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TALAT Lecture 2710: Static Design Example
1. TALAT Lectures 2710
Static Design Example
82 pages
Advanced Level
prepared during the TAS project:
TAS Leonardo da Vinci
Training in Aluminium Alloy Structural Design
Example developed with the “Mathcad” Software
Date of Issue: 1999
EAA - European Aluminium Association
2. 2710 Static Design example
Table of Contents (Active)
2710 Static Design example........................................................................................... 2
1. Introduction.......................................................................................................................6
2. Materials ............................................................................................................................7
3. Loads ..................................................................................................................................7
4. Load Combinations........................................................................................................... 9
5. Loads Effects ...................................................................................................................11
5.1. Loads per unit length and concentrated loads...........................................................................11
5.2 Finite element calculations ........................................................................................................12
5.3. Section forces for characteristic loads ......................................................................................14
5.4. Design moments, shear forces and deflections .........................................................................16
6. Code Checking.................................................................................................................28
6.1 Column A ..................................................................................................................................28
6.2 Column B ..................................................................................................................................42
6.3 Column C ..................................................................................................................................55
6.4 Floor Beam D ............................................................................................................................56
6.5 Roof Beam E .............................................................................................................................64
6.6 Roof Beam F .............................................................................................................................65
6.7 Welded Connections..................................................................................................................74
Table of Contents (Complete)
1 INTRODUCTION
1.1 Description
1.2 Sketches
1.3 References
1.4 S.I. units
2 MATERIALS
2.1 Aluminium
2.2 Other materials
3 LOADS
3.1 Permanent loads
3.2 Imposed loads
3.3 Environmental loads
3.3.1 Snow loads
3.3.2 Wind loads
TALAT 2710 2
3. 4 LOAD COMBINATIONS
4.1 Ultimate limit state
4.2 Serviceability limit state
5 LOAD EFFECTS
5.1 Loads per unit length and concentrated loads
5.1.1 Permanent loads
5.1.2 Imposed loads, uniform distributed
5.1.3 Imposed loads, concentrated
5.1.4 Snow loads
5.1.5 Wind loads
5.2 Finite element calculations
5.2.0 Nodes and elements
5.2.1 Permanent loads
5.2.2 Imposed loads, uniformly distributed
5.2.3 Imposed loads, concentrated
5.2.4 Snow loads
5.2.4 Wind loads
5.3 Section forces for characteristic loads
5.3.1 Column A
5.3.2 Column B
5.3.3 Column C
5.3.4 Floor beam D
5.3.5 Floor beam E
5.3.5 Floor beam F
5.4 Design moments, shear forces and deflections
5.4.1 Column A
5.4.2 Column B
5.4.3 Column C
5.4.4 Floor beam D
5.4.5 Floor beam E
5.4.6 Floor beam F
5.4.7 Joint A-D
5.4.8 Joint B-D
5.4.9 Joint A-E
5.4.10 Joint B-E
5.4.11 Joint B-F
5.4.12 Joint F-C
5.4.13 Column base
TALAT 2710 3
4. 6 CODE CHECKING
6.1 Column A
6.1.1 Dimensions and material properties
6.1.2 Internal moments and forces
6.1.3 Classification of the cross section in y-y-axis bending
6.1.4 Classification of the cross section in z-z-axis bending
6.1,5 Classification of the cross section in axial compression
6.1.6 Welds
6.1.7 Design resistance, y-y-axis bending
6.1.8 Design resistance, z-z-axis bending
6.1.9 Axial force resistance, y-y buckling
6.1.10 Axial force resistance, z-z axis buckling
6.1.11 Flexural buckling of beam-column
6.1.12 Lateral-torsional buckling between purlins
6.1.13 Design moment at column base
6.1.14 Deflections
6.1.15 Summary
6.2 Column B
6.2.1 Dimensions and material properties
6.2.2 Internal moments and forces
6.2.3 Classification of the cross section in y-y-axis bending
6.2.4 Classification of the cross section in z-z-axis bending
6.2,5 Classification of the cross section in axial compression
6.2.6 Welds
6.2.7 Design resistance, y-y-axis bending
6.2.8 Design resistance, z-z-axis bending
6.2.9 Axial force resistance, y-y buckling
6.2.10 Axial force resistance, z-z axis buckling
6.2.11 Flexural buckling of beam-column
6.2.12 Lateral-torsional buckling between purlins
6.2.13 Design moment at column base
6.2.14 Deflections
6.2.14 Summary
6.3 Column C
6.4 Floor Beam D
6.4.1 Dimensions and material properties
6.4.2 Internal moments and forces
6.4.3 Classification of the cross section
6.4.4 Welds
6.4.5 Bending moment resistance
6.4.6 Bending resistance in a section with holes
6.4.7 Shear force resistance
6.4.8 Deflections
6.4.8 Summary
TALAT 2710 4
5. 6.5 Roof Beam E
6.6 Roof Beam F
6.6.1 Dimensions and material properties
6.6.2 Internal moments and forces
6.6.3 Classification of the cross section
6.6.4 Welds
6.6.5 Bending moment resistance
6.6.6 Lateral-torsional buckling between purlins
6.6.7 Bending resistance in a section with holes
6.6.8 Shear force resistance
6.6.9 Concentrated transverse force
6.6.10 Deflections
6.6.11 Summary
6.7 Welded connections
6.7.1 Weld properties
6.7.2 Longitudinal weld of floor beam D
6.7.3 Base of column B
6.7.4 Connection between floor beam D and column B
Software
The example is worked out using the MathCad software in which some symbols have special meanin
according to the following
x 50.6 . mm Assign value
y 2.5 . mm Global assignment
x y = 53.1 mm Evaluate expression
a b Boolean equals
0.5 Decimal point
c (1 3 2 ) Vector
d (2 4 3 ) Vector
a ( c .d ) Vectorize (multiply the elements in vector c with corresponding elements in d)
a = ( 2 12 6 ) Result
Structure
The structure was proposed by Steinar Lundberg, who also contributed with valuable suggestions.
