12/28/21, 9:14 AM MONORAIL/RUNWAY BEAM DESIGN

ECMONORAIL
Monorail/Runway Beam Design Job : Untitled
Licensed to: mary jvnd Powered By:
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1.0   INPUT

1.1   Section
Type Type = Standard
Section Section= IPE 330
Overall Height H = 330 mm
Width of Flange B = 160 mm
Thickness of Flange T = 11.5 mm
Thickness of Web t = 7.5 mm
Root Radius r = 18 mm

1.2   Support
Simply Supported Span Ls = 2000 mm
Cantilever Span Lc = 1500 mm
Stopper Distance from Cantilever Edge Ds = 150 mm
Effective Restraint Factor Ks = 1
- Simply Supported
Effective Restraint Factor - Cantilever Kc = 2

1.3   Monorail
Monorail Capacity PL = 30 kN
Hoist Pulley Weight Sh = 2 kN
Dynamic Factor for Vertical Loads Ψv = 10 %
Dynamic Factor for Horizontal Loads Ψh = 0%

1.3.1   Wheel Arrangement

No. of Wheels N = 4
Distance between Wheels along Beam Wd = 158 mm
Distance between Wheels Wg = 100 mm

1.4   Material Properties

Steel Grade fy = 275 N/mm2
Elastic Modulus of Steel E = 200000 N/mm2
Shear Modulus of Steel G = 81000 N/mm2
Def. / Span Ratio - Simply Supported Δlims = 500
Def. / Span Ratio - Cantilever Δlimc = 300

1.5   Partial Safety Factors

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Concentrated Load γQ = 1.5

Distributed Load γG = 1.35
Bottom Flange Resistance γM0 = 1
Buckling Resistance γM1 = 1
Stress Factor γMs = 1.1

2.0   OUTPUT

2.1   Section Properties

2.1.1   I Section [From Section Table]

Cross Sectional Area A = 62.6 cm2

Moment of Inertia - Major Ixx = 11800 cm4
Moment of Inertia - Minor Iyy = 788 cm4
Radius of Gyration - Major Rxx = 13.7 cm
Radius of Gyration - Minor Ryy = 3.6 cm
Elastic Section Modulus - Major Zxx = 713 cm3
Elastic Section Modulus - Minor Zyy = 99 cm3
Plastic Section Modulus - Major Sxx = 804 cm3
Torsional Constant J = 28.1 cm4
Warping Constant Iw = 0.199 dm6

2.2   Section Classification

2.2.1   Outstand Element of Compression Flange EN 1993 - 1 - 1

Parameter e = √(235 / fy) = 0.92

Outstand c = (B - t - 2 * r) / 2 = 58.2 mm
Ratio c't = c / T = 5.1
Class 1 - Plastic

2.2.2   Web Element

Depth between Fillets C = H - 2 * (T + r) = 271 mm

Ratio C't = C / t = 36.1
Class 1 - Plastic

2.3   Calculation for Simply Supported Span

Simply Supported Span Lbs = 2000 mm

2.3.1   Design Moment

Concentrated Load Qv = (1 + Ψv / 100) * (PL + Sh) = 35.2 kN

Horizontal Load Qh = (Ψh / 100) * (PL + Sh) = 0 kN
Distributed Load - Vertical qv = A * 78.5 = 0.5 kN/m
Torsional Moment Tm = Qh * (H / 2 – T) = 0 kN-m

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Vertical Bending Moment Myed = (γQ * Qv * (Lbs/4)) + (γG * qv * (Lbs)2 / 8) = 26.7 kN-m
Horizontal Bending Moment Mzed1 = (γQ * Qh * (Lbs/4)) = 0 kN-m

