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# 9 design-of-flat-slabs

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### 9 design-of-flat-slabs

2. 2. 404 Reinforced Concrete HIT T DROP AND HITH DROP AND HITHOUT DROP AND N0H8&UHN HEAD N0 COLUMN PEAD HITH COLUMN HEAD SK 9/1 Flat slab — section. SK 9/2 Flat slab — section. SK 9/3 Flat slab — section. Column head is a local enlargement of the column at the junction with the slab. 9.2 ANALYSIS OF FLAT SLABS SK 9/4 Typical plan of ﬁat slab. Flat slabs are usually supported by a rectangular arrangement of columns. The analysis may be carried out by an equivalent frame method or by the use of a ﬁnite element computer code. When using the equivalent frame method the ratio of the longer to the shorter span should not exceed 2. The analysis for uniformly distributed vertical load may be carried out by using Tables 9.1 to 9.6. The properties of the ﬂat slab for analysis are similar to those already discussed for solid slabs in Chapter 3. 9.2.1 Effective dimension of column head 1.. = effective dimension of head 1.. ., = actual dimension of head llnrnax = [c + 2(dh '" 40) 1., = dimension of column in direction of 1.. d. . = depth of head 1.. = is taken as the lesser of 1.. ., or I. ._, ,., ,.
4. 4. 406 Reinforced Concrete FULL PAIEL WIDTH FDR VERTKAL LOADING BEl’ltEENtC UHN aerussuté um l FULL PANEIL mom 4 ETHEEN COLS HALF PAl£L HIDT . FOR . ..m. ZoNTA. _ FR SK 9/6 ‘Plan of flat slab showing ANALYSIS panel widths for analysis. The analysis should be carried out using a computer program or a moment distribution method. The analysis may also be carried out for uniformly distributed vertical loads using Tables 9.1 to 9.6. The analysis may be carried out using Table 3.13 of BS8110: Part 1: 1985"] provided the lateral stability is not dependent on the slab—column connection and loading on the ﬂat slab for the design is based on a single load case, i. e. L(. '., the ratio of Q. /G. does not exceed 1.25. Q. does not exceed 5kN/ m2. and there are at least three rows of panels. 9.3 DESIGN OF FLAT SLABS The design may be based on the negative moment at he/2 from the ccntrcline of the column. But this negative moment will have to be modiﬁed if the sum of the positive design moment and the average negative design moment is less than the following expression: Mt= <e><»e>* 8 3 L I SK 9/7 Negative moment limitation for ﬂat slabs — section.
5. 5. SK 9/8 Typical plan of ﬂat slab - negative moment limitation. Design of Flat Slabs 407 " , ,~ m u jwslég SPAN A SPAN B. [¢. SPAN 1- 93-- where I. = centre-to-centre of column in direction of span being considered 12 = centre-to-centre of column perpendicular to direction of span being considered n = total ultimate load on slab (kN/ m2). To give an example: For Span A 0.5 (M, + M2) + M-, 2 M’ For Span B + M4) + M7 2 M’ Increase negative moments Ml, M2, M3, etc. until these conditions are satisﬁed. 9.3.1 Division of panels SK 9/9 Flat slab — division of strips. COLUMN HlDOLE COLUMN STRIP STRIP STRIP
6. 6. 408 Reinforced Concrete Lx l H T P PLAN OF SLAB WITH DROP IGNORE DROP IF DROP HIDTH < Lx/3 g? _J_ , H I, ” ’ 1-2-- I I >Lx/3 I I COLU II A S RI S IP MIDDLE STRIP CDEU N COLUMN TRIP MIDDLE STRIP OLUHN STRIP SK 9/I0 Flat slab — division of strips. Panels are divided into column strips and middle strips as shown. For slab without drop the column strip is 1,, /4 wide on either side of the centreline of column, where 1, is the shorter span. For slab with drop the column strip is the size of the drop. Ignore drop if the size of the drop is less than 1,, /3. 9.3.2 Division of moments between columns and middle strips Note: The moments obtained from analysis of frames should be divided as follows (these percentages are for slabs without drops): Column strip Middle strip Negative 75% 25% Positive 55% 45% Where column drops are used and column strips are determined from the width of the drop, it may so happen that the middle strip is bigger than the middle strip in a slab without drop. In that case the moments in the middle strip will be proportionately increased and those in the column strip decreased to keep the total positive and negative moment unchanged. 9.3.3 Design of ﬂat slab panels The design is similar to the design of slabs and the worked examples are in Chapter 3.
7. 7. Design of Flat Slabs 409 COLUMN STRIP HALF. COLUMN . I STRIP 1/5A" Us A" SK 9/ll Detailing of reinforcement in flat slabs. Internal panels and edge panels Two-thirds of the negative support reinforcement in the column strip should be placed in half the width of the column strip centred over the column. 9.3.4 Moment connection to edge column I I2. = c, ,. I l'\$ COLUMN STRlP'I SK 9/12 Effective width of slab for moment connection to edge column. See sketches above to ﬁnd effective width of slab be for transfer of moment between ﬁat slab and edge column. This moment should be limited to
8. 8. 410 Reinforced Concrete MW, = 0.15b, d2fw where d = effective depth of top reinforcement in column strip. The moment M. _,, ,,, should not be less than half the design moment from an equivalent frame analysis or 70% of the design moment from a grillage or ﬁnite element analysis. The structural arrangement may be changed if M. _,, .,, , does not satisfy the above condition. g EDGE COLUMN INTERNAL COLUMN ® REa(l)5T'l: IBlTJ(T)ED ACCEEN1 pop SK 9/I3 lnsufﬁcient moment "t, max. transfer capacity at edge column. Where the design moment is larger than Mmm, redistribution of moment may be carried out to reduce the design moment to M. .,. .,, ,.. Otherwise. to transfer moments in excess of M. .,. .,, ‘ to edge column. the edge of the slab should be reinforced by an edge beam or an edge strip. The edge beam will be designed to carry the additional moment by torsion to the column. 9.3.5 Shear in ﬂat slabs Punching shear around columns should be checked according to Step 7 of Section 3.3. The shear to be considered for the punching shear calculation is increased from the calculated column shear by an amount dependent on the moment transferred to the column by frame action. For internal column connections. 1.5M. ) V. x V, “ = I/ ‘(I +
9. 9. Design of Flat Slabs 411 I — PUNCHING SHEAR PERIMETER SK 9/14 Moment diagram at an intemal column SK 9/15 Deﬁnition of dimension 2:. of a ﬂat slab. where V, = calculated shear from analysis M. = moment transferred to column by frame analysis x = length of side of perimeter considered parallel to axis of bending. Altematively, V, ;“ = 1.15 V. for simplicity For comer column connections, Veg = 1.25V. For edge column connections, Veg; = l.25V. for bending about axis parallel to free edge 1.5M Va“ = V. (l.25 + for bending about axis perpendicular to free tx edge Altematively, V, " = 1.41/I The moment M. may be reduced by 30% where the equivalent frame analysis is used and both load cases LC. and LC; have been considered. The shear reinforcement will be calculated according to Step 7 of Section 3.3.