Part 1 to 6.6 was worked out by Torsten Höglund and 6.7 by Myriam Bouet-Griffon.
TALAT 2710 5
6. 1. Introduction
1.1 Description
The industrial building contains an administration part with offices, wardrobe, meeting rooms etc.
and a fabrication hall. The load bearing system consist of frames standing at a distance of 5000 m
In the serviceability limit state max. allowable deflection is 1/250 of span.
1.2 Sketches
1.3 References
[1] ENV 1999-1-1. Eurocode 9 - Design of aluminium structures - Part 1-1: General rules. 1997
[2] ENV 1991-2-1. Eurocode 1 - Basis of design and actions on structures - Part 2-1:
Action on structures - Densities, self-weight and imposed loads. 1995
[3] ENV 1991-2-3. Eurocode 1 - Basis of design and actions on structures -
Part 2-3: Action on structures - Snow loads. 1995
[4] ENV 1991-2-4. Eurocode 1 - Basis of design and actions on structures -
Part 2-4: Action on structures - Wind loads. 1995
[5] ENV 1991-1. Eurocode 1 - Basis of design and actions on structures -
Part 1: Basis of design. 1994
1.4 S.I. units
N
kN 1000 . N MPa 1000000 . Pa kNm kN . m MPa = 1
2
mm
TALAT 2710 6
7. 2. Materials
2.1 Aluminium
[1], 3.2.2 The extrusions are alloy EN AW-6082, temper T6
The plates are EN AW-5083 temper H24
Strength of aluminium alloys
EN AW-6082 T6 fo 260 . MPa fu 310 . MPa
EN AW-5083 H24
fo 250 . MPa fu 340 . MPa
[1], 5.1.1 The partial safety factor for the members γ M1 1.10 γ M2 1.25
[1], 6.1.1 The partial safety factor for welded connections γ Mw 1.25
Design values of material coefficients
Modulus of elasticity E 700000 . MPa
Shear modulus G 27000 . MPa
Poisson´s ratio ν 0.3
α T 23 . 10
6
Coefficient of linear thermal expansion
ρ 2700 . kg . m
3
Density
2.2 Other materials
Comment: Properties of any other materials to be filled in
3. Loads
3.1 Permanent loads
[3], ?? Permanent loads are self-weight of structure, insulation, surface materials and fixed equipment
0.5 . kN . m
2
Permanent load on roof q´ p.roof
0.7 . kN . m
Permanent load on floor 2
q´ p.floor
3.2 Imposed loads
[2], 6.3.1 Office area => Category B
3 . kN . m
2
q´ k.floor
Uniform distributed load
Q k.floor 2 . kN
Concentrated load
[2], 6.3.4 Roofs not accessible except for normal maintenance etc. => Category H =>
0.75 . kN . m
2
Uniform distributed load q´ k.roof
Concentrated load Q k.roof 1.5 . kN
TALAT 2710 7
8. [?], ?? Load from crane, the crane located in the middle of the roof beam in the production hall
Concentrated load Q crane 50 . kN
3.3 Environmental loads
3.3.1 Snow loads
[3] Comment: The characteristic values of the snow loads vary from nation to nation. For simplicity
for this design example, a snow load is chosen including shape coefficient, exposure coefficient
and thermal coefficient. In a design report the calculation of the snow loads have to be shown.
2 kN . m
2
Snow load q´ snow
3.3.2 Wind loads
[3] Comment: The characteristic values of the wind loads vary from nation to nation. For
simplicity, for this design example, a wind load is chosen including all coefficients. In a design
report the calculation of the wind loads including all coefficients have to be shown.
0.70 kN . m
2
Maximum wind load on the external walls q´ w.wall
0.27 kN . m
Wind suction on leeward side of walls 2
q´ w.lee
0.70 kN . m
Wind suction on the roof 2
q´ w.roof
Max. wind suction at the lower edge of the
1.0 kN . m
2
roof and 1.6 m upwards q´ w.edge
TALAT 2710 8
9. 4. Load Combinations
4.1 Ultimate limit state
[3] To decide the section forces on the different members, the following load combinations to
.
be calculated in the ultimate limit state
LC 1: Permanent + imposed + crane + snow loads imposed load dominant
LC 2: Permanent + imposed + crane + snow loads crane load dominant
LC 3: Permanent + imposed + crane + snow loads snow load dominant
LC 4: Reduced permanent + wind loads wind load dominant
LC 5: Permanent + imposed + crane + snow + wind imposed load dominant
LC 6: Permanent + imposed + crane + snow + wind wind load dominant
[3] Comment: All possible load combinations to be calculated
[5], 9.4 ultimate limit state
Partial load factors for different load combinations in the
[5] Table 9.2 Partial factor for permanent action, unfavourable γ Gsup 1.35
Partial factor for permanent action, favourable γ Ginf 1.0
Partial factor for variable action, unfavourable γ Q 1.5
