Reference Calculations

Design of PB2
Basic Design Information

Length of PB2 = 6.00 m

Grade of Steel considered for design = S 275
material partial factor of safety M0 = 1.00
material partial factor of safety M1 = 1.00
material partial factor of safety M2 = 1.25
Elastic Modulus of Steel (E) = 210,000.00 N/mm2
Poisson's ratio of Steel = 0.30
Shear Modulus of Steel (G) = 80,769.23 N/mm2

Evaluation of Loads

Permanent action due to thickness of slab = 3.13 kN/m2

Other permanent action as per design information = - kN/m2
Total permanent action on PB2 due to slab = 3.13 kN/m
Permanent action due to wall (200mm thk block wall) = 12.00 kN/m
Permanent action due to self weight of PB2 (assume) = 0.50 kN/m
Total permanent action on PB2 = 15.63 kN/m
Variable action as per design information = 4.00 kN/m2
Variable action on PB2 due to slab = 4.00 kN/m
Minimum load combination (1.0 Gk + 1.0 Qk) = 19.63 kN/m
Maximum load combination (1.35Gk + 1.5Qk) = 27.09 kN/m

Assuming the connections are simply supported, the maximum BM

for PB2 will occur at midspan,

Design bending moment MEd = 121.92 kNm

Design shear force VEd = 81.28 kN

1.0 Initial Sizing

Assuming class 1/2 section, the required minimum plastic
modulus of PB2 = Wpl,min
 M 0  M Ed
wPl , min 
fy

wpl,min = 443.35 cm3

Hence from Steel Tables,
Select 305 x 127 x 42 UB section as PB2 with the following properties,
Reference Calculations

Steel Designer's
Manual 6th
Edition pp.1169-
1170

D = 304.4 mm
B = 123.4 mm
T = 10.7 mm
t = 7.1 mm
Area = 4720 mm2
Torsional
Constant = 148000 mm6
Warping
Constant = 7.25E+10 mm6
Iyy = 71700000 mm4
Izz = 3360000 mm4
wyy,elastic = 471000 mm3
wzz,elastic = 54500 mm3
wyy,plastic = 539000 mm3
wzz,plastic = 85400 mm3
iyy = 123 mm
izz = 26.7 mm
Buckling
parameter = 0.871

Torsional
index = 29.7
Reference Calculations

2.0 Section Classification

The beam PB2 is under flexure, with negligible axial force. The web is under
bending and bottom flange under compression.

235 = 0.92

fy
BS EN 1993-1-
1:2005 For the Web (internal) C/t = 37.40
Table 5.2 Since 37.4 < 66.56 72 

Web can be classified as Class 1

For outstand flanges

C/t = 5.77
Since 5.77 < 8.32 9 

Flange can be classified as Class 1

Hence the entire section could be classified as Class 1

3.0 Check for Shear Capacity

Cl 6.2.6 If VEd < 0.5Vc,Rd then it would be unneccessary to apply a reduction to the
yield strength of Steel owing to the impact of Shear on a flexural member.
However, it is important that VEd < Vc,Rd to prevent Shear failure.

When there is no Torsion, V c , Rd 


Av f y 3 
M0
Vc,Rd = 372.41 kN
whereas, VEd/Vc,Rd = 0.22

Hence, the section is safe from Shear failure and does not require shear
correction.

3.1 Check for Shear Buckling

 𝜀/𝜂
Shear buckling check not required if hw/tw < 72
42.87323944 < 66.56
Hence, not required
Reference Calculations
IStructE
4.0 Check for deflection

EC3 Manual for 6.2 wL3

Designing of  
Steel Structures I
= 21.99 mm
The deflection at SLS as a ratio of the length of PB2 is less than 1/250
Hence, Satisfactory

5.0 Check for Lateral Torsional Buckling

Assuming no composite action/no lateral restraint provided by concrete
Designer's Guide 0. 5
to EC3 - 2nd  2 EI z  I w L2 GI T 
Edition Eqn M cr  C1   2  Eqn D6.10
D6.10 L2  I z  EI Z 
Table 4.4 C1 = 1.13

