221.88 × ( 355 / 30.5 ) / 1
Vpl,Rd = 1275.20 kN > 152.8875 kN
Bending and Shear
Check : VED < 0.5 × Vpl,Rd
152.8875 < 0.5 × 1275.20
No Check For Bending & Shear is Required,
Shear Buckling Resistance
hw / tw = 501.9 / 10.8 = 46.47 < 72 × ε / η = 72 × 0.8136 / 1 = 58.57
No check on shear buckling required,
Deflection Check
Maximum Moment Due To Working Load Mmax = ( gk + qk ) × 3 × 9 / 2 × 6 / 2 × 3
Mmax = ( 4.5 + 3.5 ) × 3 × 9 / 2 × 2 / 2 × 3
Mmax = 324 kN m
Elastic Resistance Of The Section is Mc,RD(el) = Wel × fy / γMO
Mc,Rd(el) = 2292 × 103 × 355 / 1= 813.66 kN m
Mc,Rd(el) > Mmax
Deflection Can Be Calculated elastically,
Deflection due to point load :
w = ( gk + qk ) × 3 × 9 / 2 × 2
wtot = w × l3 / E / I / 48
wtot = 216 × 103 × ( 6×103 )3 / 210 / 103 / 61520 / 104 / 48 = 7.52 mm
Span / 300 = 30 mm
wtot < span / 300
Deflection Check is OK, The Chosen Section 533 × 210 × 101UB is suitable for the primary beams.
8
Column Design to Eurocode 3
Assume Section : 254 × 254 × 107UC
Section Properties:
Depth Of Section(h) = 266.7 mm, Width Of Section(b) = 258.8 mm, Thickness of Web(s) = 12.8 mm, Thickness of Flange(t) = 20.5 mm, Root of Radius(r) = 12.7 mm, Second Moment Of Inertia(x-x axis) = 17510 cm4, Second Moment Of Inertia(y-y axis) = 5928 cm4, Radius Of Gyration(x-x axis) = 11.3 cm, Radius Of Gyration(y-y axis) = 6.59 cm, Elastic Modulus(x-x axis) = 1313 cm3, Elastic Modulus(y-y axis) = 458 cm3, Plastic Modulus(x-x axis) = 1484 cm3, Plastic Modulus (y-y axis) = 697 cm3, Area Of Section = 136 cm2
Cross Section Classification
For Flanges
c = (b – tw – 2r) / 2 = (258.8 – 12.8 – 2 × 12.7) / 2 = 110.30
c / tf = 110.30 / 20.50 = 5.38
For Web
c = (h – 2 × tf – 2r) = (266.7 – 2 × 20.5 – 2 × 12.7) = 220.80
c / tw = 220.80 / 12.8 = 17.25
Assume S355 Steel is used
Flange thickness tf = 20.50 mm < 40 mm
hence from table 4.2 fy = 355 N/mm2
ε = (235/fy)0.5 = 0.8136
c / tf = 5.38 < 9 × ε = 7.3224
c / tw = 17.25 < 72 × ε = 58.57
The section is Class 1 As Assumed
Load on Column:
NEd = ( 11.325 × 6 × 6 × 3 ) = 1834.65 kN
Compression Resistance Of Cross Section:
Nc,Rd = Design resistance to normal forces of the cross-section for uniform compression = A × fy / ɣMO
Nc,Rd = 13600 × 355 / 1 = 4828 kN > NEd = 1834.65 kN
9
Bending Resistance Of Cross Section:
Major (y – y)
My,Ed = Maximum Bending Moment = 100 kN m
Mc,y,Rd = Design Bending Resistance Of The Cross Section = Wpl,y × fy / ɣMO
Mc,y,Rd = 1484 × 103 × 355 / 1 = 526.82 kN m > MEd = 100 kN m
Shear Resistance Of Cross Section:
VEd = Maximum Shear Force = (100 – 50) / 4 = 12.5 kN
Vpl,Rd = Av × (fy / 30.5) / ɣMO
Av = A – 2 × b × tf + (tw + 2 × r) × tf
Av = 252 × 102 – 2 × 314.5 × 31.4 + (19.1 + 2 × 15.2) × 31.4
Av = 136 × 102 – 2 × 258.8 × 20.50 + ( 12.8 + 2 × 12.7 ) × 20.50
Av = 3772.30 mm2
Av > η × hw × tw = 1 × ( 266.7 – 2 × 20.50 ) × 12.8 = 2888.96 mm2
Av 3772.30 mm2
Vpl,Rd = 3772.30 × (355 / 30.5) / 1
Vpl,Rd = 773.16 kN > 12.5 kN
Cross Section Under Bending And Axial Force:
No reduction in the major axis plastic moment resistance is necessary provide both of the following criteria are met:
NEd < 0.25 × Npl,Rd
NEd < (0.5 × hw × tw × fy) / ɣMO
NEd < 0.25 × Npl,Rd
1834.64 < 0.25 × 4828 = 1207 [ Not Satisfied ]
NEd < (0.5 × hw × tw × fy) / ɣMO
1944 < [ 0.5 × ( 266.7 – 2 × 20.50 ) × 12.8 × 355 ) / 1 ] = 512.790 kN [ Not Satisfied ]
Allowance for the effect of axial force on the major axis plastic moment resistance of the cross section must be made.