10. 10. 412 Reinforced Concrete 9.4 STEP-BY-STEP DESIGN PROCEDURE FOR FLAT SLABS Step I Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 Step 9 Step 10 Step II Step 12 Step 13 Step 14 Step I 5 Step 16 Step 17 Carry out analysis as in Section 9.2. Find moment connection to edge column as per Section 9.3.4 and redis- tribute moments if necessary. Draw bending moment diagrams and calculate moments at /2,/2 following Section 9.3. Check limitation of negative design moments following Section 9.3. Carry out division of panels as in Section 9.3.1. Divide moments between column strips and middle strips as per Section 9.3.2. Determine cover to reinforcement (see Step 3 of Section 3.3). Carry out design for ﬁexure as per Step 4 of Section 3.3. Distribute reinforcement as per Section 9.3.3. Check punching shear stress Follow Step 7 of Section 3.3. Check span/ effective depth ratio Follow Step 11 of Section 3.3 for slabs with drops. For slabs without drops follow the same step but multiply lcld from Table 11.3 by 0.9. Curtailment of bars Follow Step 12 of Section 3.3. Spacing of bars Follow Step 13 of Section 3.3. Check early thennal cracking Follow Step 14 of Section 3.3. Calculate minimum reinforcement Follow Step 9 and Step 15 of Section 3.3. Calculate ﬂexural crack width Follow Step 16 of Section 3.3. Design of connections Follow Chapter 11.
11. 11. Design of Flat Slabs 413 9.5 WORKED EXAMPLE Example 9.] Flat slab construction for a sports hall :30 I 800 deep #00 at #00 columns 9: bean t_. J L. . J . 1 r1 F1 r l"‘| " t. J n_. : I_. l t. J . 290 thick slab - r ‘I r- 1 I’ '1 r" L J 1. .1 L .1 V | - . _ . lZ3_____. .'I3_____ 3-__, _: 4---. .- Staircase and Lift Block SK 9/16 Plan on ﬁrsl ﬂoor. k00 x 600 deep > 1.00 I 500 columns adg beam +*+; +s+s +i°*%+s SK 9/17 Plan on roof.
12. 12. 414 Reinforced Concrete SK 9/18 Section through building. Two-storey building plan size: 15 m X 30m Column grid: 5 m x 6m Column size: 400 X 400 Height of building = 10m overall Topography factor, S. = 1.0 Ground roughness factor, S2 = 0.95 Statistical factor for wind, S3 = 1.0 Basic wind speed = 42 m/ s = V Design wind speed = S. S2S3 V: 40m/ s q= kV§=1kN/ m2 C, ,,_. = +0.7 and -0.3 CF. = -0.3 (four faces equally permeable) The above wind pressure coefﬁcients are obtained from CP3: Chapter V Wind loads. “"] Live load on roof= 1.5 kN/ m2 Live load on floor = 5kN/ m2 Floor slab has 2000 X 2000 drop at columns Thickness of roof and ﬂoor slab = 200 mm Thickness at drop of ﬂoor slab = 400mm Continuous perimeter edge beam 400 wide X 800 deep Centre-to-centre height of ﬂoor = 4.5 m Step I Carry out analysis Only one frame in the short direction of the building will be analysed. Column head has not been used. 4A ' Effective diameter of column, hc = 5 0.251,, A = 400 x 400 =16()000mm2 (4 x 1s0o00)% hc = ——j J! = 451mm < 0.251, I, = 5000mm 0.251, = 1250 mm
14. 14. 416 Reinforced Concrete O 0 ID ~r O 0 ID ~t SK 9/19 Frame diagram for analysis. 30|tNIm IllllIIIIIIIllllllllIlllllllllllllllllllllllIllllllllllllllllllllllllllllll 69 (D an .1 . . . -.I- . . | "“llI| IIl| IllIl| I" ""111 "1""mllllllllllllllllmn I 0 0 H LOkNIm 30kN/ m SK 9/20 Dead load on frame (B. ). lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll 30kNIrn ' 11 SK 9/21 Basic live loads B; to B3. Combinations: C] = 1.431 + '1' B3 + B4 + B5 '1' B6 *1‘ 37) C; = 1.43. + 1.6(B; + 3.. + B5 + 3,) C3 = 1.43. + 1.s(3, + B6) C‘ = 1.431 + 1.6(B; + 33 + 34)
15. 15. Design of Flat Slabs 417 _m__j Output from analysis Envelope of load cases (vertical loads) Elastic analysis — no redistribution Floor slab Member 9 Joint Maximum BM Shear Combination 2 130.80 228.5 C; 5 215.5 258.2 C4 Midspan 131.2 — C; Floor slab Member 10 Joint Maximum BM Shear Combination 5 199.9 239.0 C. 8 199.9 239.0 C. Midspan 112.5 — C3 Roof slab Member 12 Joint Maximum BM Shear Combination 3 67.8 130.0 C; 6 129.0 153.5 C. Midspan 82.0 — C; Roof slab Member 13 Joint Maximum BM Shear Combination 6 121.3 141.0 C, 9 121.3 141.0 C , Midspan 60.4 — C3 SK 9/22 Combination C. - bending moment diagram. Envelope of load cases (vertical + horizontal loads) (The analysis of hori- zontal load is carried out with half stiffness of slab) Elastic analysis — no redistribution Combinations: C5 = 1.481 + 1.43;; C6 = + + B3 ‘1' B4 ‘1' B5 + B6 '1' B7) 4'' 1.233
16. 16. 418 Reinforced Concrete Floor slab Joint 2 5 Midspan Floor slab Joint 5 8 Midspan Roof slab Joint 3 6 Midspan Roof slab Joint 6 9 Midspan Member 9 Maximum BM Shear 121.9 185.9 191.4 215.2 95.8 — Member 10 Maximum BM Shear 176.0 198.6 176.0 198.6 71.4 — Member 12 Maximum BM Shear 57.0 108 112.2 128.9 67.1 — Member 13 Maximum BM Shear 108.1 119.8 108.1 119.8 45.4 — Cany out redistribution of moment: Maximum bending moment at joint 5 = 215.5 kNm Assume 20% redistribution. Combination Combination C6 C6 Ca Set plastic moment capacity at joint 5 = 0.8 X 215.5 = 172.4 kNm Similarly maximum bending moment at joint 6 = 129 kNm Assume 20% redistribution. - SK 9/23 Shear and moment envelope for member 9.