Mcr = 332.62 kNm

Assuming lateral restraints have been provided at 2.0m regular intervals,

Wy f y
 LT 
M cr
 LT = 0.67

Eqn 6.56   
 LT  0.5 1   LT  LT  0.4    LT
2
 and h/b = 2.47 >2

Table 6.5 The buckling curve "c"shall be selected to obtain the imperfection factor

 LT = 0.49
 LT = 0.73
1
Eqn 6.57  LT 
2 2
 LT   LT    LT
XLT = 0.85
 LT
 LT ,mod 
f
where,  
f  1  0.51  k c 1  2  LT , 0  0.8
obtained from Table 6.6
2
and kc can be

f= 0.98
 XLT,mod = 0.86

Now the maximum bending moment withstandable with restraints

provided at 2.0m intervals,
Eqn 6.55 Mb,Rd = 127.93 kNm

Reference Calculations
Hence, if unrestrained lateral supports need to be provided at 2.0m intervals.

Design of Steel Column

Since the Steel beams are connected to the Columns via pin supports, there will
be no out of balance moment acting on the columns.

Consider an edge column for design,

Axial load on Column (according to SAP2000 model) = 975.00 kN

Bending moments conveyed by beams = - kN
Bending moments conveyed due to wind = - kN
Hence initial bending moment M0 = - kN

IStructE eccentricity ec is considered as 100mm.

EC3 Manual for
the design of
Steel structures
Cl 5.5 (d)

Eccentricity parallel to web e1 = 0.23 m

Eccentricity perpendicular to web e2 = 0.10 m
Hence the corresponding bending moments M1 = 30.79 kNm
M2 = 2.24 kNm

Assume 254x254x73UC section for Column

Reference Calculations

Steel Designer's
Manual 6th
Edition pp.1169-
1170

D = 254.1 mm
B = 254.6 mm
T = 14.2 mm
t = 8.6 mm
Area = 9310 mm2
Torsional
Constant = 576000 mm4
Warping
Constant = 5.62E+11 mm6
Iyy = 114000000 mm4
Izz = 39100000 mm4
wyy,elastic = 898000 mm3
wzz,elastic = 307000 mm3
wyy,plastic = 992000 mm3
wzz,plastic = 465000 mm3
iyy = 111 mm
izz = 64.8 mm
Buckling
parameter = 0.849
Torsional
index = 17.3

1.0 Section classification

The edge Column is under axial force and bending, both web and flange
are under compression
 235 = 0.92

fy
BS EN 1993-1-
1:2005 For the Web (internal) C/t = 23.29
Table 5.2 Since 23.29 < 30.51 (33𝜀)

Web can be classified as Class 1

Reference Calculations

For outstand flanges

C/t = 8.96
Since 8.96 > 8.32 9 

Flange can be classified as Class 2

Hence the entire section could be classified as Class 2

2.0 Check for effect of axial load on plastic moment resistance

Cl 6.2.9.1 In order to not reduce plastic moment resistance due to the effect of axial loads
the following two conditions are to be met
(𝐴𝑓_𝑦)/𝛾_𝑀0
〖𝐼 . 𝑁 〗 _𝐸𝑑≤0.25𝑁_(𝑃𝑙,𝑅𝑑)where NPl,Rd =

= 2,560.25 kN
n=𝑁_𝐸𝑑/ 975.00
𝑁_(𝑃𝑙,𝑅𝑑) = 2,560.25
II. 𝑁_𝐸𝑑≤(0.5ℎ_𝑤 𝑡_𝑤 𝑓_𝑦)/𝛾_𝑀0 = 0.38 >0.25

There is no necessity to check the second condition. Plastic moment resistance

reduction is required.