The moment on the minor axis is zero, so it does not matter if there is any reduction needed or not in the minor axis. That step will be skipped.
10
Reduced Plastic Moment Resistances:
Major ( y – y axis )
MN,y,Rd = MPl,y,Rd × ( 1 – n ) / ( 1 – 0.5 × aw) but MN,y,Rd < MPl,y,Rd
n = NEd / Npl,Rd = 1834.65 / 4828 = 0.38
a = ( A – 2 × b × tf ) / A
a = ( 13600 - 2 × 258.8 × 20.50 ) / 13600
a = 0.219 < 0.5
MN,y,Rd = MPl,y,Rd × ( 1 – n ) / ( 1 – 0.5 × a)
MN,y,Rd = 526.80 × ( 1 – 0.38 ) / ( 1 – 0.5 × 0.219)
MN,y,Rd = 366.77 kN m > MEd = 100 kN m
Member Buckling Resistance in Compression:
Lcr,y = Lcr,z = 1.0 × L = 4000 mm
λ1 = π × * E / fy ]0.5 = 93.9 × ε = 93.9 × 0.8136 = 76.40
λy = Lcr / iy / λ1 = 0.4633
λz = Lcr / iz / λ1 = 0.7944
Selection of buckling curve and imperfection factor α : for hot-rolled H section (h/b < 1.2, tf < 100 mm and S355 steel):
For buckling about the y-y axis, use curve b
For buckling about the z-z axis, use curve c
For curve b α= 0.34 and curve c α= 0.49
Buckling Curve : major ( y – y ) axis
Φy = 0.5 × *1 + α × ( λy – 0.2 ) + λy 2 ]
Φy = 0.5 × [1 + 0.34 × ( 0.4633 – 0.2 ) + 0.4633 2 ]
Φy = 0.6520
χy = 1 / * Φy + (Φy2 – λy
2 ) 0.5 ]
χy = 1 / [ 0.6520 + (0.65202 – 0.46332 ) 0.5 ]
χy = 0.90 < 1.0
NB,y,Rd = χy × A × fy / ɣM1
NB,y,Rd = 0.9000 × 13600 × 355 / 1
NB,y,Rd = 4345.20 > 1834.65 kN
Major Axis Flexural Buckling Is Acceptable
11
Buckling Curve : minor ( z – z ) axis
Φz = 0.5 × *1 + α × ( λz – 0.2 ) + λz 2 ]
Φz = 0.5 × [1 + 0.49 × ( 0.7944 – 0.2 ) + 0.79442 ]
Φy = 0.9611
χz = 1 / * Φz + (Φz2 – λz
2 ) 0.5 ]
χz = 1 / [ 0.9611 + (0.96112 – 0.79442 ) 0.5 ]
χz = 0.67 < 1.0
NB,z,Rd = χz × A × fy / ɣM1
NB,z,Rd = 0.67 × 13600 × 355 / 1
NB,z,Rd = 3234.76 > 1834.65 kN
Minor Axis Flexural Buckling Is Acceptable
Member Buckling Resistance In Bending
Calculation of Mcr
Length of the column between points which are laterally restrained = 4000 mm,
Lcr = 4000 mm
C1 = 1.563
Mcr = C1 × π2 × E × Iz / L2cr × [ (Iw / Iz) + ( L2
cr × G × IT / π2 / E / Iz) ]
Mcr = 1.323 × π 2 × 210000 × 5928 × 104 / 40002 × [ 898 × 109
/ 16300 / 104 + (40002 × 80769 × 1720 × 103 / π2
/ 210000 / 5928 / 104) ]0.5
Mcr = 1559.74 kN m
λLT = ( Wy × fy / Mcr )0.5 = ( 1484 × 103 × 355 / 1559.74 / 106
)0.5 = 0.58
