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BS 5950: 2000 : Part 1
Continuous construction
Multi-storey frames
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Simple Design – pin joints
(a) Web Cleats (b) End Plate (c) Fin Plates
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Continuous Design – rigid joints
Extended end plate connection
Continuous multistorey frame
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Base stiffness and strength
As in the lecture on portal framesAs in the lecture on portal frames
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Deformation of a multi-storey frame
P
P
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P - effects
P
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Stability effectsP
F
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Effective lengths and critical loads
PPE E = = 22EI/LEI/L22
PPcritcrit= 0.25= 0.2522EI/LEI/L2 2 = = 22EI/4LEI/4L22 = = 22EI/(2L)EI/(2L)22
Hence the effective length LHence the effective length LEE for a cantilever is 2L for a cantilever is 2L
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Nominal effective lengths
None Position Position Position
Position Position Position Position
Direction Direction Direction
Direction Direction Direction
1.0 L 0.85 L 0.7 L 2.0 L 1.2 L
Position
Restraint
Restraint
Practical L E
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Charts of Annex E
0.2
0.4
0.6
0.8
1.0
0.2 0.4 0.6 0.8 1.0 0 0.5
0.6
0.7
0.9 0.8
1.0
a) Non sway frames
k 2
k 1
Fixed Pinned
Pinned
0.2
0.4
0.6
0.8
1.0
0.2 0.4 0.6 0.8 1.0 0 1.0
k 2
k 1
Fixed Pinned
Pinned
1.2
1.3
1.4
1.6
2.0 3.0
1.1
b) Sway frames
K TL K TR
K u
K L
K c
K BR K BL
INFINITY
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Derivation of Charts
Model used by Scott
ku = KC / (KC + KTL + KTR) LE
k1 = Kc+ KU / (KC +KU + KTL + KTR)
kl = KC / (KC + KBL + KBR)
k2 = KC+ KL / (KC +KL + KBL + KBR)
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Use of the charts of Annex E kk11 = (K = (Kcc + K + Kuu) / (K) / (Kcc+K+Kuu+K+KTLTL +K +KTRTR))
kk22 = (K = (Kcc + K + KLL) / (K) / (Kcc+K+KLL+K+KBLBL +K +KBRBR))
The stiffness K for each member is taken The stiffness K for each member is taken as a function of I / L. as a function of I / L.
If a beam supports a floor slab, its K value If a beam supports a floor slab, its K value should be taken as I / L. should be taken as I / L.
For a beam which is not rigidly connected For a beam which is not rigidly connected to the column K should be taken as zero.to the column K should be taken as zero.
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Use of the charts of Annex E
For a beam which carries more than 90% of its For a beam which carries more than 90% of its moment capacity, a pin should be inserted at that moment capacity, a pin should be inserted at that locationlocation . .
If either end of the column carries more than 90% If either end of the column carries more than 90% of Mof Mprpr the value of k the value of k
11 or k or k22 as appropriate should as appropriate should
be taken as 1.0. be taken as 1.0. For other conditions, the appropriate values of K For other conditions, the appropriate values of K
are given in Tables E1, E2 and E3 of the code. are given in Tables E1, E2 and E3 of the code.
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Buckled mode shapes
Non Sway Frame Sway Frame
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Beam stiffness valuesbuildings without floor slabs
Provided that the frame is reasonably Provided that the frame is reasonably regular in layout:regular in layout:
For For non-swaynon-sway frames frames
KKbb = 0.5 I/L = 0.5 I/L
For For swaysway frames frames
KKbb = 1.5 I/L = 1.5 I/L
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Beam stiffness values – buildings with floor slabs
Loading condition
Non-sway mode Sway mode
Beam directly supporting concrete floor or roof slab
1.0 ( I/L ) 1.0 ( I/L )
Other beams supporting direct loads
0.75 ( I/L ) 1.0 ( I/L )
Beams with end moments only
0.5 ( I/L ) 1.5 ( I/L )
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Beam stiffness values – 2
Rotational restraint at far end of beam
Beam stiffness coefficient Kb
Fixed at far end 1.0 (I/L ) { 1 – 0.4 (Pc/PE)} Pinned at far end 0.75 (I/L ) { 1 – 1.0 (Pc/PE)} Rotation as at near end (double curvature)
1.5 (I/L ) { 1 – 0.2 (Pc/PE)}
Rotation equal and opposite to that at near end (single curvature)
0.5 (I/L ) { 1 – 1.0 (Pc/PE)}
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Frames with partial sway bracing
Two other plots exist as E4 & E5Two other plots exist as E4 & E5 Relate to kRelate to kpp =1 and k =1 and kpp =2 =2
For frames which do not satisfy the requirements For frames which do not satisfy the requirements for a non-sway frame, can still take advantage of for a non-sway frame, can still take advantage of the stiffness of the bracing using kthe stiffness of the bracing using kpp..
