APPENDIX C Slab Calculation 2

Transcription

1 APPENDIX C Slab Calculation 2 Now try calculating the same slab, with the minimum recommended balanced load of 50% of self-weight. (As opposed to 50% of DL LL used in the first calculation). The same thickness of slab will be used, with the same layout Given:-- 4 bays each way 8000 x 7000 Ult. Load factors 1.0 kpa finish 1.8 kpa partitions 2.5 kpa live load Slab will be protected from weather Assume Internal columns 500 x 500 External columns 300 x 500 Wind loads taken bv shear walls CALCULATION Assume a 210 thick slab. Floor to floor 3.0m DL 1.2 and 1.5 as SABS LL mm Dia cables, low relaxation Breaking Strength KN Area 140 sq. mm. E= 198 GPa Relaxation 1.5% at 1000 hrs Balance 50% of Permanent Load 8.05kPa /2for internal spans For external 8M spans, balance same proportion of load as previously, i.e 80/66x 0.5 =.606 Permanent load From the formula, the slab should be somewhat thicker than 230mm, but take 210 as minimum practicable. COMMENT See diagrams in first part of calculation Formula would give 230 slab then self DL Total perm = Service (1.1) Live load TOTAL 5.25 kpa Factored revised DL 12 factor to allow for extra shear in first int. Check for shear as before, 210 is a bit thin, but acceptable with capitals As contractor has column shutters with capitals, use capitals: For analysis assume E of slab = code value: i.e. ( x 30) = 26.0 GPa From the cross-sections, take the effective top cover as 47mm to C/L cable for 7m span, and ( )/2 = 47mm for 8 m span. As bottom cover is 41 for 8 m span, effective drape mm = 122 mm. For 7m span, drape is =112 mm. (See Diag. Calculation 1) From the diagrams of the cable shapes in the end spans (the cables in centre spans will have the same positions as at the ends of outer spans), the properties of the parabolas may be deduced, and then the losses due to curvature. Referring to the formula in Appendix A, the following table may be calculated, as before. Span 8.0 end 8.0 cent 7.0 end 7.0 cent a l a , b b L m n X cl c Drape CALCULATION Loads to be Balanced: For 8m span, balance x permanent load in external span: For normal external bay 7 x 8.05 x.606 = kn/m For 8 m internal span, balance 0.5 of permanent load i.e. 7 x 8.05 x 5 = kn/m For 7m span, balance 0.50x permanent load in external & internal span. For internal and external bay, 8 x 8.05 x.5 = 32.20kN COMMENT See page 1 For first internal bay, increase cables to allow for increased load at first internal support

2 SAMPLE CALCULATION 2 Page 2 CALCULATION COMMENT internal spans H>I 2 would have ~rom the loads to be balanced, the required prestress force = This is the final force after smaller prestress 8xDrape force for same losses, and assuming a loss of say 18% for 8m spans, and 16% for 7m spans, the initial prestress can be calculated, load balanced. The angle which the cable 'rotates' through is also required for the friction calculation. They are calculated from the because of formula for a parabola. larger drape. e.g. for the 8m end span O = Arctan (2 x 78.4/3268.4) x 2 + Arctan( 2 x 113.5/3931.6) x 2 =.2113 Radians The following table shows the calculation results. Span Load Drape Eff. Length Angle to C/L Angle C/L to Total angle (rad) end (Cumul) 8 mend span m 2nd span m 3rd span m last span as 1st.9073 rad 7m 1st span m 2nd span m 3rd span m 4th span as 1st span.9452 rad Losses: 8 m spans The total loss due to friction and wobble is factor = e * * =.8742 i.e % losses 7m spans The total loss due to friction and wobble is i.e % losses. factor = e-' 0025 x x a9452 =.8810 CALCULATION 8 m spans: end span The prestress required to balance 59 kn/m with a drape of.0951, and an effective length of parabola of x m is given by = 2344 kn 8 x.0944 allowing an estimated 18% losses, P, = 2859 kn If cables of 15.2mm dia (ult. strength 260 kn) are stressed to 80% of ult. the no. of cables required x 260 = 13.7 say 14 cables COMMENT Although one should calculate the losses from the exponential equation, it is quite acceptable to calculate the loss at the end, and interpolate. An even number is better if stressed from both ends. 8m spans: 2nd span The prestress required to balance kn with drape of.1122 = x = 1627 kn 8 x.1122 Allowing the same losses, P init( = 1984 kn. and no of cables =, = 9.5 cables say 10. Assume that the cables in the end span are taken to 1.5 m past the 1st support, and that half the cables are stressed from each end, the diagrams on the next page may be drawn. Although strictly one should calculate the losses from the exponential equation it is quite acceptable to use a linear method. An even number is better if stressed from both ends.