Cl 6.2.9.1 (5) Hence, the reduced plastic moment resistance

𝑀_(𝑁,𝑖,𝑅𝑑)=𝑀_(𝑃𝑙,𝑖,𝑅𝑑) (1−𝑛)/(1−0.5𝑎);𝑤ℎ𝑒𝑟𝑒 𝑎≤2𝑛

n = 0.38
a = (1 - 2btf/A) = 0.22 <2n

MN,y,Rd = 190.15 kNm

MN,z,Rd = 89.13 kNm
IstructE Manual
for EC3 Cl5.4 For simple multistorey construction braced in both directions, the columns should be
designed by applying nominal moments only at the beam to column connections.
The following should be met:
(a) columns should be effectively continuous at their splices.
(b) pattern loading may be ignored.
(c) all beams framing into the columns are assumed to be fully loaded
(d) nominal moments are applied to the columns about the two axes.
(e) nominal moments are proportioned between length above and below the beam.
(f) nominal moments are assumed to have no effects at levels above the level applied.
(g) the slenderness ratio of the columns should not exceed 180.

Reference Calculations
3.0 Check for buckling
NCCI :
Verification of For columns in simple construction, the following condition is to be met to ensure
Columns in
Simple no buckling in any direction (the slenderness ratio at both directions < 180)
Construction
SN048b-EN-GB

Where, NEd , My,Ed, Mz,Ed are the axial load and the Axial load, and Bending
moments about the major and minor axis respectively.

Nmin,b,Rd , My,b,Rd and Mz,cb,Rd which are the axial resistance, and buckling
resistance about the major and minor axis respectively, could be determined
as follows,

3.1 Axial resistance

 min Af y
N min, b , Rd 
 M1
1  93.9
235
1  93.9
fy
1  86.80
IStructE Manual
for EC3 Both ends are pin connected. Hence, (Leff /L) = 1.00
Table 5.1 Leff ,rr
rr 
irr
4.00
 yy 
0.111
𝜆_(𝑦𝑦 )= 36.04
𝜆_𝑧𝑧= 61.73
(𝜆_𝑦𝑦 ) ̅= 0.42
(𝜆_𝑧𝑧 ) ̅= 0.71
h/b = 1.00
BS EN 1993-1-
1:2005 tf = 14.2 mm
Table 6.2 Buckling curve for yy - axis = b
Buckling curve for zz - axis = c

BS EN 1993-1-
1:2005 𝜒_𝑦𝑦 = 0.94
Fig 6.4
𝜒_𝑧𝑧 = 0.73

𝜒_𝑚𝑖𝑛 = 0.73
𝑁_(𝑚𝑖𝑛,𝑏,𝑅𝑑)=(𝜒_𝑚𝑖𝑛 𝐴𝑓_𝑦)/𝛾_𝑀1
= 1,868.98 kN

Reference Calculations

Cl 6.3.2 3.2 Torsional buckling / Torsional - flexural buckling resistance

WPl f y
M y ,b , Rd   LT
 M1
0. 5
 2 EI z  I w L2 GI T 
IStructE Manual M cr  C1   2  Eqn D6.10
for EC3 L2  I z  EI Z 
Table 4.4 As a conservative approach C1 = 0.75

Mcr = 2,617.73 kNm

BS EN 1993-1-1:
2005
Wy f y
Cl 6.3.2.2  LT 
M cr
 LT = 1.00

Eqn 6.57   
 LT  0.5 1   LT  LT  0.4    LT
2
 and h/b = 1.00 <2.0
   
 LT  0.5 1   LT  LT  0.4    LT
2

The buckling curve "b"shall be selected to obtain the imperfection factor

 LT = 0.34

 LT = 0.98
1
 LT 
2 2
 LT   LT    LT

XLT = 0.70

 LT
 LT ,mod 
f
Reference Calculations

Table 6.6  
where, f  1  0.51  k c 1  2  LT , 0  0.8 
2
and kc can be
obtained from Table 6.6
f= 0.92
XLT,mod = 0.76

𝑀_(𝑦,𝑏,𝑅𝑑)=𝜒_𝐿𝑇×𝑀_(𝑁,𝑏,𝑅𝑑)= 145.40 kNm

3.3 Moment resistance of the minor axis

This was obtained under the reduced moment capacity due to the effect of
axial loads. Hence
Mz,cb,Rd = 89.13 kNm

Recalling the equation 6.44

975.00 30.79 3.36

+ + = 0.77 <1.00
1,868.98 145.40 89.13

3.4 Check for biaxial bending

0.0448512 + 0.00 = 0.05 <1.00

Hence, Satisfactory
Reference Calculations

Design of Steel Beam - Column Connection

The connection established between the Steel Beams and Columns in this case
are all Shear connections. The Shear connections are ensured by only connecting
the web of the beam to the Column.