kkpp is a measure of the stiffness of the partial sway is a measure of the stiffness of the partial sway
bracing to the stiffness of the bare frame.bracing to the stiffness of the bare frame.
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Frames with partial sway bracing
kkpp = h = h22SSpp/(80E/(80EKKcc) but ) but 22
ccis the sum of I/h for all the columns in that storeyis the sum of I/h for all the columns in that storey
SSpp is the sum of the stiffness of every panel in the storeyis the sum of the stiffness of every panel in the storey
SSpp is given by (0.6(h/b) ) t E is given by (0.6(h/b) ) t Epp/[1+(h/b)/[1+(h/b)22]]2 2
h/b is the ratio of storey height to panel width h/b is the ratio of storey height to panel width
t t is the panel thickness is the panel thickness
EEp p is the modulus of elasticity of the panel material. is the modulus of elasticity of the panel material.
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Notional horizontal loads
0.5% of (D+I)
0.5% of (D+I)
0.5% of (D+I)
0.5% of (D+I)
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Notional horizontal loads
Notional horizontal forces should Notional horizontal forces should NOTNOT::
a) be applied when considering overturninga) be applied when considering overturning
b) be applied when considering pattern loadingb) be applied when considering pattern loading
c) be combined with applied horizontal loadsc) be combined with applied horizontal loads
d) be combined with temperature effectsd) be combined with temperature effects
e) be taken to contribute to the net reactions ate) be taken to contribute to the net reactions at
the foundations.the foundations.
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Minimum horizontal forces
Factored dead load
1% of DL4
1% of DL3
1% of DL2
1% of DL1
Wind load or
Wind load or
Wind load or
Wind load or
Greater of DL4
DL3
DL2
DL1
DL 1-4 are the total dead load at each floor level
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Resistance to horizontal forces
Resistance to horizontal forces may be provided in a Resistance to horizontal forces may be provided in a number of ways as follows:number of ways as follows:
a) triangulated bracing members.a) triangulated bracing members.
b) moment resisting joints and frame action.b) moment resisting joints and frame action.
c) cantilever columns, shear walls, staircasec) cantilever columns, shear walls, staircase
and lift shaft enclosures.and lift shaft enclosures.
d) or a combination of these.d) or a combination of these.
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Classification of frames
Frames may be
1.1. Braced or unbracedBraced or unbraced –depends on how horizontal forces are transmitted to the ground.
2.2. Sway or non-swaySway or non-sway -depends on significance or otherwise of P- effects.
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Independently braced frames
a) a) The stabilizing system must have a spring The stabilizing system must have a spring stiffness at least four times larger than the total stiffness at least four times larger than the total spring stiffness of all the frames to which it gives spring stiffness of all the frames to which it gives horizontal support (i.e. the supporting system horizontal support (i.e. the supporting system reduces horizontal displacements by at least 80%).reduces horizontal displacements by at least 80%).
andand
b) Tb) The stabilizing system must be designed to he stabilizing system must be designed to resist all the horizontal loads applied including the resist all the horizontal loads applied including the notional horizontal forces.notional horizontal forces.
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Braced multi-storey frame
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Sway / non-sway categorisation
A frame can be deemed to be non-sway if,A frame can be deemed to be non-sway if,in the in the SWAYSWAY mode mode
crcr 10 10
cr cr = = 1 / 200 1 / 200 maxmax = h / 200 = h / 200 maxmax
Otherwise it is a sway frame.Otherwise it is a sway frame.
