Design concerete structures ussing this PDF to eurocode 2, pre scheme sizing and quick and dirty calculations . This is use for the chartered examinat...
Descrição: User Guide for V3 and V4 of Concrete Centre User Guide for RCC Spreadsheets
User Guide for V3 and V4 of Concrete Centre User Guide for RCC Spreadsheets
User Guide for V3 and V4 of Concrete Centre User Guide for RCC Spreadsheets
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Full description
Descripción: graded readers, penguin
Descripción: NOC Manual
Concrete, beton, pouring
Descripción completa
Simple explanation of concrete and revealed its properties
Centre Break Isolator Operation and Installation Manual - ABB MakeFull description
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?hk^phk]mhma^>nkh\h]^+^]bmbhg Since its publication in 2006, the Concrete building scheme design manual has proved a popular publication and this update is intended to assist the transition to Eurocode 2 for the design of concrete structures by showing how to carry out initial design to the Code. As before it will greatly assist candidates for the Institution of Structural Engineers chartered membership examination by drawing together in one place useful information including design aids that are specifically prepared for Eurocodes.
:\dghpe^]`^f^gmlmhma^Ëklm^]bmbhg Andrew Minson was the originator of the concept for this book, but the final format and content have evolved over time with input gratefully received by the author, Owen Brooker, from a number of people. The book contains some new material but many of the ideas and much of the data have been developed from many sources over several years. It is not possible to credit every source; indeed many are unknown, but wherever possible, references have been made to relevant documents. Particular thanks are due to Jenny Burridge and Bob Wilson who have spent considerable time reviewing and commenting on drafts and providing useful guidance on what the examiners are expecting. Thanks are also due John Brazier, Andrew Cotter, Bryan Magee and Matt Obst, who have commented or made useful contributions in more specialised areas. Gillian Bond, Issy Harvey, Sally Huish and Michael Burbridge Ltd have also made essential contributions to the production of this handbook.
*'* How to use this handbook MablaZg][hhdbleZb]hnmmhk^Ì^\mma^^qZfbgZmbhgjn^lmbhgl4mh]Zm^ma^l^aZo^_heehp^]ma^ lZf^_hkfZm'L^\mbhg*h_ma^^qZfbgZmbhgblma^]^lb`gZiikZblZeZg]mablbl\ho^k^]bgL^\mbhg+% pab\aikhob]^l^ll^gmbZebg_hkfZmbhgk^jnbk^]mhZllblmbgZglp^kbg`mabliZkmh_ma^^qZfbgZmbhg' L^\mbhg + h_ ma^ ^qZfbgZmbhg bl ]bob]^] bgmh mak^^ iZkml4 Ëklmer !l^\mbhg +\" ma^k^ bl ma^ k^jnbk^f^gm mh ik^iZk^ ln_Ë\b^gm \Ze\neZmbhgl mh ^lmZ[ebla ma^ lbs^ Zg] _hkf h_ Zee ma^ ikbg\biZelmkn\mnkZe^e^f^gml'@nb]Zg\^bl`bo^gbgL^\mbhg,h_mablaZg][hhd' L^\mbhg+]k^jnbk^lma^\Zg]b]Zm^mhik^iZk^ieZgl%l^\mbhglZg]^e^oZmbhgl%bg\en]bg`\kbmb\Ze ]^mZbel%_hk^lmbfZmbg`inkihl^l4^qZfie^lZk^`bo^gbgL^\mbhg-'?bgZeerl^\mbhg+^k^jnbk^l ]^mZbe^] f^mah] lmZm^f^gml Zg] h_m^g Zg hnmebg^ ikh`kZff^ mh [^ ik^iZk^] Zg] mabl bl \ho^k^]bgL^\mbhg.' K^Z]^kllahne]Ziik^\bZm^maZmmablin[eb\Zmbhgblpkbmm^g_hkma^lmkn\mnkZe^g`bg^^kikZ\mblbg` bgma^ND'PabelmfZgrh_ma^ikbg\bie^l\Zg[^Ziieb^]mhlmkn\mnk^lZkhng]ma^phke]%bm fnlm[^\e^Zkerng]^klmhh]maZm^gobkhgf^gmZeZg]`^hm^\agb\Ze_Z\mhkl^g\hngm^k^]fZr [^lb`gbË\Zgmer]b__^k^gm'BmblZelhpkbmm^gpbmama^]^lb`gh_[nbe]bg`lmkn\mnk^lbgfbg]kZma^k maZg[kb]`^lhkh__lahk^lmkn\mnk^l'Ma^k^jnbk^f^gml_hkjn^lmbhg1%lmkn\mnkZe]rgZfb\l%Zk^ ghmli^\bË\Zeer\ho^k^]' Text on a pale blue background provides information to help candidates prepare for the examination and is not intended for use in the examination itself.