Joints in Steel 40 mm
Construction :
Simple Joints to 2/90 x 90 x 8 L angle cleat S275
EC3 (SCI P358)
40 mm
105 mm 356 x 171 x 67kg/m I section
web thickness = 9.1mm
S275

V = 145kN M 16
Gr 8.8

EN 1993-1-8
Table 3.4 1.0 Check for Bolt Shearing resistance

A. For the bolts along the supporting Column

Capacity of one bolt

𝛼_𝑣= 0.60
Table 3.1 fub = 800.00 N/mm2
A = 157.00 mm2
 Fv,Rd = 60.29 kN
Capacity of group

𝛼=
𝛽= 0.24
VRd = 162.55 kN
VRd > 145.00 kN

B. For the bolts at supported beam

Shear force acting on one bolt Fv,Ed = 48.33 kN

Maximum shear force on bolt assembly in horizontal MEd
=
direction np1
Fh,Ed = 23.02 kN
Maximum design resultant shear force Fv,Ed = 53.53 kN
Reference Calculations

Due to double shear acting on bolt at supported beam web , Fv*,Rd = 2 x Fv,Rd
Capacity of one bolt 120.58 kN FV*,Rd > 53.53 kN

Hence, bolt shear capacity check OK

2.0 Check for Bolt Bearing resistance

A. For the bolts along the supporting Column

Bearing resistance of bolt group
𝐹_(𝑏,𝑅𝑑)=(𝑘_1 𝛼_𝑏 𝑓_𝑢 𝑑𝑡)/𝛾_𝑀2
𝛼_𝑏=min⁡{𝛼_𝑑,𝑓_𝑢𝑏/𝑓_𝑦 ,1.0}
where,
ultimate tensil strength of L angle fu,cleat = 430.00 N/mm2
thickness of L angle cleat t = 8.00 mm
fub/fu,cleat = 1.86
for end bolts ad = e1/3do
for inner bolts ad = (p1/3do) - 0.25
Hence,
ad,end = 0.74
ad,inner = 1.69

𝑘_(1,𝑒𝑑𝑔𝑒)=min⁡{(2.8𝑒_2)/𝑑_𝑜 −1.7;1.4𝑝2/𝑑_0 −1.7;𝑎𝑛𝑑 2.5}

where,
𝑘_(1,𝑖𝑛𝑛𝑒𝑟)=min{1.4 𝑝2/𝑑_0 −1.7 𝑜𝑟 2.5}

4.52
k1,edge 6.63
2.50
k1,edge,min = 2.50
6.63
k1,inner
2.50
 k1,inner,min = 2.50
minimum bearing resistance per bolt Fb,Rd,min = 81.54 kN
Bearing resistance of bolt group Vb,Rd = nxFb,Rd,min
= 489.24 kN
Hence OK

Cl 6.2.6 (2) 3.0 Check for Shear resistance of cleats connected to supporting column

Length of angle cleat = 290.00 mm

(𝐴_𝑣 𝑓_(𝑦,𝑐𝑙𝑒𝑎𝑡)/(√3))/𝛾_𝑀0
VPl,Rd =

I. Gross Section Av = 2,320.00 mm2

VPl,Rd = 575.96 kN
For 2L angle cleats = 1,151.93 kN

Reference Calculations
II. Net section Av,net = tp(hp - n1do)
Av,net = 1,888.00 mm2

(𝐴_𝑣 𝑓_(𝑦,𝑐𝑙𝑒𝑎𝑡)/(√3))/𝛾_𝑀2
VPl,Rd =

VPl,Rd = 239.81 kN
For 2L angle cleats = 479.62 kN

Hence OK

Table 3.4 Bearing resistance of beam web

𝐹_(𝑏,𝑅𝑑)=(𝑘_1 𝛼_𝑏 𝑓_𝑢 𝑑𝑡)/𝛾_𝑀2

𝛼_𝑏=min⁡{𝛼_𝑑,𝑓_𝑢𝑏/𝑓_𝑦 ,1.0}
where,
ultimate tensil strength of I section (web) fu,web = 430.00 N/mm2
thickness of web tw = 9.10 mm
fub/fu,web = 1.86
for end bolts ad = e1/3do
for inner bolts ad = (p1/3do) - 0.25
Hence,
ad,end = 0.74
ad,inner = 1.69