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Annex F - Critical load of sway frame cr = 1 / 200 max & = {n n –1}/h
0.5%(D+I)
0.5%(D+I)
0.5%(D+I)
0.5%(D+I)
1
2
3
4
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Frame design
Braced or UnbracedBraced or Unbraced
Sway or Non-swaySway or Non-sway
Elastic design or Plastic designElastic design or Plastic design
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Design of independently braced frames
Independently braced frames should be designed:Independently braced frames should be designed:
to resist gravity loads (load combination 1).to resist gravity loads (load combination 1).
the non-sway mode effective length of the columns should be the non-sway mode effective length of the columns should be obtained using Annex E. obtained using Annex E.
pattern loading should be used to determine the most severe pattern loading should be used to determine the most severe moments and forces. moments and forces.
Sub-frames may be used to reduce the number of load cases to Sub-frames may be used to reduce the number of load cases to be considered. be considered.
the the stabilizing systemstabilizing system must be designed to resist all the must be designed to resist all the horizontal loads applied including the notional horizontal forces.horizontal loads applied including the notional horizontal forces.
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Design of non-sway frames
Non-sway should be designed: Non-sway should be designed: to resist gravity loads (load combination 1). to resist gravity loads (load combination 1).
the non-sway mode effective length of the the non-sway mode effective length of the columns should be obtained using Annex E. columns should be obtained using Annex E.
pattern loading should be used to determine the pattern loading should be used to determine the most severe moments and forces. most severe moments and forces.
sub frames may be used to reduce the number of sub frames may be used to reduce the number of load cases.load cases.
the the frameframe should be checked for combined vertical should be checked for combined vertical and horizontal loads without pattern loading.and horizontal loads without pattern loading.
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Pattern Loadings : Maximum moments
Maximum beam span moments Maximum beam support moments
Maximum single curvature bending Maximum double curvature bendingin columns in columns
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Plastic design of non-sway frames
If beams are designed plastically,If beams are designed plastically,
they will contain moments in excess of 90% they will contain moments in excess of 90% of Mof Mpp..
Care must be taken when using Annex ECare must be taken when using Annex E
remembering that Kremembering that Kbb is likely to be ZERO is likely to be ZERO
in most cases,in most cases,
giving Lgiving LEE/L factors close to 1.0/L factors close to 1.0
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Plastic design of non-sway frames
Normal plastic rules apply for Normal plastic rules apply for local capacitylocal capacity and and member stabilitymember stability – as for portal frames. Usually – as for portal frames. Usually the latter will be less onerous due to the presence the latter will be less onerous due to the presence of floor slabs to provide stability.of floor slabs to provide stability.
Remember that the Remember that the frameframe must be stable must be stable perpendicular to the planes of the frames being perpendicular to the planes of the frames being designed.designed.
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Elastic design of sway sensitive frames
Sway sensitive frames should be designed as Sway sensitive frames should be designed as follows: follows:
Check in the Check in the non-swaynon-sway modemode i.e. design to resist i.e. design to resist gravity loads (load combination 1) as for independently gravity loads (load combination 1) as for independently braced frames without taking account of sway. (without braced frames without taking account of sway. (without notional horizontal forces, but with pattern loading). notional horizontal forces, but with pattern loading).
• • Check in the Check in the sway modesway mode for gravity load (i.e. load for gravity load (i.e. load combination 1) plus the notional horizontal forces without combination 1) plus the notional horizontal forces without any pattern loading. any pattern loading.
• • Check in the Check in the sway modesway mode for combined vertical and for combined vertical and horizontal loads (i.e. load combinations 2 and 3), without horizontal loads (i.e. load combinations 2 and 3), without pattern loading.pattern loading.
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Elastic design of sway framesProvided that Provided that crcr is greater than 4, the sway should be is greater than 4, the sway should be
allowed for by using one of the following methods:allowed for by using one of the following methods:
a) a) Effective length methodEffective length method. .
Use effective lengths from the Use effective lengths from the sway chartsway chart of Annex E. of Annex E.
b) b) Amplified sway methodAmplified sway method. .
The The sway momentssway moments should be multiplied by the should be multiplied by the amplification factor kamplification factor k
ampamp. . LLEE/L is /L is non-swaynon-sway value. value.