3 SAMPLE CALCULATION 2 Page 3.33 kn

4 SAMPLE CALCULATION 2 Page 4 7 m Spans Effects of friction Stressing forces in cables before other losses

5 SAMPLE CALCULATION 2 Page 5 CALCULATION For 7m external span, force to balance kn/m, with drape of.0853, and effective length of 6.3m.=1873 kn With 16% losses, no of cables = 10.8 say 12 For internal span, force to balance34.15 kn/m with drape of.1008, P fma] = 1585 kn -> 1887 kn before losses No. of cables = 9.0 say 10 Take losses due to friction and pull-in as before (see first calculation) The results are diagrammed on pages 5 and 6 COMMENT Diagrams such as shown make it easy to know what the force at any section is. Other Losses: Elastic Shortening 8m spans Average stress in end 9.5 m =( )7(7 x.21 x 2)= 1.80 MPa In Centre part (( )/2 x3.52)+ (1949)x2.98)/6.5( 7 x.21) = 1.32 MPa Losses in end span =1.80 x 1987(26 x 2) = 6.85 MPa Average loss over whole span = (1.80 x 9.5 x x 13)/(2 x 32m)x = 6.11 MPa Loss per cable = 140 x 10" 6 x 6.85 x 1000 kn =.96kN in end spans (x 4 = 3.8kN) = " x6.11 x" = 0.86 kn in centre spans (x 10=8.6kN) Total in end spans 12.4 kn Because the cables are stressed one at a time, the average loss is 0.5 times the Ratios of E moduli x stress 7m spans Average stress whole length =(( )x 8.5+( )/2 x x 1.63) /(14x8x.21)= 1.19 MPa In end average force =(( )/2= >1.35MPa Losses 10 cables per cable = 1.19x 1987(26 x 2) x 140 x 10-6 x 1000 =.63 kn Loss in 2 cables = 1.35x 140x 10' 6 x /2=0.72 kn 10 x x.72 = 7.8kN total Total in centre span = 6.3 kn The resultant forces after friction and elastic losses are shown in the diagram below.

6 SAMPLE CALCULATION 2 Page 6 Long term losses Relaxation 1.5% x 2 = 3% For 8m spans: CALCULATION end span 3% x ( )/2 kn = 79 kn centre 3% x( ( )/2x3.52 +(1940 x 2.98/6.5))=1934kN x 3%=58kN For 7m spans: end span 3% x 2252 = 67.6kN centre 3% x 1950 = 58.5 kn Creep. (Same creep factor as 1st calc) If stressed at 3 days (common for prestressed flat slabs) and with a humidity ol 45%, the creep for a 210 slab, interpolating between 150 and 300 thick slabs given in the code (BS8110) gives a creep factor of 3.7 Then loss = stress in concrete x ratio of moduli x 3.7 8m spans: Total creep loss: 7m spans End Section conc, stress = (2.578 MN )/(2*7*.21) = 1.79MPa Loss of steel stress = 1.79 * 3.7 * 198/26 = 50.5MPa Per cable 50.5 x.140(area) = 7.06 kn Centre Section( ( )x x 2.98)/6.5/(7 x.21) x 198/26 x 3.7 = 37.1 MPa Per cable 37. 1x.14 = 5.19 kn Therefore average loss in long cables =(5.19x x9.5)/16m = 6.30 kn End 7.06 x x 10 = 91.2 kn Centre 6.3x10 = 63kN End Section Conc stress after elastic losses( )/(2x 8 x.21)= 1.345MPa Centre (( )/2 x x 1.63) /(4.5 x 8 x 2.1)= MPa Average(1.154 x x 8.5)/14 = 1.27MPa Loss of steel stress : end x 3.7 x 198/26= 37.9MPa Per cable 37.9 x.14 = kn Centre loss of steel stress 1.27 x 3.7 x 198/26 = 35.78MPa loss/cable = 35.7 x.14 = 5.0kN Losses in centre 10 x 5.0 = 50kN Losses in end section 12 x = 63.7 kn 7m spans Relaxation end span 2 cables 3% x ( V2 = 11.43kN Total 10 cables (1956x x 2.63)/14--> kn x 3% = 57.4 kn Total for 12 cables = 68.8 kn Shrinkage as before: for 1 cable kn COMMENT The code gives a factor of 2.0 on the 1000 hour value for class 1 BS5896 Varies with initial prestress, temp etc. Should strictly be taken as stress at the centroid of the cables. 8m spans 14 cables >143.6 kn 10 cables > kn 7m spans 12 cables > kn 10 cables >102.6 kn Total loss Cause No of cables 8m End span 14 8m Centre 10 7m End span 12 7m Center 10 Initial force (208kN/cable) 2912 kn Elastic 12.4 kn Relaxation Shrinkage 79 kn Creep TOTAL Percentage of init. force loss after friction loss %Loss due to friction Total % loss of initial The losses are somewhat lower than with the higher prestress but still appreciably higher than the 16% (or less) assumed by some commercial designers. From the figures for long-term loss, the final prestress can be calculated

