Steel Warehouse Project

Basic Wind Velocity

= 1 x 1 x 32 m/s

= 32 m/s

Mean Wind Velocity

- Cr(z) = Kr x ln(z/z0)

-

, Terrain Category II, Table 4.1

Kr =

Kr = 0.19

Cr(z) = 0.19 x ln(10.36/0.05)

= 0.19 x ln(207.2)

= 1.013

C0(z) = 1

= 1.013 x 1 x 32m/s

= 32.416 m/s

= 32 m/s

32.416m/s

BS EN 1991-1-

4-2005

Unless

otherwise

specified

Department of Civil Engineering

CALCULATION SHEET

Module: H23S07 Sheet 1 of x

Job Title Steel Warehouse Design Calculation

Subject Wind Calculations

Made By: 009776 Date

Checked by: Date

Client:

Atkins & Partners

Wind Turbulence

= 0.19 x 32 x 1

= 6.08 m/s

Turbulence Intensity

For zmin ≤ z ≤ zmax

0.05 ≤ 10.36 ≤ 200

Peak velocity pressure

= [1+7(0.188)] x 0.5 x 1.25 x 32.4162

= 1521 Pa

6.08 m/s

(z) = 1521 Pa


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Basic Velocity Pressure

Pa

Exposure Factor

b= 60m, 2h = 20.72m, d = 45m

Hence e= 20.72m < d=45


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h/d = 10.32/45

= 0.229

Zone D: Cpe, 10 = 0.7

Zone E: Cpe, 10 = -0.3


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Wind Pressure on External Surfaces

KN/m2

KN/m2

KN/m2

KN/m2

KN/m2

+ve value denotes pressure

-ve value denotes suction


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Internal Wind Pressure

We assumed a positive internal pressure due to the limited

opening of the warehouse envelope.

We also did not take into account the uplift forces on the roof as

the loads we are using are more conservative.

= 0.2

KN/m2

KN/m2


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Dte

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Total External pressure on face D : 0.7227 KN/m2

Total External pressure on face E : - 0.7983 KN/m2

We wanted the values to be more conservative so we used the

1.0647 KN/m2 on face D and 0.7983 KN/m2 on face E.

Final External Pressure values

KN/m2

KN/m2

KN/m2

KN/m2

KN/m2


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1.0647 KN/m2

0.3042 KN/m2

0.4563 KN/m2

Plastic Analysis

Haunch length

Maximum haunch length = L/10 = 45/10 = 4.5m

Maximum Haunch Depth = 2% of Length = 0.9m

Load Combinations

Dead load + Imposed load Dead load (factored) + Transverse wind load (factored)

Dead load

Dead Load UDL = 0.167 kN/m

Self-Weight UDL = 1.251 kN/m

Total dead load UDL = 1.418 kN/m

Imposed load

Imposed load = 2.99 kN/m

Hence,

1.35Dead load + 1.5Imposed load = 1.35(1.418) + 1.5(2.99) = 6.40 kN/m

Designation

Rafter: 610x229x125

Column: 305x305x283

Mechanisms:

Haunch length = 4.5 m

Total dead load UDL:

1.418 kN/m

Imposed load:

2.99 kN/m

Dead load + Imposed =

6.40 kN/m


CALCULATION SHEET

Module: H23S07 Sheet of x


Subject


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Failure Mechanism


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Compatibility Equation

H = 7.1Φ = (x/22.5 + 0.9) θ Φ = (4x/639 + 9/71)θ

At collapse, MB = MD = 1.5MP ; MC = MP

Equilibrium equation for half of the frame:

MB(Φ+θ) + MC(-θ) = ( w x ) (

+ ( w ) (22.5-x )( ᶑ )

Vertical displacement, ᶑ = xθ

1.5Mp [θ + Φ] + Mp (θ) = (wx)(

+ (w)(22.5-x)(xθ)

Substitute Φ

1.5 MPθ + 1.5 MPθ (4x/639 + 9/71) + MPθ =

+ (22.5-x)(wxθ)