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kamp for sway sensitive frames
1) for unclad frames – or for clad structures in which the stiffening effect
of masonry infill wall panels or diaphragms of profiled steel sheeting is explicitly taken into
account in determining crcr :
kkampamp = = crcr / ( / (crcr – 1) – 1)
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kamp for sway sensitive frames
2) 2) for for clad structuresclad structures, ,
provided that the stiffening effect of provided that the stiffening effect of masonry infill wallmasonry infill wall panels or diaphragms of panels or diaphragms of profiled steel sheeting is not explicitly taken profiled steel sheeting is not explicitly taken into account:into account:
kkampamp = = crcr / (1.15 / (1.15 crcr – 1.5) but – 1.5) but 1.0 1.0
Gives smaller amplification factors than previous slide.Gives smaller amplification factors than previous slide.
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Sway effects The distortions of a frame may be divided into two The distortions of a frame may be divided into two
components:components: 1) those which arise from sway with zero 1) those which arise from sway with zero
rotation of every joint and rotation of every joint and 2) those which arise a result of joint rotations 2) those which arise a result of joint rotations
without any sway of the frame.without any sway of the frame. In the case of a symmetrical frame, with In the case of a symmetrical frame, with
symmetrical vertical loads, the sway effects can symmetrical vertical loads, the sway effects can correctly be taken as comprising the forces and correctly be taken as comprising the forces and moments in the frame due to the horizontal loads.moments in the frame due to the horizontal loads.
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Sway effects
In every other caseIn every other case need to separate need to separate moments into sway and non-sway moments into sway and non-sway components by either:components by either:
a) Deducting the non-sway effectsa) Deducting the non-sway effects
b) Direct calculationb) Direct calculation
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a) Deducting the non-sway effects:
1) Analyse the frame under the actual restraint conditions.1) Analyse the frame under the actual restraint conditions.2) Add horizontal restraints at each floor or roof level to 2) Add horizontal restraints at each floor or roof level to
prevent sway, then analyse the frame again.prevent sway, then analyse the frame again.3) Obtain the sway effects by deducting the second set of 3) Obtain the sway effects by deducting the second set of
forces and moments from the first set.forces and moments from the first set.
In 1) moments are due to sway + non-sway distortions.In 1) moments are due to sway + non-sway distortions.In 2) moments are due to non-sway distortions.In 2) moments are due to non-sway distortions.In 3) moments are thus only due to sway distortions.In 3) moments are thus only due to sway distortions.
The forces and moments from step 3 are the sway effects The forces and moments from step 3 are the sway effects which require magnifying by kwhich require magnifying by kampamp..
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b) Direct calculation:
1) Analyse the frame with horizontal restraints added 1) Analyse the frame with horizontal restraints added at each floor or roof level to prevent sway.at each floor or roof level to prevent sway.
2) Reverse the directions of the horizontal reactions 2) Reverse the directions of the horizontal reactions produced, at the added horizontal restraint produced, at the added horizontal restraint locations.locations.
3) Apply them as loads to the otherwise unloaded 3) Apply them as loads to the otherwise unloaded frame under the actual restraint conditions.frame under the actual restraint conditions.
4) Adopt the forces and moments from the second 4) Adopt the forces and moments from the second analysis (step 3) as the sway effects which require analysis (step 3) as the sway effects which require magnifying by kmagnifying by k
ampamp..
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Elastic design of sway sensitive frames
If cr 4.0 then
The frame is highly sensitive to instability effects.
The kamp approach may not be suitable and
a full second order elastic analysis should be used.
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Plastic design of sway frames
Frames may be designed using a second Frames may be designed using a second order elastic-plastic analysis or by using the order elastic-plastic analysis or by using the “ sway stability check”.“ sway stability check”.
This is derived from the Merchant-Rankine-This is derived from the Merchant-Rankine-Wood equation.Wood equation.
Origins: The Rankine strut equation isOrigins: The Rankine strut equation is
1/P1/PFF = 1/P = 1/PCRCR + 1/ P + 1/ PSQSQ
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M-R-Wood equation
Rankine’s formula proposed for struts prior to Rankine’s formula proposed for struts prior to development of Ayrton-Perry equation.development of Ayrton-Perry equation.
