Steel Warehouse Project

Department of Civil Engineering
CALCULATION SHEET
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
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
CALCULATION SHEET
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CALCULATION SHEET
Basic Velocity Pressure
Pa
Exposure Factor
b= 60m, 2h = 20.72m, d = 45m
Hence e= 20.72m < d=45
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CALCULATION SHEET
h/d = 10.32/45
= 0.229
Zone D: Cpe, 10 = 0.7
Zone E: Cpe, 10 = -0.3
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CALCULATION SHEET
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
CALCULATION SHEET
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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
CALCULATION SHEET
Client:
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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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CALCULATION SHEET
Failure Mechanism
Subject
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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
CALCULATION SHEET
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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
CALCULATION SHEET
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CALCULATION SHEET
Axial force in Rafter
Subject
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
CALCULATION SHEET
Subject
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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
CALCULATION SHEET
Subject
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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

CALCULATION SHEET
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L-Plate details
Steel grade s355 - EN 100225-2
Yield 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

CALCULATION SHEET
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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

CALCULATION SHEET
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Connection of Secondary Beam to Column

CALCULATION SHEET
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Bolt details
Grade 8.8 M20 bolt
Diameter of bolt
Tensile stress area
L-Plate details
Yield strength
Plate thickness
Taking moments about centroid of bolt group,
Resultant force,

CALCULATION SHEET
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EN 1993-1-
8:2005, T
3.4
Where,
(for class 4.6, 5.6 and 8.8)
> RA ; OK
Bearing of Bolt
Where,
is the smallest of
is the smaller of

CALCULATION SHEET
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Cl 3.10.2
Hence,
OK
Where,
kN

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T 3.4
Cl 3.10.2
Bearing of plate
is the smaller of
OK!!
Block Tearing of Web
Where,

CALCULATION SHEET
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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

CALCULATION SHEET
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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

CALCULATION SHEET
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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/mm2.
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;

CALCULATION SHEET
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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

CALCULATION SHEET
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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

CALCULATION SHEET
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BS EN
1993-1-
1:2005
Table 3.1
Connection of Primary Beam to Secondary
Bolt details
Grade 8.8 M30 bolt
Diameter of bolt
Tensile stress area
L-Plate details
Yield strength
Plate thickness
Design eccentric moment at centroid of bolt group;
But
Therefore,

CALCULATION SHEET
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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)
Therefore, the shearing resistance of bolts is acceptable.

CALCULATION SHEET
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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.

CALCULATION SHEET
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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
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

CALCULATION SHEET
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BS EN
1993-1-
8:2005
Table 3.4
Bearing Resistance of Web
Hence,

CALCULATION SHEET
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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,

CALCULATION SHEET
Bracing
Subject
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Longitudinal wind bracing
Figure 1 Side view
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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

CALCULATION SHEET
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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CALCULATION SHEET
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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CALCULATION SHEET
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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CALCULATION SHEET
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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CALCULATION SHEET
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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CALCULATION SHEET
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
Yield strength
BS steel data
Page 87
BS EN 1993-1-
8:2005
Table 3.1
Subject
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CALCULATION SHEET
Plate details
Yield strength
Plate thickness
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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CALCULATION SHEET
Where,
(for class 4.6, 5.6 and 8.8)
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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CALCULATION SHEET
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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CALCULATION SHEET
Design ultimate tensile resistance of the net cross section
Therefore, the tension resistance of the cross section is adequate.
Design for block tearing
Where,
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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CALCULATION SHEET
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
Subject
Made By: Date
Checked by: Date
Client:
Atkins & Partners

CALCULATION SHEET
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.
Subject
Made By: Date
Checked by: Date
Client:
Atkins & Partners

CALCULATION SHEET
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
Module: H23S07 Sheet of
Subject
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Checked by: Date
Client:
Atkins & Partners

CALCULATION SHEET
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
Subject
Made By: Date
Checked by: Date
Client:
Atkins & Partners
(h-2tf)/2 = (314.5-2(18.7))/2
= 138.55mm >13.18 mm
Therefore there is no overlap
between the flanges

CALCULATION SHEET
Subject
Made By: Date
Checked by: Date
Client:
Atkins & Partners
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.

CALCULATION SHEET
Subject
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
Made By: Date
Checked by: Date
Client:
Atkins & Partners
We assume force of
roof acts parallel to
plate as angle of
inclination very
small

CALCULATION SHEET
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
Subject
Made By: Date
Checked by: Date
Client:
Atkins & Partners

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
CALCULATION SHEET
Subject
Made By: Date
Checked by: Date
Client:
Atkins & Partners

CALCULATION SHEET
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
Subject
Made By: Date
Checked by: Date
Client:
Atkins & Partners

CALCULATION SHEET
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
Subject
Made By: Date
Checked by: Date
Client:
Atkins & Partners

= 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
CALCULATION SHEET
Subject
Made By: Date
Checked by: Date
Client:
Atkins & Partners

CALCULATION SHEET
EN 1993-1-8; 2005
Subject
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
Made By: Date
Checked by: Date
Client:
Atkins & Partners

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