12007-8-22 1

Continuous Construction Multi-storey frames

2007-8-22 2

Simple Frame Design pin joints

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

22007-8-22 3

Continuous Frame Design rigid joints

Rigid FrameExtended end plate connection

2007-8-22 4

Frame design

Braced or Unbraced

Sway or Non-sway

Elastic design or Plastic design

32007-8-22 5

2007-8-22 6

42007-8-22 7

Design of Independently Braced frame

Frame B is braced by frame A if lateral stiffness is KA >4KB

Stabilizing System to resist all horizontal load

Independently Braced frame designed to resist gravity load only

B A

2007-8-22 8

Design of independently braced frames (Cl: 5.6.2)

Independently braced frames should be designed: to resist gravity loads (1.4DL +1.6IL). the non-sway mode effective length of the columns

should be obtained using 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 to be considered. the stabilizing system must be designed to resist all

the horizontal loads applied including the notional horizontal forces. It can be sway or nonsway frame.

52007-8-22 9

Pattern Loadings : Maximum moments

Maximum beam span moments Maximum beam support moments

Maximum single curvature bending Maximum double curvature bendingin columns in columns

2007-8-22 10

Design of non-sway frames (5.6.3)

Non-sway frames should be designed: to resist gravity loads (load combination 1). the non-sway mode effective length of the

columns should be obtained using 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 frame should then be checked for

combined vertical and horizontal loads without pattern loading.

62007-8-22 11

Design of Sway Sensitive FramesSway sensitive frames should be designed as

follows: Check in the non-sway mode i.e. design to resist

gravity loads (load combination 1) as for independently braced frames without taking account of sway ( i.e.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.

*The sway effect should be allowed for, either by using column effective lengths by using amplified sway moments.

2007-8-22 12

Member Buckling Check for Columns in Sway Frame

FP

m MZ

m MZc

x x

x x

y y

y y+ + 10.

Effective length for sway frame

kampM, amplified moment or

1ZpMm

MMm

PF

yy

yy

b

xLT

cy

c ++

In-plane Buckling

Lateral Torsional Buckling

72007-8-22 13

Design of Sway Sensitive FramesProvided that cr is greater than 4, the sway should be allowed for by using one of the following methods (Cl. 5.6.4):a) Effective length method.

Use effective lengths from the sway chart of Annex E.

b) Amplified sway method. The sway moments should be multiplied by the amplification factor kamp. LE/L is non-sway value.

cr < 4, determine member forces by direct second-order analysis

2007-8-22 14

Effective lengths and critical loadsCritical load of a Pin-ended ColumnPcr = Pe = 2EI/L2Critical load of a column with other boundary

conditionPcr = 2EI/(KL)2 = Pe/ K2KL = Effective LengthPe = Euler buckling loadEffective Length factor, K = (Pe/Pcr)0.5

e.g., Cantilever ColumnPcr= 0.252EI/L2 = 2EI/4L2 = 2EI/(2L)2

Effective Length factor = (Pe/Pcr)0.5Hence the effective length LE for a cantilever is 2L

82007-8-22 15

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

2007-8-22 16

92007-8-22 17

2007-8-22 18

Charts of Annex E: Effective length of columns in Nonsway frames

KTL KTR

Ku

KL

Kc

KBRKBL

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Pinned

k 1

Fixed

k 2 PinnedFixed

10

2007-8-22 19

Charts of Annex E: Effective length of columns in Sway frames

k2

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Pinned

k 1

Fixed

k 2 PinnedFixed

2007-8-22 20

Buckled mode shapes

Non Sway Frame Sway Frame

11

2007-8-22 21

K

K

KTL

Ku

KTR

KBL BR

L

K1

K2

Column being consideredK

c= I/L

intJoatMembersAllofStiffnessTotalintJoatStiffnessColumnTotalk =

k k kk k k k

c u

c u TL TR1 = ++ + +

k k kk k k k

c L

c L BL BR2 = ++ + +

Effective Length of Columns in Multistorey Frame

2007-8-22 22

Derivation of Chartskl = KC / (KC + KBL + KBR)

k2 = KC+ KL / (KC +KL + KBL + KBR)

