Document Ref: SX015a-EN-EU Sheet 1 of 16 Title CALCULATION SHEET Example: Simply supported primary composite beam Eurocode Ref EN 1993-1-1, EN 1994-1-1 Made by Arnaud LEMAIRE Date August 2005 Checked by Alain BUREAU Date August 2005 Example: Simply supported primary composite beam This worked example deals with a simply supported composite beam. Two secondary beams are connected to this primary beam. 6,0 m 6,0 m 9,0 m 3,0 m 3,0 m The secondary beams are represented by two concentrated loads : 1 1 1 1 1 : Lateral restraints at the construction stage Example: Simply supported primary composite beam Created on Wednesday, February 08, 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
Model de calcul pentru o grinda metalica dubluarticluata
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Document Ref: SX015a-EN-EU Sheet 1 of 16 Title
CALCULATION SHEET
Example: Simply supported primary composite beam
Eurocode Ref EN 1993-1-1, EN 1994-1-1 Made by Arnaud LEMAIRE Date August 2005
Checked by Alain BUREAU Date August 2005
Example: Simply supported primary composite beam This worked example deals with a simply supported composite beam. Two secondary beams are connected to this primary beam.
6,0 m
6,0 m
9,0 m
3,0 m 3,0 m
The secondary beams are represented by two concentrated loads :
1 1
1 1 1 : Lateral restraints at the construction stage
Example: Simply supported primary composite beamC
reat
ed o
n W
edne
sday
, Feb
ruar
y 08
, 201
2T
his
mat
eria
l is
copy
right
- a
ll rig
hts
rese
rved
. Use
of t
his
docu
men
t is
subj
ect t
o th
e te
rms
and
cond
ition
s of
the
Acc
ess
Ste
el L
icen
ce A
gree
men
t
Document Ref: SX015a-EN-EU Sheet 2 of 16 Title
Example: Simply supported primary composite beam
Eurocode Ref EN 1993-1-1, EN 1994-1-1 Made by Arnaud LEMAIRE Date August 2005
CALCULATION SHEET
Checked by Alain BUREAU Date August 2005
The beam is a I-rolled profile in bending about the strong axis. This example includes :
- the classification of the cross-section,
- the calculation of the effective width of the concrete flange,
- the calculation of the shear resistance of a headed stud,
- the calculation of the degree of shear connection,
- the calculation of the bending resistance,
- the calculation of the shear resistance,
- the calculation of the longitudinal shear resistance of the slab,
- the calculation of the deflection at serviceability limit state.
This example does not include any shear buckling verification of the web.
Eurocode Ref EN 1993-1-1, EN 1994-1-1 Made by Arnaud LEMAIRE Date August 2005
Checked by Alain BUREAU Date August 2005
Basic data Design a composite beam of a multi-storey building according to the data given below. The supporting beams are not propped. The profiled steel sheeting is parallel to the primary beam.
• Span length : 9,00 m
• Bay width : 6,00 m
• Slab depth : 14 cm
• Partitions : 0,75 kN/m2
• Secondary beams (IPE 270) : 0,354 kN/m
• Imposed load : 2,50 kN/m2
• Construction load : 0,75 kN/m2
• Reinforced concrete density : 25 kN/m3
Try IPE 400 – Steel grade S355
Depth ha = 400 mm
Width b = 180 mm
Web thickness tw = 8,6 mm
Flange thickness tf = 13,5 mm
Fillet r = 21 mm
Mass 66,3 kg/m
z
z
y y
tf
tw
b
ha
Euronorm
19-57
Section area Aa = 84,46 cm2
Second moment of area /yy Iy = 23130 cm4
Elastic section modulus /yy Wel,y = 1156 cm3
Plastic section modulus /yy Wpl.y = 1307 cm3
Radius of gyration /zz iz = 3,95 cm
Modulus of elasticity of steel Ea = 210 000 N/mm2
Example: Simply supported primary composite beamC
reat
ed o
n W
edne
sday
, Feb
ruar
y 08
, 201
2T
his
mat
eria
l is
copy
right
- a
ll rig
hts
rese
rved
. Use
of t
his
docu
men
t is
subj
ect t
o th
e te
rms
and
cond
ition
s of
the
Acc
ess
Ste
el L
icen
ce A
gree
men
t
Document Ref: SX015a-EN-EU Sheet 4 of 16 Title