Merchant suggested its use for frames with:Merchant suggested its use for frames with: Elastic critical load replacing Euler load.Elastic critical load replacing Euler load. Plastic collapse load replacing squash load.Plastic collapse load replacing squash load.
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Plastic design of sway framesMerchant-Rankine EquationMerchant-Rankine Equation
1 = 1 + 1
PF Psq
Pcr
Psq/Pcr
PF/Psq
Merchant-Rankine
1.0
1.0
0.5
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Development of M-R-Wood Eqn1/P1/PFF = 1/P = 1/PCRCR + 1/ P + 1/ PSQSQ
1/1/FF = 1/ = 1/ CRCR + 1/ + 1/ PP
={ ={ P P + + CRCR }/ }/CRCR P P
F F = = CRCR P P / { / { P P + + CRCR } }
But But FF = 1 = 1
P P + + CR CR = = CRCR P P
CRCR = = CRCR P P - - P P = = P P {{CRCR –1} –1}
P P = = CRCR /{ /{CRCR –1} –1}
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Development of M-R-Wood Eqn
But for CR / P >10
stray composite action &
strain hardening means
no need to reduce P .
For 10<CR/ P <4.0
change 1 to 0.9 for
continuity at CR/ P =10
r = 0.9CR /{CR –1}
P P
1.0
P /CR 0.1 0.25
Merchant –Rankine-Wood
Merchant-Rankine
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Combined plastic collapse mechanism
Determine plastic collapse load using sway mode shown
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Plastic collapse load
1.1. The bases of the columns should be fixed.The bases of the columns should be fixed.
2.2. Mechanism as on previous slide.Mechanism as on previous slide.
3.3. Ensure that columns remain elastic at assumed Ensure that columns remain elastic at assumed plastic moment.plastic moment.
4.4. Check that no localised beam or storey Check that no localised beam or storey mechanism is more dangerous.mechanism is more dangerous.
5.5. Storey height less than mean column spacing.Storey height less than mean column spacing.
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Development of M-R-Wood EqnFor actual unclad frames or clad frameswhere cladding stiffness is utilised
For CR / P >20
strain hardening means
no need to reduce P .
For 20<CR/ P <5.75
change 1 to 0.95 for
continuity at CR/ P =20
r = 0.95CR /{CR –1}
P
1.0
P /CR 0.1 0.25
0.05
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Summary of design procedures
1. Select member sizes for initial analysis 1. Select member sizes for initial analysis bases on experience.bases on experience.
2. Choose elastic or plastic design (usually 2. Choose elastic or plastic design (usually elastic)elastic)
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Elastic Design
Classify frames as independently braced (5.1.4), sway or non sway (2.4.2.6)
Independently braced frames (5.6.2):
Design to resist gravity loads (load combination 1)
The non-sway mode effective length of the columns should be found from Annex E.
Pattern loading should be used to determine the most severe moments and forces.
Sub-frames may be used to reduce the number of load cases
The stabilizing system must be designed to resist all the horizontal loads applied including the notional horizontal forces.
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Elastic Non-sway frames (5.6.3):
Design to resist gravity loads (load combination 1) Design to resist gravity loads (load combination 1)
The non-sway mode effective length of the columns should The non-sway mode effective length of the columns should be found from Annex E. be found from Annex E.
Pattern loading should be used to determine the most Pattern loading should be used to determine the most severe moments and forces. severe moments and forces.
Sub-frames may be used to reduce the number of load Sub-frames may be used to reduce the number of load cases.cases.
The The frameframe should be checked for combined vertical and should be checked for combined vertical and horizontal loads without pattern loading.horizontal loads without pattern loading.
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Elastic sway sensitive frames Check in the non-sway mode i.e. design to resist gravity
loads (load combination1) as for independently braced frames without taking account of sway. (without notional horizontal forces, but with pattern loading).
Check in the sway mode for gravity load (i.e. load combination 1) plus the notional horizontal forces without any pattern loading.
Check in the sway mode for combined vertical and horizontal loads (i.e. load combinations 2 and 3) without pattern loading.
Allow for sway using the effective length method Annex E or the amplified sway method.