Conservative Formulae for the curvesFor non sway

( ) 22121 )(55.014.05.0/ kkkkLLE ++++

LE

For sway frames

++

2121

2121

6.0)(8.0112.0)(2.01

/kkkkkkkkLLE

1

2

12

2007-8-22 23

Use of the charts of Annex E k1 = (Kc + Ku) / (Kc+Ku+KTL +KTR) k2 = (Kc + KL) / (Kc+KL+KBL +KBR) The stiffness K for each member is

taken as a function of I / L If a beam supports a floor slab, its K

value should be taken as I / L For a beam which is not rigidly

connected to the column K should be taken as zero.

2007-8-22 24

Use of the charts of Annex E For a beam which carries more

than 90% of its moment capacity, a pin should be inserted at that location .

If either end of the column carries more than 90% of Mp, the value of k1 or k2 as appropriate should be taken as 1.0.

For other conditions, the appropriate values of K are given in Tables E1, E2 and E3 of the code.

90%Mp

Ib/|Lb = 0

k2 = 1

13

2007-8-22 25

Beam stiffness values

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 )

Table E.1 BS5950:Part1

Non Sway Frame Sway Frame

Beam in double curvature

Beam in single curvature

2007-8-22 26

Effect of Axial Force on Beam stiffness Kb -Non sway frames

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)}

Far end of the beam

Table E.2 BS5950:Part1

PE = 2EI/L2

PcPc

14

2007-8-22 27

Columns in a Mixed Frame (E.5)

srV suV

Effective length to be increase by f

Storey buckling amplification factor

= total vertical load in that storey in the columns that resist sway in that plane

= total vertical load in that storey in the columns not resisting sway in that plane

Rigid frame Simple frame

2007-8-22 28

Frames with partial sway bracing Two other plots exist as E4 & E5 Relate to kp =1 and kp =2

For frames which do not satisfy the requirements for a non-sway frame, can still take advantage of the stiffness of the bracing using kp.

kp is a measure of the stiffness of the partial sway bracing to the stiffness of the bare frame.

15

2007-8-22 29

Frames with partial sway bracing(Annex E.3)

kp = h2Sp/(80EKc) but 2

c = sum of I/h for all the columns in that storey Sp = sum of the stiffness of every panel in the storey

Sp = (0.6(h/b) ) t Ep/[1+(h/b)2]2

h/b = ratio of storey height to panel width t = panel thickness Ep = modulus of elasticity of the panel material.

To account for stiffening effect of infill wall panels in determining the effective length ratio for frame with partial wall bracing of relative stiffness kp

Spring stiffness Sp

h

2007-8-22 30

Kp = 1

Kp =2

16

2007-8-22 31

Amplified Sway Method

Sway moment should be multiplied by the amplification factor kamp

Non-sway mode in-plane effective length should be used.

2007-8-22 32

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 cr :

kamp = cr / (cr 1)

17

2007-8-22 33

kamp for sway sensitive frames2) for clad structures,

provided that the stiffening effect of masonry infill wall panels or diaphragms of profiled steel sheeting is not explicitly taken into account:kamp = cr / (1.15 cr 1.5) but 1.0

Gives smaller amplification factors than previous slide.

2007-8-22 34

Sway effects

In the case of a symmetrical frame, with symmetrical vertical loads, the sway effects can correctly be taken as comprising the forces and moments in the frame due to the horizontal loads.

18

2007-8-22 35

Sway Moments

In every other case it is necessary to separate moments into sway and non-sway components by either:

a) Deducting the non-sway effects.

b) Direct calculation.

2007-8-22 36

a) Deducting the non-sway effects:

1) Analyse the frame under the actual restraint conditions.

2) Add horizontal restraints at each floor or roof level to prevent sway, then analyse the frame again.