[5] Table 9.3 ψ factor for imposed loads ψ 0i 0.7
ψ factor for snow loads ψ 0s 0.6
ψ factor for wind loads ψ 0w 0.6
[5] Eq In load combinations where the imposed load is
(9.10b) dominating ξ in Eq (9.10b) is less than 1.0, say ξ 0.9
Load combinations
ξ . γ Gsup ξ . γ Gsup ξ . γ Gsup γ Ginf ξ . γ Gsup ξ . γ Gsup
Load case
1 permanent loads
γ Q ψ 0i . γ Q ψ 0i . γ Q 0 γ Q ψ 0i . γ Q
2 distributed loads
ψ u ψ 0i . γ Q γ Q ψ 0i . γ Q 0 ψ 0i . γ Q ψ 0i . γ Q
3 crane load
ψ 0s . γ Q ψ 0s . γ Q γ Q 0 ψ 0s . γ Q ψ 0s . γ Q4 snow loads
0 0 0 γ Q ψ 0w . γ Q γ Q 5 wind loads
Resulting load factors in the ultimate limit state
1.215 1.215 1.215 1 1.215 1.215
1.5 1.05 1.05 0 1.5 1.05
ψ u = 1.05 1.5 1.05 0 1.05 1.05
0.9 0.9 1.5 0 0.9 0.9
0 0 0 1.5 0.9 1.5
TALAT 2710 9
10. 4.2 Serviceability limit state
[5], 9.5.2 Partial load factors forfrequent load combinations in the serviceability limit state
and 9.5.5
LC 1: imposed load dominant
LC 2: crane load dominant
LC 3: snow load dominant
LC 4: wind load dominant
LC 5: wind load only (for comparison)
LC 6: simplified, [5] (9.20) (for comparison)
[5] Table 9.3 ψ factor for imposed loads ψ 1i 0.5 ψ 2i 0.3 ( = 0 for roof)
ψ factor for crane loads ψ 1c 0.5 ψ 2c 0.3
ψ factor for snow loads ψ 1s 0.2 ψ 2s 0
ψ factor for wind loads ψ 1w 0.5 ψ 2w 0
Load combination
1 2 3 4 5 6
Load case
1 1 1 1 0 1
permanent loads
ψ 1i 0 0 0 0 0.9
imposed distributed loads
ψ s ψ 2c ψ 1c ψ 2c 0 0 0.9 imposed crane load
ψ 2s ψ 2s ψ 1s ψ 2s 0 0.9 snow loads
ψ 2w ψ 2w ψ 2w ψ 1w 1 0 wind loads
Resulting partial load factors in theserviceability limit state
Load combination
1 2 3 4 5 6
1 1 1 1 0 1
0.5 0 0 0 0 0.9
ψ s = 0.3 0.5 0.3 0 0 0.9
0 0 0.2 0 0 0.9
0 0 0 0.5 1 0
For the floor, the load combination 1 is valid
For roofs, the load combinations 2, 3 and 4 are valid
Comment: Load combinations 5 and 6 for comparison only
TALAT 2710 10
11. 5. Loads Effects
5.1. Loads per unit length and concentrated loads
Distance between all frames c frame 5000 . mm
Because of continuous purlins and secondary floor beams the load on a beam in a frame is mo
than the distance between the beam times the load per area. Therefore, for the second frame, t
load is increased with a factor ofkf where
k f 1.1
5.1.1 Permanent loads
k f . c frame . q´ p.floor q p.floor = 3.85 kN . m
1
Permanent load on floor q p.floor
k f . c frame . q´ p.roof q p.roof = 2.75 kN . m
1
Permanent load on roof q p.roof
5.1.2 Imposed loads, uniform distributed
k f . c frame . q´ k.floor q k.floor = 16.5 kN . m
1
Distributed load on floor q k.floor
k f . c frame . q´ k.roof q k.roof = 4.125 kN . m
1
Distributed load on roof q k.roof
5.1.3 Imposed loads, concentrated
Concentrated load on floor Q k.floor 2 . kN
Concentrated load on roof Q k.roof 1.5 . kN
Concentrated load from crane P crane 50 . kN
5.1.4 Snow loads
k f . c frame . q´ snow q snow = 11 kN . m
1
Snow load on roof q snow
TALAT 2710 11
12. 5.1.5 Wind loads
k f . c frame . q´ w.wall q w.wall = 3.85 kN . m
1
Maximum wind load on the q w.wall
external walls
k f . c frame . q´ w.lee q w.lee = 1.485 kN . m
1
Wind suction on leeward q w.lee
side of walls
k f . c frame . q´ w.roof q w.roof = 3.85 kN . m
1
Wind suction on the roof q w.roof
Max. wind suction at the k f . c frame . q´ w.edge q w.edge = 5.5 kN . m
1
q w.edge
lower edge of the roof and
1.6 m upwards
5.2 Finite element calculations
Nodes and elements
TALAT 2710 12
13. Moment diagrams
5.2.1 Permanent loads
Values of moments and shear
forces for separate columns
and beams are given in 5.3
5.2.2 Imposed loads,
uniformly distributed
5.2.3 Imposed loads,
concentrated
5.2.4 Snow loads
5.2.5 Wind loads
TALAT 2710 13
14. 5.3. Section forces for characteristic loads
5.3.1 Column A
(FE-cal- Bending moments, section 1, 2, 3 and 4 2.90 4.40 2.14 1.88
culation)
Sections in columns, load cases in rows 9.43 15.6 12.1 5.96
row 1, permanent loads MA 3.65 4.60 5.22 5.81 . kNm
row 2, distributed loads
row 3, crane load 4.28 4.63 4.22 3.04
row 4, snow loads 12.8 6.45 2.52 1.95
row 5, wind loads
18.8 7.32
Axial force, part 1-2 and 3-4
60.5 12.0
Parts in columns, load cases in rows
NA 1.42 2.84 . kN
0.56 28.5 27.8
0.50 15.6 12.0
Deflection in section 4 δ A 2.48 . mm
4.19
8.36
5.3.2 Column B
(FE-cal- Bending moments, section 1, 2, 3, 4,5 and 6 0.07 2.73 4.51 4.35 7.46 9.07
culation) 4.26 15.4 18.2 4.69 8.29 15.5