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Note This graph assumes solar shading used and air-conditioning will be installed in homes when temperatures become uncomfortable, but will be used with natural ventilation whenever possible.
a) Flat soffit Duct leaving riser / core consider universal columns as beams
VAV terminal box
A B C
500
400
E
D
F Below deep beams final terminal device only
Main duct run-outs below normal maximum beam size G
FFL H
b) Downstand beams Key A B C D E F G H
Structural zone 50 mm deflection and tolerance Approx. 500 mm HVAC duct or terminal device 50 mm support and tolerance 5 150 mm sprinkler zone (if required) 150 mm lighting and ceiling zone Floor to ceiling height Raised floor
Engineering Consultants 99 High Street Newtown NW9 9AA Your ref: Our ref: 1099/OB/1.01/21-11-06
21 November 2006
FAO Mr J Client Dear Mr Client
Call Centre – Implications of Recent Minor Seismic Activity As we are sure you are aware there has recently been a minor earthquake in the region of the recently completed Call Centre. We would like to address the concerns you may have regarding the strength of your building if faced with a similar event. Initially we would like to confirm that the structure has been designed to the latest Building Regulations and Codes of Practice. There is no requirement to explicitly consider the effects of seismic activity because historically the most significant horizontal forces on structures in this region are wind forces, rather than forces arising from seismic activity. We would suggest that, before making any decisions about strengthening the structure, we wait for a consensus amongst the experts as to whether this type of event is likely to occur more frequently in the area and what they recommend as the design requirements. We would expect that, if changes are necessary, the Building Regulations will be revised and that clear guidance will emerge on whether existing buildings should be strengthened. The Call Centre is just one of many buildings that could be affected by a similar earthquake in this region. In the meantime, we would like to reassure you that the building as it stands will have reasonable strength to resist to a minor seismic event. It has been designed with ties to ensure that ‘progressive collapse’ does not occur and these will act to strengthen the building in the event of an earthquake. There are, however, a number of specific areas where localised damage could occur as a result of a small earthquake; these are detailed below. s !N EARTHQUAKE RESISTANT BUILDING SHOULD BE ABLE TO CARRY THE LATERAL LOADS IMPOSED TO A STABILITY CORE )N THIS CASE THE precast units and topping screed are designed to act as a stiff diaphragm. However, depending on the intensity of the earthquake, this system may have insufficient strength to carry the forces. A design check can be carried out when there is agreement on the