3) Obtain the sway effects by deducting the second set of forces and moments from the first set.

In 1) moments are due to sway + non-sway distortions.In 2) moments are due to non-sway distortions.In 3) moments are thus only due to sway distortions.

The forces and moments from step 3 are the sway effects which require magnifying by kamp.

Sway Moments

19

2007-8-22 37

Deducting the non-sway effects

Step 1Step 2

Sway moment = Moment in Step 1 Moment in Step 2

Design moment = nonsway moment + kamp.x Sway moment(step 2)

2007-8-22 38

b) Direct calculation:

1) Analyse the frame with horizontal restraints added at each floor or roof level to prevent sway.

2) Reverse the directions of the horizontal reactions produced, at the added horizontal restraint locations.

3) Apply them as loads to the otherwise unloaded frame under the actual restraint conditions.

4) Adopt the forces and moments from the second analysis (step 3) as the sway effects which require magnifying by kamp.

Sway Moments

20

2007-8-22 39

Direct calculation

Step 2: Sway MomentStep 1 Nonsway moment

R3

R2

R1

R3

R2

R1

Design moment = nonsway moment + kamp.x Sway momentStep 1 Step 2

2007-8-22 40

Elastic design of sway sensitive frames

If cr < 4.0 then The frame is sensitive to instability effects.

The kamp approach is not suitable and a full second order elastic analysis should be used.

21

2007-8-22 41

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

16kN/m72kN/m

Factored dead plus live load

Beams are supporting concrete slab, Kb = 1.0 I/L

2007-8-22 42

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 kNat each floor F2 = 0.5 (72 x 28.8)/100 = 10.4 kN

16kN/m72kN/m

Factored dead plus live load

Inter-storeyF1

F2

F2

F2

deflections9mm

8mm

6mm

4mm

22

2007-8-22 43

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!)

2007-8-22 44

Using Appendix E

Column 1 Beams KTL = KTR = KBL = KBR = I/L =21500/720 = 29.9Columns KU = KC = KL = I/L = 6090/360 = 16.9Restraint factorsTop k1 = (KC+KU) / (KC + KU +KTL+KTR) = 0.36Bottom k2 = (KC+KL) / (KC + KL +KTL+KTR) = 0.36

The frame is a sway frame; use Figure E.2LE/L = 1.27 i.e. LE = 1.27 x 3.6 = 4.57m

If 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 i.e. LE = 0.625 x 3.6 = 2.25m

3.6m

3.6m

3.6m

3.6m

7.2m 7.2m 7.2m 7.2m

1

2

3

23

2007-8-22 45

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Pinned

k 1

Fixed

k 2 PinnedFixed

2007-8-22 46

Column 2Beams KTL = KBL = I/L =21500/720 = 29.9Columns KC = KL = I/L = 6090/360 = 16.9

End Restraint factorsTop k1 = (KC) / (KC +KTL) = 0.36Bottom k2 = (KC+KL) / (KC + KL +KTL) = 0.53

Therefore as the frame is a sway frame From Figure E.2

LE/L = 1.4 i.e. LE = 1.4 x 3.6 = 5.04m

3.6m

3.6m

3.6m

3.6m

7.2m 7.2m 7.2m 7.2m

1

2

3

24

2007-8-22 47

Column 3Beams KTL = KTR = I/L =21500/720 = 29.9

Columns KU = KC = I/L = 6090/360 = 16.9Restraint factorsTop k1 = (KC+KU) / (KC + KU +KTL+KTR) = 0.36Bottom k2 = (KC) / (KC + 0.1 x KC ) = 0.91

Therefore as the frame is a sway frame from Figure E.2LE/L = 2.0 i.e. LE = 2.0 x 3.6 = 7.20m

The design would then proceed as normal using the effective lengths calculated above.