𝑘_(1,𝑒𝑑𝑔𝑒)=min⁡{(2.8𝑒_2)/𝑑_𝑜 −1.7;1.4𝑝2/𝑑_0 −1.7;𝑎𝑛𝑑 2.5}

where,
𝑘_(1,𝑖𝑛𝑛𝑒𝑟)=min{1.4𝑝2/𝑑_0 −1.7 𝑜𝑟 2.5}

4.52
k1,edge
 k1,edge -
2.50
k1,edge,min = 2.50
-
k1,inner
2.50
k1,inner,min = 2.50

Fb,Rd = 92.75 kN

Since Fb,Rd > FV,Ed the web resistance of the beam is satisfactory

Reference Calculations

Cl 3.10.2(2) 4.0 Check for block tearing resistance

For symmetric blot group subjected to concentric loading

𝑉_(𝑒𝑓𝑓,𝑙,𝑅𝑑)=(𝑓_𝑢
𝐴_𝑛𝑡)/𝛾_𝑀2 +(𝑓_𝑦
𝐴_𝑛𝑣/√3)/𝛾_𝑀0

Ant = (a2-0.5d0)tw
Ant = 364.00 mm2

Anv = {Lv + (a1-(n1-0.5)d0)tw

Anv = 2,184.00 mm2

Veff,l,Rd = 471.97 kN

Since Veff,l,Rd > VEd the block tearing resistance of the beam web is satisfactory
Reference Calculations

Design of Plate girder with Columns

Grade of Steel used for plates = S 275

Span at Grid 2 = 18,000.00 mm
Depth to be maintained between 1/12 < D < 1/8; D = 1,600.00 mm
Breadth to be maintained between D/5 < B < D/3; B = 350.00 mm
Flange thickness = 25.00 mm
Web thickness = 12.00 mm
end distance for rigid end post = 350.00 mm
end post type = NON RIGID
equally spaced transverse stiffener interval = 3,300.00 mm
BM at midspan = 3,724.00 kNm
Maximum shear force = 766.00 kN

1.0 Section classification

2.0 Section Classification

The beam PB2 is under flexure, with negligible axial force. The web is under
bending and bottom flange under compression.

235 = 0.92

fy
 235

fy
BS EN 1993-1-
1:2005 For the Web (internal) C/t = 129.17
Table 5.2 Since 129.17 > 114.63 (124𝜀)

Web can be classified as class 4

For outstand flanges

C/t = 6.76
Since 6.76 < 8.32 9 

Flange can be classified as Class 1

3.0 Determination of bending resistance

Using the flanges only method, the moment resistance of the section
could be expressed as, 𝑀_(𝑓,𝑅𝑑)=2𝑏_𝑓 𝑡_𝑓
(ℎ_𝑤/2+𝑡_𝑓/2) 𝑓_𝑦

Mf,Rd = 3,789.84 kNm

Reference Calculations

BS EN 1993-1-5: check for flange induced buckling

2005 Cl8 (1) ℎ_𝑤/𝑡_𝑤 ≤𝑘 𝐸/𝑓_𝑦𝑓
to prevent flange induced buckling, it should suffice √(𝐴_𝑤/𝐴_𝑓𝑐 )

k = 0.30
Aw = 18,600.00 mm2
Afc = 8,750.00 mm2
(hw/tw)max = 334.01
(hw/tw)actual = 129.17

4.0 Resistance to shear

4.1 Shear Lag effect

The section does not have to be checked for shear lag if b o < Lc/50
Cl 3.1 bo = 350.00
Lc/50 = 360.00
 Hence, section does not need to be checked for shear lag

