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Concrete Filled CHS Composite Column
The following Eurocodes and pre-Eurocodes have been used for this worked example:
BS EN 1990, Basis of Structural Design, July 2002, with UK National Annex, March 2004
BS EN 1991-1-1, Eurocode 1 Actions on structures Part 1.1: General actions Densities, self-weight,
imposed loads for buildings, July 2002
prEN 1992-1-1, Eurocode 2 Design of concrete structures Part 1.1: General rules and rules for
buildings, April 2003
prEN 1993-1-1, Eurocode 3 Design of steel structures Part 1.1: General rules and rules for buildings,
December 2003
prEN 1994-1-1, Eurocode 4 Design of composite steel and concrete structures Part 1.1: General rules
and rules for buildings, January 2004
prEN 10025-2, Hot rolled products of non-alloy structural steels Part 2: Technical delivery conditions for
flat products, March 1998
The following design guidance document has been used for this worked example:
T.T.Lie and V.K.R. Kodur (1996), Fire Resistance of Steel Columns Filled with Bar-Reinforced Concrete,
Journal of Structural Engineering, Vol. 122, Jan 1996, ASCE
Notes on European Standards
BS EN denotes a European Standard that has been published by BSi
prEN denotes a draft European standard that is not publicly available
Note on values contained in this worked example
The computer software used to calculate the expressions given in this worked example does not
round the values at intermediate stages during the calculation. Therefore some values given on thefollowing sheets may appear to be incorrect when determined using rounded input values.
BRE and Buro Happold have made every effort to ensure the accuracy and quality of all the information
in this document when first published. However, they can take no responsibility for the subsequent use
of this information, nor for any errors or omissions it may contain.
Queen's Printer and Controller of Her Majesty's Stationery Office 2005
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NG+NQ
4 . 5
0 m
MG+MQ
Figure 0: Overview of Composite Column
Design Example of a concrete filled composite column to EN1994-1-1 Tables and clause
numbers relate to
Consider the 4.5m high column shown EN1994-1-1
subject to an axially applied point unless stated
load, and an applied moment. This otherwise
mimics the common situation where a
column is subject to axially applied
load from storeys above and a
moment induced by a floor beam at
the storey considered.
The column section is circular andtherefore doubly symmetric. It is
assumed to be pinned top and
bottom.
It has been designed using the
simplified method of analysis in
accordance with clause 6.7.3EN1994.
The design is based on first order
analysis with appropriate amplificationto the design moments to account for
second order effects.
The required fire resistance is 60minutes.
1. Design Data
L = 4.50 m
Effective or buckling length, L e = 4.50 m
Note: The Eurocodes do not give any guidance regarding effective or buckling lengths ofmembers subject to compressive axial load. Reference should be made to textbooks for
this information.
1.1. Loading
1.1.1. Permanent Actions (G)
NG = 3000 kN
MG = 50.00 kNm
1.1.2. Variable Actions (Q)
NQ = 1300 kNMQ = 45.00 kNm
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1.2. Loading factors
Partial loading factor for permanent actions, !G = 1.35 EN 1990 TablePartial loading factor for variable actions, !Q = 1.50 A1.2(B) & N.A
1.3. Materials
1.3.1. Steelwork
Grade S355, nominal yield strength, f y = 355 N/mm2
Partial Safety Factor, !M0 = 1.00 6.1(1) EN1993-1-1Note: Recommended value used for ! M0 . This value may be altered by the UK National
Annex to EN1993-1-1.
Design strength, f yd = f y/!M0 = 355 N/mm2
Modulus of elasticity, E a = 210 kN/mm
2 3.2.6(1)EN1993-1-1
1.3.2. Concrete
Normal Weight concrete strength class C40/50
Characteristic cylinder strength, f ck = 40 N/mm2
Partial Safety Factor, !c = 1.5 Table 2.1NNote: Recommended value for ! c , given in Table 2.1N, EN1992-1-1. This value may be EN1992-1-1 altered by the UK National Annex to EN1992-1-1.
Design value of concrete cylinder strength, f cd = f ck /!c = 26.7 N/mm 2 Secant modulus of elasticity, E cm = 35.0 kN/mm 2 Table 3.1
Note: alternatively the secant modulus can be calculated from, E cm = 22[(f cm /10) 0.3 ], EN1992-1-1 where f cm is the mean value of concrete c ylinder compressive strength and is equal to
f ck +8(MPa.)
1.3.3. Reinforcement
Grade S460, yield strength, f sk = 460 N/mm2
Partial Safety Factor, !s = 1.15 Table 2.1NNote: Recommended value for ! c , given in Table 2.1N, EN1992-1-1. This value may be EN1992-1-1 altered by the UK National Annex to EN1992-1-1.
Design strength, f sd = f sk /!s = 400.0 N/mm 2 Modulus of elasticity, E s = 210 kN/mm
2 Note: for composite structures, the design value of the modulus of elasticity E s may be
taken as equal to the value for structural s teel given in EN1993-1-1, 3.2.6 (from clause
3.2(2), EN1994-1-1).
2. Global Analysis
The effects of deformed geometry (second-order effects) must be considered. In
addition, appropriate allowances must be incorporated within the structural analysis
to cover the effects of imperfections, including geometrical imperfections such as
lack of verticality.
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Second-order effects will be included indirectly by using a first-order analysis 5.2.2(2)
modified with appropriate amplification. The amplification considered in this example
is in accordance with clause 6.7.3.4(5). See section 12 for details.
In accordance with clause 5.3.2.3(1), a design value for initial bow for the composite
column has been taken from Table 6.5. Assuming a reinforcement ratio of less than
3%, the design value of initial bow should be taken as L/300. Therefore, the member
imperfection should be taken as 15.00 mm
The effects of the applied support moment and the moment due to the initial member
imperfection have been combined and the maximum combined moment at either the
support or mid-span has been used as the design bending moment.