3.6m

3.6m

3.6m

3.6m

7.2m 7.2m 7.2m 7.2m

1

2

3

Column based is pinned

2007-8-22 48

If the column bases were rigid base stiffness is taken as column stiffness : cl. 5.1.3.2

Column 3Beams KTL = KTR = I/L =21500/720 = 29.9Columns KU = KC = I/L = 6090/360 = 16.9

End restraint factorsTop k1 = (KC+KU) / (KC + KU +KTL+KTR) = 0.36Bottom k2 = (KC) / (KC + KC ) = 0.50

From Figure E.2 LE/L = 1.35 i.e. LE = 1.35 x 3.6 = 4.86m

The effective length is much reduced and the column will be smaller but the cost of providing moment resisting foundations may out-weight the cost of the savings in steelwork. The fixity would also be beneficial in controlling sway deformations.

1

2

3

25

2007-8-22 49

If all floor beams carrying load exceeding 90% of their moment capacity

Column 1 Beams KTL = KTR = KBL = KBR = 0Columns KU = KC = KL = I/L = 6090/360 = 16.9Restraint factors

Top k1 = (KC+KU) / (KC + KU + 0+ 0) = 1Bottom k2 = (KC+KL) / (KC + KL + 0+ 0) = 1

From Figure E.2, the effective length ratio would be equal to infinity.

1

2

3

2007-8-22 50

Using the amplified sway method.Alternatively consider the design of the original frame using the amplified sway method.The maximum value of the sway index is in the lower storey and is given by

= (u-l)/h = 4/3600 = 1.1 x 10-3The elastic critical load factor cr is then given by :

cr = 1/(200) = 4.55 > 4.0 (Note this suggests a very flexible frame)Amplification factor = cr /(cr-1) = 4.55/(4.55-1) = 1.28The design can proceed using effective length ratios of nonsway frame and all the moments due to horizontaldeformations multiplied by 1.28.

26

2007-8-22 51

EXAMPLE 2

Notional Horizontal Load = 0.5% floor gravity load applied in one direction at a time.The frame is a sway frame in both direction.Column designed will be governed by buckling about the minor axis.Hence, design the member for buckling about the minor axis

NHLx

x

y

100 kN

150 kN

100 kN

100 kN100 kN

NHLy

150 kN150 kN

150 kN

NHLy NHLx

All members UC 152 x 152 x 37 S 275 steel

2007-8-22 52

Example 2 : Continue

Alternate designProvide bracing to prevent side sway in the x direction as shown.

Design the column as nonsway in the x direction, but consider sway effect in the y direction.

NHLyNHLy

x

y

100 kN

100 kN

NHLx150 kN

100 kN

100 kN

100 kN100 kN

27

2007-8-22 53

100kN 100kN 1kN

1kN

1.5kN

1.5kN

Lateral Deflection at the storey level (cm)

1.503

1.305

0.942

0.412

100 100

150 150

150 150

A

B 3.5m

3.5m

3.5m

3.5m

3.5m

Frame Classification for sway or nonsway Limiting value for uncladded frame h/2000 = 350/2000 = 0.175cmsince interstorey sway deflections due to notional load at all storey level are greater than the limiting value of 0.175cm, the frame is classified as sway frame.

Example 2 : Continue. Consider sway effect in the y direction.

Y direction

NHL

2007-8-22 54

Compute Elastic Buckling Load FactorUse Annex E.6 and F.2Notional horizontal load in each storey = 0.5%(factored gravity loads).Conduct a linear elastic analysis to give sway deflection . F.2Compute sway indices for each storey s u Lstorey height=

(Storey height = 350cm)

0.0005640.0010430.0015120.001181

Sway index, sStorey No.

= 0.00151

0.4311.300151.0200

1200

1

maxscr

28

2007-8-22 55

Column Effective Length

Axial load in column AB, F = 504.7 kN

Euler buckling load of column = 2EI/L2

=

Effective length factor for column AB

48.17.504x311.3

3667LEI

LL

2cr

2E ==

=

A

B

504.7 kN

504.7 kN

2.7 kNm

5.973 kNm

Forces in column AB(from structural analysis

Alternatively, we may use the limited frame method,Beam stiffness kb = 1.5I/L (Sway mode, Table E.1 in Annex E)kbottom = 0.5 (Nominal rigid base, Cl 5.1.3.2)

5.05.1011

11ktop =++++= 53.1

LLE =

Let us proceed with the calculation!