CALCULATION SHEET
Example: Simply supported primary composite beam
Eurocode Ref EN 1993-1-1, EN 1994-1-1 Made by Arnaud LEMAIRE Date August 2005
Checked by Alain BUREAU Date August 2005
Profiled steel sheeting Thickness of sheet t = 0,75 mm
Slab depth h = 140 mm
Overall depth of the profiled steel sheeting excluding embossments
hp = 58 mm
b1 = 62 mm b2 = 101 mm e = 207 mm
Connectors Diameter d = 19 mm
Overall nominal height hsc = 100 mm
Ultimate tensile strength fu = 450 N/mm2
Number of studs n = 74, 1 row
(Stud at beam mid-span ignored)
0,5hp
hp
hsc
h
b0
b1 b2
e
Concrete class : C 25/30 Value of the compressive strength at 28 days fck = 25 N/mm2
Secant modulus of elasticity of concrete Ecm = 31 000 N/mm2
Reduction factor for lateral torsional buckling To determine the design buckling resistance moment of a laterally unrestrained beam, the reduction factor for lateral torsional buckling must be determined. The restraint provided by the steel sheet is in this case quite small and it is neglected. The following calculation determines this factor by a simplified method for lateral torsional buckling. This method avoids calculating the elastic critical moment for lateral torsional buckling.
Non-dimensional slenderness
The non-dimensional slenderness may be obtained from the simplified method for steel grade S355:
Interaction between bending moment and shear force If Vz,Ed < Vpl,Rd / 2 then the shear force may be neglected.
So, Vz,Ed = 90,73 kN < Vpl,Rd / 2 = 874,97 / 2 = 437,50 kN OK
EN 1993-1-1
§ 6.2.8 (2)
Final stage
Effective width of concrete flange The effective width is constant between 0,25 L and 0,75 L, where L is the span length. From L/4 to the closest support, the effective width decreases linearly. The concentrated loads are located between 0,25 L and 0,75 L.
Eurocode Ref EN 1993-1-1, EN 1994-1-1 Made by Arnaud LEMAIRE Date August 2005
Checked by Alain BUREAU Date August 2005
Degree of shear connection The degree of shear connection is defined by :
fc,
c
NN
=η
Where : Nc is the design value of the compressive normal force in the concrete flange
Nc,f is the design value of the compressive normal force in the concrete flange with full shear connection
EN 1994-1-1
§ 6.2.1.3 (3)
At the load location:
The compressive normal force in the concrete flange represents the force for full connection.
Ac is the cross-sectional area of concrete, so at the load location: Ac = beff hc
with hc = h - hp = 140 – 58 = 82 mm
Ac = 2250 × 82 = 184500 mm2
So, =1051
25×184500×850=850=850= 3
C
ckccdcfc, ,
,f
A,fA,Nγ
2614 kN
Since the maximum moment is nearly reached at the load location, the studs should be placed between the support and the concentrated load. However studs should also be placed between the concentrated loads.
e2
e1
3,0 m 1,5 m
31 studs spaced at e1 = 95 mm and 6 studs spaced at e2 = 220 mm
The design value of the normal force in the structural steel section is given by :
2998 kN =0110×355×8446== 3M0yaapl, ,/ /fAN γ
So, =2614×5370=×=> fc,capl, ,NNN η 1403 kN
EN 1994-1-1
§ 6.2.1.2 and
§ 6.2.1.3
With the ductile shear connectors and the cross-section of the steel beam in Class 1, the resistance moment of the critical cross-section of the beam MRd at the load location is calculated by means of rigid-plastic theory except that a reduced value of the compressive force in the concrete flange Nc is used in place of the force Ncf.
Here, the plastic stress distribution is given below:
Eurocode Ref EN 1993-1-1, EN 1994-1-1 Made by Arnaud LEMAIRE Date August 2005
Checked by Alain BUREAU Date August 2005
The value for Δx is half the distance between the section where the moment is zero and the section where the moment is a maximum, and we have two areas for the shear resistance.