Sections in columns, load cases in rows
MB 3.27 0.96 2.15 19.27 15.7 11.2 . kNm
row 1, permanent loads
row 2, distributed loads 6.57 5.24 1.59 20.0 34.6 33.5
row 3, crane load
row 4, snow loads 17.0 13.5 0.49 10.95 11.4 12.7
row 5, wind loads
35.1 23.5 9.18
Sections in columns, load cases in rows 84.4 33.9 12.8
Axial force, parts 1-2, 3-4 and 5-6 NB 31.7 31.1 4.34 kN
0.57 95.0 95.7 38.2
0.53 31.2 34.8 13.8
Deflection in section 6 δ B 2.57 . mm
4.30
8.32
TALAT 2710 14
15. 5.3.3 Column C
Comment: To reduce the extent of this example, this column is left out. It can be given,
conservatively the same section as column B
5.3.4 Floor beam D
(FE-cal- Bending moments, section 1, 2 and 3 6.56 10.4 7.25
culation) 27.8 43.6 33.6
Sections in columns, load cases in rows
MD 0.62 4.86 3.11 . kNm
row 1, permanent loads
row 2, distributed loads 0.40 1.62 3.65
row 3, crane load
row 4, snow loads 8.97 2.00 13.0
row 5, wind loads
11.7 11.4
Sections in columns, load cases in rows 48.5 50.5
Shear force, section 1 and 3 VD 1.4 0.6 . kN
3.21 0.67 0.67
13.1 3.66 3.66
Deflection in section 2 δ D 1.63 . mm
0.92
0.91
5.3.5 Roof beam E
Comment: To reduce the extent of this example, this column is left out
5.3.6 Roof beam F
(FE-cal- Bending moments, section 1, 2 and 3 11.8 26.4 4.08
culation) 13.0 42.4 5.28
Sections in columns, load cases in rows
MF 35.0 102.0 10.94 kNm
row 1, permanent loads
row 2, distributed loads 54.1 101.7 17.5
row 3, crane load
row 4, snow loads 22.4 38.0 0.26
row 5, wind loads
TALAT 2710 15
16. Shear force, section 1 and 3 14.3 13.2
Sections in columns, load cases in rows 21.1 20.1
VF 43.4 23.3 . kN
7.71 57.6 52.5
12.6 21.1 17.4
Deflection in section 2 δ F 22.2 . mm
29.3
11.1
5.4. Design moments, shear forces and deflections
5.4.1 Column A
Bending moments
For the load cases 1 - 5 the bending moment in section 1 of column B is:
2.9
9.43
<1>
(5.3.1) MA = 3.65 kNm
4.28
12.8
The load factor matrix is
1.215 1.215 1.215 1 1.215 1.215
1.5 1.05 1.05 0 1.5 1.05
(4.1) ψ u 1.05
= 1.5 1.05 0 1.05 1.05
0.9 0.9 1.5 0 0.9 0.9
0 0 0 1.5 0.9 1.5
TALAT 2710 16
17. The values in the moment vector shall be multiplied with the corresponding load factor for every lo
combination
i 1 .. cols ψ u (cols (ψu) is the number of columns in the matrix ψu)
<i > <1> . <i >
M MA ψ u
3.524 3.524 3.524 2.9 3.524 3.524
14.145 9.901 9.901 0 14.145 9.901
M = 3.832 5.475 3.832 0 3.832 3.832 kNm
3.852 3.852 6.42 0 3.852 3.852
0 0 0 19.2 11.52 19.2
M
The moments in the columns of the matrix (= load combination) are added
<i >
M sum M
i
T
M sum = ( 25.353 22.752 23.677 16.3 13.833 1.909 ) kNm
Maximum and minimum of moment are
M Amax max M sum M Amax = 25.353 kNm
1 1
M Amin min M sum M Amin = 16.3 kNm
1 1
<i > <s> . <i > <i >
(5.3.1) Moments in section 2 s 2 M MA ψ u M sum M
i
T
M sum = ( 37.743 32.793 33.501 5.275 31.938 21.048 ) kNm M Amax max M sum
s
M Amin min M sum
s
<i > <s> . <i > <i >
(5.3.1) Moments in section 3 s 3 M MA ψ u M sum M
i
T
M sum = ( 11.471 3.677 3.494 1.64 9.203 2.246 ) kNm M Amax max M sum
s
M Amin min M sum
s
<i > <s> . <i > <i >
(5.3.1) Moments in section 4 s 4 M MA ψ u M sum M
i
T
M sum = ( 7.86 2.563 7.002 1.045 6.105 2.253 ) kNm M Amax max M sum
s
M Amin min M sum
s
TALAT 2710 17
18. Resulting maximum moments and minimum moments in section 1 to 4 are
T
M Amax = ( 25.353 5.275 11.471 1.045 ) kNm
T
M Amin = ( 16.3 37.743 1.64 7.86 ) kNm
Axial force
Axial force in part 1-2 s 1
18.8
60.5
<s>
(5.3.1) NA = 1.42 kN <i > <s> . <i > <i >
N NA ψ u N sum N
28.5 i
15.6 N Amax max N sum
s
T
N sum = ( 137.751 109.887 127.626 4.6 123.711 87.126 ) kN N Amin min N sum
s
<i > <s> . <i > <i >
(5.3.1) Axial force in part 3-4 s 2 N NA ψ u N sum N
i
T
N sum = ( 48.932 42.254 60.212 10.68 38.132 25.532 ) kN N Amax max N sum
s
N Amin min N sum
s
Resulting maximum and minimum axial forces in part 1, 2 and 3 are:
T T
N Amax = ( 4.6 10.68 ) kN N Amin = ( 137.8 60.2 ) kN
Deflection
Deflection in section 6 s 1
0.56
0.5
<s> <i > <s> . <i > <i >
(5.3.1) δ A = 2.48 mm δ δ A ψ s δ sum δ
i
4.19
δ Amax max δ sum δ Amin min δ sum
8.36 s s
T
δ sum = ( 1.054 1.8 2.142 4.74 8.36 6.113 ) mm
Resulting maximum and minimum deflection in section 6 δ Amax= ( 8.36 ) mm
δ Amin ( 1.054 ) mm
=
TALAT 2710 18
19. 5.4.2 Column B
Bending moments
For the load cases 1 - 5 the bending moment in section 1 of column B is:
0.07