2.3.2   Capacity of Bottom Flange Resistance EN 1993 - 6

Stress due to Vertical Moments fbv = Myed / Zxx = 37.492 N/mm2

Stress due to Horizontal Moments fbh = Mzed1 / Zyy = 0 N/mm2
Lever Arm m = 0.5 * (B - t) - 0.8 * r – (B / 2 – Wg / 2) Cl 6.7 Eq (6.3)
= 31.8 mm
Effective Length leff = 2√2 * (m + (B / 2 – Wg / 2)) + 0.5 * Wd Table 6.2
= 253.9 mm
Design Resistance of Bottom Flange Ffrd = leff * T2 * fy / γM0 * (1 - (fbv * γM0 / fy)2) Cl 6.7 Eq (6.2)
/ (4 * m) = 71.1 kN
Vertical Crane Wheel Load Fzed = γQ * Qv / N = 13.2 kN
Since Fzed ≤ Ffrd, Design Resistance of Bottom
Flange is Satisfactory

2.3.3   Design Values of Torsional Effects

Torsional Bending Constant a = √((E * Iw) / (G * J)) = 1322.3 mm

Torsional Moment Td = γQ * Qh * (H - 2 * T) / 2 = 0 kN-m
Portion of Length α = 0.5
Distance along Member z = α * Lbs = 1000 mm
Hyperbolic Sine of Alpha L/a t1 = sinh(α * Lbs / a) = 0.8
Hyperbolic Tangent of L/a t2 = tanh(Lbs / a) = 0.9
Hyperbolic Cosine of Alpha L/a t3 = cosh(α * Lbs / a) = 1.3
Hyperbolic Sine of z/a t4 = sinh(z / a) = 0.8
Angle of Twist Φ = Td * a / (G * J) * ((1 - α) * z / a + ((t1 / t2 - t3) * t4)) = 0 rad
Second Derivative of Φ Φ" = -Td / (G * J * a) * ((t1 / t2 - t3) * t4) = 0
Minor Axis Moment Mzed2 = Φ * Myed = 0 kN-m
Design Warping Torsional Moment Mwed = E * Iyy / 2 * (H - T) * Φ" / 200 = 0 kN-m

2.3.4   Torsional Moment

Hyperbolic cosine of 0 t4' = cosh(0) = 1

Term for Φ' Φ' = (1 - α) + (t1 / t2 - t3) * t4 = 0.1
St Venant's Torsional Moment Tted = Td * Φ' = 0 kN-m

2.3.5   Moment Resistance EN 1993 - 1 - 1

Moment Resistance Mcrd = Sxx * fy / γM0 = 221.1 kN-m Cl 6.2 Eq (6.13)

Unity Factor Ucm = Myed / Mcrd = 0.12 Cl 6.2 Eq (6.12)
Since Ucm < 1, Section Chosen is Suitable

2.3.6   Maximum Shear Force due to Vertical Load

Shear Force due to Concentrated Loads Vc = γQ * Qv = 52.8 kN

Shear Force due to Distributed Loads Vd = γG * qv * Lbs / 2 = 0.7 kN
Maximum Shear Force Vzed = Vc + Vd = 53.5 kN

2.3.7   Shear Plastic Resistance EN 1993 - 1 - 1

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Shear Area Avz = A - 2 * B * T + (t + 2 * r) * T = 3080.2 mm2 Cl 6.2 (3)

Shear Plastic Resistance VplRd = Avz * (fy / √3) / γM0 = 489.1 kN Cl 6.2 Eq (6.18)
Unity Factor Ucs = Vzed / VplRd = 0.11 Cl 6.2 Eq (6.17)
Since Ucs < 1, Section Chosen is Suitable

2.3.8   Reduced Shear Resistance in the Presence of Torsion EN 1993 - 1 - 1

Torsional Shear Stress tted = Tted * t / J = 0 N/mm2 Cl 6.2 Eq (6.20)

Reduction Factor tr = √(1 - tted * √3 / (1.25 * fy) / γM0) = 1
Reduced Shear Plastic Resistance Vtrd = tr * VplRd = 489.1 kN Cl 6.2 Eq (6.26)
Since Vzed < Vtrd, Section is OK