ΔFd = Nc / 2 = 1403 / 2 = 701,5 kN
hf = h - hp = 140 – 58 = 82 mm
=4500×82
10×5701==
3
f
d ,xh
FvEd Δ
Δ1,9 N/mm2
To prevent crushing of the compression struts in the concrete flange, the following condition should be satisfied :
ffcdEd cossin θθν fv < with [ ]250/16,0 ckf−=ν and θf = 45 °
5,45,05,1
252502516,0Ed =××⎥⎦
⎤⎢⎣⎡ −×<v N/mm2 OK
The following inequality should be satisfied for the transverse reinforcement :
Asf fyd / sf ≥ vEd hf / cot θf where fyd = 500 / 1,15 = 435 N/mm2
Assume the spacing of the bars sf = 200 mm and there is no contribution from the profiled steel sheeting
Asf ≥ =01×435200×82×91,
,71,6 mm2
We can take 10 mm diameter bars (78,5 mm2) at 200 mm cross-centres could be used for this design.
Serviceability Limit State verifications
Formula for the calculation of the deflection due to G + Q :
Qy
22
Q
Gy
22
y
4G
G
24)4(3w
24)4(3
384 5
FIE
aLa
FIE
aLaIE
Lqw
−×=
−×+=
So, w = wG + wQ
Example: Simply supported primary composite beamC
reat
ed o
n W
edne
sday
, Feb
ruar
y 08
, 201
2T
his
mat
eria
l is
copy
right
- a
ll rig
hts
rese
rved
. Use
of t
his
docu
men
t is
subj
ect t
o th
e te
rms
and
cond
ition
s of
the
Acc
ess
Ste
el L
icen
ce A
gree
men
t
Document Ref: SX015a-EN-EU Sheet 15 of 16 Title
CALCULATION SHEET
Example: Simply supported primary composite beam
Eurocode Ref EN 1993-1-1, EN 1994-1-1 Made by Arnaud LEMAIRE Date August 2005
Checked by Alain BUREAU Date August 2005
Construction stage
SLS Combination during the construction stage : FG + FQ = 49,28 + 13,5 = 62,78 kN
qG = 0,65 kN/m
EN 1990
§ 6.5.3
Deflection during the construction stage : Iy is the second moment of area of the steel beam.
492801023130 210000 24
)30004-90003(30001023130210000 384
900065,0 5 4
22
4
4
G ××××
×××+
×××××
=w
mm 3,272,261,1 G =+=w
mm 7,2135001023130 210000 24
)30004-90003(3000w 4
22
Q =××××
×××=
So, w = wG + wQ = 27,3 + 7,2 = 34,5 mm
The deflection under (G+Q) is L/261
Final stage
SLS Combination
FG + FQ = 62,78 + 45,0 = 107,78 kN
qG = 0,65 kN/m
EN 1990
§ 6.5.3
Deflection at the final stage : Iy depends on the modular ratio (n) depending on the type of loading. By simplification, we can take :
n0 = Ea / Ecm = 210000 / 31000 = 6,77 for short-term effects (Q)
Eurocode Ref EN 1993-1-1, EN 1994-1-1 Made by Arnaud LEMAIRE Date August 2005
Checked by Alain BUREAU Date August 2005
Note : It may be used for both short-term and long-term loading, a nominal modular ratio (n) corresponding to an effective modulus of elasticity for concrete Ec,eff taken as Ecm / 2.
EN 1994-1-1
§ 5.4.2.2 (11)
wG = 27,3 mm
13500×10×62919×210000×24
3000×4-9000×3×3000= 4
2
partitions )(
w2
= 2,6 mm
45000×10×82458×210000×24
3000×4-9000×3×3000= 4
2
)(
w2
Q = 6,7 mm
So, w = wG + wpartitions + wQ = 27,3 + 2,6 + 6,7 = 36,6 mm
The deflection under (G + Q) is L/246
Note 1 : The limits of deflection should be specified by the client. The National Annex may specify some limits. Here the result may be considered as fully satisfactory.
Note 2 : Concerning vibrations, the National Annex may specify limits concerning the frequency. Here the total deflection is low and the mass fairly high and by experience there is no problem of vibration.