2.1. Design Value Actions
NEd = N G" !G + N Q" !Q = 6000 kNMEd = 0.5 " (MG" !G + M Q" !Q) + N Ed" 0.015 = 158 kNm (at mid height)VEd = (M G" !G + M Q" !Q)/L = 30 kN
3. Trial Column size
Trial column size assumed considering axial load only, given the required fire
resistance is (from Lie and Kodur (1996)):
Trial diameter = [(t fire" (Le -1000)) /(0.08 " (f ck + 20)) " # (NEd )]0.4 where: t fire = required fire resistance in minutes = 60 minutes
! trial column diameter = 409 mm
Try CHS 406.4 section
determine minimum wall thickness to prevent local buckling:
maximum (d/t) = 90 " ( # (235/f y)) = 73.225 Table 6.3minimum t = 5.550 mm therefore min 6 mm thick wall required
Try CHS 406.4x10.0 section
d = 406.4 mm
t = 10.0 mm A a = 124.5 cm2
Ia = 24476 cm4 W pa = 1572 cm
3
Assume 1.5% reinforcement, based on concrete area. Approximate reinforcementrequired is:
minimum reinforcement = (( $" (d-2 " t)2)/4) " 1.5/100 = 1759 mm 2
A minimum of 6 bars should be adopted, therefore try 10 No. 16 dia bar(s)
As = 2010 mm2 d rebar = 16 mm A bar = A s /10 = 201 mm
2
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4. Composite Cross Section Properties
Aa = 124.5 cm2 As = 20.1 cm
2 Ac = (( $" d2)/4)-A a-As=1153 cm 2 concrete diameter, d c = d - 2 " t = 386 mm
Second Moment of Area of Steel Section, I a= 24476 cm 4
Second Moment of Area of Reinforcement,
Is given by:
Assume 8mm links around main
reinforcement and 50mm cover.
therefore R = d c/2-50d rebar /2 - 8 = 127 mm
y1 = 121 mm y 2 = 75 mm
therefore,
Is = 4 " Abar " y12 + 4 " Abar " y22 = 1629 cm 4
Second Moment of Concrete, I c = ($" (d-2 " t)4)/64 = 109425 cm 4
4.1. Reinforcement RatioThe ratio of reinforcement area to concrete area should not exceed 6%. In concrete 6.7.3.1(3)
filled hollow sections no longitudinal reinforcement is normally necessary unless
required for fire resistance.
actual reinforcement ratio = (A s /Ac) = 1.744 %Therefore reinforcement ratio is less than 6%
5. Simplified Method of Design
A simplified axial load - moment (N-M) interaction curve is produced in order to
determine the resistance of a composite cross-section to combined compression andbending. The simplified interaction curve is illustrated in Figure 6.19. A modified
version, indicating a CHS column example, has been reproduced below.
R y1 y2
Figure 0: Composite column cross-section
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Figure 1: Simplified Interaction Curve and Corresponding Stress Distributions
In order to produce the N-M interaction curve the cross-sectional capacities at points
A to D should be determined assuming the stress distributions indicated.
In addition, it should be noted that the Simplified Method of design is subject to the
following limitations:
The column cross-section must be prismatic and symmetric about both axes
over its whole height
The relative contribution of the steel section to the design resistance of the
composite section must be between 0.2 and 0.9
The relative slenderness of the composite column, %, must not be greaterthan 2.0
The limitations noted have been considered at the appropriate location within this
calculation.
The following sections outline the calculation of points A to D on the simplified N-M
interaction curve for the cross-section considered.
5.1. Point A on Simplified N-M Interaction Curve
The plastic resistance to compression N pl.Rd of a composite cross-section should be
calculated by adding the plastic resistances of its components: 6.7.3.2
Npl.Rd = A a" f yd + 0.85 " Ac" f cd + A s" f sd equation (6.30)
Note: for concrete filled sections the 0.85 factor can increase up to 1.0 (as shown in
Figure3). A further increase in concrete strength may also be taken in accordance withclause 6.7.3.2(6) with a concrete filled tube of circular cross section if " does not exceed
0.5 and e/d is less than 0.1 (where e is the eccentricity of loading given by M Ed /N Ed and d
N
M
A
C
D
B
A
B
C
D
f cd
f cd
f cd
f cd
f yd
f yd
f yd
f yd
hn
hn
2h n
hn
hn
2h n
f sd
f sd
f sd
f sd
Npl.Rd
Npm.Rd
Npm.Rd /2
Mpl.Rd
Mpl.Rd
MMax.Rd
+
+
+ +
+
+
--
-
-
-
- -
- -
-
--
-
-
-
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is the external diameter of the column). An iterative process is therefore required to
determine whether this additional increase in concrete strength can be used.
The plastic resistance to axial force, N pl.Rd , of a concrete filled section is therefore:
Npl.Rd = A a" f yd + A c" f cd + A s" f sd = 8298 kN
5.1.1. Check Applicability of Simplified Method
At this point we have sufficient data to check the steel section contribution ratio and
the relative slenderness of the column to confirm that the simplified method is
applicable. In addition, the value of relative slenderness may allow a further increase
in concrete strength to be taken:
The steel section contribution ratio, &, should fufill the following condition: 0.2< &
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where ' t is the creep coefficient according to clause 5.4.2.2(2), taken as 1.5 fromFigure 3.1, prEN 1992-1-1, for loading at 28 days
NEd is the total design normal force
NG.Ed is the part of this normal force that is permanent, therefore, N G.Ed = !G"NG
! E c.eff = 17.4 kN/mm2
and EI eff = E a" Ia + E s" Is + K e" Ec.eff " Ic = 66239 kNm 2
The elastic critical normal force, N cr , for a pin ended column is given by:Ncr = $2" EIeff /Le 2 = 32284 kN
Therefore the relative slenderness, % is:%= # ( N pl.Rk /N cr )%=0.555 < 2.0 therefore it is OK to use simplified method of design however, as % is now greater than 0.5 no further enhancement of the concretestrength due to confinement is allowed.