Cl 5.1 4.2 Shear buckling

For stiffened web if the following requirement ℎ_𝑤/𝑡_𝑤 ≤31/𝜂 𝜀√(𝑘_𝜏 )
is met, the web shall be "stocky" for shear buckling
Note2 𝜂 = 1.20
a/hw = 2.13
𝑘_𝜏=5.34+4(ℎ_𝑤/𝑎)^2
APPENDIX A
= 6.22
Hence, (hw/tw)max = 59.57
(hw/tw)actual = 129.17

Hence, section is to be checked for shear buckling resistance

Cl 5.2 𝑉_(𝑏,𝑅𝑑)=𝑉_(𝑏𝑤,𝑅𝑑+) 𝑉_(𝑏𝑓,𝑅𝑑)

𝑉_(𝑏𝑤,𝑅𝑑)=(𝜒_𝑤 𝑓_𝑦𝑤 ℎ_𝑤 𝑡)/(√3

Cl 5.3 contribution from the web against shear buckling,
𝛾_𝑀1 )

Note 3 (b) (𝜆_𝑤

) ̅=ℎ_𝑤/(37.4𝑡𝜀√
(𝑘_𝜏 ))
(𝜆_𝑤 ) ̅ 1.50
=

Reference Calculations

Fig 5.2

𝜒_𝑤
= 0.62
Vbw,Rd = 1,830.95 kN
 The contribution from flanges could be neglected hence Vb,Rd = Vbw,Rd

Eq 5.1 the code imposes a maximum value for shear buckling resistance which is
(𝑉_(𝑏𝑤,𝑅𝑑))𝑚𝑎𝑥=(𝜂
ℎ_𝑤 𝑡)/(√3 𝛾_𝑀1 )
= 3,543.78 kN

Hence Vb,Rd = 1831kN could be considered as the shear buckling

resistance which is significantly higher than VEd.

5.0 Interaction of bending and shear

VEd/Vb,Rd = 0.42
0.42 is lesser than 0.5, no further checks for moment - shear interaction are required.

6.0 Design of end post & transverse stiffeners

3,300.00

18000

Reference Calculations

6.1 Design of stiffener - initial sizing

15𝜀𝑡 15𝜀𝑡
Cl 9.1 Fig 9.1 _𝑤 _𝑤

𝑑_1
𝑡_𝑤

𝐴
_
𝑠
width of stiffener, d1 = 162.00 mm
〖
thickness_ of stiffener t = 12.00 mm
〗
EC 1993-1-5: 〖
2006 Cl 9.3.3(3) a/hw =_ 2.13 > √2
〗
Eqn 9.6 Ist > 0.75hwt3
(Ist)min = 200.88 cm4
(Ist)actual = 425.15 cm4
 6.2 Check for capacity of intermediate stiffener
Recalling Fig 9.1, part of the web also contributes to the stiffener
15𝜀𝑡 = 166.39 mm

The stiffener coupled with the contribution from the web forms an imaginery column
this column, should be designed to resist the axial load at the stiffener location
in this case, the most onerous case is for the stiffener adjacent to the support
column. Hence, as a conservative approach it is assumed that
the maximum shear force VEd shall be acting on the imaginery column as an axial load.
stiffeners are to be designed for an axial load if the following condition is not met
Critical shear resistance > VEd
NOTE 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑠ℎ𝑒𝑎𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒= [(𝑓_𝑦𝑤 ℎ_𝑤
𝑡)/(√3×(𝜆_𝑤^2 ) ̅𝛾_𝑀1 )]

= 1,316.51 kN
Hence, design for axial load not required.

6.3 Check for lateral torsional buckling

Structural Steel
Work: Design to For the stiffener plate IT = 8.90 cm4
Limit State IP = 1,700.61 cm4
Theory, 4th Edn. IT/TP = 0.01
Eqn 5.24 (IT/TP)min = 5.3fy/E
= 0.007

Reference Calculations
Hence, no lateral torsional buckling
There is no requirement to check for load bearing stiffener since
the connection with the column is through a fin plate / angle cleat

6.4 Design of end post

In this case the end post is non rigid

6.5 Design of support column

 The design axial force acting on Column = 766.00 kN
Assume 533x210x101UB
Hence eccentricity about major axis = 0.37 m
Myy = 282.16 kNm
the moments about Mzz are negligible due to equally loaded spans

