Post on 04-Apr-2018
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CSE 362 Design of Steel StructuresDesign philosophy, allowable stress design, design examples,limit stress design, comparison
Professor K.F. ChungProfessor K.F. Chung
Department of Civil and Structural EngineeringThe Hong Kong Polytechnic University
Hong Kong SAR, China
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Overall stability / equilibrium considerations / Linear elastic design
ACTION RESISTANCE
Section capacityAxial forces
Tension
Compression
where F is the applied forceA is the cross-sectional areaDesign is the design strength
Shear force
Bending moment
DesignA
F =
Design
A
F =
DesigntI
yAS =
DesignZ
M =
Allowable stress design
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F = 100 kN
DataDesign strength of mild steely,ASD = 165 N/mm
2
Consider
DesignAF =
A
FDesign
165
10x100
FA
3
Design
required =
Use a 25 x 25 mm square rod with the following cross-sectional area:
A = 625 mm2 Arequired = 606 mm2
Design of a tie
= 606 mm2
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DataP = 10 kNDesign strength of mild steelDesign = 165 N/mm
2 or 0.6 x 275 N/mm2
Density of steel = 7850 kg/m3
Unit cost= $20 /kg
Assume no instability.
Truss 1
AB
C
P1.0 m
1.0 m
-P
P2+
LABLAC
Truss 2
A
B
C
P
1.0 m
1.0 m
+P
P2LABLAC
2.0 m
Truss 3
A
C
P
LABLAC
B
2
P+
2P
1.0 m
Design of a two-member truss
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Member Length (m) Total length (m) Force (kN)
Tensionmembers
AB 2.8
7.6
141
BC 2.0 200
CE 2.8 283
Compressionmembers
AD 4.0
10.8
100
BD 2.8 141CD 2.0 200
ED 2.0 200
2.0 m 4.0 m
P = 100 kN
P
2.0 m
-2P -P
-2P
+2P
P2+P2
P22+
A
BC
DE
2.0 m
Design of a cantilever truss
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(N/mm2)
Design = 275 N/mm2
t = 183 N/mm2 with load factor 1.5 tension
c = 100 N/mm2 compression due to buckling
Spare strength / reserved strength
PlasticElastic
Simplified stress-strain curve
Design of a cantilever truss
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For S275 steel, the yield strength is 275 N/mm2. With a factor of safetyequal to 1.5, the design strength of the members under tension is
For those members under compression, buckling may occur, and hence,the design strength is further reduced to
c = 100 N/mm2 for I section
No reduction in the design strength for square hollow section incompression.
tensionmm/N1835.1
275 2t ==
SchemeS275 steel
Design strength (N/mm2)
Tension Compression
I Solid bar 183 100
II Joist 183 100
III Square hollow section 183 183
Design of a cantilever truss
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Design for member under tension (Scheme I)
F = 283 kN (largest tension force in member CE)
t = 183 N/mm2
The cross sectional area, A, of the member required is
Arequired = F /t = 283 x 103 / 183
= 1546 mm2
Use 40 x 40 mm solid bar
Aprovided = 40 x 40
= 1600 mm2 > Areq = 1546 mm2
or the stress,= F / Aprovided= 283 x 103 / 1600
= 176.9 N/mm2
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F = 200 kN (largest compression force in members CD & ED)
c = 100 N/mm2
The cross sectional area, A, of the member required is
Arequired = F /c = 200 x 103 / 100= 2000 mm2
Use 45 x 45 mm solid bar
Aprovided = 45 x 45
= 2025 mm2 > Areq = 2000 mm2
or the stress,= F / Aprovided= 200 x 103 / 2025
= 98.8 N/mm2 Areq = 1546 mm2
or the stress,= F / Aprovided= 283 x 103 / 1620
= 174.7 N/mm2
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F = 200 kN (largest compression force in members CD & ED)
c = 100 N/mm2
The cross sectional area, A, of the member required is
Arequired = F /c = 200 x 103 / 100= 2000 mm2
Use 89 x 89 x 19.4 kg/m Joist
Aprovided = 24.9 cm2
= 2490 mm2 > Areq = 2000 mm2
or the stress,= F / Aprovided= 200 x 103 / 2490
= 80.3 N/mm2 Areq = 1546 mm
2
or the stress,= F / Aprovided= 283 x 103 / 1690
= 167.5 N/mm2
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F = 200 kN (largest tension force in member CE)
c = 183 N/mm2 (=t =y / 1.5)
No reduction for buckling is necessary.