17. 17. Design of Flat Slabs 419 SK 9/24 Shear and moment envelope for member 10. Set plastic moment capacity at joint 6 = 0.8 X 129 = 103.2 kNm The following steps of reanalysis of frame are carried out: Step I: For one combination at a time increase live load on span until plastic moment is reached at a joint in a member. Plastic moment capacity of members on first floor is 172.4 kNm and member on roof is 103.2 kNm. Step 2: Release joint where plastic moment is reached and increase loading until plastic moment capacity is reached at another joint. Step 3: Progressively release joints and increase live load until full comp- lement of live load is on structure. Step 4: Find cumulative effect of all incremental live load on structure. The following tables become useful if a non-linear ﬁnite element computer package is not available. Frame types: F. = no member end releases F; = member in F. nos 12 and 14 ends released at joints 6 and 9 F3 = member in F; nos 9 and 11 ends released at joints 5 and 8 F4 = member in F3 no. 13 ends released at joints 6 and 9 F5 = member in F. no. 10 ends released at joints 5 and 8 Plastic Hinges ‘ 9 12 SK 9/25 Frame type I«‘, _ SK 9/26 Frame type F; .
18. 18. 420 Reinforced Concrete Plaitic Hinges 10 SK 9/27 Frame type F. » SK 9/28 Frame type F. .. Plastic Hinges B5 6 B6 9 B7 12 Imm. .uuuu-mu-m-ummuumm 0~3|(NIm SK 9/29 Frame type F5. SK 9/30 Unit live load on frame. Combination C7 = 1.4 X dead load or 1.418. The method is illustrated for combinations C] and C; only. C} = l.6(B§ + B5 + BI, + B; + B}, + 13;) C5 = l.6(B§ + B], + B; + B5) B5, B5. B1. = 1kN/ m 3;. Ba. 3; = 0.3 kN/ m Combination Frame Member end bending moments (kNm) WP‘? ’ Member 9 Member 10 Member 12 Member 13 Joint 2 Joint 5 Joint 5 Joint 8 Joint 3 Joint 6 Joint 6 Joint 9 C7 F1 62.7 102.5 95.5 95.5 44.7 97.5 91.6 91.6 C; F. 2.0 3.7 3.4 3.4 0.7 1.1 1.0 1.0 Ci F2 2.] 3.6 3.5 3.5 0.9 0 0.5 0.5 1 F3 3.] 0 2.4 2.4 1.1 0 0.8 0.8 . ’| F4 3.] 0 2.2 2.2 l. l 0 0 0 C i F; 3.1 0 0 0 1.1 0 0 0 C5 F. 2.3 2.7 0.8 0.8 0.8 0.8 0.2 0.2 C5 F2 2.3 2.6 0.8 0.8 0.9 0 -0.2 —0.2 (Ii F: 3.1 0 0 0 1.0 0 0 0 Cg F4 3.1 0 0 0 1.0 0 0 0 C5 F5 3.1 0 0 0 1.0 0 0 0
19. 19. Design of Flat Slabs 421 T_ C; = 1 unit of live load combination in combination C. i. e. C; = 1kN/ m of B2, B3 and B4 and 9/30kN/ m of B5, B5 and B7. Full compliment of B2, B3 and B4 is 30kN/ m and of B5, B6 and B7 is 9kN/ m. Plastic moment at joint 5 is ﬁxed at 172 kNm Plastic moment at joint 6 is ﬁxed at 103.2 kNm Dead load moment at joint 6= 97.5 kNm Each unit of combination C. produces 1.1kNm at joint 6 for frame type 1']. Therefore units of live load in combination C 1 required to form first plastic hinges at joint 6 and joint 9 in members 12 and 14 _ 103.2 — 97.5 1-1 = 5 units, say Frame type F; has joints released at joints 6 and 9 for members 12 and 14. After 5 units of combination C. the bending moments at joints are as follows: Frame type F, Member 9 Joint 2 62.7 + 5 X 2.0 = 72.7kNm Joint 5 102.5 + 5 X 3.7 = 121.0kNm Member 10 Joint 5 95.5 + 5 X 3.4 = 112.5 kNm Member 12 Joint 3 44.7 + 5 X 0.7 = 48.2kNm Joint 6 97.5 + S X 1.1 = 103kNm *plastic Member 13 Joint 6 91.6 + 5 X 1.0 = 96.6kNm Units of live load in combination C. to form second plastic hinges at joints 5 and 8 in members 9 and 11 _ 172 -— 121 3.6 = 14 units of combination C 1 Total number of units of C . to cause plastic hinges at joints 5 and 8 in members 9 and 11 is 19. After 19 units of combination C, , the bending moments at joints are as follows: Frame type F2 Member 9 Joint 2 72.7 + 14 X 2.1 1l)2.1kNm Joint S 121 + 14 x 3.6 = 171.4kNm *plastic Member 10 Joint 5 112.5 + 14 X 3.5 161.5 kNm Member 12 Joint 3 48.2 + 14 X 0.9 60.8 kNm Joint 6 plastic = 103 kNm *p| astic Member 13 Joint 6 96.6 + 14 X 0.5 = 103.6kNm *plastic ll Joint 5 of member 9 and joint 6 of member 13 have gone plastic simul- taneously at 19 units of combination C, . Therefore frame type F3 is not considered.
