onw way slab design
- 1. DESIGN OF REINFROCED CONCRETE STRCUTURE: ONE WAY SLAB PREPARED BY : PALAK PATEL CIVIL ENGINEERING MAHAMTA GANDHI INSTITUTE OF TECHNICAL EDUCATION AND RESEARCH CENTER, NAVSARI
- 2. SLABS Slab ― A slab is a plate element having depth (D), very small as compared to it’s length and width. ― Slabs form floors and roofs of buildings. ― They are assumed to carry uniformly distributed loads. ― Usually slabs are horizontal except in the case of stair cases and ramps for car parks. 2
- 3. SLABS TYPES OF SLABS SLABS TYPE • Simply supported slab • Continues slabs • Cantilever slabs SLABS TYPE BASED ON SUPPORT CONDITIONS • One way slabs • Two way slabs • Flat slabs • Grid slabs 3
- 4. SLABS One way slabs ― If the slab is supported on two opposite sides. ― If the slab is supported on four sides, and Ly/Lx ≥ 2. ― Both of condition called one way slab. ― In one way slab main reinforcement is provided along short span and distribution reinforcement provided along long span. Two way slabs ― If the slab is supported on all four edge, and Ly/Lx < 2. ― Above condition called two way slab. ― In two way slabs main reinforcement is provided along both the direction. Ly = longer span Lx = shorter span 4
- 5. SLABS Flat slabs ― When slab is directly supported on columns, without beams. ― It’s provided to increase the floor height and to permit large amount of light. Grid slab ― When slab is supported on beams with columns only on the periphery of the hall. Continuous slab ― In case of large halls, auditoriums, marriage halls, etc. the length is divided into equal bays by providing beams. ― The slab provided over such area is called, continuous slab. 5
- 6. SLABS Design considerations → Effective depth (d) (IS: 456-2000,Pg-37, cl 23.2.1) ― 𝐿 𝐷 = 20 × 𝑀𝑜𝑑𝑖𝑓𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝐹𝑎𝑐𝑡𝑜𝑟 (IS: 456-2000, Pg-38, Fig.4) → Effective span (le) (IS: 456-2000, Pg-34,cl 22.2) ― A) simply supported slab i.Clear span + effective depth (d) ii.Centre to centre of support Which ever is smaller 6
- 7. SLABS ― B) continuous slab ― If width of support < 1 12 × clear span, then effective span is taken as per simply supported beam. ― If width of support > 1 12 × clear span (OR) width of support > 600 mm, then i. For end span one end fixed and the other continuous OR for intermediate span: effective span = clear span ii. For end span with one end free other continuous: a. Clear span + d/2 b. Clear span +( ½ x width of dis-continuous support ) Which ever is smaller 7
- 8. SLABS Cover to reinforcement ― Provided for the protect steel reinforcement from corrosion → Nominal cover ( Clear cover ) ― The thickness of concrete from the surface of reinforcement bar to the nearest edge of the concrete is called Nominal cover. → Effective cover ― The thickness of concrete from the centre of bar to the nearest edge of concrete is called effective cover. → IS code provision ( IS: 456-2000, Pg-46, cl 26.4 ) ― Nominal cover should not be less than the diameter of bar. 8
- 9. SLABS Spacing of reinforcement ( IS : 456-2000, Pg-45, cl 26.3) → Minimum horizontal distance ― Shall not be less of these values ― The diameter of the bar if the diameter are equal ― The diameter of the larger bar if the diameter are unequal. ― 5 mm more than the nominal maximum size of course aggregates. 9
- 10. SLABS → Maximum distance between bars in tension ( IS : 465-2000, Pg-46, cl 26.3.3 ) ― Main reinforcement i. 3d, where, d = effective depth ii. 300 mm Which ever is smaller 10
- 11. SLABS ― Distribution steel i. 5d ii. 450 mm Which ever is smaller. ― Maximum spacing of shear reinforcement i. 0.75d ii. 300 mm Which ever is smaller 11