4.26
<1>
(5.3.2) MB = 3.27 kNm
6.57
17
The load factor matrix is
1.215 1.215 1.215 1 1.215 1.215
1.5 1.05 1.05 0 1.5 1.05
(4.1) ψ u 1.05
= 1.5 1.05 0 1.05 1.05
0.9 0.9 1.5 0 0.9 0.9
0 0 0 1.5 0.9 1.5
The values in the moment vector shall be multiplied with the corresponding load factor for every l
combination
i 1 .. cols ψ u (cols (ψu) is the number of columns in the matrix ψu)
<i > <1> . <i >
M MB ψ u
0.085 0.085 0.085 0.07 0.085 0.085
6.39 4.473 4.473 0 6.39 4.473
M= 3.433 4.905 3.433 0 3.433 3.433 kNm
5.913 5.913 9.855 0 5.913 5.913
0 0 0 25.5 15.3 25.5
M
The moments in the columns of the matrix (= load combination) are added
<i >
M sum M
i
T
M sum = ( 3.042 6.43 8.901 25.43 12.258 20.541 ) kNm
Maximum and minimum of moment are
M Bmax max M sum M Bmax = 25.43 kNm
1 1
M Bmin min M sum M Bmin = 8.901 kNm
1 1
TALAT 2710 19
20. <i > <s> . <i > <i >
(5.3.2) Moments in section 2 s 2 M MB ψ u M sum M
i
T
M sum = ( 20.693 13.331 10.619 22.98 32.843 34.013 ) kNm M Bmax max M sum
s
M Bmin min M sum
s
<i > <s> . <i > <i >
(5.3.2) Moments in section 3 s 3 M MB ψ u M sum M
i
T
M sum = ( 31.953 22.796 24.717 5.245 32.394 24.498 ) kNm M Bmax max M sum
s
M Bmin min M sum
s
<i > <s> . <i > <i >
(5.3.2) Moments in section 4 s 4 M MB ψ u M sum M
i
T
M sum = ( 36.484 47.266 50.594 12.075 26.629 22.169 ) kNm M Bmax max M sum
s
M Bmin min M sum
s
<i > <s> . <i > <i >
(5.3.2) Moments in section 5 s 5 M MB ψ u M sum M
i
T
M sum = ( 69.124 72.458 86.153 9.64 58.864 48.293 ) kNm M Bmax max M sum
s
M Bmin min M sum
s
<i > <s> . <i > <i >
(5.3.2) Moments in section 6 s 6 M MB ψ u M sum M
i
T
M sum = ( 76.18 74.245 89.305 9.98 64.75 50.155 ) kNm M Bmax max M sum
s
M Bmin min M sum
s
Resulting maximum moments and minimum moments in section 1 to 6 are
T
M Bmax = ( 25.43 10.619 32.394 50.594 9.64 9.98 ) kNm
T
M Bmin = ( 8.901 34.013 5.245 12.075 86.153 89.305 ) kNm
Axial force
Axial force in part 1-2 s 1
35.1
84.4
<s>
(5.3.2) NB = 31.7 kN <i > <s> . <i > <i >
N NB ψ u N sum N
95 i
31.2 N Bmax max N sum
s
T
N sum = ( 288.031 264.317 307.051 11.7 259.951 203.251 ) kN N Bmin min N sum
s
TALAT 2710 20
21. <i > <s> . <i > <i >
(5.3.2) Axial force in part 3-4 s 2 N NB ψ u N sum N
i
T
N sum = ( 198.187 196.927 240.352 28.7 166.868 130.732 ) kN N Bmax max N sum
s
N Bmin min N sum
s
<i > <s> . <i > <i >
(5.3.2) Axial force in part 5-6 s 3 N NB ψ u N sum N
i
T
N sum = ( 69.291 65.484 86.451 11.52 56.871 42.831 ) kN N Bmin min N sum
s
N Bmax max N sum
s
Resulting maximum and minimum axial forces in part 1, 2 and 3 are:
T T
N Bmax = ( 11.7 28.7 11.52 ) kN N Bmin = ( 307.1 240.4 86.5 ) kN
Deflection
Deflection in section 6 s 1
0.57
0.53
<s> <i > <s> . <i > <i >
(5.3.2) δ B = 2.57 mm δ δ B ψ s δ sum δ
i
4.3
δ Bmax max δ sum δ Bmin min δ sum
8.32 s s
T
δ sum = ( 1.076 1.855 2.201 4.73 8.32 6.276 ) mm
Resulting maximum and minimum deflection in section 6 δ Bmax ( 8.32 ) mm
=
δ Bmin ( 1.076 ) mm
=
5.4.3 Column C
Comment: To reduce the extent of this example,
calculation of this column is left out. It can,
conservatively, be given the same dimensions
as column B
TALAT 2710 21
22. 5.4.4 Floor beam D
Bending moments
For the load cases 1 - 5 the bending moment in section 1 of beam D is:
6.56
27.8
<1>
(5.3.4) MD = 0.62 kNm
0.4
8.97
The load factor matrix is
1.215 1.215 1.215 1 1.215 1.215
1.5 1.05 1.05 0 1.5 1.05
(4.1) ψ u 1.05
= 1.5 1.05 0 1.05 1.05
0.9 0.9 1.5 0 0.9 0.9
0 0 0 1.5 0.9 1.5
The values in the moment vector shall be multiplied with the corresponding load factor for every l
combination
i 1 .. cols ψ u (cols (ψu) is the number of columns in the matrix ψu)
<i > <1> . <i >
M MD ψ u
7.97 7.97 7.97 6.56 7.97 7.97
41.7 29.19 29.19 0 41.7 29.19
M = 0.651 0.93 0.651 0 0.651 0.651 kNm
0.36 0.36 0.6 0 0.36 0.36
0 0 0 13.455 8.073 13.455
M
The moments in the columns of the matrix (= load combination) are added
<i >
M sum M
i
T
M sum = ( 49.379 36.59 37.109 6.895 41.306 23.414 ) kNm
Maximum and minimum of moment are
M Dmax max M sum M Dmax = 6.895 kNm
1 1
M Dmin min M sum M Dmin = 49.379 kNm
1 1
TALAT 2710 22
23. <i > <s> . <i > <i >
(5.3.4) Moments in section 2 s 2 M MD ψ u M sum M
i
T
M sum = ( 84.597 67.164 65.949 13.4 86.397 67.977 ) m kN M Dmin min M sum
s
M Dmax max M sum
s
<i > <s> . <i > <i >
(5.3.4) Moments in section 3 s 3 M MD ψ u M sum M
i
T
M sum = ( 52.658 36.139 35.348 26.75 64.358 57.038 ) m kN M Dmin min M sum