7 SAMPLE CALCULATION 2 Page 7 Initial prestress (after elastic losses) Loss Final prestress 8m end span 2578 kn kn 8m 2nd span m end span m 2nd span ANALYSIS The slabs are analysed by the equivalent frame method, using Long's method to calculate the equivalent stiffness of the columns. Three loading cases are needed: Dead load on all spans, Line load on even spans, and live load on odd spans. These may be combined together with the load factors to give the desired bending moments diagrams. Any method of analysis may be used: moment distribution if a computer is not available, or a frame analysis if one is available. The hogging moments for the ultimate limit state may then be reduced by 15%, and the sagging moments increased accordingly to maintain equilibrium. The design moment is at the face of the column or capital, but the the total statically required moment is: W(L-2D/3) /8 If a computer program is used, there is an advantage in arranging a node at 1/3 of the column or capital dimension from the centreline of the column, as the moment at that point will not be less than the moment given for statically required moment. Loads (As for previous calculation) LOAD Finishes Partitions Self. Wt Total DL Live load Service (1.1DL + LL) 1.l0kPa Ultimate (1.2DL + 1.6LL) 1.20kPa Ultimate Dead Load (1.5 DL) l.skpa 2.7 kpa 7.87kPa 12.07kPa Total loads on spans Service Ultimate Self only 8m Spans DL 61.95kn/m 67.62kN/m m spans LL m spans DL m spans LL NOTE: As the Ultimate Deadload case of 1.5 DL at kpa is less than the Ultimate DL + LL of kpa, it may be effectively ignored, although there may be some places where the moments could be fractionally higher. A construction loading at initial prestress may need to be calculated if the propping is not adequately arranged. Loads due to prestress For a parabola, the equivalent uniform load caused by a tension in the cable is given by w L 2 /8 = P h where h is the drape for the cables. The drapes can be read from the table on page 1 The loads will not be uniform, but will be trapezoidal if the variation of prestress along the beam is taken into account In addition the moments due to eccentricity at the ends must be taken into account in the analysis if the cables are not exactly central. In our case we have assumed a 20mm eccentricity upwards at the end' The moment at the end of the 8m spans will be 2267 x.02 =56.7 knm The moment at the end of the 7m spans will be 1959 x.02 = 49.0kNm The loads given are for final prestress. Initial prestress (after friction and elastic losses) combined with deadload only may be a critical state, but it is probably sufficient to take the stresses due to final prestress and multiply them by an average factor.

8 SAMPLE CALCULATION 2 Page 8 Note: The forces at intermediate points are interpolated 8m 1st. span L end 8m 1st start of sag 8m 1st. end of sag 8m 1st span RH end 8m 2nd span L end end of cable beyond end 8m 2nd RH end 7m 1st span LH 7m 1st: start of sag 7m 1st end of sag 7m 1st RH 7m 2nd LH end 7m 2nd end of cable 7m 2nd beyond end cable 7m 2nd Rh end Prestress Drape Length Lateral load kn/m down 33.1kN/m up kn/m down down up 29.1 up kn/m down down up up down down up up kn/m down 272 kn/m Loads on 8m spans due to final prestress kn/m Loads on 7m spans due to final prestress Columns: Stiffness by Long's method (See Sample Calculation 1) 7m spans Exterior: E (30 MPa)=26GPa f e =.5 x.3 3 /12 = x 10-3 m units K c = 4EI/L C K c = K c /( K C L/Eh J c) K e = K c /(I (4I c L/L c h 3 c)