The plastic resistance to axial force, N pl.Rd , remains 8298 kN
5.2. Point D on Simplified N-M Interaction Curve
Plastic Bending Capacity, M maxRd , is given by:
MmaxRd =W pa " f yd + W pc " f cd + W ps " f sd
Plastic modulus of steel section is W pa = 1572 cm3
Plastic modulus of the reinforcement is W ps = 4 " y1" Abar + 4 " y2" Abar = 158 cm 3 Effective plastic modulus of concrete is W pc = (d c
3)/6 W ps = 9458 cm3
! Mmax.Rd = W pa " f yd + 0.5 " Wpc " f cd + W ps " f sd
Mmax.Rd = 747 kNm
The axial load at the point of maximum bending is ! Npm.Rd
therefore 0.5 " Npm.Rd = 1564 kN
5.3. Point C on Simplified N-M Interaction Curve
The plastic resistance moment of the composite section, M pl.Rd , is given by:
Determination of position of neutral axis depth, h n , when axial load is zero:
hn = (N pm.Rd A sn " (2" f sd f cd )) /(2 " d" f cd + 4 " t" (2" f yd f cd )) = 57.463 mm
-
-
-
-
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where:
Asn is the reinforcement area within h n, therefore (initial guess) A sn = 2 " Abar = 402 mm 2
and N pm.Rd is the compressive resistance of the whole area of concrete (see below).
Generally, W psn = W ps but as the worst case is where only two bars occur within h n,
and these are on the centre line; W psn = 0 cm3
Wpcn = (d-2 " t)"hn2 W psn = 1276 cm 3 Wpan = d " hn2 W pcn W psn = 66 cm 3
Determine the plastic resistance of composite section, M pl.N.Rd , taking into accountthe compressive normal force:
Mpl.N.Rd = W pan " f yd + 0.5 " Wpcn " f cd + W psn " f sd Mpl.N.Rd = 40.456 kNm
Mpl.Rd = M max.Rd M pl.N.Rd
Mpl.Rd = 707 kNm
The design value of the resistance of the concrete to compression, N pm.Rd , is given
by:
Npm.Rd = ( $" d c2)/4 " f cd = 3127 kN
5.4. Point B on Simplified N-M Interaction Curve
The value of M pl.Rd has previously been determined in order to define point C on the
N-M interaction curve.Mpl.Rd = 707 kNm
6. Plastic Resistance to Vertical Shear, V Rd
As a simplification, shear may be assumed to be resisted by the steel section alone. 6.7.3.2(4)
In the absence of torsion, the design plastic shear resistance, V pl.a.Rd , is given by: 6.2.6 (2)
Vpl.a.Rd = A v" (f y/ # (3))/ !M0 EN1993-1-1
where A v is the shear area given by:
Av = 2 " Aa /$ = 79.28 cm 2 6.2.6(3)EN1993-1-1
! Design shear resistance of composite section,V pl.a.Rd = 1625 kN 6.2.2.2
Therefore as the applied shear is less than half the shear resistance, the effect of shear on
the plastic moment resistance can be ignored 6.2.2.4(1)
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7. Resistance of Column to Buckling under Axial Loads
For simplification for composite members in axial compression, the design value of
the normal force, N Ed , should satisfy the following:
NEd / x Npl.Rd ! 1.0; 6.7.3.5(2)
where:
Npl.Rd is the plastic resistance of the composite section determined using the material
factor !M1 instead of !M0 to determine the steel strength, f yd (Note: in this case !M1 =!M0 therefore there is no change in the value of N pl.Rd ).
x is the reduction factor for the relevant buckling mode given in clause 6.3.1.2
EN1993-1-1, in terms of the relative slenderness %. The relevant buckling curve forcross-sections of composite columns are given in Table 6.5, EN1994, where ( s is thereinforcement ratio A s /Ac = 1.744 % therefore from Table 6.5, buckling curve a
should be used.
For the determination of the internal forces the design value of the effective flexural
stiffness, EI eff.II , used to determine the relative slenderness of the member, should be
determined from: 6.7.3.4(2)
EIeff.II = Ko" (E a" Ia + E s" Is + K e.II"E c.eff " Ic) equation (6.42)where K e.II is a correction factor which should be taken as 0.5
and K o is a calibration factor which should be taken as 0.9
Note: The value E c.eff has been used in place of E cm in equation (6.42) in order to allow for
long term effects (in the same way as calculated in section 6.).
therefore EI eff.II = 579. "10 6 kN/cm-2
and the elastic critical normal force, N cr , for a pin ended column is then given by:
Ncr.II = $2" EIeff.II/Le 2 = 28221 kN
Therefore the relative slenderness, % is:% = # ( N pl.Rk /N cr.II ) = 0.594
The reduction factor x is given by:
x = 1/( ) + # () 2-%2)); where: ) = 0.5 " [1 + *" (%-0.2) + %2] where * =0.21 for buckling curve a ! x = 0.89 ;
and x " Npl.Rd = 7404 kN
therefore N Ed /(x" Npl.Rd ) = 0.81 < 1.0 therefore buckling resistance is OK
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8. Resistance of Column to Bending Moments
For non-sway columns the use of the 1 st order moments calculated may be permitted
by multiplying the design bending moment, M Ed , by a correction factor, k, given by: 6.7.3.4 (5)
k = +/(1-N Ed /Ncr.eff ) ! 1.0 equation (6.43)where:Ncr.eff is the critical normal force for the relevant axis, corresponding to the effective
flexural stiffness given in clause 6.7.3.4 (2), with the effective length taken as the
column length. In this case, N cr.eff = N cr.II = 28221 kN
+ is an equivalent moment factor given in Table 6.4, which is equal to 0.66, for a
column with a moment diagram corresponding to end moments and 1.0 for momentdiagram produced from lateral load or member imperfection.
Two values of k must therefore be calculated, corresponding to the applied moment
and the moment due to member imperfections. The design moment, M Ed , should
then be modified if appropriate.
when considering the applied end moment, k 1 = +1 /(1-N Ed /Ncr.eff ) = 0.838 < 1.0therefore k 1 = 1.0
for the moment due to member imperfection, k 2 = +2 /(1-N Ed /Ncr.eff ) = 1.270
! the design moment at mid-height, M Ed = 0.5 " k1" (MG" !G + M Q" !Q) +k2" (NEd"0.015) = 181.80 kNm
9. Plot Simplified Interaction Curve (Figure 6.19 EN1994).
The points A-D, calculated above should now be plotted to produce the simplified N-
M interaction curve (Figure 6.19, EN1994):
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Figure 2: Simplified N-M Interaction Curve.