20. 20. 422 Reinforced Concrete After 24 units of combination C1, the bending moments at joints are as follows: Frame type F, Member 9 Joint 2 102.1 + 5 X 3.1 = l17.6kNm Joint 5 plastic = 171.4 kNm *plastic Member 10 Joint 5 161.5 + 5 X 2.2 172.5 kNm ‘plastic Member 12 Joint 3 60.8 + 5 X 1.1 = 66.3kNm Joint 6 plastic = 103kNm *plastic Member 13 Joint 6 plastic = 103.6 kNm *plastic Frame type F 5 After 30 units of combination C . , the bending moments at joints are as follows: Member 9 Joint 2 117.6 + 6 X 3.1 = 136.2 kNm Joint 5 = 171.4 kNm Member 10 Joint 5 = 172.5 kNm Member 12 Joint 3 66.3 + 6 X 1.1 = 72.9kNm Joint 6 = 103.0 kNm Member 13 Joint 6 = 103.6 kNm Formula for calculating midspan bending moment and shear C7 + 5 units of C’. (F. ) + 14 units of C’, (F2) + 5 units of C; (F4) + 6 units Of Cl (F5) Member 9 — combination C , (20% redistribution) Midspan moment = 56.3 + 2.2 X 5 + 2.2 X 14 + 5 X 3.5 + 6 X 3.5 = 136.6kNm End shear, joint 2 = lll.1+ 3.7 X 5 + 3.7 X14 + 4.6 X 5 + 4.6 X 6 = 232.0kN End shear, joint 5 = 126.9 + 4.3 X 5 + 4.3 X 14 + 3.4 X 5 + 3.4 X 6 = 246.0kN Similarly Member 10 — combination C, (20% redistribution) Midspan moment = 115.7 kNm End shear = 239 kN Member [2 - combination C 1 (20% redistribution) Midspan moment = 90.9 kNm End shear 3 = 134.9 kN End shear 6 = 147.1kN Member 13 — combination C, (20% redistribution) Midspan moment = 72.7kNm End shear 6 = 141 kN
21. 21. Design of Flat Slabs 423 NA N; m; o. _ v. _ 3 3. 3 ca 3” 0.. . on an U N; N; 3 3 v. _ A: 3. 3 ma 3 0.». on Q _ NA NA >6 3 Y. 3 3. 3. 3 em a. .. Wm at _ N. _ N; 3 3 v. _ _. _ 3. 3. 3 3. En Nd «K W N. _ N. _ no n. _ _. _ ed 3. 3. 3 3. 2 N. ~ _. _~ 5 ads 98. won 3: 13 W3 3.: 3:. 3.. .32 _. _: new _. .~ 6 a o . =oEoE o m : .uEcE w m . :uEoE n N . =oEoE Roam = «%_. ._2 . _ao__w = a%E2 . _ao: m 5&3} . _ao. _m . _.a%_2 2:. 2 .3502 E . _2_EoS E . _unEo2 a 59802 oﬁﬁm : o_: E_eEoU . m:= u«o_ :5. . .o_: :. womb 25.. .. m_. o_. _a> 5.. A25 Esuﬁ can AEZJV m: _oEoE : mqmEE . _onEo_2 ; =&= e = ..a. _.8 EO. _.. ..—
22. 22. 424 Reinforced Concrete 1036 ltNm l03€| tNrn SK 9/3| Bending moment diagram combination C, (20% redistribution). Note: Only one combination C, has been fully analysed to demonstrate the procedure for redistribution of moments in a frame structure. in practice all combinations of loads should be similarly processed to get an envelope of moments and shears. For all combinations of loads the plastic hinges will form at the same moment, i. e. 172 kNm at first floor level and 103.2 kNm at roof level. Step 2 Check moment connection to edge column Edge of slab 400 b, = 800mm Column SK 9/32 Effective width of slab for moment transfer. M. ‘ , .,, , = 0.15b. d2f. , be = C1. + (7,. = 400 + 400 = 800mm (1 = 175 mm assumed fa, = 40N/ mmz M. _ W = 0.15 x 800 x 1752 x 40 = 147kNm > 136.2 kNm (member 9. joint 2) The column slab connection at the edge can transfer the applied moment and no further redistribution is necessary. It is conservatively assumed in this analysis that the depth of the slab at the column is 200mm, ignoring the drop. The moment M, _ , n,, , is greater than the design moment obtained from an equivalent frame analysis.
23. 23. Design of Flat Slabs 425 Step 3 Find bending moments at h, /2 and at edge of drop SK 9/33 Bending moments at critical points — combination C , (20% redistribution). Member 9 -— combination C, (20% redistribution) Joint 5: Bending moment = 172 kNm Shear = 246 kN Dead load: 1.4 X 40 = 56 kN/ m near support Live load: 1.6 X 30 = 48kN/ m near support hc/2 = 0.225 m (56 + 48) x 0.2252 2 Bending moment at ht/2 = 172 — 246 X 0.225 + = 119.3 kNm (top tension) Edge of drop = 1000mm from centreline of column + (56 + 48) x 12 2 = -22 kNm (bottom tension) ll Bending moment at edge of drop 172 — 246 x 1 Joint 2, similarly: (56 + 48) x 0.2252 Bending moment at he/2 = 136.2 — 232 X 0.225 + 2 = 86.6 kNm (top tension) Bending moment at edge of drop = -43.8 kNm (bottom tension) Member 10 — combination C, (20% redistribution)
24. 24. 426 Reinforced Concrete Step 4 Joint 5: Bending moment at he/2 = 121.4 kNm (top tension) Bending moment at edge of drop = -14.5 kNm (bottom tension) Member 12 - combination C, (20% redistribution) Joint 3: Bending moment at hc/2 (1.4 x 30 +1.6 x 9) x 0.2252 = 72.9 — 134.9 x 0.225 + 2 = 44.0kNm Joint 6: Bending moment at he/2 = 71.3 kNm (top tension) Member 13 — combination C, (20% redistribution) Joint 6: Bending moment at he/2 = 73.3 kNm (top tension) Check limitation of negative design moment 0-225 he/2-0-225 25kNm ; I72-O kNm I36-2 kNm l ' 141-3 kNm 111-6|tNrn ‘ "93 "N'“ 86-6 kNm ' Column ‘35'5 mm column JOINT 2 JOINT 5 SK 9/34 Member 9 — combination C, (20% redistribution): limitation of negative design moment. -'2>< er M"(8 1‘ 3 where n = loading per unit area on slab. Average n on span on ﬁrst ﬂoor = 1.4 x 5 + 1.6 X 5 = 15 kN/ m2 I, = 5.0m I2 = 6.0m h, _. = 0.225m M- = (g) (5.0 _ = 264.6 kNm 2 X 0.225)2 3 (floor slab)
25. 25. Design of Flat Slabs 427 Average n on span on roof = 1.4 X 5 + 1.6 x 1.5 = 9.4kN/ m’ 2 M, = (9.4 >; 6.0) (5.0 _ 2 X 3.225) = 165.8 kNm (roof slab) Cheek negative moment limitation Member 9 Joint 2 at hc/2 = M2 = 86.6kNm (see Step 3) Joint 5 at he/2 = M5 = 119.3 kNm Midspan moment = 136.6kNm (see Step 1) Average of M2 and M5 plus midspan moment = 0.5 (86.6 + 119.3) + 136.6 = 239.6kNm < M’ = 264.6kNm The negative moments will have to be increased by (264.6 - 239.6) 25.0kNm Revised negative moments: Joint 2: 86.6 + 25.0 = 1I1.6kNm Joint 5: 119.3 + 25.0 = 144.3 kNm Member 10 Joint 5 at hc/2 = M5 = 12l.4kNm Midspan moment = 1l5.7kNm Average of negative and positive = 121.4 + 115.7 = 237.lkNm < M’ 264.6 kNm The negative moments will have to be increased by (264.6 - 237.1) 27.5 kNm Revised negative moments: Joint 5: 121.4 + 27.5 = 148.9kNm Member 12 Joint 3 at he/2 = M3 = 44.0kNm Joint 6 at h. /2 = M6 = 71.3 kNm Midspan moment = 90.9kNm Average of negative and positive = 0.5 (44.0 + 71.3) + 90.9 = 148.55 kNm < M’ = 165.8 kNm The negative moments will have to be increased by (165.8 — 148.6) = 17.2 kNm Revised negative moments: Joint 3: 44.0 + 17.2 = 6l.2kNm Joint 6: 71.3 + 17.2 = 88.5 kNm Member 13 Joint 6 at hc/2 = M, , = 73.3 kNm Midspan moment = 72.7kNm