The cross sectional area, A, of the member required isArequired = F /c = 200 x 10
3 / 183
= 1093 mm2
Use 90 x 90 x 3.6 SHS, 9.3 kg/m
Aprovided = 1240 mm2 > Areq = 1093 mm
2
or the stress,= F / Aprovided= 200 x 103 / 1240
= 161.3 N/mm2
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Scheme
Section designationMemberlength
(m)
*Weight
(kg)
Totalweight
(kg)
**Unitcost
($/kg)
Totalcost
($)DimensionsUnit
weight(kg/m)
IT
C
40 x 40 Solid bar
45 x 45 Solid bar
12.6
15.9
7.6
10.8
96
172268 30 8040
IIT
C
76 x 76 x 12.7 Joist
89 x 89 x 19.4 Joist
12.7
19.4
7.6
10.8
97
210307 15 4605
IIIT
C
90 x 90 x 5.0 SHS
90 x 90 x 3.6 SHS
13.3
9.3
7.6
10.8
102
101203 18 3654
* Minimum amount of materials (weight/tonnage) does NOT always meansmost economical design.
** The unit cost also includes cost of fabrication / welding / erection.
Design of a cantilever trussTwo section sizes
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Scheme One sectionsize
Two sectionsize
Ratio
One : Two
I 8790 (2.0) 8040 (2.2) 1.09 : 1.0
II 5355 (1.2) 4605 (1.3) 1.16 : 1.0
III 4410 (1.0) 3654 (1.0) 1.21 : 1.0
Scheme III is always preferred based on the above assumption and
data according to design strength unit cost (material / fabrication / welding / erection) availability of materials
Although there is saving of about 20% in using two section sizes,only one section size may be used in practice due to small quantity.The normal unit of purchase is one ton.
Design of a cantilever truss
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As demonstrated in the above example Design of a truss, thefollowing data are very important:
section properties of members, such as cross-sectional area
design strength of materials under different types of internal forces- t tension- c compression (buckling ?)- s shear- b bending- tor torsion
partial safety factor for materials under different types of internalforces
unit cost of material / fabrication / welding / erection / time
Structural design
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Linear elastic design, i.e. all members of a structure operate withinthe linear elastic range of the stress-strain curve of the steelmaterial. Moreover, the structural response of all the membersoperate within the linear elastic range of the load-deformation curvesof the structure.
DataLoading, F obtained from loading allowance
Strength of material, fy obtained from tests or codesSafety factor, = 1.5 ~ 2.0
Design strength ft =
Section properties of the members, A
fy
Linearelasticrange
LOAD RESISTANCE
Applied stress Design strength
F/A ft = fy/
Allowable stress designWorking stress design, or permissible stress design
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Partial safety factors to allow for extreme situations
-f on loading G for permanent actions
Q for variable actions
Loads U.K. H.K. EuropeDead load (D)Imposed load (L)Wind load (W)
1.41.61.4
1.41.61.4
1.31.51.3
D + L + W 1.2 1.2 1.1
Dead load restraining upliftand overturning
1.0 1.0 1.0
-M on materials/resistance
Steel M = 1.0 or 1.05 (tension)
= 1.1 (local buckling)
= 1.2 (connection)
Concrete M = 1.5 (compression)
Partial safety factors for extreme situations
Limit state designUltimate limit state
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DataLoad = FPartial safety factor,f = 1.0 ~ 1.6Factored design load =f x F (increased)Strength of material = fy from tests or codes (y ormax)Partial safety factor,M = 1.0 ~ 1.5Section properties, A
Fullrange
LOAD RESISTANCE
Internal forces due toapplied load
Section capacityMember resistance
ftA = fyA /
ULS - strength (yielding, buckling, rupture)- stability against overturning & sway- fracture due to fatigue- brittle fracture
Ff M
Limit state designUltimate limit state
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Long term behaviour (not extreme situation) tends to happenduring the life time of the structure.
deflection / deformation (most common)
vibration, fatigue, corrosion & durability
From deflection, the design criterion is
Expected deflection < Limits of deflection
Compared to load/resistance calculation, the deflection is always veryapproximate. It is the order of magnitude other than the exact value moremeaningful.
Typical deflection limits
deflection due to unfactored imposed load
beams carrying plaster or other brittle finish
all other beams
180
Span
360
Span
200
Span
Limit state designServiceability limit state
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Allowable stress design Limit state design
Stress Cross-section capacity
Axial force F P = pyA
Shear force Fv P = pvAv
Bending moment M Mc = py S or 1.2 py Z
Interaction
yc pA
Fp =
[ ]vp
tI
yAS =
yb pzMp =
ypz
M
A
P+ 1
M
M
P
F
c
+
Allowable stress design vs Limit state design
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ASDDesign on the allowable
LSDDesign on the limit
One overall factor of safety on each loading type and load combinationon each material, failure mode and otherconstruction effects such as connection.
All the members of a structureoperate within the linearelastic range.
Design for failureSome materials may be yielded in service.
Deflection check Deflection check in serviceability limit state withdifferent set of partial safety factor (usually equal to1.0)
M
f
Allowable stress design vs Limit state design