design forces that should be used for an earthquake in this region. s 4HE CURRENT BUILDING IS DIVIDED INTO TWO DISCRETE STRUCTURES WITH A SMALL MOVEMENT JOINT BETWEEN THEM $URING AN earthquake the structures will move laterally and the width of the joint is unlikely to be sufficient to prevent the two structures from making contact. We think there are two options for overcoming this; either the joint could be made wider by making alterations to the existing structure, or the lateral stiffness of the building could be increased by incorporating more stability walls. The latter would reduce the sideways sway during a seismic event. It may also be the preferred solution if the floor diaphragm requires strengthening. We will need to have a discussion so we can find an acceptable solution that provides minimal disruption to your business and the minimum effect on the architecture. s 4HE BUILDING IS FOUNDED ON WEATHERED ROCK WHICH SHOULD BE SUFlCIENTLY ROBUST IN THE EVENT OF AN EARTHQUAKE (OWEVER we will need to review the strength of the foundations to resist lateral forces, for which they were not designed. The building has a suspended ground floor, which means that there are beams tying the foundations together. This is beneficial in the event of an earthquake, because it prevents the foundations moving apart possibly causing major damage to the frame. s 4HE BEAM TO COLUMN CONNECTIONS ARE A CAUSE FOR CONCERN 4HERE ARE MANY WAYS TO FORM A CONNECTION AND THIS IS generally left to the precast manufacturer to design. This is usually beneficial because he can carry out the work more efficiently and to suit his working methods. On this project the manufacturer has used a ‘billet’ type connection, which is perfectly adequate for the job it was designed to do. However, it offers less spare capacity to resist the forces from an earthquake and strengthening may be required. s 4HE TWO STOREY GLAZED ENTRANCE AREA WILL REQUIRE SOME STRENGTHENING MEASURES 4HE LATERAL MOVEMENT THAT WILL OCCUR during an earthquake is almost certain to damage the glazing. The current glazing is laminated glass, which means that it will not break into small pieces. However, there is a risk that whole panes will come away from their fixings with the potential to cause loss of life. To overcome this, strengthening of the column–to-roof beam connections will be necessary to reduce lateral sway. We cannot be certain about how design guidance will change as a result of the earthquake, but the above comments should give you an appreciation of the preventive measures that are likely to be necessary. If you would like to discuss the situation further then please contact us.
Yours sincerely
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38
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,Design calculations (section 2c) ,'*Expectations of the examiners