26. 26. 428 Reinforced Concrete Average of negative and positive = 73.3 + 72.7 = 146kNm < M’ 165.8kNm The negative moments will have to be increased by (165.8 — 146) 19.8kNm Revised negative moments: Joint 6: 73.3 + 19.8 = 93.lkNm Step 5 Cony out division of panels 20(1) Gohrm Strip 4000 Middle Strip 2&1) Column Strip 4 Middle strip Column Strip SK 9/35 Plan of ﬂoor slab - division of strips, First floor slab Size of drop = 2000 mm %= %0 = 1667mm < 2000mm Column strip = 2000 mm With drop middle strip = 5000 — 2000 = 3000mm (y-y) and = 6000 — 2000 = 4000mm (x-x) Middle strip in a slab without drop = 6000 — I, /2 = 6000 - 2500 = 3500mm 4000 Proportion of middle strip with drop and without drop = ﬂ = 1.14 Roof slab
27. 27. Design of Flat Slabs 429 2500 Column Strip 35% Mid1l¢Str'p 25(1) Column Strip 3500 Middle Strip 2500 Colimn Strip SK 9/36 Plan of roof slab - division of strips. Column strip = Middle ‘x _ M _ §— 2 — 2500mm strip = 6000 - 2500 = 3500mm (x—x) = 5000 — 2500 = 2500mm (y-y) Step 6 Divide moments between column strip and middle strip For slabs without drops, Negative moments - 75% column strip 25% middle strip Positive moments - 55% column strip 45% middle strip Floor slab: design moments Member 9: negative moments Joint 2: Joint 5: 111.6 kNm (see Step 4) l44.3kNm (see Step 4) Middle strip moments Joint 2: 111.6 X 0.25 X 1.14 (see Step 4) = 3l.8kNm (top tension) Edge of drop = 43.8 X 0.25 X 1.14 = 12.5 kNm (top tension) Joint 5: 144.3 X 0.25 X 1.14 = 41.1kNm (top tension) Edge of drop = 22.0 X 0.25 X 1.14 = 6.3 kNm (top tension)
28. 28. 430 Reinforced Concrete Note: Column strip moments Joint 2: 111.6 - 31.8 = 79.8 kNm (top tension) Edge of drop: 43.8 - 12.5 = 31.3 kNm (top tension) Joint 5: 144.3 — 41.1 = 103.2 kNm (top tension) Edge of drop: 22 - 6.3 = 15.7 kNm (top tension) Member 9: positive moments Design midspan moment = 136.6 kNm (bottom tension) Middle strip moments 136.6 X 0.45 X 1.14 = 70.1kNm (bottom tension) Column strip moments Midspan: 136.6 - 70.1 = 66.5 kNm (bottom tension) Member 10: negative moments Joint 5: 148.9 kNm (see Step 4) Middle strip moments Joint 5: 148.9 x 0.25 x 1.14 = 42.4 kNm (top tension) Edge of drop: 14.5 X 0.25 X 1.14 = 4.1l(Nm (top tension) Midspan: Column strip moments Joint 5: 148.9 — 42.4 = 106.3kNm (top tension) Edge of drop: 14.5 — 4.1 = l0.4kNm (top tension) Member 10: positive moments Design midspan moment = l15.7kNm (bottom tension) Middle strip moments Midspan: 115.7 X 0.45 X 1.14 = 59.4 kNm (bottom tension) Column strip moments Midspan: 115.7 — 59.4 = 56.3 kNm (bottom tension) Similarly calculate moments in column strips and middle strips in roof slab. Step 7 Determine cover to reinforcement See Step 3 of Section 3.3. Step 8 Design for ﬂexure Note: See Step 4 of Section 3.3. The increased slab thickness at drops may be considered for the determi- nation of reinforcement provided all reinforcement is properly anchored. Check reinforcement also at edge of drop. ln this example the reinforcement is found for the ﬂat slab spanning in the short direction only. Exactly the same method of analysis and design should be used to ﬁnd the reinforcement in the long direction.
29. 29. Design of Flat Slabs 431 Step 9 Detailing of reinforcement Two-thirds of the negative support reinforcement in the column strip should be placed in half the width of the column strip centred over the column. Step 10 Calculate punching shear and shear stress Punching shear at ﬂoor slab Check joint 5. 215-5kNrn [ 199-9 kNm t. ... mlll| mm ‘ Wllnt &NT 5 MEMBER 9 FEMBER 10 SK 9/37 Moment transfer to column :0, punching Shea, ELASTIC ANALY§l-S - N0 QDISTRIWTIQ mlculatim-| _ CL ' t check ' ,1.. . Mm ﬁ: :% 1"’? Perimeter to check Punching Shear in 400 slab ToJoirll8 Drop inslab SK 9/38 Punching shear perimeters — plan of ﬂoor slab.
30. 30. 432 Reinforced Concrete Use results of elastic analysis of frame before redistribution. Maximum column moment at joint 5 = 215.5 — 199.9 = 15.6kNm M, = 15.6 kNm A 30% reduction is allowed if frame analysis is carried out. M, = 0.7 x 15.6 = 10.9kNm V, = 258.2 + 239.0 = 497.2kN Punching shear perimeter at 1.Sd from face of column. cl = 400 - 0 = 370mm 1.5d = 1.5 X 370 = 555mm x = 400 + 2 X 555 = 1510mm = l.5l0m 1.5M. ) ( 1.5 x 10.9 ) v = V =497.2 1 —: °‘‘ '( 1 + v, x + 497.2 x 1.510 = 508 kN Maximum shear stress at column perimeter (U0 = 4 X 400) Va, 503 x 103 _U°d_4x400x370 Veff Ud U=4x15l0=604(l _ 508x103 V_6040x370 reinforcement = 0.86N/ ml < 5N/ m’ OK Shear stress. v = 0.23Nlmm2 < vc for minimum percentage of No shear reinforcement is required in slab with drop. For slab outside drop consider that loaded area is perimeter of drop. Vt" = 508 kN — (load on the area ()f drop) = 508 — 4 X 22 = 420 kN Perimeter of slab at 1.5d (d = 170mm) = 4 x (2000 + 3 X 170) = l004()mm 420 x 103 Shear stress, v = -T10 040 X 170 = 0.25 N/ mmz < v¢ for minimum percentage of reinforcement No shear reinforcement is required at internal columns of floor slab. Similarly check for an extemal column and a comer column.