Using this simplified N-M interaction curve, the design value of the plastic moment ofthe composite member taking into account the relevant compression normal force,termed M pl.N.Rd , can be determined. The value of M pl.N.Rd , is defined as: 6.7.3.6(1)
Mpl.N.Rd = d" Mpl.Rd
and the following inequality should be satisfied:
MEd /Mpl.N.Rd = M Ed / d" Mpl.Rd ! * M
where the coefficient * M is taken as 0.9 for steel grades between S235 and S355and 0.8 for steel grades between S420 and S460.
when N Ed = 6000 kN, d is:
d = (N pl.Rd N Ed )/(N pl.Rd N pm.Rd ) = 0.444
! MEd /( d" Mpl.Rd )= 0.579 < 0.9 therefore the bending resistance taking into account thenormal force N Ed is OK
A
C
D
B
d" Mpl.Rd
Npl.Rd =8298kN x Npl.Rd =7404kN
MMax.Rd = 747kNm
N
M
Npm.Rd =3127kN
! Npm.Rd =1564kN
Mpl.Rd = 707kNm
NEd =6000kN
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Continuous Steel and Concrete Composite Beam
The following Eurocodes and pre-Eurocodes have been used for this worked example:
BS EN 1990, Basis of Structural Design, July 2002, with UK National Annex, March 2004
BS EN 1991-1-1, Eurocode 1 Actions on structures Part 1.1: General actions Densities, self-weight,
imposed loads for buildings, July 2002
prEN 1992-1-1, Eurocode 2 Design of concrete structures Part 1.1: General rules and rules for
buildings, April 2003
prEN 1993-1-1, Eurocode 3 Design of steel structures Part 1.1: General rules and rules for buildings,
December 2003
prEN 1994-1-1, Eurocode 4 Design of composite steel and concrete structures Part 1.1: General rules
and rules for buildings, January 2004
prEN 10025-2, Hot rolled products of non-alloy structural steels Part 2: Technical delivery conditions for
flat products, March 1998
Notes on European Standards
BS EN denotes a European Standard that has been published by BSi
prEN denotes a draft European standard that is not publicly available
Note on values contained in this worked example
The computer software used to calculate the expressions given in this worked example does not
round the values at intermediate stages during the calculation. Therefore some values given on the
following sheets may appear to be incorrect when determined using rounded input values.
BRE and Buro Happold have made every effort to ensure the accuracy and quality of all the information
in this document when first published. However, they can take no responsibility for the subsequent use
of this information, nor for any errors or omissions it may contain.
Queen's Printer and Controller of Her Majesty's Stationery Office 2005
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Design Example of a 20m span continuous composite beam to EN1994-1-1. Table and clause
Consider the internal composite beam A-A between edge columns and a numbers relate to
central support. Beam is subject to uniform floor loading and is assumed to EN1994-1-1
be fully continuous. unless stated
otherwise
Figure 1: Floor Layout
b = 3.00m b = 3.00mb = 3.00m b = 3.00m
L = 10.00m
Composite Beam
CompositeDeck Span
L = 10.00m
A
A
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where f cm is the mean value of concrete cylinder compressive strength and is equal to
f ck +8(MPa.)Dry Density = 24.0 + 1.0 kN/m 3 (for normal percentage reinforcement) Table A.1 Annex AWet Density = 24.0 + 1.0 + 1.0 kN/m 3 (for normal percentage reinforcement) EN 1991-1-1
1.2.3. Shear Connectors
19mm diameter stud, d stud = 19.00 mmNote: EN1994-1-1 uses d to denote the clear depth of the web of the structural section,
the overall diameter of a circular hollow section and the minimum transverse dimension of
a column, in addition to the diameter of a stud connector. The variable d stud has been used
in this example to differentiate between these various definitions.
95mm length after welding h sc = 95mmUltimate tensile strength of stud, f u = 450.00 N/mm
2
Partial Safety Factor, !V = 1.25 2.4.1.2(5) andNote: Recommended value used for ! V . This value may be altered by the UK National 6.6.3.1(1)
Annex to EN1994-1-1.
1.2.4. Reinforcement
Grade S460, yield strength, f sk = 460 N/mm2
Partial Safety Factor, !s = 1.15 Table 2.1NNote: Recommended value for ! c , given in Table 2.1N, EN1992-1-1. This value may be EN1992-1-1 altered by the UK National Annex to EN1992-1-1.
Modulus of elasticity, E s = 210 kN/mm2
Note: for composite structures, the design
value of the modulus of elasticity E s may be
taken as equal to the value for structural steel
given in EN1993-1-1, 3.2.6 (from clause
3.2(2), EN1994-1-1).
2. Selection of Trial Beam Size
An approximate span to depth ratio for
the steel section for a continuouscomposite secondary beam is 25.
Therefore given a 10.0 m span, try at
least a 400 mm deep beam.
Try UB 457x191x89
h a = 463.4 mm b = 191.9 mm
d = 407.6 mm t w = 10.5 mm
tf = 17.7 mm r = 10.2 mm
Aa = 114 cm 2 Iyy = 41015 cm 4 Wel.y = 1770 cm
3 Wpl.y = 2014 cm3
iz = 4.29 cm
z
z
y y
t w
t f
b
h d
r
Figure 2: Steel Beam Cross-section
a
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3. Section Classification
For t f = 17.700 mm
Yield strength, f y = 275.0 N/mm2 EN 10025
" = # ((235)/ f y) = 0.924 7.3 & Table 4
During the execution stage the neutral axis will be at mid depth of the steel section
and classification of the section is based on sheet 1 of Table 5.2, EN1993-1-1:
c = (bt w 2 $r)/2 = 80.50 mm
Flange: c/t f = 4.55 < 9$" = 8.32 Therefore flange is class 1 EN1993-1-1
Web: d/t w = 38.82 < 72 $" = 66.56 Therefore web is class 1 Table 5.2Therefore section is class 1 during execution
In the composite stage it is possible for a distance c of the web to be in
compression over the central support. Assuming the neutral axis is at least 100mm
below the top surface of the upper flange of the beam, c = d-100 = 307.60 mm
Web: c/t w = 29.30 < 33 $" = 30.51 Therefore web is class 1 EN1993-1-1Table 5.2
Therefore the web of section is class 1 during the composite stageNote: This classification assumes a neutral axis depth. A more detailed check should be
carried out once the depth of the web in compression is calculated (see section 6.3.3).