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Plastic design
Classify frames as independently braced Classify frames as independently braced (5.7.2), sway or non sway (5.7.3)(5.7.2), sway or non sway (5.7.3)
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Plastic braced frames
Independently braced frames (5.7.2)
Design to resist gravity loads (load combination 1). Design to resist gravity loads (load combination 1).
The effective length The effective length LLE E of the columnsof the columns in the plane in the plane
of the frame should generally be taken as equal to of the frame should generally be taken as equal to the storey height the storey height L.L.
Check columns under pattern loading using an Check columns under pattern loading using an effective length from Annex E. effective length from Annex E.
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Plastic design- unbraced frames
Check for possible non-sway modes of failure Check for possible non-sway modes of failure as recommended for independently braced frames.as recommended for independently braced frames.
Satisfy the simplified M-R-Wood frame Satisfy the simplified M-R-Wood frame stability check (given in clause 5.7.3.2), stability check (given in clause 5.7.3.2),
Or design to resist sway mode failure using a Or design to resist sway mode failure using a
second order elastic-plastic analysis.second order elastic-plastic analysis.
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Example : Use of Annex E : Effective lengths in a continuous multi-storey frame
3.6m
3.6m
3.6m
3.6m
7.2m 7.2m 7.2m 7.2m
1
2
3
Ix beams = 21500cm
Ix columns = 6090cm
4
4
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Check if the frame is a sway frame
The notional horizontal force
= 0.5% factored dead plus imposed load =
at roof level F1 = 0.5 (16 x 28.8)/100 = 2.3 kN
at each floor F2 = 0.5 (72 x 28.8)/100 = 10.4 kN
16kN/m run 72kN/m run
Factored dead plus live load
F1
F2
F2
F2
deflections Absolute
9mm
8mm
6mm
4mm
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Sway or non-sway frame
The deflection in the lower storey exceeds h/2000 = 3600/2000 = 1.8mm and the frame is a sway frame.
Only the upper storey does not violate the non-sway limit.
(*these deflection values have been guessed and are not the result of an analysis!)
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Using Appendix E
Column 1 Column 1
BeamsBeams KKTLTL = K = KTRTR = K = KBLBL = K = KBRBR = I/L =21500/720 = 29.9 = I/L =21500/720 = 29.9
Columns Columns KKUU = K = KCC = K = KL L = I/L = 6090/360 = 16.9= I/L = 6090/360 = 16.9
Restraint factorsRestraint factors
TopTop kk11 = (K = (KCC+K+KUU) / (K) / (KCC + K + KUU +K +KTLTL+K+KTRTR) = 0.36) = 0.36
BottomBottom kk22 = (K = (KCC+K+KLL) / (K) / (KCC + K + KLL +K +KTLTL+K+KTRTR) = 0.36) = 0.36
The frame is a sway frame; use Figure E.2The frame is a sway frame; use Figure E.2
LLEE/L = 1.27 i.e. L/L = 1.27 i.e. LEE = 1.27 x 3.6 = 4.57m = 1.27 x 3.6 = 4.57m
IFIF bracing were provided and the frame became a non-sway frame, the bracing were provided and the frame became a non-sway frame, the effective length ratio from Figure E.1 of the code would be 0.625 effective length ratio from Figure E.1 of the code would be 0.625
i.e. Li.e. LEE = 0.625 x 3.6 = 2.25m = 0.625 x 3.6 = 2.25m
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Using Appendix E
Column 2 Column 2
Beams Beams KKTLTL = K = KBLBL = I/L =21500/720 = 29.9 = I/L =21500/720 = 29.9
Columns Columns KKCC = K = KL L = I/L = 6090/360 = 16.9= I/L = 6090/360 = 16.9
Restraint factorsRestraint factors
TopTop kk11 = (K = (KCC) / (K) / (KCC +K +KTLTL) = 0.36) = 0.36
BottomBottom kk22 = (K = (KCC+K+KLL) / (K) / (KCC + K + KLL +K +KTLTL) = 0.53) = 0.53
Therefore as the frame is a sway frame from Figure E.2Therefore as the frame is a sway frame from Figure E.2
LLEE/L = 1.4 i.e. L/L = 1.4 i.e. LEE = 1.4 x 3.6 = 5.04m = 1.4 x 3.6 = 5.04m
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Using Appendix EColumn 3Column 3
Beams Beams KKTLTL = K = KTRTR = I/L =21500/720 = 29.9 = I/L =21500/720 = 29.9
Columns Columns KKUU = K = KCC = I/L = 6090/360 = 16.9 = I/L = 6090/360 = 16.9
Restraint factorsRestraint factors
TopTop kk11 = (K = (KCC+K+KUU) / (K) / (KCC + K + KUU +K +KTLTL+K+KTRTR) = 0.36) = 0.36
BottomBottom kk22 = (K = (KCC) / (K) / (KCC + 0.1 x K + 0.1 x KCC ) = 0.91 ) = 0.91
Therefore as the frame is a sway frame from Figure E.2Therefore as the frame is a sway frame from Figure E.2
LLEE/L = 2.0 i.e. L/L = 2.0 i.e. LEE = 2.0 x 3.6 = 7.20m = 2.0 x 3.6 = 7.20m
The design would then proceed as normal using the effective The design would then proceed as normal using the effective lengths calculated above.lengths calculated above.