31. 31. Design of Flat Slabs 433 Rules for calculation of pcrimeters of external and corner columns l‘5d C1 150 C; 1'5d P=2Cx*Cy*5d F=2(x+C, )oC, o6d DGE COLIMN ON EDGE COLUMN INSIDE EDGE OF SLAQ SLAB X cl l'5d Cr l5d P: C, ,+ Cyo3d P: ICSSCI’ Ol : - 2(x+C, ,) ¢C, +6d NER UHN on 2(c. . c, )n2a . QN_£D_G£_QE. §l. AB. EDGE COLUMN INSIDE SLA§ P: C. ¢C, + xoyoad (RNER COLUMN INSIDE P: l¢SS¢f of : - SLAB . Cxo C, ¢xoyo3d s: ::e= .,: *;. ~.é: :v- on 2<c. »c. >+=2a CORNER COLUMN INSIDE SK 9/39 Punching shear pcrimeters for flat slab. The illustrations show the different column conﬁgurations with respect to a free edge and the corresponding pcrimeters for the calculation of punching shear stresses.
32. 32. 434 Reinforced Concrete When the column face is more than 1.5d away from a free edge of slab, then there are two altcmative pcrimeters possible as illustrated. Take the least of these two altematives for the calculation of punching shear stress. Punching shear at roof slab Check joint 3 — edge column. v. = 130kN The frame action considered is in the x—x direction as explained in Section 9.3.5. V”, = 1.25v, = 1.25 x 130 = 162.5kN d = 170mm assumed 1.5:! = 1.5 X 170 = 255mm Shear stress at column perimeter (U. , = 3 X 400 = 1200) V, “ 162.5 x 103 = —=j=0.sN/ 2 SNI 20K Uod 1200x170 '"'" < ""“ Sh 15d — — V°" ear SITCSS at . — V — U = [2 x (400 + 255)] + (400 + 510) = 2220 162.5 x 103 v = T 2220 x 170 = 0.43 N/ mmz Assume minimum percentage of tensile reinforcement in slab. v, . = 0.48 N/ mmz for Grade 40 concrete and an effective depth of 170 mm. No shear reinforcement is necessary. Note: The punching shear check should also be carried out for the Hat slab spanning in the long direction and the worst result should be used. Step 11 Check span/ eﬂective depth ratio Follow Step 11 of Section 3.3 Step 12 Curtailment of bars Follow Step 12 of Section 3.3. Step 13 Spacing of bars Follow Step 13 of Section 3.3. Step 14 Check early thermal cracking Follow Step 14 of Section 3.3. Step 15 Calculate minimum reinforcement Follow Step 9 and Step 15 of Section 3.3.
33. 33. Design of Flat Slabs 435 Step 16 Calculate / lexural crack width Follow Step 16 of Section 3.3. Step I 7 Design of connections Follow Chapter 11 9.6 TABLES AND GRAPHS FOR CHAPTER 9 How to use Tables 9.1 to 9.6 km} Intermediate f| aor. Typical section h column A C>C=267 First floor . Typical section Plan shovlm points for which coefficients are in tables 9-1 to 96 SK 9/40 Sketches to be used in conjunction with Tables 9.1-9.6.
34. 34. 436 Reinforced Concrete , __ r-—-- —-v r—1 ---- 1-51 — | ' ' BIIIIIIIIB l l-3ly— C= ' Fa‘ C- . - 1 _ n 0.1ly : 0 Sly — L I I o-3ty— l‘ l o 01,- ] C. ' SK 9/41 Zones of stiffness °'°lV 0'3” 0&3’ 0'71)’ "ow "3.7 "SW correction factors to be applied to C3 points of interest. Points to note: (1) The ﬂat slab system should have at least 3 spans in the 1, direction and 3 spans in the 1,. direction. (2) The coefﬁcients are valid for equal spans in the I, and 1, direction. They may be used up to a maximum variation of 20% in the span lengths. (3) The tables can be used only for uniformly distributed loads with all spans loaded simultaneously with the maximum load. (4) To account for the possible increase in moment due to variation of live loads in different panels of the ﬂat slab, no redistribution of moments should be carried out. (5) For horizontal loading a separate frame analysis should be carried out and the appropriate moments will be combined with the venical load moments. In general, ﬁat slab construction should be fully braced and the horizontal load should be carried entirely by a shear-wall system. (6) The coefﬁcients are applicable to a corner panel, an edge panel with a free edge in the 1, direction, an edge panel with a free edge in the I, direction and an internal panel. (7) The moment triad (M, ,, M, and M, y) obtained by this method of analysis should be combined using the Wood—Armer method as described in Section 1.12. Sign convention for bending Banting moment Bending momerl moments (+ vc in directions 85°“ Y'3’"9 about Maxis Torsional moment shown) "'V_ ‘ ’“' = My - unnwsamooeiemem °""“"‘“'°'°'°' '°“' Note: A positive moment denotes hogging. This sign convention is opposite to the Wood—Armer convention. Reverse the signs of the moments before carrying out Wood-Armer combination as per Section 1.12.
35. 35. —Design of Flat Slabs 437 Note: Step-by-step analysis procedure Step I : Determine value of 01 (see SK9/40 and ﬁnd I, /l_, .). Step 2: Assume d is the thickness of slab and h is the dimension of a side of a square column. Step 3: From available L, as shown in SK 9/40, determine S = otd3L/ It“. Step 4: Select a point of interest from SK9/40 or SK9/41 where the moments have to be found. Corresponding to the zone of inﬂuence, ﬁnd appropriate stiffness connection factor K depending on S from Graphs 9.1 to 9.18. Step 5: Find moment coefﬁcients from Tables 9.1 to 9.3 corresponding to I, /!_, , and the location of the point of interest. Step 6: Find the ultimate unifonnly distributed load on the ﬂat slab = n kN/ m2. Step 7: Find moment triads: M, = nC, K,l§kNm/ m M, = nC, ,K_, .l_§kNm/ m M, _,. = nC, ‘,, K,_, .l_f, kNm/ m Step 8: Carry out Wood—Armer combination. Step 9: Find column reactions corresponding to I, /l_. , (see SK9/41 for column locations). Step I0: The moments obtained using these coefﬁcients are in kNm/ m. Find the effective width be as in Section 9.3.4. Multiply the moment obtained by analysis at edge and corner columns with the effective width be to ﬁnd the slab-to-column connection moment. This transfer moment should be less than M. _,. ,,. ,,‘ as deﬁned in Section 9.3.4. Reaction at column C1 = nC1I, l_v Reaction at column C2 = nC2I, ,l, , etc. If in the zone of K A (as shown in SK9/41) the point of interest is located, then to ﬁnd M, use stiffness correction factor K Ax corresponding to S and I, /l_. , as in Graph 19.1. Similarly if the point of interest lies in the zone of K; then to ﬁnd M, use KFX as in Graph 9.16. The beneﬁts of using these tables and graphs are that the analysis can be done very quickly and the necessity of carrying out the two analyses for the two orthogonal directions may be avoided. These tables can also be used for the analyses of raft foundation where the loading n may be assumed to be uniformly distributed over an inverted ﬂat slab. The total loads from a structure will be assumed uniformly distributed at the under- side of the raft. Step II: Calculate total column moments using Table 9.7 and divide the total moment between the columns at the junction depending on their relative stiffness. The stiffer the column, the more moment it will carry. The stiffness of a column may be calculated as II! where I is the moment of inertia and I is the effective height.
