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Continuous Construct

### Transcript of Continuous Construct

• BS 5950: 2000 : Part 1Continuous constructionMulti-storey frames

• Simple Design pin joints

• Continuous Design rigid joints

Continuous multistorey frame

• Base stiffness and strength

As in the lecture on portal frames

• Deformation of a multi-storey frame

• P - effects

• Stability effects

• Effective lengths and critical loads PE = p2EI/L2

Pcrit= 0.25p2EI/L2 = p2EI/4L2 = p2EI/(2L)2

Hence the effective length LE for a cantilever is 2L

• Nominal effective lengths

• Charts of Annex E

• Derivation of ChartsModel 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 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.

• Use of the charts of Annex EFor 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 Mpr 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.

• Buckled mode shapes

• Beam stiffness valuesbuildings without floor slabs Provided that the frame is reasonably regular in layout:For non-sway framesKb = 0.5 I/LFor sway framesKb = 1.5 I/L

• Beam stiffness values buildings with floor slabs

Non-sway mode

Sway mode

Beam directly supporting concrete floor or roof slab

1.0 ( I/L )

1.0 ( I/L )

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 bracingTwo other plots exist as E4 & E5Relate 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.

• Frames with partial sway bracing kp = h2SSp/(80ESKc) but 2

SKc is the sum of I/h for all the columns in that storeyS Sp is the sum of the stiffness of every panel in the storeySp is given by (0.6(h/b) ) t Ep/[1+(h/b)2]2 h/b is the ratio of storey height to panel width t is the panel thickness Ep is the modulus of elasticity of the panel material.

• Notional horizontal loads Notional horizontal forces should NOT: a) be applied when considering overturning b) be applied when considering pattern loading c) be combined with applied horizontal loads d) be combined with temperature effects e) be taken to contribute to the net reactions at the foundations.

• Minimum horizontal forces

• Resistance to horizontal forcesResistance to horizontal forces may be provided in a number of ways as follows: a) triangulated bracing members. b) moment resisting joints and frame action. c) cantilever columns, shear walls, staircase and lift shaft enclosures. d) or a combination of these.

• Classification of frames Frames may be Braced or unbraced depends on how horizontal forces are transmitted to the ground.Sway or non-sway -depends on significance or otherwise of P- effects.

• Independently braced framesa) The stabilizing system must have a spring stiffness at least four times larger than the total spring stiffness of all the frames to which it gives horizontal support (i.e. the supporting system reduces horizontal displacements by at least 80%).andb) The stabilizing system must be designed to resist all the horizontal loads applied including the notional horizontal forces.

• Braced multi-storey frame

• Sway / non-sway categorisationA frame can be deemed to be non-sway if,in the SWAY mode lcr 10

lcr = 1 / 200 max = h / 200 max Otherwise it is a sway frame.

• Annex F - Critical load of sway frame cr = 1 / 200 max & = {n n 1}/h

• Frame design Braced or Unbraced Sway or Non-sway Elastic design or Plastic design

• Design of independently braced frames

Independently braced 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 to be considered. the stabilizing system must be designed to resist all the horizontal loads applied including the notional horizontal forces.

• Design of non-sway frames Non-sway 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 be checked for combined vertical and horizontal loads without pattern loading.

• Plastic design of non-sway framesIf 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

• Plastic design of non-sway framesNormal plastic rules apply for local capacity and member stability as for portal frames. Usually the latter will be less onerous due to the presence of floor slabs to provide stability.

Remember that the frame must be stable perpendicular to the planes of the frames being designed.

• Elastic design of sway framesProvided that lcr is greater than 4, the sway should be allowed for by using one of the following methods: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.

• kamp for sway sensitive framesfor 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)

• kamp for sway sensitive frames 2) 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.

• Sway effects The distortions of a frame may be divided into two components: 1) those which arise from sway with zero rotation of every joint and 2) those which arise a result of joint rotations without any sway of the frame.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.

• Sway effects In every other case need to separate moments into sway and non-sway components by either:

a) Deducting the non-sway effects b) Direct calculation

• 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.

• b) Direct calculation:

1) Analyse the frame with horizontal restraints added at each floor or roof level to prevent sway.2) Reverse