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
If the column bases were fixed
Beams Beams KKTLTL = K = KTRTR = I/L =21500/720 = 29.9 = I/L =21500/720 = 29.9
Columns Columns KKUU = K = KCC = I/L = 6090/360 = 16.9 = I/L = 6090/360 = 16.9
Restraint factorsRestraint factors
TopTop kk11 = (K = (KCC+K+KUU) / (K) / (KCC + K + KUU +K +KTLTL+K+KTRTR) = 0.36) = 0.36
BottomBottom kk22 = (K = (KCC) / (K) / (KCC + 1.0 x K + 1.0 x KCC ) = 0.50 ) = 0.50
From Figure E.2 LFrom Figure E.2 LEE/L = 1.35 i.e. L/L = 1.35 i.e. LEE = 1.35 x 3.6 = 4.86m = 1.35 x 3.6 = 4.86m
The effective length is much reduced and the column will be smaller The effective length is much reduced and the column will be smaller but the cost of providing moment resisting foundations may outweigh but the cost of providing moment resisting foundations may outweigh the cost of the savings in steelwork. The fixity would also be the cost of the savings in steelwork. The fixity would also be beneficial in controlling sway deformations.beneficial in controlling sway deformations.
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Beams carrying >90% of their moment capacity
Column 1 Column 1
Beams Beams KKTLTL = K = KTRTR = K = K
BLBL = K = KBRBR = 0 = 0
Columns Columns KKUU = K = KCC = K = K
L L = I/L = 6090/360 = 16.9= I/L = 6090/360 = 16.9
Restraint factorsRestraint factors
TopTop kk11 = (K = (KCC+K+K
UU) / (K) / (KCC + K + K
UU + 0+ 0) = 1 + 0+ 0) = 1
BottomBottom kk22 = (K = (KCC+K+K
LL) / (K) / (KCC + K + K
LL + 0+ 0) = 1 + 0+ 0) = 1
From Figure E.2, the effective length ratio would be From Figure E.2, the effective length ratio would be equal to equal to infinityinfinity..
Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute
Using the amplified sway method.AlternativelyAlternatively consider the design of the original frame consider the design of the original frame using the amplified sway method.using the amplified sway method.
The maximum value of the sway index The maximum value of the sway index is in the lower is in the lower storey and is given by storey and is given by
= (= (uu--ll)/h = 4/3600 = 1.1 x 10)/h = 4/3600 = 1.1 x 10-3-3
The elastic critical load factor The elastic critical load factor crcr is then given by : is then given by :
crcr = 1/(200 = 1/(200) = 4.55 ) = 4.55
(Note this suggests a very flexible frame)(Note this suggests a very flexible frame)
Amplification factor = Amplification factor = crcr /( /(crcr-1) = 4.55/(4.55-1) = 1.28-1) = 4.55/(4.55-1) = 1.28
The design can proceed using effective length ratios of The design can proceed using effective length ratios of oneone and and allall the moments due to the moments due to horizontalhorizontal deformations deformations multiplied by 1.28.multiplied by 1.28.