s
M Dmax max M sum
s
Resulting maximum moments and minimum moments in section 1 to 3 are
6.895 49.379
M Dmax = 86.397 kNm M Dmin = 13.4 kNm
26.75 64.358
Shear force
Shear force in section 1 s 1
11.7
48.5
<s>
(5.3.4) VD = 1.4 kN <i > <s> . <i > <i >
V VD ψ u V sum V
0.67 i
3.66 V Dmax max V sum V Dmin min V sum
s s
T
V sum = ( 89.038 67.844 67.615 6.21 85.745 61.723 ) kN
Shear force in section 2 s 2
11.4
50.5
<s>
(5.3.4) VD = 0.6 kN <i > <s> . <i > <i >
V VD ψ u V sum V
0.67 i
3.66 V Dmax max V sum V Dmin min V sum
s s
T
V sum = ( 88.368 65.373 65.241 16.89 91.662 71.133 ) kN
Resulting maximum and minimum shear forces in section 1 and 3 are
89.038 6.21
V Dmax = kN V Dmin = kN
91.662 16.89
Deflection
TALAT 2710 23
24. 89.038 6.21
V Dmax = kN V Dmin = kN
91.662 16.89
Deflection
Deflection in section 2 s 1
3.21
13.1
<s>
(5.3.4) δ D = 1.63 mm <i > <s> . <i > <i >
δ δ D ψ s δ sum δ
0.92 i
0.91 δ Dmax max δ sum δ Dmin min δ sum
s s
T
δ sum = ( 10.249 4.025 3.883 2.755 0.91 17.295 ) mm
Resulting maximum and minimum deflection in section 2
δ Dmax= ( 17.295 ) mm δ Dmin ( 0.91 ) mm
=
5.4.5 Roof beam E
Comment: To reduce the extent of this example, calculation of this beam is left out. It can be given
the same dimensions as floor beam D
5.4.6 Roof beam F
Moment
For the load cases 1 - 5 the bending moments in section 1 to 3 of the beam F are
<i > <s> . <i > <i >
(5.3.6) Moments in section 1 s 1 M MF ψ u M sum M
i
T
M sum = ( 119.277 129.177 145.887 21.8 99.117 79.827 ) m kN M Fmin min M sum
s
M Fmax max M sum
s
<i > <s> . <i > <i >
(5.3.6) Moments in section 2 s 2 M MF ψ u M sum M
i
T
M sum = ( 294.306 321.126 336.246 30.6 260.106 218.226 ) m kN M Fmin min M sum
s
M Fmax max M sum
s
<i > <s> . <i > <i >
(5.3.6) Moments in section 3 s 3 M MF ψ u M sum M
i
T
M sum = ( 40.114 42.661 48.238 3.69 39.88 37.348 ) m kN M Fmin min M sum
s
M Fmax max M sum
s
TALAT 2710 24
25. Resulting maximum moments and minimum moments in section 1 to 3 are
21.8 145.887
M Fmax = 336.246 kNm M Fmin = 30.6 kNm
3.69 48.238
Shear force
Shear force in section 1 s 1
14.3
21.1
<s>
(5.3.4) VF = 43.4 kN <i > <s> . <i > <i >
V VF ψ u V sum V
57.6 i
21.1 V Fmax max V sum V Fmin min V sum
s s
T
V sum = ( 146.435 156.47 171.5 17.35 127.444 105.289 ) kN
Shear force in section 3 s 2
13.2
20.1
<s>
(5.3.4) VF = 23.3 kN <i > <s> . <i > <i >
V VF ψ u V sum V
52.5 i
17.4 V Fmax max V sum V Fmin min V sum
s s
T
V sum = ( 117.903 119.343 140.358 12.9 102.243 82.758 ) kN
Resulting maximum and minimum shear force in section 1 and 3 are
171.5 17.35
V Fmax = kN V Fmin = kN
140.358 12.9
Deflection
Deflection in section 2 s 1
<s>T <i > <s> . <i > <i >
(5.3.4) δ F = ( 7.71 12.6 22.2 29.3 11.1 ) mm δ δ F ψ s δ sum δ
i
δ Fmin min δ sum
s
T
δ sum = ( 20.67 18.81 20.23 2.16 11.1 65.4 ) mm δ Fmax max δ sum
s
[5] (9.20) Simplified verification δ Fmax = ( 65.4 ) mm δ Fmin ( 11.1 ) mm
=
[5] (9.16) Load combination 3 δ sum = 20.2 mm
3
TALAT 2710 25
26. 5.4.7 Joint A-D
(5.4.1) and Moment M Amax = 11.5 kNm
(5.5.4) 3
M Dmin = 49.4 kNm
1
M Amin = 37.7 kNm
2
Shear V Dmax = 89 kN
1
Check: M Amax M Dmin M Amin = 0.17 kNm
3 1 2
5.4.8 Joint B-D
(5.4.2) and Moment M Bmax = 32.4 kNm M Bmin = 34 kNm
(5.5.4) 3 2
M Dmin = 64.4 kNm V
3
Check: M Bmax M Dmin M Bmin = 2.05 kNm V
3 3 2
Shear V Dmax = 91.7 kN
2
N
V B3 = 2.306 kN V B2 = 18.184 kN
Axial N B3 = 240.4 kN N B2 = 307.1 kN
5.4.9 Joint A-E and joint B-E
5.4.10 Comment: To reduce the extent of this example, calculation of this joint is left out
5.4.11 Joint B-F
(5.4.2) and Moment M Bmin = 86.153 kNm
(5.5.6) 5
M Fmin = 145.9 kNm
1
M Bmax = 50.6 kNm
4
Shear V Fmax = 171.5 kN
1
Check: M Bmin M Fmin M Bmax = 9.14 kNm
5 1 4
(The reason why the sum of the moments is not = 0 is the fact that all the moments does not belong to
the same load combination)
TALAT 2710 26
27. 5.4.12 Joint F-C
(5.4.6) Moment M Fmin = 48.238 kNm
3
M Fmax = 3.7 kNm
3
Shear V Fmin = 12.9 kN
2
V Fmax = 140.4 kN
2
5.4.13 Column bases See 6.1.13 and 6.2.13
TALAT 2710 27
28. 6. Code Checking
6.1 Column A
6.1.1 Dimensions and material properties
Section height: h 160 . mm
Flange depth: b 150 . mm
Web thickness: tw 5 . mm
Flange thickness: tf 14 . mm
Overall length: L1 3 .m
Distance between purlins: cp 1 .m
[1] Table 3.2b Alloy: EN AW-6082 T6 EP/O t > 5 mm
newton newton
f 0.2 260 . fu 310 .