,'*'* Principal elements Ma^_heehpbg`blZeblmh_lmkn\mnkZef^f[^klmaZm\hne][^\hglb]^k^]mh[^ikbg\biZe^e^f^gml' BmfZrghm[^Z_neeeblmZg]_hklhf^[nbe]bg`lma^l^^e^f^gmlfb`amghmZiier3 ■ LmZ[bebmrlrlm^f!bg\en]bg`Zll^llf^gmh_ma^ehZ]l" ■ ?hng]Zmbhgl!bg\en]bg`Zll^llf^gmh_ma^\hf[bg^]^__^\mlh_`kZobmZmbhgZeZg]eZm^kZeehZ]l%
Percentage of tension reinforcement (As,req /bd) Notes 1 For two-way spanning slabs, the check should be carried out on the basis of the shorter span. 2 This graph assumes simply supported span condition (K = 1.0). K = 1.5 for interior span condition K = 1.3 for end span condition K = 1.2 for flat slabs K = 0.4 for cantilevers 3 Compression reinforcement, r', has been taken as 0.
4 Curves based on the following expressions: l 1.5 fck r 0 + 3.2 = K 11 + r d
[
fck
1.5 r0 –1 r
( )]
where r ≤ r0 and l 1.5 fck r 0 + = K 11 + d ( r – r ')
c2 = 0.2, gG = 1.35 To determine stress in the provided reinforcement (ss ), calculate the ratio Gk/Qk, read up the graph to the appropriate curve and read across to determine ssu. ss can be calculated from the expression: ss = ssu f
,'1 Detailing ,'1'* Maximum and minimum areas of reinforcement Ma^fZqbfnfZk^Zh_^bma^kma^m^glbhghk\hfik^llbhgk^bg_hk\^f^gmbgZg^e^f^gmbl- h_ ma^`khll\khll&l^\mbhgZeZk^Zh_ma^\hg\k^m^4mabl\Zg[^bg\k^Zl^]mh1 ZmeZil'Ma^fbgbfnf i^k\^gmZ`^lZk^`bo^gbgMZ[e^;'-!l^^:ii^g]bq;"'
,'1'+ Minimum spacing of bars Ma^fbgbfnfliZ\bg`h_ma^[Zklblma^fZqbfnflbs^h_ma^\hZkl^Z``k^`Zm^ienl.ffhkma^ [Zklbs^%pab\a^o^kblma^`k^Zm^k'?hk+)ffZ``k^`Zm^Zg][Zklh_+.ffbg]bZf^m^kZg]ho^k% ma^fZqbfnfgnf[^kh_[ZklbgZeZr^kbl3
,'1', Maximum spacing of bars Ma^fZqbfnfliZ\bg`bl`bo^gbgmZ[e^0',h_>nkh\h]^+!l^^MZ[e^;',bg:ii^g]bq;"' La^Zkebgdllahne][^ZmZliZ\bg`h_ghfhk^maZg)'0.]%Zg]ghehg`bmn]bgZe[Zklahne][^fhk^ maZg*.)ffhk]_khfZo^kmb\Zee^`'
,'2 Design of beams ,'2'* Governing criteria NlnZeerma^`ho^kgbg`\kbm^kbhgblma^[^g]bg`lmk^g`ma'=^Ì^\mbhgfZr[^\hf^\kbmb\Ze_hkehg` liZglhk\Zgmbe^o^kl%Zg]la^Zkblebd^ermh[^\kbmb\Ze_hkmkZgl_^k[^Zfl%^li^\bZeer_hka^Zober ehZ]^]lahkmliZgl'
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50
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Worked example 1 Transfer beam
CCIP - 018 La^^mgh'
MS
TB1 =Zm^
TCC
Aug 09
Gk = 1500 kN, Qk = 1000 kN
9.0 m
1.5 m
Initial sizing
Shear stress not to exceed 4 N/mm2 (to avoid reinforcement congestion). Ultimate load = 1.35 x 1500 + 1.5 x 1000 = 3525 kN (ignoring self-weight) Take b = 600 3 V =d = Ed = 3525 x 10 = 1469 mm vEdb 4 x 600
Section 2.10.2
Take overall depth as 1650 mm (d = 1550) Bending
Mmax = 3525 x 1.5 = 5288 kNm For Mmax = 5288, b = 600, d = 1550, fck = 32 N/mm2 K = 0.115, As = 8858 mm2 (z = 1297 mm) Use 8 H40 (10100 mm2) in 2 layers. vEd = 3525 x 103/(0.9 x 600 x 1550) = 4.21 MPa
Figure 3.1
Assuming the strut is at 45° Asw s
≥
4.21 x 600 435 x cot 45°
≥
5.81 mm
8H40bars
Slab reinforcement
Table 3.7
1650
Shear
Figure 3.1
Try H12 links s = 452/5.81 = 78 mm Say 4 H12 links @ 75 mm ctrs 4 H12 links @ 75 ctrs
600
Comments