4. Actions
4.1. Execution Stage
4.1.1. Permanent Actions, g k
Concrete Slab area (per m width) A c = [(h-h p)$1000 + (140 $hp$(1000/153))] Ac =145667 mm
2
Weight of Wet Concrete slab = 3.79 kN/m 2
Weight of Steel deck (allow) 0.17 kN/m 2
Weight of Reinforcement (allow) 0.04 kN/m 2 Weight of Steel beam (allow) 0.25 kN/m 2
Therefore, g exe = 4.25 kN/m2
4.1.2. Variable Actions, q k
Execution loading = 0.50 kN/m 2
Therefore, q exe = 0.50 kN/m2
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4.2. Composite Stage
4.2.1. Permanent Actions, g k
Weight of Cast Concrete slab = 3.64 kN/m 2
Weight of Steel deck (allow) 0.17 kN/m 2
Weight of Reinforcement (allow) 0.04 kN/m 2
Weight of Steel beam (allow) 0.25 kN/m 2
Services and Ceilings = 0.50 kN/m 2
Therefore, g comp = 4.60 kN/m2
4.2.2. Variable Actions, q k
Occupancy, (for category C2 ocupancy) = 4.00 kN/m 2 Table 6.1EN1991-1-1
Moveable Partitions = 0.5 kN/m 2 6.3.1.2(8)
Note: Variable actions are not independent of each other. EN1991-1-1
Therefore, q comp = 4.50 kN/m2
4.3. Partial factors for Actions
Partial factor for permanent actions (where unfavourable), !G.sup = 1.35 Table A1.2(B)
Partial factor for permanent actions (where favourable), !G.inf = 1.00 EN 1990-1-1Partial factor for variable actions (where unfavourable), !Q = 1.50Partial factor for variable actions (where favourable), !Qi = 0.00
Note: As the variable actions considered in this example are not independent " factors
need not be considered.
5. Execution Stage Design
5.1. Global Analysis
The moments and forces within the section are determined using elastic analysis
methods.
It has been assumed that the rotational stiffness and moment resistance of thebeam/column connection will provide full continuity.
The resistance of the steel beam is determined using plastic section analysis.
Maximum uniformly distributed design load per span
F exe.max = 3 $(!G.sup $g exe + !Q$qexe ) = 19.45 kN/mMinimum uniformly distributed design load per span, F exe.min = 0.00 kN/m
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5.1.1. Load Case 1 Maximum Design Load on both spans
Figure 3: Design loading
Figure 4: Bending Moments with full design load on both spans.
5.1.2. Load Case 2 Maximum Design Load on one span only
Figure 5: Design loading
Figure 6: Design Bending Moments with maximum load on one span only.
Considering both loadcases, the maximum design moments on the steel section
during the execution stage are:Design negative moment, M a.Ed.negative = 243.28 kNm
Design positive moment, M a.Ed.positive = 186.26 kNm
19.45 kN/m
10.00 m10.00 m
-243.28 kNm
136.85 kNm
10.00 m10.00 m
19.45 kN/m
-121.64 kNm
186.26 kNm
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In addition, the maximum shear force on the steel section is:
Design shear, V a.Ed = 121.64 kN
5.2. Plastic Resistance to Vertical Shear, V Rd 6.2.2.2
The design plastic shear resistance of the structural steel section, V pl.a.Rd , in the
absence of torsion is given by: 6.2.6 (2)
Vpl.a.Rd = A v$(f y/ # (3))/ !M0 EN1993-1-1
Where A v is the shear area given by:
Av = A
a (2 $ b $ t
f ) + ((t
w + (2 $ r)) $ t
f ) = 5130 mm 2 6.2.6(3)
But not less than % $ h w $ tw = 4494 mm 2 EN1993-1-1where h w is the clear web depth between flanges h w = h a - (2 $ tf ) = 428 mmand % is conservatively taken as 1.0. 6.2.6(3)
EN1993-1-1
Av = 5130 mm2
! Design shear resistance V pl.a.Rd = 814.53 kN 6.2.2.2
Therefore as the applied shear is less than half the shear resistance, the effect of shear on
plastic moment resistance can be ignored 6.2.2.4(1)
5.3. Plastic Moment Resistance of Steel Section
It is assumed that the beam is fully restrained against lateral torsional buckling
under positive moments, since the decking spans perpendicular to the beam and is
directly attached to the top flange.
! Positive moment Resistance of steel section, M pl.a.Rd = W pl.y$f yd = 553.75 kNm
Beam is Satisfactory for positive moment resistance during execution of the
Structure
5.4. Lateral Torsional Buckling resistance of the Steel Beam 6.3.2.1
Considering the negative design moment, the design buckling resistance of a
laterally unrestrained beam is taken as: EN1993
Mb.Rd = <$ W pl.y $f y/!M1 equation (6.55)where:
Wy= W pl.y for a class 1 section
< is the reduction factor for lateral torsional buckling, which for a rolled section, isgiven by:
< = 1/( ' LT+ # (' LT2-($) LT2)) 6.3.2.3(1)
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where:
! LT = 0.5 " [1+# LT" ( $LT $LT.0 )+ %"$LT2] the parameters $LT.0 and % may be given in the National Annex. Recommendedvalues of, $LT.0 = 0.4 and % = 0.75 are given in EN1993-1-1.
# LT is given in Table 6.3, EN1993 using the appropriate buckling curve, determinedusing Table 6.5, EN1993. For h a /b = 2.415 , use buckling curve c and # LT = 0.49
The non-dimensional slenderness, $LT, is defined as & (Wpl.y" f y/Mcr ) where M cr is theelastic critical moment for lateral torsional buckling. Unfortunately EN1993-1-1 does
not give expressions for calculating M cr and therefore designers must use textbooksto obtain the most appropriate equation for M cr . However, the non-dimensional
slenderness, $LT , can more conveniently be found using the following relationship:
$LT = [$LT/$1]"%0.5 where $1 = '" [Ea /f y]0.5
Note: This is equation (F.12) in Appendix F.2 of the draft Eurocode DD ENV1993-1-1,
published in 1992. This information has been removed from the latest draft EN1993-1-1,
presumably as it is considered to be textbook information.