Eliminate θ

1.5 MP + 1.5 MP (4x/639 + 9/71) + MP =

+ 22.5wx–wx2

2.5Mp +

+ =

Mp (2x/213 + 191/71) =

MP =

(GENERAL EQUATION)

For dead load + imposed load (Mechanism 1)

1.35Dead load + 1.5Imposed load = 1.35(1.418) + 1.5(2.99) = 6.40 kN/m

MP =

Maximum MP with respect with x is when

=

,

x = 21.68m

Dead load + Imposed =

6.40 kN/m

X=21.68 m


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Therefore, Mp = 559.09 kNm < Mpl,Rd = 1304.98 kNm OK

Plastic moment in column = 1.5Mp = 838.64 kNm

RA (Reaction at A) =

Therefore, 1.5Mp = 838.64 = 7.1HA, where HA = 118.12 kN

Eaves moment = 8HA = 944.96 kN

Bending moment in rafter

M + 6.4x2/2 + (8+x/22.5)118.12 – 144x = 0

M + 3.2x2 + 944.96 + 5.25x – 144x = 0

At x = 22.5, at apex

M= 556.92 kNm

Shear force in Rafter

Converting Vertical UDL to be in rafter’s plane,

6.4cos6 = 6.36 kN/m (rafter plane)

V + 6.36x = 144cos6

V = 143.21 – 6.36x

Mpl,Rd =1304.98kNm

Mp=559.09 kNm

1.5Mp =838.64 kNm

RA = 144 kN

HA = 118.12 kN


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Axial force in Rafter

UDL to point load = 6.4 x 22.5 = 144 kN

Axial force = 144sin6 = 15.05 kN

Shear Force in Column = HA = -118.12kN

Wind load - Mechanism 2

Wind Load UDL = 6.39 kN/m


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At collapse:

MB = -1.5MP, MD = 1.5MP

Equilibrium Equation:

MB(-θ) + MD(θ) = (9.59)(7θ)+(6.39)(6θ)+(6.39)(5θ)+(6.39)(4θ)+(6.39)(3θ)+(6.39)(2θ)+ 9.59θ

-MB θ + MD θ = 204.52θ

1.5MP θ + 1.5MP θ = 204.52θ

3MP = 204.52

MP = 68.17 kNm < Mpl,Rd = 1304.98 kNm

Plastic Moment in Column = 1.5MP = 102.26 kNm

Reaction at A, VA = -4.54 kN

Horizontal reaction at A, HA = -22.61 kN

Combined Mechanism: (1) + (2)

Dead Load + Imposed Load + Wind Load

MB( θ+ Φ) + MC(- θ) = 1617.85 θ

Φ = 0.26θ

MB(1.26θ) + MC(-θ) = 1617.85 θ

1.26MBθ - MC(θ) = 1617.85 θ

Combination: (1) + 1.26(2)

1.26MBθ - MC(θ) = 1617.85 θ

+1.26[-MBθ +MDθ = 204.52θ]

-MCθ + 1.26MDθ =1875.55θ

MPθ + 1.26MPθ = 1875.55θ

2.26MP = 1875.55

MP = 829.89 kNm < Mpl,Rd = 1304.98 kNm

OK

Mpl,Rd =

1304.98kNm

Mp = 68.17kNm

HA = -22.61kN

Mp = 829.89kNm

OK


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Plastic Moment in Column = 1.5MP = 1244.84 kNm

Reaction at A, VA = 139.46 kN

Therefore, 1.5MP = 1244.84 = 7.1HE

HE = 175.33 kN

HA = 124.21 kN

Bending moment in Rafter:

M + 6.4x2/2 + 6.39(8)(4 + x/22.5) + 124.21(x/22.5 +8) -139.46(x) = 0

M + 3.2x2 +204.48 +2.272x + 5.52x + 993.68 – 139.46x = 0

At x = 22.5m

M = 144.37 kNm

Axial Force in Rafter:

UDL to point load = 6.4 x 22.5 = 144 kN

Axial Force = 144sin6 = 15.05 kN

Shear force in column = -HA = -124.21 kN

VA = 139.46kN

HA = 124.21kN

Moment in Rafter

=144.37kNm


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CALCULATION SHEET

Based on SCI calculation sheet

Job No. H23S07 Rev NA Sheet of

Job Title Steel Warehouse Design Calculations

Subject Load Analysis

Client:

Atkins & Partners Made by: Ruchika-009508 Date

Checked by: Date

EN 1993-1-8: 2005

Primary Beam to Column Connection

Bolt details Grade 8.8 M30 bolt Diameter of bolt Tensile stress area Clearance hole diameter For class 8.8 non preloaded bolts Ultimate tensile strength


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L-Plate details Steel grade s355 - EN 100225-2 Yield strength

Ultimate tensile strength Plate thickness

Taking moments about centroid of bolt group- 3FB = FA

Resultant force :

RA = 165.23 kN

RB = 73.89 kN Therefore is the highest force on a bolt


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EN 1993-1-8: 2005, T 3.4

Shear resistance of bolts Category A bearing type The resistance of a single bolt in shear,

Where, (for class 4.6, 5.6 and 8.8)

No. of shear planes,

> RA ; OK

Bearing of Bolt

Where,

is the smallest of

is the smaller of


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Cl.3.10.2

Hence,

OK

Block Tearing of Plate

Where, is net area subjected to tension is net area subjected to shear


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Table 3.4

Bearing of plate

is the smaller of

OK


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EN 1993-1-8 2005 ( 3.10.2) Table 3.4

Block tearing of web

OK

Bearing of web

is the smaller of

OK


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Critical Design force= 165.23 kN Shear of bolt Bearing Of bolt

Block tearing of plate kN

Bearing of plate


Bearing of web


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Connection of Secondary Beam to Column


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Taking moments about centroid of bolt group,

Resultant force,


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EN 1993-1-8:2005, T 3.4




> RA ; OK

Bearing of Bolt

Where,

is the smallest of

is the smaller of


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Cl 3.10.2

Hence,

OK



kN


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T 3.4 Cl 3.10.2

Bearing of plate

is the smaller of

OK!!

Block Tearing of Web



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T 3.4 Bearing of Web

is the smaller of

OK!!

Summary Critical Design force= 58.94 kN Shear of bolt Bearing Of bolt Block tearing of plate kN > 58.94kN OK!!!!

Bearing of plate


Bearing of web


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Connection of Primary Beam to Secondary Beam


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(EN 1994-1-1)

Design of composite slab and Secondary beam

The composite office slab is to be designed atop the secondary beam. It is assumed to be propped during construction so the secondary beam does not need to support the weight of concrete while it hardens. The general description of the composite slab is as follows- Height of Slab- 150mm Direction of steel decking- Parallel to the beam Height of steel decking-50mm

Calculation of

Determination of Neutral Axis Resistance of the concrete flange Rcf = 0.567 fck beff ( h - hp ) Resistance of the steel section Rs = fy Aa

Resistance of the steel flange Rsf = fy b tf

Resistance of overall web depth Rw = Rs – 2 Rsf


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Therefore Neutral Axis is in the Steel Flange.

Plastic Analysis-Ultimate Limit State Verification

Shear Connectors Diameter d = 19 mm Overall nominal height hsc = 100 mm Ultimate tensile strength fu = 450N/mm2 Number of shear connector studs n = le/e = 9000/225=40 Number of studs per rib nr = 1


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Clause 6.6.3.1 (EN 1994-1-1)

Shear Resistance of a stud

Limited by Concrete

Limited by Stud

With; α = 0.2(hsc/d +1); (for 3<hsc/d<4) or α = 1; (hsc/d > 4) Where; γv is the partial factor 1.25. d is the diameter of the stud. 19 mm fu is the ultimate tensile strength of the stud material 450 N/mm

2.

fck is cylindrical compressive strength of concrete at age consider- 25Mpa hsc is the overall height of the studs. Ecm is the modulus of elasticity of the concrete; Ecm = 22000((fck +8)/10)

0.3.