2.3.9   Shear Buckling

Depth between the Flanges hw = H - 2 * T = 307 mm

Buckling Ratio hw't = hw / t = 40.93 Cl 6.2 (6)
Since hw't ≤ 72 * e / 1.2, No Check for Shear
Buckling is Required

2.3.10   Lateral Torsional Buckling EN 1993 - 1 - 1

Effective Length of Beam Le = Lbs * Ks = 2000 mm

2.3.10.1   Terms for Critical Moment

Euler Term Et = π2 * E * Iyy / (Le2) = 3888624.1 N

Load to Shear Center Distance Zg = H / 2 = 165 mm
Shear Modulus Term Gt = (Le2 * G * J) / (π2 * E * Iyy) = 5853.2 mm2
Square Root Term Sr = √(Iw / (Iyy + Gt)) = 176.4 mm
Coefficient C1 C1 = 1.4
Critical Moment Mcr = C1 * Et * Sr = 927.3 kN-m
Non Dimensional Slenderness λlt = √(Sxx * fy / Mcr) = 0.5
Limiting Slenderness Value λlt0 = 0.4 Cl 6.3.2.3
Ratio h'b = H / B = 2.1
Imperfection Value αlt = 0.49 Table 6.5
Factor Φlt = 0.5 * (1 + αlt * (λlt - λlt0) + 0.75 * λlt2)
= 0.61
Modification Factor Ψlt 2 2
= 1 / (Φlt + √(Φlt - 0.75 * λlt )) = 0.95 Cl 6.3.2.2 Eq (6.56)
Correction Factor kc = 1 / √C1 = 0.86 Table 6.6
Moment Distribution Factor f = 1 - 0.5 * (1 - kc) * (1 - 2 * (λlt - 0.8)2) = 0.94
Modified Ψlt Factor Ψlt' = Ψlt / f = 1.01 Cl 6.3.2.3 Eq (6.58)
Design Buckling Resistance Moment Mbrd = Ψlt' * Sxx * fy / γM1 = 222.7 kN-m Cl 6.3.2.3 Eq (6.55)
Unity Factor Ucb = Myed / Mbrd = 0.12
Since Ucb < 1, Section Chosen is Suitable

2.3.11   Local Bending Stresses in the Bottom Flange due to Wheel Loads EN 1993 - 6

Ratio μ = 2 * (B /2 - Wg / 2) / (B - t) = 0.39 Cl 5.8 Eq (5.7)

2.3.11.1   Stress Points are at i = 0, 1, 2

Coefficient 1 Cx0 = 0.05 - 0.58 * μ + 0.148e3.015μ = 0.31 Table 5.2

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Coefficient 2 Cx1 = 2.23 - 1.49 * μ + 1.39e-18.33μ = 1.64 Table 5.2

Coefficient 3 Cx2 = 0.73 - 1.58 * μ + 2.91e-6.33μ = 0.35 Table 5.2
Coefficient 4 Cy0 = -2.11 + 1.977 * μ + 0.0076e6.530μ = -1.23 Table 5.2
Coefficient 5 Cy1 = 10.108 - 7.408 * μ - 10.108e-1.364μ = 1.28 Table 5.2
Coefficient 6 Cy2 = 0 Table 5.2
Local Longitudinal Bending Stress σoxed = Cx * Fzed / T2 = 164.169 N/mm2 Cl 5.8 Eq (5.5)
Local Transverse Bending Stress σoyed = Cy * Fzed / T2 = 128.08 N/mm2 Cl 5.8 Eq (5.6)

2.3.11.2   Serviceability Limit State Stress Check EN 1993 - 6

Global Shear Stress ved = Vzed / (H * t) = 21.601 N/mm2

Shear Resistance vRd = fy / √3 * γMs = 144.338 N/mm2 Cl 7.5 Eq (7.2b)
Since Ved ≤ VRd, Shear Stress is OK
Global Bending Stress σxed = fbv = 37.492 N/mm2 Cl 7.5 Eq (7.2a)
Since σxed ≤ fy / ƳMs, OK
Reversible behaviour due to Secondary effects induced by deformations are limited by the following expressions :
Expression 7.2c exp7.2c = √(σxed2 + 3 * ved2) = 52.967 N/mm2 Cl 7.5 Eq (7.2c)
Since exp7.2c ≤ fy / ƳMs, OK
Expression 7.2e exp7.2e = √(((σxed + σoxed)2 + σoyed2) - ((σxed + σoxed)
* σoyed)+ (3 * ved2)) = 180.673 N/mm2 Cl 7.5 Eq (7.2e)
Since exp7.2e ≤ fy / ƳMs, OK