For a beam with uniform doubly symmetric cross-sections $LT can be obtained from:$LT = 0.9 " (L/iz)/[C10.5 " [1+0.05 " [(L/iz)/(h a /tf )]2]0.25 ] = 124.067 where C 1 = 1.285 (corresponding to the shape of the bending moment diagram withload on both spans)
Note: This is equation (F.20) in Appendix F.2 of the draft Eurocode DD ENV1993-1-1,
published in 1992. This information has been removed from the latest draft EN1993-1-1,
presumably as it is considered to be textbook information.
Therefore, $LT = 1.238 and ! LT = 1.280
which gives ( LT = 0.505 ! Mb.Rd = 279.86 kNm > M a.Ed.negative therefore OK Beam is stable without further restraint during the execution stage
6. Composite Stage Design
6.1. Global Analysis
The moments and forces within the section are determined using linear elastic
global analysis methods. Moment redistribution, in accordance with clause 5.4.4
has been carried out.
The cross section over the support is class 1, and therefore the support moments
(determined for the uncracked section) have been reduced by 40%, the maximum
permitted for this class in Table 5.1.
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The resistance of the steel beam and composite section are determined using
plastic section analysis.
Maximum uniformly distributed design load on beam, F comp.max = 3 $(!G.sup $g comp +!Q$q comp ) = 38.89 kN/mMinimum uniformly distributed design load on beam, F comp.min = 3 $!G.inf $g comp =13.80 kN/m
6.1.1. Load Case 1 Maximum Design Load on both spans
Figure 7: Design loading
Figure 8: Design Bending Moments with full design load on both spans (with no
moment redistribution)
6.1.2. Load Case 2 Maximum Design Load on one span only
Figure 9: Design loading
38.89 kN/m
10.00 m10.00 m
-486.00 kNm
273.38 kNm
13.80 kN/m
10.00 m10.00 m
38.89 kN/m
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Figure 10: Design Bending Moments with load on one span only (with no moment
redistribution)
Therefore considering both loadcases and adopting the maximum 40% reduction insupport moment permitted, the maximum moments on the composite section are:
Design hogging moment, M Ed.negative = 486.00 $0.6 = 291.600 kNm (loadcase 1)Design sagging moment, M Ed.positive = 335.32+0.5 $(0.4 $329.25) = 401.170 kNm(loadcase 2)
Note: This redistributed design positive moment is an approximate value, assuming the
positive moment at mid-span increases by approximately ! the decrease in negative
moment over the support.
The maximum elastic shear force on the section is 242.00kN. The redistributed
shear force is given by:
VEd = F comp.max $ L/2 + M Ed.negative /L = 223.594 kN
! Design shear, V Ed = 223.59 kN
6.2. Plastic Resistance to Vertical Shear, V Rd 6.2.2.2
In this example, because the contribution of the reinforced concrete part of thebeam has not been established, the design plastic shear resistance of the
composite section, V Rd , is conservatively taken as that for the structural steel
section alone, V pl.a.Rd , in accordance with clause 6.2.6, EN1993-1-1.
! Design shear resistance (as previously calculated in section 5.2), V Rd = V pl.a.Rd =
814.53 kN 6.2.2.2
Therefore as the applied shear is less than half the shear resistance, the effect of shear on
plastic moment resistance can be ignored 6.2.2.4(1)
6.3. Moment Resistance with full Shear Interaction, M pl.Rd 6.2.1.2 6.3.1. Effective Widths of Compression Flange, b eff 5.4.1.2
Effective length of beam span between supports, L e.1 =0.85 $L = 8500 mm Figure 5.1
-329.25 kNm
335.32 kNm
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Effective length of beam over support, L e.2 = 0.25 $(2$L) = 5000 mm Figure 5.1
Midspan effective breadth, b eff.1 = 2 $ Le.1 /8 = 2125 mm < beam spacing and support effective breadth, b eff.2 = 2 $ Le.2 /8 = 1250 mm < beam spacing
6.3.2. Midspan Moment Resistance
In the midspan region of the beam, the moment resistance of the composite section
is calculated in the same way as for a simply-supported composite beam.
Compressive Resistance of Slab, N c.f , neglecting the contribution of any
reinforcement in compression in accordance with clause 6.2.1.2(1)c, is:Nc.f = 0.85 $f ck$b eff.1 $h c/!c = 4768.50 kN
Tensile Resistance of Steel Section, N pl.a , is:
Npl.a = f yd$ Aa = 3128.56 kN
Since N pl.a M Ed.positive therefore OK
6.3.3. Support Moment Resistance
At the internal support the negative (hogging) moment resistance is obtained by
considering the tensile resistance of the reinforcement within the slab.
Assume T16 bars at 200mm centres (1005mm 2/m) with 25mm cover to the top of
the slab.
dbar = 16mm A bar = +$dbar 2/4 = 201 mm 2 As = b eff.2 $ 1005 = 1256 mm 2
Axial capacity of the steel section, N pl.a , is:
Npl.a = f yd$ Aa = 3128.56 kN
Axial capacity of the slab reinforcement, N s , is:
Ns = f sk $ As /!s = 502.50 kN
Depth of the neutral axis below the top surface of upper flange , z cw , is given by:
z cw = h a /2 - N s /(2$tw$f yd) = 145 mm
therefore, M pl.Rd.negative is given by:
Mpl.Rd.negative = M pl.a.Rd + N s$(h a /2 + z) N s 2/(4$tw$f yd)
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where z is the distance from the top of the steelwork to the centreline of the slab
reinforcement, z = 117 mm
Therefore M pl.Rd.negative = 707.11 kNm > M Ed.negative , therefore OK
6.3.4. Lateral Torsional Buckling of Composite Section
In accordance with guidance outlined in clause 6.4.3, the lateral torsional buckling
resistance of this continuous beam (within a building structure), has not been
checked. This is because the relevant conditions outlined in clause 6.4.3(1) have
been satisfied.