In this instance: hsc/d = 100/19 = 5.263 > 4, so α= 1 Ecm = 22000((25 + 8)/10)

0.3 = 31476

Resistance of studs limited by concrete;

Resistance of studs limited by studs;


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Clause 6.2.1.3 (EN 1994-1-1) Clause 5.4.1.2 (EN 1994-1-1-2004) Clause 6.6.3.1 (EN 1994-1-1) Table 6.2 (EN 1994-1-1)

So, PRd = min{ PRd limited by studs, PRd limited by concrete} = 74.29kN

Number of shear connectors for full interaction

Where,

Calculation of Reduction factor Ribs are perpendicular to the supporting beam

PRd=74.29kN


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Clause 6.6.1.2 (EN 1994-1-1)

For the full span, the number of studs required for full interaction

Partial shear connection Degree of interaction between steel and concrete deck is;

Limitation on the use of partial shear connection

=


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Moment Reduction in Partial Shear Connection The reduced moment of the section

for when neutral axis is in the steel flange,

Moment Resistance in partial shear connection,

OK!!!


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Elastic Analysis-Serviceability Limit State Verification

Transformation Properties Concrete Transform into steel; For short term loading; 0r

For Long term loading

Ecm is the modulus of elasticity of concrete, Ecm = 22000((fck +8)/10)0.3

. is the creep coefficient according to the age of concrete at the moment of consideration. Normally assumed as 1.5 for concrete at 28 days is the creep multiplier depending on type of loading. Normally assumed as 1.1.

is the modular ratio

for short - term loading.

For short term loading

For Long term loading

We use the lower value of the two, For long term loading

Determination of neutral axis


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The Neutral axis is in the steel beam.

Determination of moment of inertia


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Deflection of composite beam

Deflection Limits

Deflection Check


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BS EN 1993-1-1:2005 Table 3.1

Connection of Primary Beam to Secondary




Design eccentric moment at centroid of bolt group;

But Therefore,


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BS EN 1993-1-8:2005 3.4.1(1) Table 3.2 Table 3.4

Vertical component at bolts,

Horizontal component at bolts,

Resultant force at A and B,

Resultant force at C and D,

Taking the larger of the two values, we design for


Where, (for class 4.6, 5.6 and 8.8) No. of shear planes,

Therefore, the shearing resistance of bolts is acceptable.


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BS EN 1993-1-8:2005 Table 3.4

Bearing resistance of bolts Bearing resistance per bolt,

Hence,

Therefore, bearing resistance per bolt is acceptable.


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BS EN 1993-1-8:2005 Clause 3.10.2 Table 3.4


Where, is net area subjected to tension=0 is net area subjected to shear

Therefore, it is acceptable.

Bearing Resistance of Plate


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BS EN 1993-1-8:2005 Clause 3.10.2

Hence,

Therefore, bearing resistance of plate is acceptable.

Block tearing resistance of web

Where, is net area subjected to tension=0 is net area subjected to shear



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BS EN 1993-1-8:2005 Table 3.4

Bearing Resistance of Web

Hence,



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Secondary Beam to Primary Beam Connection Summary Critical Design force Shear resistance of bolt,

Bearing resistance of bolt, Block tearing of plate,

Bearing resistance of plate, Block tearing of web,

Bearing resistance of web,

Bracing

Longitudinal wind bracing

Figure 1 Side view

Wind braces are design to resist wind loads acting on the structure. Wind load acting on the transverse face is transferred to the bracings and are resisted. Circular Hollow Sections (CHS) are chosen for the design. There are two bays of K-bracings at the extremities of the structure on each longitudinal side. Both the braces are considered to be in compression. Wind pressure acting on the front face (45m side) The projected area of the vertical front face Characteristic value of total wind load acting

Since there are two bays of bracings on each face,

Wind load value acting on front face


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Factors on actions

Partial factors: Permanent actions

Reduction factor Variable actions

Factors on accompanying actions: Imposed loads for storage areas

Actions

Permanent actions

Variable action

Combination of actions for ULS, using equation 6.10

Figure 2 Forces acting

P1 and P2 are point loads acting on the brace members. A1 and A2 are the forces in the brace members due to P1 and P2.