2.3.12   Combined Bending and Torsion Check EN 1993 - 6

Total Minor Axis Design Moment Mzed = Mzed1 + Mzed2 = 0 kN-m

Minor Axis Resistance Moment Mzrd = fy * Zyy / γM1 = 27.2 kN-m
Warping Torsional Resistance Moment Mwrd = Mzrd / 2 = 13.6 kN-m
Coefficient 7 kw = 0.7 - 0.2 * Mwed / Mwrd = 0.7 Annex A
Coefficient 8 kzw = 1 - (Mzed / Mzrd) = 1
Coefficient 9 ka = 1 / (1 - Myed / Mcr) = 1.03
Equivalent Uniform Moment Factor Cmz = 0.9
Unity Factor Uccm = Myed / Mbrd + (Cmz * Mzed) / Mzrd + (kw *
kzw * ka * Mwed)/ Mwrd = 0.12 Annex A Eq (A1)

2.3.13   Check for Deflection EN 1993 - 6

Deflection due to Concentrated Loads δ1 = (Qv * Lbs3) / (48 * E * Ixx) = 0.2 mm

Deflection due to Distributed Loads δ2 = 5 / 384 * qv * Ls4 / (E * Ixx) = 0 mm
Total Deflection δ = δ1 + δ2 = 0.3 mm
Allowable Deflection δallow = Ls / Δlims = 4 mm Cl 7.3 Table 7.2
Since δ ≤ δallow, Deflection OK

2.3.14   Check for Vibration EN 1993 - 6

In order to limit the lateral vibration of the bottom flange the slenderness ratio is calculated as below :
Bottom Flange Inertia lzf = 3940000 mm4
Radius of Gyration of the Flange iz = (lzf / (B * T))0.5 = 46.3 mm
Slenderness of Bottom Flange Sbf = Lbs / iz = 43.2 Cl 7.6 (2)

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Allowable Slenderness Ratio of Bottom

Sbfallow = 250 Cl 7.6 (2)
Flange
Since Sbf ≤ Sbfallow, Vibration is Satisfactory

2.4   Calculation for Cantilever Span

Cantilever Span LbcC = Lc - Ds = 1350 mm

2.4.1   Design Moment

Concentrated Load QvC = (1 + Ψv / 100) * (PL + Sh) = 35.2 kN

Horizontal Load QhC = (Ψh / 100) * (PL + Sh) = 0 kN
Distributed Load - Vertical qvC = A * 78.5 = 0.5 kN/m
Torsional Moment TmC = QhC * (H / 2 – T) = 0 kN-m
Vertical Bending Moment MyedC = (γQ * QvC * LbcC) + (γG * qvC * (LbcC)2 / 2) = 71.9 kN-m
Horizontal Bending Moment Mzed1C = (γQ * QhC * 0.9 * LbcC) = 0 kN-m

2.4.2   Capacity of Bottom Flange Resistance EN 1993 - 6

Stress due to Vertical Moments fbvC = MyedC / Zxx = 100.8 N/mm2

Stress due to Horizontal Moments fbhC = Mzed1C / Zyy = 0 N/mm2
Lever Arm mC = 0.5 * (B - t) - 0.8 * r – (B / 2 – Wg / 2) Cl 6.7 Eq (6.3)
= 31.8 mm
Effective Length leffC = 2 * (mC + (B / 2 – Wg / 2)) Table 6.2
= 123.7 mm
Design Resistance of Bottom Flange FfrdC = leffC * T2 * fy / γM0 * (1 - (fbvC * γM0 / fy)2) Cl 6.7 Eq (6.2)
/ (4 * mC) = 30.6 kN
Vertical Crane Wheel Load FzedC = γQ * QvC / N = 13.2 kN
Since FzedC ≤ FfrdC, Design Resistance of
Bottom Flange is Satisfactory