6.5.1 Check for effect of axial load on plastic moment resistance

Cl 6.2.9.1 In order to not reduce plastic moment resistance due to the effect of axial loads
the following two conditions are to be met
(𝐴𝑓_𝑦)/𝛾_𝑀0
〖𝐼 . 𝑁 〗 _𝐸𝑑≤0.25𝑁_(𝑃𝑙,𝑅𝑑)where NPl,Rd =

= 3,547.50 kN
n=𝑁_𝐸𝑑/ 766.00
𝑁_(𝑃𝑙,𝑅𝑑) = 3,547.50
II. 𝑁_𝐸𝑑≤(0.5ℎ_𝑤 𝑡_𝑤 𝑓_𝑦)/𝛾_𝑀0 = 0.22 <0.25

There is no necessity to check the second condition. Plastic moment resistance

reduction is required.

Cl 6.2.9.1 (5) Hence, the reduced plastic moment resistance

𝑀_(𝑁,𝑖,𝑅𝑑)=𝑀_(𝑃𝑙,𝑖,𝑅𝑑) (1−𝑛)/(1−0.5𝑎);𝑤ℎ𝑒𝑟𝑒 𝑎≤2𝑛

n = 0.22
a = (1 - 2btf/A) = 0.43 <2n

Reference Calculations

MN,y,Rd = 718.50 kNm

MN,z,Rd = 109.73 kNm

EN 1993-1-1
Eqn 6.62

the selected section is Class 1

𝜆_𝑧𝑧= 87.53
(𝜆_𝑧𝑧 ) ̅= 1.01
check for curve b, in accordance with Table 6.2
Fig 6.4
 Xzz = 0.60
Hence, buckling resistance to axial loads = 2,128.50 kN

0. 5
 2 EI z  I w L2 GI T 
IStructE Manual M cr  C1   2  Eqn D6.10
for EC3 L2  I z  EI Z 
Table 4.4 As a conservative approach C1 = 0.75

Mcr = 787.61 kNm

BS EN 1993-1-1:
2005
Wy f y
Cl 6.3.2.2  LT 
M cr
 LT = 0.95
Eqn 6.57   
 LT  0.5 1   LT  LT  0.4   
2
LT  and h/b = 2.56 >2.0

Reference Calculations
The buckling curve "c"shall be selected to obtain the imperfection factor

 LT = 0.49

 LT = 0.98

1
 LT 
2 2
 LT   LT    LT

XLT = 0.67

 LT
 LT ,mod 
f
Table 6.6 where,
obtained from Table 6.6
 
f  1  0.51  k c 1  2  LT , 0  0.8 
2
and kc can be

f= 0.92
 XLT,mod = 0.73

Lateral torsional buckling resistance = 523.16 kNm

This I section satisfies the combined effect of axial load and bending

Reference Calculations

Design of wind bracings

The wind bracing shall be designed for the section under compression as
steel is stable under tension.

The maximum axial load acting on the bracings in Grid 5 = 60.00 kN

Let's consider a hot finished circular hollow section, where d = 114.30 mm

t= 5.00 mm
L= 7,210.00 mm
1.0 Section classification
Table 5.2
Sheet 3 235 = 0.92

fy
 235

fy
d/t = 22.86
22.86 < 50

Hence the section could be classified as class 1

A= 1720 mm2
I= 2570000 mm4
i= 38.7 mm
welastic = 45000 mm3
wplastic = 59800 mm3

2.0 Check for flexural buckling

Cl 6.3.1.3 𝜆 ̅=𝐿_𝑐𝑟/(𝑖𝜆_1 )

Table 6.2 the buckling curve type is "a"

𝜆 ̅= 2.15
𝜙= 3.01
𝜒= 0.20

design buckling resistance of member = 92.48 kN

Hence the selected section of 4.5"diameter, 5mm thickness is adequate.

272.81
0.697016
60288
3.20
8.00
707602.5
0.709522

1.4094
221.2758

kn
knpm2 m
 8.55 321.4286
0.297619

16.5

1.929825