9 where L=8m, L C =3, h=.21. and c=.3 Then equivalent stiffness = stiffness x.645 orequiv. I = 725 x 10" 6 8m spans interior : I c =.5 4 /12 = x 10-3, k =.0564 c= 0.5 Then equiv. stiff = stiff x.596 Equiv I = 3.10 x m exterior I c =.3 x -5 3 /12 = :-.125 x 10-3 c=0.5 k=.1272 Then equiv stiff = stiff x.555 Equiv I = 1.73 x m interior I = 5.20?, x c=.5. k=.0564 Then equiv. stiff = stiff x.628 Equiv I = 3.27 x 1-3 SAMPLE CALCULATION 2 Page 9 Slab Stiffness The moment of Inertia of the 8m spans is 7 x.21 3 /12 =.0054 m The moment of inertia of the 7m spans is 8 x.21 3 /12 = m The E is taken as the same as the columns, and the creep factor (for calculation deflections due to long term loads) as 3.5 The results of the analysis are given below: Because the structure is symmetrical, only the first two spans are shown. It should be realised that as the loads are calculated for an interior span, the first interior column band should be reinforced for approximately a 7% greater load. i.e. for this design, 1 extra prestressing cable, and some additional reinforcement over the columns. MOMENTS 8m 1st span Left ULT (1.2DL +1.6L1) knm ULT (adjusted 15%) Service (1.1DL +LL) DL self only Final Prestress m 1st span centre m 1st span right m 2nd span left m 2nd span centre m 2nd span right m 1st span left m 1 st span centre m 1st span right m 2nd span left m 2nd span centre m 2nd span right m 1st span 8m 2nd span 7m 1st span 7m 2nd span DEFLECTIONS (allowing 3.5 creep factor on permanent loads) Assuming an uncracked section (see later for correction) Service Load (1.1 DL +LL) 28.7 mm 9.0 mm 14.9 mm 5.0 mm Serviceability Limit State Final Prestress mm mm mm mm Net Deflection 18.0 mm 6.3 mm 10.3 mm 3.0 mm For controlling cracking, it has been traditional to limit tensile stresses. For the sake of completeness, this will be done, but the incremental stress method is considered better.. (Use the program described in Appendix E) The permissible tensile stress is given as 0.45Ö f c or 2.46 MPa 4.O4 Mfti \^ \ T = 4.O4x75.2/2 \ =151.9 kn ^ / / \ 752. mm t» h = 21O mm 4.P4 \ / 1 x 210 mm/ V^ i / / = 75.2 mm 7.24 MRi Elastic interacted etraeeee- equivalent tension \

10 SAMPLE CALCULATION 2 Page 10 Serviceability Tensile stress 8m 1st L 8m 1st C Net moment (service-prestress) Comp. kn Stress (moment) ± Stress (comp) Net tensile stress (MPa) 2.36 Z of section (M units).0514 Slab Area (M units) m 1st R 8m 2nd L * 3.23* 8m 2nd C 8m 2nd R 7m 1st L 7m 1st C m 1st R * 7m 2nd L m 2nd C m 2nd R It may be seen that 3 of the stresses are more than the Report 25 stresses, and reinforcement is required. The tension to be taken by reinforcement is calculated by simply taking the force from the tensile stress diagram as in the sketch at a stress of 0.58 f v It should be noted that it is assumed that the moments are evenly distributed across the section. This is clearly incorrect kN /(.58 x 450) ->582sq.mm/m 180 To calculate the steel required to control crackwidth by a more logical method, the formula in BS8007 is used. Moments are taken about the level of the tension steel reinforcement, i.e. The required area of steel is given by calculating the tension required for a moment of M + P(d-h/2), and then reducing the tension by P. i.e. transferring the compressive force to the reinforcement level.. 75% of the total hogging moment and 55% of the total sagging moment should be taken in the column band for the purpose of calculating crackwidths The following table was calculated using a computer program for crackwidths to BS8007 It may be seen that areas are similar to the areas required for ultimate load, which supports the Report 25 stresses. SABS 0100 seems to give slightly higher areas if the tensile stress in the concrete is taken into account. Hogging moments are taken as 75% of total in column bands, and 55% for sagging moments The result is slightly more reinforcement in the position which requires more than nominal reinforcement Required area of reinforcement for cracking (0.15% = 315 sq.mm/m) Net. Moment (service-prestress) Moment /m in band Prestress MPa Steel stress MPa Bar Dia. d Area (sq.mm/m) 8m 1st L 8m 1st C 8m 1st R 8m 2nd L 8m 2nd C 8m 2nd R 7m 1st L 7m 1st C 7m 1st R 7m 2nd L 7m 2nd C 7m 2nd R 46.4 knm ! @ @ 120 nom 10@250 12@180 nom nil nil nil inal(297) nil inal 625 inal inal (Calculated using SABS0100 with 0.2mm crackwidth. See Appendix E) The areas are slightly greater than for the Report 25 Calculation (e.g Y16 at 200 or 1?0 as compared with Y12 at 180) Ultimate Load Limit State. 15% reduction of hogging moments ( with corresponding increase of sagging moments) is allowed and has been taken. The decompression moment, equal to the prestress x Z, is the moment required to reduce the compression on the extreme fibre to zero. If the applied moment (Ultimate dead + Live + prestress) is less than the decompression moment, no reinf. is required. If it is greater, an equivalent moment M' is calculated, by displacing the prestress to the level of the reinforcement, and adding a moment of P(d-h/2). The tension calculated from this moment then has the prestress force deducted, to obtain the net tension on the reinforcement M7Jd-P. The required steel area, is then (M;/Jd-P)/f s The ultimate M.R. of concrete = 4651 bd 2 = 126 knm for the 7m spans and 154 knm for the 8m spans. None of the design moments