2.4.3   Design Values of Torsional Effects

Torsional Bending Constant aC = √((E * Iw) / (G * J)) = 1322.3 mm

Torsional Moment TdC = γQ * QhC * (H - 2 * T) / 2 = 0 kN-m
Portion of Length αC = 0.9
Distance along Member zC = αC * LbcC = 1215 mm
Hyperbolic Sine of Alpha L/a t1C = sinh(αC * LbcC / aC) = 1.1
Hyperbolic Tangent of L/a t2C = tanh(LbcC / aC) = 0.8
Hyperbolic Cosine of Alpha L/a t3C = cosh(αC * LbcC / aC) = 1.5
Hyperbolic Sine of z/a t4C = sinh(zC / aC) = 1.1
Angle of Twist ΦC = TdC * aC / (G * J) * ((1 - αC) * zC / aC + ((t1C / t2C
- t3C) * t4C)) = 0 rad
Second Derivative of Φ Φ"C = -TdC / (G * J * aC) * ((t1C / t2C - t3C) * t4C) = 0
Minor Axis Moment Mzed2C = ΦC * MyedC = 0 kN-m
Design Warping Torsional Moment MwedC = E * Iyy / 2 * (H - T) * Φ"C / 200 = 0 kN-m

2.4.4   Torsional Moment

Hyperbolic cosine of 0 t4C' = cosh(0) = 1

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Term for Φ' Φ'C = (1 - αC) + (t1C / t2C - t3C) * t4C = 0

St Venant's Torsional Moment TtedC = TdC * Φ'C = 0 kN-m

2.4.5   Moment Resistance EN 1993 - 1 - 1

Moment Resistance McrdC = Sxx * fy / γM0 = 221.1 kN-m Cl 6.2 Eq (6.13)

Unity Factor UcmC = MyedC / McrdC = 0.33 Cl 6.2 Eq (6.12)
Since UcmC < 1, Section Chosen is Suitable

2.4.6   Maximum Shear Force due to Vertical Load

Shear Force due to Concentrated Loads VcC = γQ * QvC = 52.8 kN

Shear Force due to Distributed Loads VdC = γG * qvC * LbcC = 0.9 kN
Maximum Shear Force VzedC = VcC + VdC = 53.7 kN

2.4.7   Shear Plastic Resistance EN 1993 - 1 - 1

Shear Area AvzC = A - 2 * B * T + (t + 2 * r) * T = 3080.2 mm2 Cl 6.2 (3)

Shear Plastic Resistance VplRdC = AvzC * (fy / √3) / γM0 = 489.1 kN Cl 6.2 Eq (6.18)
Unity Factor UcsC = VzedC / VplRdC = 0.11 Cl 6.2 Eq (6.17)
Since UcsC < 1, Section Chosen is Suitable

2.4.8   Reduced Shear Resistance in the Presence of Torsion EN 1993 - 1 - 1

Torsional Shear Stress ttedC = TtedC * t / J = 0 N/mm2 Cl 6.2 Eq (6.20)

Reduction Factor trC = √(1 - ttedC * √3 / (1.25 * fy) / γM0) = 1
Reduced Shear Plastic Resistance VtrdC = trC * VplRdC = 489.1 kN Cl 6.2 Eq (6.26)
Since VzedC < VtrdC, Section is OK

2.4.9   Shear Buckling

Depth between the Flanges hwC = H - 2 * T = 307 mm

Buckling Ratio hw'tC = hwC / t = 40.93 Cl 6.2 (6)
Since hw'tC ≤ 72 * e / 1.2, No Check for Shear
Buckling is Required