Remember to check headroom beneath the beam H40 bars will be heavy; if there is no reasonable alternative, ensure that the contractor is aware so he may take steps to safeguard the health and safety of the steel fixers.
,'*)', Fire resistance Ma^leZ[mab\dg^llZg]Zqbl]blmZg\^lahne][^\hfiZk^]pbmama^k^jnbk^f^gmlbgMZ[e^,'*.% pab\abl[Zl^]hg>nkh\h]^+%IZkm*&+'Ma^Zqbl]blmZg\^%Z%blma^]blmZg\^_khfma^_Z\^h_ma^ \hg\k^m^mhma^\^gmk^h_ma^fZbg[ZkZg]ma^]^lb`g^klahne]^glnk^maZm3 Z6\ghf$febgd$f[Zk(+' MZ[e^,'*. Fbgbfnf]bf^glbhglZg]Zqbl]blmZg\^l_hkhg^&pZrliZggbg`leZ[l LmZg]Zk]_bk^k^lblmZg\^ REI 60 REI 90 REI 120 REI 240
Imposed load = 2.5 kN/m2 Superimposed dead load = 1.5 kN/m2 Concrete class C28/35 cnom = 25 mm
Initial sizing
From Economic concrete frame elements – 167 or 6000/32 = 187.5, say 200 mm
Loading
ULS = 1.35 (0.2 x 25 + 1.5) + 1.5 x 2.5 = 12.5 kN/m2
Bending
Check first support from end M = – 0.086Fl = – 0.086 x 12.5 x 62 = – 38.7 kNm
Section 2.10
Table 3.14
For b = 1000, d = 200 − 25 − 10 = 165, fck = 28 Then K = 0.051, As, req = 567 mm2 (z = 157 mm) Use Shear
H12 @ 175 ctrs (As,prov = 646 mm2)
VEd = 0.6F = 0.6 x 12.5 x 6 = 45.0 kN V 45.0 x 103 vEd = Ed = = 0.27 N/mm2 bd 1000 x 165
Table 3.14
100 As = 100 x 646 = 0.39 bd 1000 x 165
Deflection
= vRd,c = 0.57 vEd < vRd,c = no shear links required
Table 3.6
Maximum sagging moment = 0.075 Fl = 0.075 x 12.5 x 62 = 33.75 kNm
Table 3.14
For fck = 28, d = 165, b = 1000 Then K = 0.044, As,req = 494 mm2 (z = 157 mm) r = As,req/bd = 494 x 100/(1000 x 165) = 0.30% Basic l/d = 35.2 x 1.3 = 45.8 (K = 1.3 for end span condition) actual l/d = 6000/165 = 36.4 =OK
From Economic concrete frame elements: 213 mm or 9000/36 = 250 mm say 250 mm
Loads
n = 1.35 (1.5 + 6.25) + 1.5 x 5 = 18.0 kN/m2 (ULS)
Check short span
d = 250 − 25 − 8 = 217 mm
Section 2.10
ly 9.0 = = 1.3 lx 7.2 bsx = – 0.069 and bsx = 0.051 Support moment critical in bending
Table 3.17
= – 64.4 kNm/m msx = bsx n lx = – 0.069 x 18.0 x For msx = 64.4, b = 1000, d = 217, fck = 28 2
7.22
Then As = 718 mm2/m (K = 0.049 & z = 206 mm) Use H12s @ 150 ctrs (As,prov = 754 mm2/m) Shear
bvx = 0.50
Table 3.18
Vsx = bvx n lx = 0.5 x 18.0 x 7.2 = 64.8 kN/m width 3 V vEd = Ed = 64.8 x 10 = 0.30 N/mm2 bd 1000 x 217 100 As = 100 x 754 = 0.35 bd 1000 x 217 = vRd.c = 0.57 N/mm2 > 0.30 = no shear links required Deflection
Table 3.6
Maximum sagging moment = 0.051 x 18.0 x 7.22 = 47.6 kN/m For b = 1000, d = 217, fck = 28 Then K = 0.036, As req = 531 mm2 (z = 206 mm) r = As,req/bd = 531 x 100/(1000 x 217) = 0.24% Basic l/d = 35.2 x 1.3 = 45.8 (K = 1.3 for end span condition) actual l/d = 7200/217 = 33.2 =OK
,'*,'+ Geometry Ma^pb]mah_ma^kb[bl]^m^kfbg^][r\ho^k%[ZkliZ\bg`Zg]Ëk^k^jnbk^f^gml'Mrib\ZeerliZ\bg`l _hkkb[lZk^/))ff%0.)ff%2))ffZg]*+))ff%pbmaZkb[pb]mah_*.)ff':o^kZ`^ l^e_&p^b`amh_ma^kb[[^]leZ[\Zg[^]^m^kfbg^]_khf?b`nk^,'1%hkZgZllnf^]l^e_&p^b`am\Zg [^h[mZbg^]_khfMZ[e^+'*)\' T H
Using Economic concrete frame elements: 284 mm Or 8000/26 = 307 mm, say 300 mm