6.4. Moment Resistance with Actual Shear Interaction
6.4.1. Shear Connector Resistance 6.6.3.1
The design shear resistance of a single shear connector welded in accordance with
EN14555 is given by:
P Rd = [0.8 $f u$(+$d stud2)/4]/!V equation (6.18)
or
P Rd = [0.29 $*$ d stud 2$# (f ck$Ecm )]/!V equation (6.19)
whichever is the smaller, with:* =0.2 $(h sc /d stud + 1) if 3 < h sc / d stud < 4 equation (6.20)or
* =1 if h sc / d stud > 4 equation (6.21)h sc /d stud = 5.00 therefore * =1
therefore:
P Rd = [0.8 $f u$(+$d stud 2)/4]/!V = 81.66 kN equation (6.18)
or
P Rd = [0.29 $*$ d stud 2$# (f ck$Ecm )]/!V = 99.10 kN equation (6.19)
The deck spans perpendicular to the beam and is therefore transverse. The effectof a reduction factor, k t, on the shear connector resistance should be checked. 6.6.4.2
There is one stud per trough of the deck, therefore n r = 1
kt = (0.7 $bo)/( # (n r )$hp)$(h sc /hp 1) = 1.66 equation (6.23)> 1.0 therefore no reduction in the shear connector resistance
therefore P Rd = 81.66 kN
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6.4.2. Shear Stud Layout Loadcase 1
Figure 12: Shear Stud Layout corresponding to loadcase 1
The maximum positive moment is at a point approximately 3.75m away from the far
left hand support. Therefore, 24 troughs are available for the positioning of the
shear stud connectors between this support and the point of maximum negative
moment. Between the internal support and the point of maximum negative moment,
there are 41 shear stud positions.
6.4.3. Shear Stud Layout Loadcase 2
Figure 13: Shear Stud Layout corresponding to loadcase 2
The point of maximum sagging moment is approximately 4.0m away from the far
left hand support. Therefore 26 troughs are available for the positioning of the shear
stud connectors. There are 39 shear stud positions between the internal support
and the point of maximum sagging moment.
6.4.4. Degree of Shear Connection, 6.6.1
The longitudinal shear force transfer, R q , between the left hand support and thepoint of maximum positive moment, is:
R q.positive = 24 $P Rd = 1959.84 kN (loadcase 1)
24 studs at 153mm centres
10.0 m
ColumnCL
250 41 studs at 153mm centres
ColumnCL
26 studs at 153mm centres
10.0 m
ColumnCL
250 39 studs at 153mm centres
ColumnCL
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R q.positive = 26 $P Rd = 2123.16 kN (loadcase 2)
Note: It has been assumed that the area between the left hand support and the point of
maximum positive moment is critical as there are significantly more shear studs between
the point of maximum positive moment and the internal support, even if those within the
hogging region of the beam are ignored. This should be checked when the number of
shear studs required to yield the slab reinforcement in the hogging region of the beam is
determined (see section 6.4.5).
Minimum degree of shear connection, %min , for beam of length less than 25.00m isgiven by:
%min = 1-[(355/f y)$(0.75-0.03 $L)] = 0.419 or 0.4 (whichever is greater) 6.6.1.2
actual degree of shear interaction, %, is given by:%positive = N c/Nc.f = R q.positive /Npl.a = 0.626 (loadcase 1)%positive = N c/Nc.f = R q.positive /Npl.a = 0.679 (loadcase 2)both of which are greater than %min and are therefore OK
6.4.5. Moment Resistance, M Rd 6.2.1.3
The negative plastic moment resistance in hogging bending should be determined
in accordance with clause 6.2.1.2, assuming there is full interaction between the
structural steel, reinforcement and concrete. Appropriate shear connection should
be provided to ensure yielding of the reinforcement in tension.
In other words, sufficient shear studs should be provided in the hogging region of
the beam to yield the reinforcement in tension prior to failure of the shear
connectors.
The number of studs required to yield the reinforcement, N s /P Rd = 6
There are therefore sufficient shear studs in the hogging region of the beam to yieldthe reinforcement in tension. Therefore, M Rd.negative is given by:
MRd.negative = M pl.Rd.negative = 707.11 kNm > M Ed.negative , therefore OK
Beam is satisfactory for negative moment resistance in service
The positive moment resistance of the composite beam is obtained using the linear
interaction method in accordance with clause 6.2.1.3(5) as follows:
MRd = M pl.a.Rd + (N c/Nc.f )$(Mpl.Rd -Mpl.a.Rd ) equation (6.1)where:
Nc is the compressive normal force in the concrete flange
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Nc.f is the compressive normal force in the concrete flange if full shear interaction is
provided, where the ratio % = N c/Nc.f is the degree of shear interaction.
The degree of shear interaction varies between loadcase 1 and loadcase 2, as the
number of shear studs between the supports and the point of maximum positive
bending moment varies between the two loadcases. The separate values of
positive moment resistance corresponding to the two loadcases should,theoretically, be considered.
In practice, however, it would normally be acceptable to conservatively calculate the
positive moment resistance of the composite section using the minimum shear
interaction (corresponding to loadcase 1) and compare this with the maximumpositive design moment (corresponding to loadcase 2 in this example). This
approach has been adopted.
Therefore the moment resistance with partial shear interaction, for the sagging
regions of the composite beam (using the minimum shear interaction) is:
MRd.positive = M pl.a.Rd + %positive $(Mpl.Rd.positive M pl.a.Rd ) = 891.28 kNm > M Ed.positive ,therefore OK
Beam is satisfactory for positive moment resistance in service
6.5. Transverse Reinforcement and Longitudinal Shear Resistance Check
Check resistance of the concrete flange to splitting. To prevent concrete failure the
longitudinal shear stress should satisfy the following inequality:
vEd < ,$ (f ck/!c)$sin( - f )$cos( - f ) equation (6.22)EN1992-1-1
where:
, = 0.6 $(1-f ck/250) 6.2.3(3) & 6.2.2(6)
- f is the angle between the diagonal strut assumed in the Eurocode 2 model andthe longitudinal axis of the slab, which is chosen (within limits) by the designer.