Buckling Factor χLT
Imperfection factor αLT = 0.21
ΦLT = 0.5 × *1 + αLT × ( λLT – 0.2 ) + λLT 2 ]
ΦLT = 0.5 × [1 + 0.21 × ( 0.58 – 0.2 ) + 0.58 2 ]
ΦLT = 0.7081
χLT = 1 / * ΦLT + (ΦLT2 – λLT
2 ) 0.5 ]
χLT = 1 / [ 0.7081 + (0.70812 – 0.582 ) 0.5 ]
χLT = 0.8974
12
Lateral Torsional Buckling Resistance
MB,Rd = χLT × WPl, y × fy / γM1
MB,Rd = 0.8974 × 1484 × 103 × 355 / 1
MB,Rd = 472.76 kN m
MEd = 100 kN m
MB,Rd > MED
Member Buckling Resistance In Combined Bending And Axial Compression
Member subjected to combine bending and axial compression must satisfy both of following equations :
NEd / ( χy × NRk / γM1 ) + kyy × My,Ed / ( χLT × My,Rk / γM1 ) + kyz × Mz,Ed / ( Mz,Rk / γM1 ) ≤ 1
NEd / ( χz × NRk / γM1 ) + kzy × My,Ed / ( χLT × My,Rk / γM1 ) + kzz × Mz,Ed / ( Mz,Rk / γM1 ) ≤ 1
My,Ed = 100 kN m, Mz,Ed = 0, above equations become :
NEd / ( χy × NRk / γM1 ) + kyy × My,Ed / ( χLT × My,Rk / γM1 ) ≤ 1
NEd / ( χz × NRk / γM1 ) + kzy × My,Ed / ( χLT × My,Rk / γM1 ) ≤ 1
Determination of Interaction Factors kij
The equivalent uniform moment factors Cmy and CmLT
Cmy = 0.6 + 0.4 × Ψ = 0.6 + 0.4 × 0.50 = 0.8 > 0.4
CmLT = 0.6 + 0.4 × Ψ = 0.6 + 0.4 × 0.50 = 0.8 > 0.4
λy = Lcr / iy / λ1 = 0.4633
λz = Lcr / iz / λ1 = 0.7944
For Class 1 and 2 I Sections :
kyy = Cmy × * 1 + ( λy – 0.2 ) × NEd / χy / ( NRk / γM1 ) ] < Cmy × [ 1 + 0.8 × NEd / χy / ( NRk / γM1 ) ]
kyy = 0.8 × [ 1 + ( 0.4633 – 0.2 ) × 1834.65 / 0.90 / ( 4345.20 / 1 ) ] = 0.8988 < 0.8 × [ 1 + 0.8 × 1834.65 / 0.90 / ( 4345.20 / 1 ) ] = 1.10
kyy = 0.8988
kzy = 1 – ( 0.1 × λz ) / ( Cm,LT – 0.25 ) × ( NEd ) / * χz × ( NRk / γM1 ) ]
kzy = 1 – ( 0.1 × 0.7944 ) / ( 0.8 – 0.25 ) × ( 1834.65 ) / [ 0.67 × ( 4345 / 1 ) ]
kzy = 0.9089
13
Check Compliance with Interaction Formulae:
NEd / ( χy × NRk / γM1 ) + kyy × My,Ed / ( χLT × My,Rk / γM1 ) ≤ 1
1846.65 / ( 0.90 × 4345.20 / 1 ) + 0.8988 × 100 / ( 0.8974 × 366.77 / 1 ) ≤ 1
0.7452 ≤ 1 [ Satisfied ]
NEd / ( χz × NRk / γM1 ) + kzy × My,Ed / ( χLT × My,Rk / γM1 ) ≤ 1
1846.65 / ( 0.67 × 8946 / 1 ) + 0.9089 × 100 / ( 0.8974 × 366.77 / 1 ) ≤ 1
0.5842 ≤ 1 [ Satisfied ]
Therefore a hot – rolled 254 × 254 × 107UC is acceptable for the column
Secondary Beams: Light Gray
Primary Beams: Darker Gray
Column: Black