Loading
ULS = 1.35 (1.5 + 0.3 x 25) + 1.5 x 5 = 19.65 kN/m2
Bending
Check long span end bay condition M = – 0.086Fl
Centre strip critical
= – 0.086 x 19.65 x 7 x 92 = – 958 kNm
Design moment = 0.75 x 958 = 718 kNm
Section 2.1
Table 3.20 Table 3.21
For b = 3500, d = 300 − 30 − 10 = 260 mm, fck = 28 Then K = 0.108, As = 7110 mm2 Centre column strip = 4740 mm2 (2708 mm2/m) Use T20 @ 100 ctrs (As,prov = 3140 mm2/m) Outer column strip = 2370 mm2 (1354 mm2/m) Use T20 @ 200 ctrs (As,prov = 1570 mm2/m)
Punching shear
VEd = 19.65 x 7.0 x 9.0 = 1238 kN bV 1.15 x 1238 x 103 At column face, vEd = u d Ed = = 4.07 < 4.98 4 x 350 x 250 i eff Shear resistance without links:
100 As 100 x 3140 = = 1.26% bd 1000 x 250
=vRd,c = 0.75 MPa Area of steel at first control perimeter, 2d = 2 x 250 = 500 mm from column face. Length of perimeter = 4 x 350 + 2Q x 500 = 4542 mm Radial spacing sr = 0.75 x 250 = 188 mm Asw = (vEd – 0.75 vRd,c)sr u1/(1.5 fywd,ef) 3
= d1.15 x 1238 x 10 – 0.75 × 0.75m x 188 x 4542 /(1.5 x 313) 4542 x 250 =1257 mm2/perimeter Using 10 mm dia. links, no of links = 1257/78.5 = 16, spacing = 4542/16 = 284 < 1.5d Deflection
62
Maximum sagging moment = 0.075 Fl = 0.075 x 19.65 x 7 x 92 = 836 kNm
Table 3.22
=^lb`g\Ze\neZmbhgl
Centre strip critical
Design moment = 0.55 x 836 = 460 kNm For b = 3500, d = 260 and fck = 28 K = 0.069, As = 4351 mm2 Or As = 1243 mm2/m r = As,req/bd = 1243 x 100/(1000 x 260) = 0.48% Basic l/d = 20.3 x 1.2 = 24.4 (K = 1.2 for flat slab condition) Figure 3.2 actual l/d = 9000 / 260 = 34.6 Increase area of steel to reduce steel stress, assume H20 @ 225 ctrs (1400 mm2) Figure 3.3 Approximate steel stress at SLS = 236 MPa Therefore approximate steel stress, ss, when As,prov = 1400 mm2 ss = 236 x 1243/1400 = 210 MPa Increased basic l/d = 24.4 x 310/210 = 36 > 34.6 =OK =Use H20 @ 225 ctrs As,prov = 1400 mm2
Comments
1. The design in the orthogonal direction has not been included. 2. For the design of the strips across two panels only, coefficients from Table 3.20 are not appropriate.
Using Economic concrete frame elements: 255 mm or 7200/29 = 248 mm, say 250 mm Assume self-weight = 4.0 kN/m2
Table 2.9c
Loading
ULS = 1.35 (4.0 + 1.5) + 1.5 (2.5) = 11.2 kN/m2
Bending
Check sagging moment in end bay d = 250 – 26 – 6 – 8 = 210 mm Effective length = 6.2 + 2 x 1/2 x 0.210 = 6.41 m
BS EN 1992-1-1: Fig 5.4
M = 0.075Fl = 0.075 x 11.2 x 6.412 x 0.75 = 25.9 kN/m per rib
Table 3.20
For b = 750, d = 210, fck = 28 K = 0.030, As,req = 291 mm2 (z = 200 mm) x = 2.5 (210 – 200) = 25 mm
= N.A. in flange – design as rectangular section
Deflection
r = As,req/Ac = 291 x 100/(125 x 110 + 750 x 100) = 0.33% Reduction factor for flanged section = 1 – 0.1 x ((750/125) – 1) = 0.5 = use 0.8. Basic l/d = 0.8 x 33 x 1.3 = 34.3 (K = 1.3 for end span condition) actual l/d = 7200 / 210 = 34.3 =OK
Using Economic concrete frame elements: 425 mm thick (i.e. 325 moulds + 100 topping) or 9000/20 = 450, say 425 thick