2007-8-22 56

For UC 152 x 152 x 37 S 275 steel, use design table Page 265Local Capacity Check (Simplified formula)

Overall Buckling Check (simplified formula)

(In-Plane Buckling )

(Lateral Torsional Buckling)

OK0.146.00703.039.085973.5

13007.504

MM

AF

cx

x

y

29

2007-8-22 57

By Amplified Sway Method

Af = =

Column effective length for nonsway framekbottom = 0.5 (Nominal rigid base, Cl 5.1.3.2)Beam stiffness = 0.5I/L (Non-sway mode, Table E.1 in Annex E)Ktop = (1 + 1)/(1+ 1 + 0.5) = 0.8In-plane Column effective length factor = 0.77 (Fig. E.1, Annex E for nonsway frame)Lex=3.5m x 0.77 =2.7m Pcx = 1180kN In planeLey = 3.5m Pcy = 668kN out-of-plane = -2.7/5.973 = - 0.452 mLT = 0.45 (Table 18 of BS5950)mx = 0.51 (Table 26 of BS5950)In Plane Buckling

Lateral Torsional Buckling (out of plane buckling; no amplification for moment)

OK

1crcr

43.1

131.331.3 =

0.180.075

97.5x51.0x43.1668

7.504Z

MmkPF

yy

xxamp

c

4 cycles/second(including interaction between primary and secondary beams)

Building deflection (< H/500) Stability, Overturning & Uplift (foundation and continuity of column) Building frequencies (Period < 2 seconds) Continuity of floor diaphragms Continuity of vertical bracing Location of lateral bracing

30

2007-8-22 59

Disproportionate Collapse

2007-8-22 60

Structural Integrity

Deflected shapeR

1 Tying force = R or 75 kN

31

2007-8-22 61

Avoidance of Disproportionate Collapse2 Tying of columns

2007-8-22 62

3.Continuity of columns. all columns should be carried through at each beam-to-column connection. All column splices should be capable of resisting a tensile force equal to the largest factored vertical dead and imposed load reaction applied to the column at a single floor level

4.Resistance to horizontal forces. Braced bays or other systems for resisting horizontal forces should be distributed throughout the building such that, in each of two directions approximately at right angles.

5.Heavy floor units. Where precast concrete or other heavy floor or roof units are used they should be effectively anchored in the direction of their span

32

2007-8-22 63

If the five criteria are not satisfied

Check disproportionate collapse: to ensure that removal of single element does not cause too large an area (15% of floor area or 70m2) to collapse

If large area collapse occurs, design the key element to withstand a blast of 34 kN/m2

2007-8-22 64

Check disproportionate collapse: to ensure that removal of single element does not cause too large an area (15% of floor area or 70m2) to collapse

If large area collapse occurs, design the key element to withstand the blast pressure

Alternate Load Path Method

Core

Exterior columns

Belt and outrigger

33

2007-8-22 65

Progressive Collapse Resistant Design

Local Resistance Method:

This method seeks to provide sufficient strength to resist an extreme loading event. It requires that a specific collapse-initiating event be identified so that the local resistance can be referenced to a specific limit state.

Alternate Load Path Method

This method requires that the structure is able to provide alternate load path to one lost column. The collapse-initiating event and threat level is not considered.

2007-8-22 66

Design SummaryElastic Design

Classify frames as braced or unbraced frame (5.1.4), and sway or nonsway (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.

34

2007-8-22 67

Elastic Non-sway frames (5.6.3):

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 frame should be checked for combined vertical

and horizontal loads without pattern loading.

Design SummaryElastic Design

2007-8-22 68

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 methodAnnex E or the amplified sway method.