Note: the recommended range of # f may be found within the National Annex to EN1992,
but in the absence of more rigourous calculation, the limits are 45 o># f >26.5 o, for 6.2.4(4)
compression flanges or 45 o># f >38.6 o for tension flanges. EN1992-1-1
The design shear stress, v Ed , for one stud per trough, is given by:
vEd = P Rd /(2$0.153 $h f ) = 2.70 N/mm 2 6.6.6.1(5) using h f = h c = 99 mm 6.6.6.4(1)
If we choose the angle, - f = 26.5 (which leads to a lower bound) ,$ (f ck /!c)$sin( - f )$cos( - f ) = 5.37 N/mm 2 therefore v Ed < ,$ (f ck/!c)$sin( - f )$cos( - f )is satisfied
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In addition, the area of transverse reinforcement within the slab, A sf , should satisfy
the following:
Asf $(f sk /!s)/s f > v Ed $h f /cot( - f ) equation (6.21)EN1992-1-1
where:
Asf is the area of the transverse reinforcing bars at spacing s f Note: in this example the contribution of the profiled metal sheeting has been
conservatively neglected, although clause 6.6.6.4(4) allows the contribution of sheeting
with ribs transverse to the beam and continuous over it to be taken.
The minimum slab reinforcement, A s.min = k s$kc$k$f ct.eff $ Act /. s 7.4.2(1)where:
f ct.eff is the mean value of tensile strength which can be taken as f ctm = 2.60 N/mm2
k is a coefficient which allows for the effect of non-uniform self-equilibriating
stresses and is equal to 0.8
ks is a coefficient which allows for the effect of the reduction of the normal force of
the concrete slab due to initial cracking or local slip and is taken as 0.9
kc is a coefficient which takes account of the stress distribution within the section
immediately prior to cracking and is given by:
kc=1/(1+h c/(2$zo)) + 0.3;
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Try T12-200 reinforcement to the top face of the slab, so A sf = 113.1 mm 2 and s f =
200mm
Hence, A sf $(f sk /!s)/s f = 226 N/mmand v Ed $h f /cot( - f ) = 133 N/mm
therefore A sf $(f sk /!s )/s f > v Ed $h f /cot( - f ) is satisfied and no additional reinforcement isrequired.
7. Serviceability Limit States7.1. Elastic Stress Check
No stress checks are required for normal conditions and consequently no limits are
outlined in EN1994. 7.2.2(1)
7.2. Deflections
7.2.1. Execution Stage Deflections
Deflection of the bare steel beam during execution is given by:Second moment or area of the steel section, I yy = 41015 cm
4
The deflection due to variable actions, w var.exe , during execution is given by:Unfactored variable action on beam, F var.exe = 15.00 kN
The critical loadcase for deflection is where only one span is fully loaded, the
deflection (determined using a commercial software package) is:wvar.exe = 1.63 mm
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7.2.2. Composite Stage Deflections
The deflection of the composite section due to variable and total actions has been
determined using commercially available software, using the second moment of
area of the composite section based on elastic (uncracked) properties to the
sagging regions of the beam. Appropriate allowance must be made for the effects
of cracking of concrete and in accordance with clause 5.4.2.3(3), a reduced flexural
stiffness has been used over 15% of the span on each side of the internal support.
Second moment of area of the composite section based on elastic (uncracked )
properties, I c , is given, from first principles, by:
Ic = [Aa$(h a+2 $hp+h c)2]/[4$(1+n L$R)] + (b eff.1 $h c3)/(12 $nL) + I yy where:
R = A a /(b eff.1 $h c) = 0.054 The effects of creep may be taken into account by using a modular ratio, n L, given 5.4.2.2(2)
by:
nL = n 0$(1+ / L$0 t) = 15.90 where:n0 = E a /E cm where E cm is the secant modulus of elasticity for short term loading
0 t is the creep coefficient taken as 1.5 from Figure 3.1, EN 1992-1-1, for loading at
28 days / L is a creep multiplier depending upon the type of loading, taken as 1.1 forpermanent loads.
Note: For simplification in building structures, the effects of creep may alternatively be 5.4.2.2(11) taken into account by replacing the concrete area, A c , by effective steel areas A c /n for
both short and long term loading, where n is the nominal modular ratio corresponding to
an effective modulus of elasticity for concrete of E c,eff taken as E cm /2.
Appropriate allowance should also be made for the effects of concrete shrinkage in 7.3.1(8)
accordance with clause 5.4.2.2(1), but, unless specifically required by the client, the
effect of curvature due to the shrinkage of normal weight concrete need not beincluded when the ratio of span to overall depth of composite beam is not greater
than 20.
The overall depth of composite beam is 613 mm and the ratio of span to overalldepth is therefore 16.303 and the effect can be ignored.
In addition, the effects of incomplete interaction may be ignored for most cases, 7.3.1(4)
provided that the shear interaction is greater than 0.50.
Therefore I c = 109601 cm4
Over the central support a reduced flexural stiffness, E a I2, is used, where E a is the
Youngs modulus of steel and I 2 is the second moment of area of the effective steel 1.5.2.12
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Project Title:EC3 & EC4 Worked Examples
Project Number: Sheet 21 of 21 Rev:07
Subject:Continuous steel and concrete composite beam
Made by/date:CGR/September 2004
Client: Checked/date:GHC/October 2004
provide any guidance regarding the magnitude of this value, but does suggest that
the value be agreed with the client and/or the relevant authority.
A natural frequency limit of 4Hz has been used in this example, assuming that
walking is the main source of the vibration.
Consider the weight of the floor in the dynamic calculations to include the self
weight of the slab and beam, 10% of the imposed load (excluding partitions) and
ceilings and services.
therefore, F vib = 150.05 kN
It is necessary to consider the mode shape of vibration when calculating the naturalfrequency of a continuous beam. Due to the influence of the asymmetric inertial
forces, the natural frequency is approximately the same as that of a simply-
supported beam.
! The deflection of the composite beam subject to instantaneously applied self
weight is:
wvib = (5 $Fvib$L3)/(384 $Ea$Ic) = 8.49 mm
and the natural frequency of the beam, f , (subject to uniformly distributed loading) isgiven by:
f = 18/ # (wvib) = 6.178 Hz > 4.0Hz therefore OK
Note: Whilst the 4Hz natural frequency limit is an almost universally accepted industry
standard for vibrations, satisfying the limit will not guarantee that the element or structure
as a whole will perform adequately.