Section 2.9
Assume self-weight = 7.3 kN/m2 Ultimate load
n = 1.35 (7.3 + 1.5) + 1.5 (2.5) = 15.6 kN/
Table 2.9c m2
There is a substantial beam along the column strips, which can therefore resist torsion at the corners. Design as a two-way spanning slab. ly 9.0 = = 1.25 lx 7.2 Interpolating for a corner panel Hogging moments bsx = − 0.066, Sagging moments bsx = 0.049
Table 3.17
Bending at support Msx = bsx nlx2 = − 0.066 x 15.6 x 7.22 = − 53.4 kN/m width For ribs @ 900 ctrs M = 53.4 x 0.9 = 48.0 kNm/rib For d = 425 − 26 − 6 − 8 = 385 mm, b = 900, fck = 28 Then As,req = 299 mm2 (Taking z = 0.95d and therefore N.A. is in flange) Deflection
r = As,req/Ac = 299 x 100/(180 x 285 + 900 x 100) = 0.21% Deflection not critical by inspection
Figure 3.2
Use 2 x H16 per rib (As,prov = 402)
,'*/', Stresses Ma^lmk^llbgma^^e^f^gm\Zg[^\Ze\neZm^]Zl_heehpl3 F Lmk^llZmma^mhih_ma^l^\mbhg%sm6I$ [ :\Sm F Lmk^llZmma^[hmmhfh_ma^l^\mbhg%s[6I$ [ :\S[ pa^k^ :\ 6 Zk^Zh_ma^\hg\k^m^ F[ 6 [ZeZg\^]fhf^gm!b'^'bg\en]bg`ihlm&m^glbhgbg`^__^\ml" Sm 6 l^\mbhgfh]nenl&mhi S[ 6 l^\mbhgfh]nenl&[hmmhf I 6 ik^lmk^ll_hk\^ 69
Area of concrete, Ac = 7000 x 250 = 1750 x 103 mm2 Second moment of area, Ic = 7000 x 2503/12 = 9.11 x 109 mm4 Distance to extreme fibres from neutral axis, yb = yt = 125 mm Section modulus, Zb = Zt = 9.11 x 109/125 = 72.9 x 106 Strand diameter = 12.9 mm 125 Minimum cover = 20 mm 125 Distance to centre of strand = 60 mm Tendon profile for end span (most critical):
Table 2.22
60 a
60 Idealised tendon a = 250 – 60– 60 – 65 = 98 shape for initial 2 design only
Number of
Characteristic value of maximum force, 186 kN
strands required
Initial prestress = 0.8 x 186 = 149 kN (allow for 80% of characteristic force)
Table 3.29
Prestress in service condition = 0.7 x 149 = 104 kN (allow for 10% loss at transfer and 20% loss at service, check in detailed design) Balance dead loads with prestressing P = ws2/(8a) = 7 x 6 x 92/(8 x 0.098) = 4339 kN
Section 3.16.2
=No. of tendons required = 4339/104 = 41.7, try 9 x 5 strands per duct Total force = 4680 kN Moments (SLS)
Applied loads wa = 7 x (6.25 + 1.25 + 5) = 87.5 kN/m Balancing load wb = 8aP/s2 = 8 x 0.098 x 4160 / 92 = 40.2 kN/m Balanced moment M ≈ (wa – wb) l2/10 = (87.5 – 40.2) x 92/10 = 383.1 kNm
Mid-span stresses
Comments
st =
P + M = 4680 x 103 + 383.1 x 106 = 2.7 + 5.3 = 8.0 N/mm2 Act Zt 1750 x 103 72.9 x 106
For class C32/40 allowable compressive stress is 12.8 N/mm2 = OK
Table 3.30
P – M = 4680 x 103 – 383.1 x 106 = 2.7 – 5.3 = – 2.6 N/mm2 sb = Act Zb 1750 x 103 72.9 x 106
Section 3.16.3
For class C32/40 allowable tensile stress is 2.7 N/mm2 = OK
Table 3.30
Stresses at the support and punching shear should also be checked at this stage for a flat slab (see 3.12.3) It is assumed that deflection, the ULS requirements and transfer requirements can be met with passive reinforcement in detailed design, as is usually the case.