Design SummaryElastic Design

35

2007-8-22 69

Reading assignment:

BS5950:Part1 Cls 5.1, 5.2 & 5.6

R Liew note: Chapter 3 Sections 3

HOMEWORK 4: CONTINUOUS FRAMES

2007-8-22 70

Take Home Design Project

DL =24kN/mIL= 30 kN/m

6m

Storey height =4m

6m 6m 6m

Five framesSpacing 6m

Comment on the structural integrity

36

2007-8-22 71

PLASTIC DESIGN

2007-8-22 72

Plastic design

Classify frames as independently braced (5.7.2), sway or non sway (5.7.3)

37

2007-8-22 73

Plastic braced frames

Independently braced frames (5.7.2) Design to resist gravity loads (load

combination 1). The effective length LE of the columns in the

plane of the frame should generally be taken as equal to the storey height L.

Check columns under pattern loading using an effective length from Annex E.

2007-8-22 74

Plastic design of non-sway frames

If beams are designed plastically,they will contain moments in excess of 90% of Mp.Care must be taken when using Annex Eremembering that Kb is likely to be ZERO in most cases,giving LE/L factors close to 1.0

38

2007-8-22 75

Plastic design of non-sway frames

Normal plastic rules apply for local capacityand member stability as for portal frames. Usually the latter will be less onerous due to the presence of floor slabs to provide stability to the beams.

Remember that the frame must be stable in the plane perpendicular to the plane of the frame being designed.

2007-8-22 76

Plastic design- unbraced frames

Check for possible non-sway modes of failure as recommended for independently braced frames.

Satisfy the simplified M-R-Wood frame stability check (given in clause 5.7.3.2),

Or design to resist sway mode failure using either elastic analysis or second order elastic-plastic analysis.

39

2007-8-22 77

Plastic design of sway frames

Frames may be designed using a second order elastic-plastic analysis or by using the sway stability check.

This is derived from the Merchant-Rankine-Wood equation.

Origins: The Rankine strut equation is1/PF = 1/PCR + 1/ PSQ

2007-8-22 78

M-R-Wood equation

Rankines formula proposed for struts prior to development of Ayrton-Perry equation.

Merchant suggested its use for frames with: Elastic critical load replacing Euler load. Plastic collapse load replacing squash load.

40

2007-8-22 79

Plastic design of sway framesMerchant-Rankine Equation

1 = 1 + 1PF Psq

Pcr

Psq/Pcr

PF/Psq

Merchant-Rankine1.0

1.0

0.5

2007-8-22 80

Development of M-R-Wood Eqn 1/PF = 1/PCR + 1/ PSQ

1/F = 1/ CR + 1/ P ={ P + CR }/CR P

F = CR P / { P + CR } But F = 1 P + CR = CR P CR = CR P - P = P {CR 1} P = CR /{CR 1}

41

2007-8-22 81

Development of M-R-Wood Eqn

But for CR/ P >10stray composite action &strain hardening means noneed to reduce r below P.

For 10 < CR/ P < 4.0change 1 to 0.9 for continuity at CR/ P =10

r = 0.9CR /{CR 1}

r P

P /CR0.1 0.25

Merchant Rankine-Wood

Merchant-Rankine

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Combined plastic collapse mechanism

Determine plastic collapse load using sway mode shown

42

2007-8-22 83

Plastic collapse load

1. The bases of the columns should be fixed.2. Mechanism as on previous slide.3. Ensure that columns remain elastic at

assumed plastic moment.4. Check that no localised beam or storey

mechanism is more dangerous.5. Storey height less than mean column

spacing.

2007-8-22 84

Development of M-R-Wood EqnFor actual unclad frames or clad frames

where cladding stiffness is utilised

For CR / P > 20strain hardening means noneed to reduce r below P . For 20 < CR/ P < 5.75change 1 to 0.95 for continuity at CR/ P = 20

r = 0.95CR /{CR 1}

r P

P /CR0.1 0.25

0.05