Axial force in bracing,

is the design value of the axial force

BS EN 1990 Table (2.1) NA. A1.2(B) Table 2.2


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Partial factors for resistance

Trail section

Steel grade: EN 10025-2 - S355 for t < 40mm Nominal value of yield strength

Nominal value of ultimate tensile strength

Dimension and properties

Outside diameter Thickness Mass per meter Area of section

Ratio for local buckling

Second moment of area Radius of gyration Elastic modulus Plastic modulus Torsional constant Surface area per meter per tonne

BS EN 1993-1-1:2005 6.1(1) Table 3.1 BS steel data (pg 16)


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Material properties

As t < 16mm, for steel type S355 Yield strength

Modulus of elasticity Section classification - Tubular sections

Class 1 limit for section in compression,

,

Therefore,

, the section is class 1 for axial compression.

Design of member in compression

Cross-sectional resistance to axial compression

Basic requirement

Is the design resistance of the cross-section for uniform compression. Equation (6.10)

Therefore, the resistance of the cross section is adequate.

BS EN 1993-1-1:2005 3.2.6(1) Table 5.2 (sheet 3 of 3) 6.2.4(1) Equation (6.9) 6.2.4(2)


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Flexural buckling resistance

For uniform member under axial compression the basic requirement is,

is the design buckling resistance

For class 1, 2 and cross section)

is reduction factor for buckling and determined from figure 6.4: Buckling curve Selection of buckling curve, For flexural buckling slenderness is determined from,

(for class 1, 2 and 3)

Where, is the buckling length in the buckling plane considerd.

i is the radius of gyration about the relevant axis, determined using the properties of the gross cross section

Using figure 6.4: Buckling curve for and buckling curve "a"

Therefore,

Therefore, the flexural buckling resistance of the section is adequate.

BS EN 1993-1-1:2005 6.3.1 6.3.1.1(1) Equation (6.46) 6.3.1.1(3) Equation 6.47 6.3.1.2(1) Table 6.2 Figure 6.4 6.3.1.1(3) Equation 6.47 6.3.1.1(1) Equation 6.46


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Design of member in tension

Due to the negative pressure Design value of the tension force at each cross section should satisfy,

Therefore, the resistance of the cross section is adequate.

BS EN 1993-1-1:2005 6.2.3(1) Equation 6.5 6.2.3(2) Equation 6.6


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Connection design

Figure 3 Connection

Circular hollow section is connected to the portal frame by the use of gusset plates. Flat end plates are fillet welded to the slots in the CHS. Bolts in clearance holes transfer the load the end plate and gusset plates. Shear plane is assumed to pass through the thread of the bolt. Connection design resistance for force.

Bolt selection

Bolt details

Grade 8.8 M20 bolt - none preloaded Diameter of bolt Tensile stress area Clearance hole diameter For class 8.8 non preloaded bolts Yield strength

Ultimate tensile strength

BS steel data Page 87 BS EN 1993-1-8:2005 Table 3.1


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Plate details

Steel grade s275 - EN 100225-2 Yield strength


Positioning of holes for bolts

Minimum and maximum spacing, end and edge distances Minimum End distance Edge distance Spacing distance Spacing distance Maximum End distance Edge distance Spacing distance Spacing distance Hence the bolt spacing's and distances, End distance Edge distance Spacing distance Spacing distance

Figure 4 End plate and CHS

BS EN 1993-1-1:2005 Table 3.1


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Total shearing resistance for 4 bolts,

Therefore, the shearing resistance of bolts is acceptable.