2 2
mm mm
heat_treated 1 (if heat-treated then 1 else 0)
[1] (5.4), (5.5) f o f 0.2 fa fu
fo newton newton newton
[1] (5.6) fv f v = 150 E 70000 . G 27000 .
2 2 2
3 mm mm mm
Partial safety factors: γ M11.10 γ M2 1.25
Inner radius: r 5 . mm
Web height: bw h 2 .t f 2 .r b w = 122 mm
S.I. units: kN 1000 . newton kNm kN . m MPa 1000000 . Pa
TALAT 2710 28
29. 6.1.2 Internal moments and forces
(5.4.2) Bending moments and axial forces for load case LC1, LC3 and LC6 in section 1, 2, 3 and 4
0 25.4 23.8 1.91
3.0 37.7 33.5 21.1
x M LC1 . kNm M LC3 . kNm M LC6 . kNm
3.1 11.5 3.49 2.24
5.5 7.86 7.00 2.25
138 127 87
138 127 87
N LC1 . kN N LC3 . kN N LC6 . kN
49 60 26
49 60 26
Bending moment kNm Axial force kN
4 4
2 2
0 0
40 20 0 20 40 0 50 100 150
Load case 1 Load case 1
Load case 3 Load case 3
Load case 6 Load case 6
Load case 1 Load case 3 Load case 6
Moment in section 2 M LC1 = 37.7 kNm M LC3 = 33.5 kNm M LC6 = 21.1 kNm
2 2 2
Moment at column base 1 M LC1 = 25.4 kNm M LC3 = 23.8 kNm M LC6 = 1.91 kNm
1 1 1
Axial force in part 1-2 N LC1 = 138 kN N LC3 = 127 kN N LC6 = 87 kN
1 1 1
Preliminary calculations show that load case 1 is governing. Study part 1-2 from column base
to floor beam. Moment in top of part 1-2 (section 2) is larger than at column base (section 1) why
M 1.Ed (below) correspond to section 2 andM 2.Ed to section 1 of the column.
TALAT 2710 29
30. Load case 1
Bending moment in section 2 M 1.Ed M LC1 M 1.Ed = 37.7 kNm
2
Bending moment at column base (1) M 2.Ed M LC1 M 2.Ed = 25.4 kNm
1
Axial force in part 1-2 (compression) N Ed N LC1 N Ed = 138 kN
1
6.1.3 Classification of the cross section in y-y-axis bending
β w bending
a) Web
b1
[1] 5.4.3 b1 bw t1 tw β w 0.40 . β w 9.76
=
t1
250 . newton
[1] Tab. 5.1 ε β 1w 11 . ε β 1w 10.786
=
fo 2
mm
Heat treated,
unwelded = no β 2w 16 . ε β 2w 15.689
=
longitudinal
weld β 3w 22 . ε β 3w 21.573
=
class w if β w β 1w , if β w > β 2w , if β w > β 3w , 4 , 3 , 2 , 1
> class w = 1
[1] 5.4.5 Local buckling
β w 32 220
ρ cw if 22 , 1.0 , ρ cw 1
=
ε β w β w
2
ε ε
t w.ef.b if class w 4 , t w . ρ cwt w
, ( b = bending) t w.ef.b = 5.0 mm
b) Flanges
[1] 5.4.3 ψ 1
0.8
[1] (5.7.),(5.8.) g if ψ > 1 , 0.7 0.3 . ψ , g=1
1 ψ
b tw 2 .r b2
b2 t2 tf β f g. β f 4.821
=
2 t2
[1] Tab. 5.1 ε = 0.981 β 1f 3 . ε β 1f 2.942
=
β 2f 4.5 . ε β 2f 4.413
=
β 3f 6 . ε β 3f 5.883
=
class f if β > β 1f , if β f > β 2f , if β f > β 3f , 4 , 3 , 2 , 1
f class f = 3
TALAT 2710 30
31. [1] 5.4.5 Local buckling:
β f 10 24
ρ cf if 6 , 1.0 , ρ cf 1
=
ε β f β f
2
ε ε
t f.ef if class f 4 , t f . ρ cft f
, t f.ef = 14.0 mm
Classification of the cross-section in y-y axis bending
class y if class f > class w , class f , class w class y = 3
6.1.4 Classification of the cross section in z-z-axis bending
Cross section class of web: No bending stresses class w 1
Cross section class for flanges: According to above class f = 3
class z if class f > class w , class f , class w class z = 3
6.1.5 Classification of the cross section in axial compression
β wc compression
a) Web
b1
b1 bw t1 tw β wc β wc = 24.4
t1
[1] Tab. 5.1 β 1w 10.786
=
β 2w 15.689
=
β 3w 21.573
=
class wc if β wc β 1w , if β wc > β 2w , if β wc > β 3w , 4 , 3 , 2 , 1
> class wc = 4
[1] 5.4.5 Local buckling
β wc 32 220
ρ cw if 22 , 1.0 , ρ cw 0.931
=
ε β wc β wc
2
ε ε
t w.ef if class wc 4 , t w . ρ cwt w
, t w = 5 mm t w.ef = 4.7 mm
b) Flanges
Same as in bending t f.ef = 14.0 mm class f = 3
Classification of the total cross-section in axial compression
class c if class f > class wc , class f , class wc class c = 4
TALAT 2710 31
32. 6.1.6. Welds
[1] 5.5 HAZ softening at column ends
[1] Tab.5.2 ρ haz 0.65
[1] Fig.5.6 Extent of HAZ (MIG-weld) t1 tf
b haz if t 1 > 6 . mm , if t 1 > 12 . mm , if t 1 > 25 . mm , 40 . mm , 35 . mm , 30 . mm , 20 . mm
b haz = 35 mm
6.1.7 Design resistance, y-y-axis bending
[1] 5.6.1 W
Elastic modulus of the gross cross section el:
2 .b .t f 2 .t f .t w A g = 4.86 . 10 mm
3 2
Ag h
1. . 3
tw . h 2 .t f
3
Ig bh b
12
I g = 2.341 . 10 mm
7 4
I g .2
W el = 2.926 . 10 mm
5 3
W el
h
W
Plastic modulus of the gross cross section pl:
1.
