Date post: | 13-Dec-2015 |

Category: |
## Documents |

Upload: | adnan-jadoon |

View: | 230 times |

Download: | 0 times |

Share this document with a friend

Description:

Continuous Construct

Embed Size (px)

Popular Tags:

of 67
/67

BS 5950: 2000 : Part 1 Continuous construction Multi-storey frames

Transcript

BS 5950: 2000 : Part 1

Continuous construction

Multi-storey frames

Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute

Simple Design – pin joints

(a) Web Cleats (b) End Plate (c) Fin Plates

Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute

Continuous Design – rigid joints

Extended end plate connection

Continuous multistorey frame

Department of Civil and Structural Engineering, University of SheffieldIn association with the Steel Construction Institute

Base stiffness and strength

As in the lecture on portal framesAs in the lecture on portal frames

Deformation of a multi-storey frame

P

P

P - effects

P

Stability effectsP

F

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

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

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

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)

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.

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.

Buckled mode shapes

Non Sway Frame Sway Frame

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

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 )

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)}

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.

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.

Notional horizontal loads

0.5% of (D+I)

0.5% of (D+I)

0.5% of (D+I)

0.5% of (D+I)

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.

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

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.

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.

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.

Braced multi-storey frame

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.

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

Frame design

Braced or UnbracedBraced or Unbraced

Sway or Non-swaySway or Non-sway

Elastic design or Plastic designElastic design or Plastic design

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.

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.

Pattern Loadings : Maximum moments

Maximum beam span moments Maximum beam support moments

Maximum single curvature bending Maximum double curvature bendingin columns in columns

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

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.

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.

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.

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)

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.

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.

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

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..

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..

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.

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

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.

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

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}

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

Combined plastic collapse mechanism

Determine plastic collapse load using sway mode shown

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.

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

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)

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.

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.

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.

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)

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.

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.

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

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

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!)

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

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

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.

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.

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..

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.

Recommended