Design value per bolt

Bearing resistance of bolts Bearing resistance per bolt,

Where,

is the smallest of

In the direction of the load transfer: For end bolts

Design value per bolt

BS EN 1993-1-8:2005 3.4.1(1) Table 3.2 Table 3.4 Table 2.1 Table 3.4


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For inner bolts

Hence, Choosing the smallest, Perpendicular to the direction of load transfer:

Edge bolts is the smaller of

Inner bolts is the smaller of

Therefore, for both edge and inner bolts Hence,

Therefore, bearing resistance per bolt is acceptable.

End plate resistance

Tension resistance

Design value of the tension force should satisfy,

For sections with holes , Design plastic resistance of the gross cross-section

BS EN 1993-1-8:2005 Table 3.4 6.2.3(1) 6.2.3(2) Equation 6.6 Equation 6.7


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Design ultimate tensile resistance of the net cross section

Therefore, the tension resistance of the cross section is adequate.

Design for block tearing



Design for bearing resistance

Where,

is the smallest of

In the direction of the load transfer: For end bolts

For inner bolts

Hence, Choosing the smallest, Perpendicular to the direction of load transfer:

BS EN 1993-1-8:2005 6.2.3 3.10.2.(2) Equation 3.9 Table 3.4


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Edge bolts is the smaller of

Inner bolts is the smaller of

Therefore, for both edge and inner bolts Hence,


Welding For S257 steel

Simplified method for design resistance of fillet weld. Considering leg length fillet on both the sides, top and bottom.

Correlation factor for S275 steel,

Design shear resistance of weld per unit length;

Effective weld length,

The shear resistance is

Therefore, design shear resistance of weld is adequate.

The shear resistance

BS EN 1993-1-8:2005 Table 3.4 4.5.3.3(3) Table 4.1 Equation 4.4 Equation 4.3


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Summary Critical design force CHS member, Member compression,

Flexural buckling, Connection design, Shear resistance bolts, Bearing resistance of bolts, End plate, Tension resistance, Block tearing,

Bearing resistance,

Welding, Shear resistance,

All checks have passed.


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Column Base Connection for Office

Design condition for column C2

The column is pin-ended. However, the column must be stable during erection phase therefore 4 bolts outside the column profile will be used.

Plan of base Plate

Characteristic force due to permanent action, FGk = 304.6kN

Characteristic force due to variable action, FQk = 225kN

Ultimate Limit State (ULS)

Partial factors for actions

Permanent action, G = 1.35

Variable action, Q = 1.5

Combination actions for ULS

NEd = 1.35(340.6) + 1.5(225) = 797.31kN

Column Details

305x305x118 in s355 steel

b = 307.4mm

h = 314.5mm

d = 246.7mm

tw = 12.0mm

tf = 18.7mm

NEd = 797.31kN

EN 1993-1-1


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r = 15.2mm

A = 15000mm2

Section perimeter, P = 1834.6mm

Base Plate Details

Strength of concrete C20/30

fck = 30N/mm2

fcd = αcc fck / c

c is partial safety for concrete

c = 1.5

αcc = 1.0

fcd = (1.0 x 30)/1.5 = 20N/mm2

Area required = (797.31x103)/20 = 39865.5mm2

Effective area = 4c2 + (section perimeter x c) + section area

where c is the cantilever outstand of the effective area as shown below.

39865.5 = 4c2 + 1834.6c + 15000

c = 13.175mm

c = 1318mm

Thickness of base plate

tp = c(3fcd mo /fy)0.5

tp = 13.18((3x20x1.0)/355)0.5 = 5.42mm

fcd =20N/mm2

EN 1992-1-1

T3.1

Eq: 3.15

EN 1991-1-1

2.4.2.4


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(h-2tf)/2 = (314.5-2(18.7))/2

= 138.55mm >13.18 mm

Therefore there is no overlap

between the flanges

tp < 40mm, therefore nominal design strength = 355N/mm2

Adopt 10mm plate

Connection of base plate to column; it is assumed that the axial force is

transformed by direct bearing, which is achieved by normal fabrication

process. Only nominal welds are required to connect the baseplate to

the column though in practice full profile 6mm fillet are often used.