b .h tw . h 2 .t f W pl = 3.284 . 10 mm
2 2 5 3
W pl b
4
TALAT 2710 32
33. W
Elastic modulus of the effective cross section effe:
t f = 14 mm t f.ef = 14 mm
bw
As tf.ef = tf then bc b c = 61 mm
2
t w = 5 mm
t w.ef.b = 5 mm
bf 0.5 . b tw 2 .r b f = 67.5 mm
2 .b f . t f b c. t w A eff = 4.86 . 10 mm
3 2
A eff Ag t f.ef t w.ef.b
Shift of gravity centre:
2
h tf bc 1
e ef 2 .b f . t f t f.ef . . t
w t w.ef.b . e ef = 0 mm
2 2 2 A eff
Second moment of area wiht respect to centre of gross cross section:
2 3
h tf bc
2 .b f . t f t f.ef . . t I eff = 2.341 . 10 mm
7 4
I eff Ig w t w.ef.b
2 2 3
Second moment of area wiht respect to centre of effective gross section:
e ef . A eff I eff = 2.341 . 10 mm
2 7 4
I eff I eff
I eff
W eff = 2.926 . 10 mm
5 3
W eff
h
e ef
2
[1] Tab. 5.3 Shape factor α
- for class 1 or 2 cross-sections:
W pl
α 1.2.w α 1.2.w = 1.122
W el
- for welded, class 3 cross-sections:
TALAT 2710 33
34. [1] Tab. 5.3 Shape factor α
- for class 1 or 2 cross-sections:
W pl
α 1.2.w α 1.2.w = 1.122
W el
- for welded, class 3 cross-sections:
β 3w β w W pl W el
[1] (5.16) α 3.ww 1 . α 3.ww 1.245
=
β 3w β 2w W el
β 3f β f W pl W el
[1] (5.16) α 3.wf 1 . α 3.wf= 1.088
β 3f β 2f W el
β, β2, β3 are the slenderness parameter and the limiting values for the most critical
element in the cross-section, so it is the smaller value of3.ww and α3.wf
α
α 3.w if α 3.ww α 3.wf , α 3.ww , α 3.wf α 3.w 1.088
=
W eff
- for class 4 cross-sections: α 4.w α 4.w = 1
W el
class y = 3
α y if class y > 2 , if class y > 3 , α 4.w , α 3.w , α 1.2.w α y 1.088
=
M
Design moment of resistance of the cross section c,Rd
f o . α .y el
W
[1] (5.14) M y.Rd M y.Rd = 75.3 kNm
γ M1
6.1.8 Design resistance, z-z-axis bending
Cross section class class z = 3
t f .b
3
2. I z = 7.875 . 10 mm
6 4
Gross cross section: Iz
12
t f.ef . b
3
2. I z.ef = 7.875 . 10 mm
6 4
Effective cross section: I z.ef
12
I z.2 I z.ef . 2
Section moduli: Wz W z.ef
b b
Wz
Shape factor: α z α z=1
W z.ef
f o . α .z z
W
Bending resistance: M z.Rd M z.Rd = 24.818 kNm
γ M1
TALAT 2710 34
35. 6.1.9 Axial force resistance, y-y buckling
[1] 5.8.4 Cross section area of gross cross sectionAgr
b .h tw . h 2 .t f A gr = 4.86 . 10 mm
3 2
A gr b
Cross section area of effective cross sectionAef t w.ef = 4.653 mm
2 .b 2 . t f b w. t w A ef = 4.818 . 10 mm
3 2
A ef A gr t f.ef t w.ef
( t f = 14 mm t w = 5 mm 2 . b 2 = 135 mm t f.ef = 14 mm t w.ef = 4.653 mm )
A ef
Effective cross section factor η η = 0.991
A gr
Second moment of area of gross cross sectionIy:
2
2. . 3 h tf 1.
2 .b .t f . 2 .t f .t w
3
Iy btf h
12 2 12
[1] Table 5.7 Buckling length factor Ky 1.5 L1=3 m
Case 5
l yc K y .L 1 l yc = 4.5 m
π .E .I y
2
Buckling load N cr
2
l yc
N cr = 798.639 kN
A gr . η . f o
[1] 5.8.4.1 Slenderness parameter λ y λ y 1.252
=
N cr
[1] Table 5.6 α if ( heat_treated 1 , 0.2 , 0.32 ) α = 0.2
λ o if ( heat_treated 1 , 0.1 , 0 ) λ o 0.1
=
0.5 . 1 α . λ y
2
φ λ o λ y φ = 1.399
1
χ y χ y = 0.494
2 2
φ φ λ y
[1] Table 5.5 Symmetric profile k1 1
[1] Table 5.5 No longitudinal welds k2 1
fo
Axial force resistance N y.Rd χ .y . k 1 . k 2 .
η .A
gr N y.Rd = 562.6 kN
γ M1
TALAT 2710 35