2.4.10   Lateral Torsional Buckling EN 1993 - 1 - 1

Effective Length of Beam LeC = LbcC * Kc = 2700 mm

2.4.10.1   Terms for Critical Moment

Euler Term EtC = π2 * E * Iyy / (LeC2) = 2133675.8 N

Load to Shear Center Distance ZgC = H / 2 = 165 mm
Shear Modulus Term GtC = (LeC2 * G * J) / (π2 * E * Iyy) = 10667.5 mm2
Square Root Term SrC = √(Iw / (Iyy + GtC)) = 189.5 mm
Coefficient C1 C1C = 1.4
Critical Moment McrC = C1C * EtC * SrC = 546.7 kN-m
Non Dimensional Slenderness λltC = √(Sxx * fy / McrC) = 0.6
Limiting Slenderness Value λlt0C = 0.4 Cl 6.3.2.3
Ratio h'bC = H / B = 2.1
Imperfection Value αltC = 0.49 Table 6.5
Factor ΦltC = 0.5 * (1 + αltC * (λltC - λlt0C) + 0.75 * λltC2) = 0.71

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Modification Factor ΨltC = 1 / (ΦltC + √(ΦltC2 - 0.75 * λltC2)) = 0.86 Cl 6.3.2.2 Eq (6.56)
Correction Factor kcC = 1 / √C1C = 0.86 Table 6.6
Moment Distribution Factor fC = 1 - 0.5 * (1 - kcC) * (1 - 2 * (λltC - 0.8)2) = 0.93
Modified Ψlt Factor ΨltC' = ΨltC / fC = 0.93 Cl 6.3.2.3 Eq (6.58)
Design Buckling Resistance Moment MbrdC = ΨltC' * Sxx * fy /γM1 = 204.7 kN-m Cl 6.3.2.3 Eq (6.55)
Unity Factor UcbC = MyedC / MbrdC = 0.35
Since UcbC < 1, Section Chosen is Suitable

2.4.11   Local Bending Stresses in the Bottom Flange due to Wheel Loads EN 1993 - 6

Ratio μC = 2 * (B /2 - Wg / 2) / (B - t) = 0.39 Cl 5.8 Eq (5.7)

2.4.11.1   Stress Points are at i = 0, 1, 2

Coefficient 1 Cx0C = 0.05 - 0.58 * μC + 0.148e3.015μC = 0.31 Table 5.2

Coefficient 2 Cx1C = 2.23 - 1.49 * μC + 1.39e-18.33μC = 1.64 Table 5.2
Coefficient 3 Cx2C = 0.73 - 1.58 * μC + 2.91e-6.33μC = 0.35 Table 5.2
Coefficient 4 Cy0C = -2.11 + 1.977 * μC + 0.0076e6.530μC = -1.23 Table 5.2
Coefficient 5 Cy1C = 10.108 - 7.408 * μC - 10.108e-1.364μC = 1.28 Table 5.2
Coefficient 6 Cy2C = 0 Table 5.2
Local Longitudinal Bending Stress σoxedC = CxC * FzedC / T2 = 164.169 N/mm2 Cl 5.8 Eq (5.5)
Local Transverse Bending Stress σoyedC = CyC * FzedC / T2 = 128.08 N/mm2 Cl 5.8 Eq (5.6)

2.4.11.2   Serviceability Limit State Stress Check EN 1993 - 6

Global Shear Stress vedC = VzedC / (H * t) = 21.695 N/mm2

Shear Resistance vRdC = fy / √3 * γMs = 144.338 N/mm2 Cl 7.5 Eq (7.2b)
Since VedC ≤ VRdC, Shear Stress is OK
Global Bending Stress σxedC = fbvC = 100.82 N/mm2 Cl 7.5 Eq (7.2a)
Since σxedC ≤ fy / ƳMs, OK
Reversible behaviour due to Secondary effects induced by deformations are limited by the following expressions :
Expression 7.2c exp7.2cC = √(σxedC2 + 3 * vedC2) = 107.595 N/mm2 Cl 7.5 Eq (7.2c)
Since exp7.2cC ≤ fy / ƳMs, OK
Expression 7.2e exp7.2eC= √(((σxedC + σoxedC)2 + σoyedC2) - ((σxedC + σoxedC)
* σoyedC)+ (3 * vedC2)) = 232.585 N/mm2 Cl 7.5 Eq (7.2e)
Since exp7.2eC ≤ fy / ƳMs, OK