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40. 40. 442 Reinforced Concrete 3.9.. ud . .m. .9.. .uG m. ... ._3.. ..u~u . .:. .~. ..u. u ‘| 'n § 0 3 8E. mu 3.. .. 5 3.. .. K 3.. ... . 3 2.. ... . No «S. .. no SE. .0 R. ... ... T on . ._. ... T 2§. ... .+ m. .:. ... T 3 : ... .T Q. 2.. ... - m. .S. ... .+ «U _m: ... T 3 2.. ... - on . ..&. ..+ na. ... ... T E . ... ~.. T an 92.. .: ~: ... ~.. .+ .3 5.9 on 8.. .. 2 £. ... . 8 .8.. . e. ... ... ... 83.. .. mm 3.. .. . ... m.. ... ... .U R: ... .T §. ... ... T 3m. ... ..T . ... \$m. ..: x. ... ... ..T E £. ... T 3.o8.. T . .U 5.8. . .T . ._. ... ... ..T M: .. :.. .+ 2 m. ... ..T 33.. ..» 93: . .T 3 Cool .2.. . . ... ..U . ,a. ... ... .. 8 BE. 9 2.. ... . . ... m.. ... .. . ... _. .E. 8.. ... ... . 55.. ... . .9 mm 9.. ... .» mm ~m. ... T . ... 9.. ..T N» . ... ... .T R 9.. ... .T 3 8.. ... - 2 SET «U . ... ... ... .T 3 mm. ... T on : ... .+ 8.. ... ... T 8 §. .T Q _: ..T . §§. .T . .n. .. 5.9 R 2.. ... . R : E. . ... Rod an .8.. . 3.8.. .. S Ma. .. 3 8.. .. . ...2.. ..T mu m~. ... T . .. . .NET m~. ..m. ... .u . .. \$. ... T 8\$. ... .: 8.. m.. ..T . ..: ... ... T Q . ..m. ... T . .m3.. ..+ am n. ... .T \$. .2.. T . ... ..: ..T . ... .8.. .: . ... .. a. ... ... .. . a.. ... ... . 9. wood 32.. ... . t. ~.. ... .. . ... x.. ... . Z 8.. .. \$2.. ..» on . ._. ... .+ \$. .8.. .+ m. :.. ... ... + . .. n8.. T . .2:. ... T ~. .§. ..+ £_: ..T . ... 3.. ... - 8 R. ... ..+ 33.. ..» . ... m. .~. .T . .~. ... ... .T \$. ... ... .+ 5 SE. 23.. ... . .38.. .. 2 BE. _t. ... ... .E. _.. ... .. S.. ... ... .. ... .._. ... T §. ... ... T am . ... ET 3:. ... T S .8.. .: £. .m. ... T . .N : ... ..T Re. .. . .T on . .m. ... T S m. .N. ..+ . ... 9.. ... - mm . ... ~.. T . ... NM: .. T . .. NNo. oI .3. 8.. ... ... . 32.. ... . I 2.. ... . N3.. ... .. 8 . ... E. on 8.. .. . ... ... ... ... nS_. ... T §: ... .T _. ... n.. ..T E ~8.. T mo : ... ..T 88.. ... - ~m£. ... .T K > GLFLY L3‘o“L3‘ n5‘u“§ do“ . .~. ... ..T BN8.. - o2.: ... + R . E.. .T omR. ...1 . ... ... :.. .- . .. . .8.. T .5 i U 323.. . . .:o_uEooo he . :_on_ we . =oEoE .5 .3. «. ... _ . ... .. .5. km. .. . ... :o_.8o. _ Aw. — H . ..: ... : 3.2» 3: a0 : M_mOU . _O. — mEo_uEuou ~: UEOE m: __u: om W5 o_. _a. —.
41. 41. Design of Flat Slabs 443 8.. .:. _ H6 8e&. .on, .u vgntoud ms3~. on_o D 8 83. 25.: _3..8.o 82.3. 823. Sumo. .. : .m: .o G 383.: 383: 885+ E _ S61 8 _ 55.. 2 237 n: .8.o+ G B «E 6: R 3:7 882+ So: .cI 8 SET Semmou \$S~. o+ AB 56 G 83. %. ..s. o 9:8.. . 3:86 03.8.: 8:3. 3 Be. .. .0 338.. .: S 337 : .\$o. .T on mad: 8 3561 2 Rod: 8 .8.. .» U «\$2.. .- ~3x: T 3 32+ 882.: omommdl 8 :3: 8 mad: Sud 56 38:6 3 :5.. . R 83. mm 82. E Sod E X56 G _ 5.: .. .U Q 3.37 8 95.? 3 ~85: E \$3.: hm ms. ..» R \$5.9: am 8. 6! «U E. 2: . o.. 9. Scan 5 S~. o+ mm wood: 8 £3: 95. ad: mo .8.. .: am. .. to 883. em as. .. 8 N86 83.. .. _m 83. 2. _ 8.: F 58.. . AU mweedu vm mmodl as ~85: _%8.. T mu mmodl mm mad: .. .83.ou RD 8925: S 837 ~§. ~.o+ 833: 832.: e. ..2~. .T 8 .8.. .: AR. .. 5.0 8 .86 on :5.. . E E3 mm 83. 2. N8.. . S 83. on as . .. A0 «\$8.? 8 _ :3: en 83+ 2 _8.o+ So3.c. . 83:. » 8 8. . ..+ «U R we . cl 8 %o. .T _. §wn. o+ N3. _ . c| Q. SN. ..» on 327 %m: .o+ S: :9 BS9: . .£: o.o .385 W385 2 So. .. 53:6 2:55 9 _£s. .T 8337 3 Sod: 88.27 mm SST 339:: E ~3.. T G N. 82.: «\$3.? 8a. .~. o+ R827 89.~. o: R. 83: Sass: AR; .30 \$586 8 83. a 8.8 8 Bed 3. Sod 8 83. R m8.o ab 3 ass: 3 23.: 85°61 o\$8.o: mm 33.: mm R37 383T U S. _2 6| 3 mad: _\$-. c+ R 83.: mm ~3.. T w2.: ..T um mmodl .3; "‘ U . mu. _o. :_ . .=o_oEooo we .52. _o . =oEo_2 \$4 .3; as; \$6 uﬁd «Rd «E = o_.83 .8.N N \$.13 32w 3: . _o cwwov 5.. m. :u_uEooo : _oEoE wﬁucom e. a o_au. _.