11 SAMPLE CALCULATION 2 Page 11

12 SAMPLE CALCULATION 2 Page 12 exceeds this. The lever arm as a proportion of effective depth is given approximately by d-h/2 = 65mm for the 8m spans, and 49mm for the 7m spans. The approximate moment /metre taken by inal reinforcement is 391 MPa x 0.75 d x.15% x.21m= 8m 1st L 8m 1st C 8m 1st R 8m 2nd L 8m 2nd C 8m 2nd R 7m 1st L 7m 1st C 7m 1st R 7m 2nd L 7m 2nd C 7m 2nd R Net. Moment (Ult-prestress) Required Area of reinforcement for Ultimate Limit State Net Moment /m in band (A) Mom. /m P(d-h/2) (B) ' DecomMom/m Prestress x Z Design Moment/m (A+B) Net Tension kn/m knmford=.182 mm 15.3kNm/m for d=. 166 Bar Dia & spacing Y10@250 Y10@250 Y10@200 Y10@250 Area (sq.mm/m) This method does not take account of the moments due to prestress, but effectively assumes the cables are bonded, and that the moments are taken over the full width of the section. At the first interior support, of the 8m spans f, = 2.358lMN/( 14 x 140. x 10-6 ) = MPa (Effective pe prestress) f pu = MN/140. x 10-6 = MPa -. Aps = 14x 140 x10-6 =l.96x 10-6 Effective depth of cables =.163m (see page 3 of calcs) From equation 52 of BS8110 for l/d =16/0.163 =98, f pb = = 1264 MPa For end 8m spans (SABS 0100) 1=32/2= f = 1302 MPa therefore use 1264MPa For internal spans, 1= 32/3 = For 7m end spans, 1=28/2=14, and for 7m internal spans, 1=28/3=9.33 (for 8m int 1295MPa, for 7m end For prestress only, T ult = MN for 7m internal f b = 1302MPa) Then N.A. depth = 2.478/(7 x.45 x 30) = 26.2 mm and lever arm = /2 =.150 m M.R. Prestress =.150x 2.478= knm Actual ultimate moment 441.7kNm. Therefore additional steel is required. d reinf =.182and d cable =.163 Say approximately Tension reinff = ( ) x 2478kN = 469 At 391 MPa (450/1.15), the area required = 1392 sq.mm over 3.5 m width = 343 sq. mm/m 220 ) Try 200 : Area = 1374 sq.mm Tensile capacity =537 kn See diagram, Tension = kn = 3015kN. N.A depth = 3015/C x 30)=31.9mm MR =537x x.147 = 453kNm O.K. The other positions are checked in the table below 0.15% 250 8m 1st L 8m 1st C 8m 1st R 8m 2nd C Ult. M f pb x!40. x 10-6 x No. cables MN MN Eff.d N.A. depth 27mm The reinforcement areas required for ultimate load by this method are a bit smaller than those required for crack control Compare the above calculation with the Report 25 method of calculating the reinforcement required for ultimate moments:- 19.3mm Lever arm M.R DM As reqd. 0 (.15% min) inal