2.4.12   Combined Bending and Torsion Check EN 1993 - 6

Total Minor Axis Design Moment MzedC = Mzed1C + Mzed2C = 0 kN-m

Minor Axis Resistance Moment MzrdC = fy * Zyy / γM1 = 27.2 kN-m
Warping Torsional Resistance Moment MwrdC = MzrdC / 2 = 13.6 kN-m
Coefficient 7 kwC = 0.7 - 0.2 * MwedC / MwrdC = 0.7 Annex A
Coefficient 8 kzwC = 1 - (MzedC / MzrdC) = 1
Coefficient 9 kaC = 1 / (1 - MyedC / McrC) = 1.15
Equivalent Uniform Moment Factor CmzC = 0.9
Unity Factor UccmC = MyedC / MbrdC + (CmzC * MzedC) / MzrdC + (kwC *
kzwC * kaC * MwedC) / MwrdC = 0.35 Annex A Eq (A1)

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2.4.13   Check for Deflection EN 1993 - 6

Deflection due to Concentrated Loads δ1C = ((QvC * (LbcC)2 * ((LbcC) + Ls)) / (3 * E * Ixx)) = 3 mm
Deflection due to Distributed Loads δ2C = (qvC * (LbcC)4) / (8 * E * Ixx) = 0 mm
Total Deflection δC = δ1C + δ2C = 3.1 mm
Allowable Deflection δallowC = LbcC / Δlimc = 4.5 mm Cl 7.3 Table 7.2
Since δC ≤ δallowC, Deflection OK

2.4.14   Check for Vibration EN 1993 - 6

In order to limit the lateral vibration of the bottom flange the slenderness ratio is calculated as below :
Bottom Flange Inertia lzfC = 3940000 mm4
Radius of Gyration of the Flange izC = (lzfC / (B * T))0.5 = 46.3 mm
Slenderness of Bottom Flange SbfC = LbcC / izC = 29.2 Cl 7.6 (2)
Allowable Slenderness Ratio of Bottom
SbfallowC = 250 Cl 7.6 (2)
Flange
Since SbfC ≤ SbfallowC, Vibration is Satisfactory

3.0   SUMMARY

Critical Section : Cantilever Span

3.1   Bottom Flange Resistance

Description Actual Allowable Status
Vertical Crane Load (kN) FzedC = 13.2 FfrdC <= 30.6 PASS

3.2   Moment and Shear Resistance

Description Actual Allowable Status
Vertical Moment (kN-m) MyedC = 71.9 McrdC <= 221.1 PASS
Torsional Buckling Moment (kN-m) MyedC = 71.9 MbrdC <= 204.7 PASS
Shear Force (kN) VzedC = 53.7 VtrdC <= 489.1 PASS

3.3   Serviceability Limit State Check

Description Actual Allowable Status
2
Global Bending Stress (N/mm ) σxedC = 100.82 <= 250 PASS
Reversible Behaviour Stresses - exp 7.2c (N/mm2) exp7.2cC =107.595 <= 250 PASS
2
Reversible Behaviour Stresses - exp 7.2e (N/mm ) exp7.2eC =232.585 <= 250 PASS

3.4   Combined Bending and Torsion Check

Description Actual Allowable Status
Unity Factor UccmC = 0.35 <= 1 PASS

3.5   Deflection

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Description Actual Allowable Status

Total Deflection (mm) δC = 3.1 δallowC <= 4.5 PASS

3.6   Vibration Check

Description Actual Allowable Status
Slenderness of Bottom Flange SbfC = 29.2 SbfallowC <= 250 PASS

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