Total ultimate axial load = 4 x 19.7 x 7 x 9 N = 4952 kN Initial sizing
Using Economic concrete frame elements – 450 mm square
Column design
NEd/bhfck = 4952 x 103/(4502 x 32) = 0.76 Take minimum moment as 4952 x 0.02 = 99 kNm (assume column is not slender) MEd/bh2fck = 99 x 106/(4503 x 32) = 0.03 d2 = 35 + 8 +16 = 59 mm d2/h = 59/450 = 0.13 Use column graph for d2/h = 0.15 Asfyk/bhfck = 0.35 As = 0.35 x 4502 x 32/500 = 4536 mm2 Use 6 H32 (As,prov = 4830 mm2)
Assume critical combination is 1.0 Gk + 1.5 Qk,w where Qk,w is wind load. Assuming the floor load is 4000 kN supported on 8 columns, the horizontal action due to imperfections = 0.25% x 4000 = 10 kN
3.0 m gk = 35 kN/m
75 kN 3.0 m
gk = 35 kN/m
75 kN
ft
=N − M Lt (tL2/6)
N
= 1.0 x 35 x 4 x 2.5 + 0.200 x 12 x 2.5 x 25 = 500 kN
MEd = 1.5((40 + 10) x 12 + (75 + 10)(9 + 6 + 3)) = 3195 kNm ft
3.0 m gk = 35 kN/m
75 kN
=
500 x 103 − 3195 x 106 2500 x 200 (200 x 25002/6) ?b`nk^,'**
= 1.0 − 15.3
La^Zk\^gmk^_hk lmZ[bebmrpZeel
= − 14.3 N/mm2
3.0 m
200 mm thick wall
2.5 m
Assume that the tension is resisted by 1 m at the end of the wall. As =
0.5 ft Lt t fyd
= 0.5 x 14.3 x 1000 x 200 = 3287 mm2 or 1644 mm2/face 435 Use T20s @ 175 ctrs Comment
(1800 mm2/face)
This design assumes that the base is sufficient to resist overturning. When designing the base for overturning the EQU category should be used rather than STR as used here.
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Ch[gh'
OB
Worked example 10 Shear walls with varying stiffness
CCIP - 018 La^^mgh'
JB
SW2 =Zm^
TCC
Dec 06
60.0 y
Floor-to-floor height = 3.6 m
5.0 W4 W2
Shear centre
8.0 25.0
W1 10.0
W3 5.0
e Wind force, take as 1.0 kN/m2
The relative stiffness of the walls can be calculated as follows (note as the walls are rectangular and the same thickness we can use the depth only) W1 W2
= 103 = 83 Total
= 1000 m3 = 512 m3 = 1512 m3
y = 0.1 x 1000 + 59.9 x 512 = 20.3 m 1512 Eccentricity, e = 30 – 20.3 = 9.7 m Twisting moment, Mt = 9.7 x 1.0 x 60 x 3.6 = 2095 kNm per floor In the worst case this moment can be resisted by the walls W3 and W4 only, the force (F) in each wall is: F = 2095/24.8 = 84.5 kN per floor To check for the critical case the direct design force on the walls W3 and W4 from the wind parallel to these walls = 1.0 x 3.6 x 25/2 = 45 kN, and therefore the forces imposed by the twisting action are more onerous and should be used for design. Comments
This layout is not particularly eccentric and yet still imposes large torsional forces.
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Ch[gh'
OB
Worked example 11 Bored pile in clay
CCIP - 018 La^^mgh'
JB
BP1
=Zm^
TCC
0m
c u = 70 kN/m
2
Dec 06
Find safe working load of a 450 mm diameter bored pile in clay Qf = ac As + cu Nc Ab gs gb Qf = 0.45 (70 + 170)/2 p x 0.45 x 10 + 170 x 9 x p x 0.2252 3.0 2.5 = 254 + 97.3
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