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Connection of Purlins to Rafter

Grade S355 L plate, M8, 8.8 Bolt,

M10 Area = 58mm2

Plate 5mm S355

Roofing sheet load = 0.027kN/m2

Load = 0.027x7.2

= 0.1944kN

Purlin = 0.317kN/m

Load = 1.902kN

Total Dead Load imposed on plate = 2.09kN

Factored ULS x1.35 = 2.83kN

Live Load = 1kN/m2

Load = 7.2kN

Factored ULS x1.5 = 10.8kN


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We assume force of

roof acts parallel to

plate as angle of

inclination very

small

Therefore:

Total load imposed DL+LL = 12.89kN = Fbolt

Force per bolt = 12.89/4 = 3.2225kN

Shear of Bolt

Fv,Rd = (0.6x800x58)/1.25

= 22.272kN > Fbolt OK

Bearing of Bolt

b = min

= min

= min {1 ; 0.69 ; 1}

b = 0.69

k1 = min {2.8(e2/d0)-1.7 ; 2.5}

= min {4.13 ; 2.5}

k1 = 2.5

Fb,Rd = (2.5x0.69x800x10x5)/1.25

= 55.2 > Fbolt OK

Force per bolt

= 3.2225kN

Fv,Rd

= 22.272kN

Fb,Rd

= 55.2

EN 1993-1-8:

2005,Table

3.4


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Veff,Rd =

Anv = 2x25x5 = 250

Veff,Rd = (1/ ) x 355 x 250


Bearing of Plate

b =

= 800/510 = 1.57

b = 0.69

k1 = 2.5

Fb,Rd = (2.5x0.69x510x10x5)/1.25


Shear of Bolt = 22.272kN

Bearing of Bolt = 55.2kN > Fbolt = 12.89 OK

Block tearing of Plate = 51.239kN

Bearing of Plate = 35.19kN

Veff,Rd

= 51.239kN

Fb,Rd

= 35.19kN


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Side Rail Bolt Connection

Self-weight of cladding 1m x 6m spacing = 0.0471kN/m2 x 6m2

= 0.283kN

Self-weight of side rail 6m length = 0.0296kN/m x 6m = 0.1776kN

Total Dead Load = 0.460kN

= 1.35

ULS Combination

1.35(0.46) = 0.621kN

FEd = 0.621kN

VA = VB = FEd/2 =0.3105kN

Design force per bolt = 0.3105kN

Fbolt = 0.3105kN


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Shear of Bolt EN1993-1-8:2005 Table 3.4

= (0.6x800x14.2)/1.25

= 5.453kN

> Fbolt

Bearing of Bolt Table 3.4

=

= 1

= 2.38

= 1.0

k1 = 2.8(50/7) – 1.7 = 18.3

Therefore k1 = 2.5

tp = 5mm

=

= 40kN

> Fbolt


= (800x0)/1.25 + (800x2x50x5)/

= 230.94kN > Fbolt

Bearing of Plate

=

= 1.569

OK

OK

OK

Equation 3.9


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= 2.381

= 1.0

k1 = 2.5

=

= 25.5kN

> Fbolt

Block Tearing of Flange

= (510x0)/1.25 + ((2x50x16.4) x 355)/

= 315.64kN > Fbolt

Bearing of Flange Table 3.4

=

= 1.569

= 2.381

k1 = 2.5

=

= 25.5kN > Fbolt

Fbolt = 0.3105kN

OK

OK

OK


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EN 1993-1-8; 2005

Plate Bearing, Fb,Rd = 25.5kN

>Fbolt

Plate Block Tearing, Veff,Rd = 230.94kN

Flange Bearing, Fb,Rd = 25.5kN

Flange Block Tearing, Veff,Rd = 315.64kN

Bolt Shear, Fv,Rd = 5.453kN

Bolt Shearing, Fb,Rd = 40kN


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Steel Warehouse Project

Documents

date client

h23s07 sheet

load imposed load

atkins partners wind

knm2 total external

date total external

kr x lnzz0

basic wind velocity