13 SAMPLE CALCULATION 2 Pase 13 Ult. M f pb x!40. x 10-6 x No. cables Eff.d N.A. depth Lever arm M.R. DM As reqd. 8m 2nd R 7m 1st L MN MN mm m 1st C m 1st R 7m 2nd C 7m 2nd R MN 16.8mm Y10@250 Y10@250 The reinforcement required for ultimate moments by this method are slightly less than required for cracking, and slightly less than the reinforcement required by the suggested method.. Ultimate MR of section = 537x x.147= 453 knm It is also desirable to check the ultimate load case with half of the cables in the end span removed, with DL +.25 LL. in the end spans only. The MR due to half the number of cables may be taken as half the M.R. in the table above if Report 25 method is used, or the moments calculated for prestress may be halved and deducted from the moments for DL +.25LL The second method will be used Ult. M (DL +.25LL) PSMom/2 Moment/m PS (d-h/2)/2 Ult Mom/m Net. M/m for tension Required Area (per m col band) 8m 1st L 8m 1st C 8m 1st R knm sq mm 622. sq mm Norn Y12 at 180 7m 1st L 7m 1st C 7m 1st R sq mm It may be seen that the reinforcement for the 1.2DL + 1.6LL ultimate load case is adequate everywhere except the first interior support of the 8m spans Deflections: The deflections in the table above were calculated for a non-cracked slab If the sections are cracked, the Moment of Inertia would be 2.56 x 10-3 ' instead of the uncracked value of 5.4 x 10-3 Using the ACI formula, and taking moment at first cracking as the moment at which the tensile stress reaches 2MPa, the cracking moment in the 8m 1st span is ( ) x 7 x.21 2 /6 = 182 knm. Taking the design moments as the serviceability moments, and taking the moments due to prestress into account, the moment at midspan of the 8 m 1st span is knm. i.e. cracked, and at the first internal support 290 knm Then I e midspan = 4.13 xl0-3 I Support = 3.26 x 10-3 using the service load moments calculated above The actual moment over the support is more concentrated, and there may be more cracking. The net equivalent M. of I. is then.85 x x 3.26 =4.0 The calculated deflection will then be 5.4/4.0 x the calculated one. Now the calculated deflection assumes that the slab acts as a band spanning in one direction, and neglects the span in the other direction. If one adds half the calculated deflection in the short direction to the calculated deflection in the long direction, this will probably be a reasonable estimate.

14 SAMPLE CALCULATION 2 Page 14 The total deflection is then ( ) x 5.4/4.0 = 31mm This is 1/223 of the short span, and may be acceptable if there are no rigid partitions. It should be noted that if there are reasons the slab might be more cracked, e.g. temperature stresses or shrinkage, the deflections could be considerably greater. Shear: The calculated ultimate shear at the first interior support ( at the face of the supports) is (from the computer calculations) kn (as compared to the simply supported load of kn). The shear assumed in the preliminary calculations was 860 kn The design shear is 1.15 times this, or 968. kn Alternatively, from the code, V eff = V t ( M/V t x) M t, the moment transmitted to the column, is knm, and x is x 1.5 (The column capital +1.5 slab depths) then V eff = x = kn From this may be deducted the vertical component of the prestressing cables At 1.06 m from the c/1 of column, the slope of the cables is.0529 from the properties of the parabola in the 8m span vertical component = 4/14x 2267 x.0529 =34 kn x 2 = 68 kn In the 7m span the slope is.0575 Component = 5/12 x.0525 x 2056 kn =44.9 kn x 2 =89.8 kn Total vert. component on 4 sides =157.8 kn and net design shear = =875 kn Area of reinforcement (Y10@200) in one direction, and 180 in the other. Average is 510 sq mm/m Average % = 100(1860 x 4,5 x 140.x 10" x1.71 x 10-6 x 450)/(1.71 x.21x 450) = 0.97% Permissible shear stress =.0.70 MPa Actual shear stress = 851/(4 x 2.04 x.167) =.63 MPa. Therefore no shear reinf. reqd At other columns, shear is less, and moment transferred to column is smaller. No shear reinforcement required. One should also calculate that the width of slab at the external columns is adequate to transfer the moment, but with Long's method, the moment transferred is quite small, and with the column capitals, it is not necessary.