42. 42. 444 Reinforced Concrete L: average storey heigrl T SLAB S. ECI| QN_Il: l81.111itL\$LAB SK 9/42 Column number identiﬁcation. Table 9.7 Bending moment coefﬁcient for design of columns in ﬁat slab construction. I, /I, 1 1.2 1.4 1.6 1.8 2 Moment T coefﬁcients Column no. 1 0.01175 0.01345 0.01526 0.01717 0.01916 0.02123 Cc, 1 -0.011 75 -0.017 92 -0.025 94 -0.036 03 -0.048 36 -0.063 07 C . .. 2 -0.002 44 -0.002 33 -0.002 24 -0.00218 -0.002 16 -0.002 18 Ct, 2 -0.022 80 -0.032 71 -0.045 25 -0.060 73 -0.079 39 -0.101 47 Cc, 3 0.022 80 0.027 65 0.03301 —0.0388S 0.045 13 0.05184 Cc, 3 0.00244 0.00441 0.00706 0.01040 0.01445 0.01919 C. .. 4 -0.003 76 —0.00393 -0.00418 -0.00454 —0.00506 -0.00571 Cc, 4 0.003 76 0.006 59 0.010 28 0.014 84 0.020 31 0.026 67 CE. Column total moment = C. X R X X n where 1] = load per unit area on slab. See Graphs 9.19-9.26 for factors R, and R1. Note: Divide total moment in column to top and bottom column in proportion to their stiffness.
43. 43. Design of Flat Slabs 445 . N can _ Hat. ..‘ .2 mo>. =.o . < o: o.N mo. 8:o_uEooo L . N van — u .5.. . new 323 . < o. ._oN . .o. « £= o_o_. ._. «ooo : c_6oE8 Soeuum L2 .8 EM ad . _.. _a. _U = o_6o. :8 age. ..» as: .8. EM .5 . .._. :o
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45. 45. Design of Flat Slabs 447 Graph 9.6 K3,, for M, ,,. . Sti ness correction coefﬁcients for Zone 8, cu for I, /I, =1. 1.2. 1.4. 1.6, 1.8 and 2. 1 Q‘) L1 _J 0 B 0.9 - 0 1 0.6 Graph 9.5 KB, for My: Stiffness correction coefﬁcients for Zone B. curves for I, /l_, = 1. 1.4, 1.6, 1.8 and 2
46. 46. .. .3.E_. .u «no 3.. . Scan A . ~ wen _ H5.. . .8 mo>. =o .0 2.oN .8 £= o_uEuoo :88o. .ou mmoctzm . ... < .8 CM ad . _.. _.: U 448 Reinforced Concrete 0.6 No.0 3.0 85 . m use _ H >. =.. ~ .8 mo>. =o .0 o_. _oN .8 m: .o_uEooo . .8.. ..o. .oo mmoctzm . .€< .8 .68 ha . _._. ... o
47. 47. .. :._. ua. n 0.9 Design of Flat Slabs 449 . ... 3 X DEN — ">. ~k~ hO. — . ..o>. =o . Q 05 .8 m. =o_oEooo nan : o=oo. .oo \$o: .._. m ax. .8 .58 and E30 e. _ . v._ A H. ..~. .~ .8 wotau .0 u. _oN .8 m. ..o_uEoeo 58:8 a2_. ._. m . ..: 5. 5.. .. .. . .._. =o
48. 48. 450 Reinforced Concrete 23.. .: ca . .~ E. .. o. _ . N.— . _ H .58 .8 mo>. _._o .9 o: QN .8 m= .o. _uEooo = o_. uu. =oo mmuctzw . ..: .8 . .5¥ and . _.. _a. _0 . .:. ..uu. a . . V 0 N — 9 9.0 0.0 N. _.mN iv . .. 3 . ~ .25 _ H _. .:¢ .8 mo>. =o . Q o= oN .8 u. :o_uEuou : o=oo. .oo mmoctzm A2 .8 EM :6 . _._. ..O
49. 49. Design of Flat Slabs 451 u - . '- ° 18 ' ' :7. f P Q - ' C — 0!’ — . I-0 = . —. 75 > " 1:? ‘ 8-4‘ h V \$ O V ﬁt; 5 8 n" '-E «iv .5 a ">8 . ' co : >‘l‘-'11 v , _o 28 -I 248 33 ~ 85 . ='5- gesg - gas. on 9 3 Q Q 5 a 4 8-ad Lin 0 coefﬁcients for zone B. curves for I, /l_, = 1, 1.2, 1.4 Graph 9.13 K5,. for M, . Stiffness correcti n oi C N u - O In F in 1- G -n I‘ s 2 8 3 3 § 5 8 2 V’ F 9“ O Q
50. 50. 452 Reinforced Concrete aglnuuuu r T ‘ . ~ .25 v. _ . N._ . _ H 5». .2 mo>. :.u . ... o. .oN .2 2=u_uEuco = o=uo. .8 \$u= .=_. m . ..: .2 £2 £5 EEO as 8.. . 8. F . _:. ... .a-» . ~ E3 o. _ . m._ . _ n :3 .2 noise . m 25M .2 m. :o_uEuco : o=uo. .3 \$o= .c_. w . :v< .2 ink m_. o _= _a.5 5.. . 2.9.0 .33 23
51. 51. EH. Design of Flat Slabs 453 A . _u. V . uux E Ur .2 V. .. 3.0%! . ‘ K‘ L . ..‘ v «S. W A _ N . N E; .. .. .3 .3 .3 . _ u .3 3. 3:8 . . Eon 3. a_a_2.. .8o . .o_. uo. ..8 mmo. ..c_. m . ..: .2 . ... .. and .320 . ~ Ea . ... 6.. .v. _ . N.— J Hatu. .2 meta . ... BEN .02 m= .o_oEooo . _o_.3:8 m%= .._. m . ... 3. . ... :5 . _.. Ec
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53. 53. Design of Flat Slabs 455 . ~ . .E_. ._oO .2 8.. u_oEoou . .o_. oo. ..co \$o. E_. m LE .2 at and .320 . N _= ..= _oD .2 m= .o_uEo8 . .o_. oo. .oo 3u. ..c_. m . ... .< .2 J. 3.. . .320
54. 54. «#18 v. — Q. . .. . n . .:. :_oU .2 . .. . . .E= _oO .2 £: o_oc. oou . .o_. uo. .oo 8o: .._. m L2 .2 .1 was .320 m. ..o_uEooo . .o_. oo. .ou 8u. E_. m ax. .2 . .< and .3s. ¢ 456 Reinforced Concrete
55. 55. Design of Flat Slabs 457 . . . .. E=_oU .2 3.. u_uEoou . .o= oo. ..ou 80.5.6 LE . .. 3.: .395 : _o m. ..o_oEooo . .o_. oo. .ou m. ..o. E_. m . ..}. .2 . . . a .3 C