- 12. SLABS Requirement of reinforcement for slab ( IS : 456-2000, Pg-48, cl 26.5.2 ) → Minimum reinforcement ― For mild steel bar, ― Reinforcement shall not be less than 0.15% of the total c/s area. ― For high strength deformed bars ( tor bar) ― Reinforced shall not be less than 0.12% of the total c/s area. → Maximum diameter ― The diameter of bar shall not exceed ,1/8 of total slab thickness. 12
- 13. SLABS Slabs design checks → Check for cracking ( IS : 456-2000, Pg-46 ) ― For Main steel i. 3d ii. 300 mm Spacing should not exceed of this two values. ― For distribution steel i. 5d ii. 450 mm Spacing should not exceed of the two values. 13
- 14. SLABS Check for deflection ( IS: 456-2000, Pg-38, Fig-4 ) ― Allowable 𝐿 𝑑 = 20 × 𝑀𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝐹𝑎𝑐𝑡𝑜𝑟 ― Find actual 𝐿 𝑑 ― Actual 𝐿 𝑑 < Allowable 𝐿 𝑑 Check for development length (Ld) ( IS: 456-2000, Pg-44, cl 23.2.3.3 (c) ) ― Ld should be ≤ 1.3 × 𝑀1 𝑉 + 𝐿0 ― Here Ld = ∅×𝜎 𝑠 4×𝜏 𝑏𝑑 14 𝜎 𝑠 = 0.87 × 𝑓𝑦 M1 = Moment resistant 50% steel at support V = shear force at support L0 = sum of anchorage beyond centre of support L0 = d = 12∅ take smaller value of two
- 15. ONE WAY SLABS DESIGN ( EXAMPLE ) Design the slab for the room for office building 3.2 m x 9.2 m . The slab is resting on 300 mm thick wall and resting live load of 2.5 kN/m2. ― Use M-20 concrete mix and Fy-415 steel as reinforcement ― Check the slab for cracking, check for deflection, check for development length and check for shear. Assume 3.2 m x 9.2 m as clear dimension Fck = 20 N/mm2 Fy = 415 N/mm2 15
- 16. 16 Ly/Lx = 9.2/3.2 = 2.875 > 2 Design the slab as one-way simply supported slab. Effective depth : ( IS 456:2000, Pg-38, table 4) Consider shorter span as l, l = 3200 mm l/d = 20 x M.F Assume 0.6% steel Fy = 415 N/mm2, Fs = 240 N/mm2 M.F = 1.15 l/d = 20 x 1.15 d = l/(20 x 1.15) = 3200/(20 x 1.15) = 139.13 mm Provided d = 150 mm Assume 10 ∅ mm bars Over all depth = 150 + ∅/2 + clear cover
- 17. 17 = 150 + 10/2 + 20 = 150 +5 + 20 = 175 mm Effective span: ( IS 456:2000, Pg-34, Cl-22.2.9 ) i. 3200 + 150 = 3350 mm ii. c/c of support = 3200 + 300 = 3500 mm Whichever is smaller l = 3350 mm = 3.35 m Load calculation Assume 1 m strip of slab Dead load of slab = ( 1 x 0.175 ) x 25 = 4.375 kN/m Floor finish = 1 x 1 = 1.0 kN/m Live load = 2.5 x 1 = 2.5 kN/m Total load = DL + LL + FL = 4.375 + 2.5 + 1 = 7.875 kN/m
- 18. 18 Factor load = 1.5 x 7.875 = 11.812 kN/m Bending moment Mu = 𝑤𝑙2 8 = 11.812 ×3.352 8 = 16.58 kN.m Effective depth for flexure For Fy = 415 N/mm2 Mu = 0.138 Fck b d2 16.58 x 106 = 0.138 x 20 x 1000 x d2 d2 = 16.58 𝑥 106 0.138 𝑥 20 𝑥 1000 d = 77.50 mm < 150 mm……(OK) Main steel (SP 16, Pg-48, Table 2 ) 𝑀𝑢 𝑏𝑑2 = 16.58 𝑥 106 1000 𝑥 1502 = 0.73
- 19. 19 Fy = 415 N/mm2 Fck = 20 N/mm2 Pt = 0.218 Ast = 𝑃𝑡 100 × 𝑏 × 𝑑 = 0.218 100 × 1000 × 150 = 327 mm2 Consider 10 mm ∅ bar ast = 𝜋 4 × 𝑑𝑖𝑎2 = 𝜋 4 × 102 = 78.53 mm2 Spacing = 𝑎 𝑠𝑡 𝐴 𝑠𝑡 × 1000 = 78.53 327 × 1000 = 240.15 mm Provided spacing 230 mm c/c
- 20. 20 Ast = 𝑎 𝑠𝑡 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 × 1000 = 78.53 230 × 1000 = 341.43 mm2 Distribution steel ( IS: 456:2000, Pg-48, Cl 26.5.2.1 ) Provide minimum 0.12% of total cross section area. Ast = 0.12 100 × 1000 × 175 = 210 mm2 Consider 8 mm diameter bar, ast = 𝜋 4 × 𝑑𝑖𝑎2 = 𝜋 4 × 82 = 50.26 mm Spacing = 𝑎 𝑠𝑡 𝐴 𝑠𝑡 × 1000 = 50.26 210 × 1000 = 239.33 mm Provided spacing 230 mm c/c
- 21. 21 Ast = 𝑎 𝑠𝑡 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 × 1000 = 50.26 230 × 1000 = 218.52 mm2 Check for cracking ( IS: 456:2000, Pg-46 ) For main steel 3d = 3 x 150 = 450 300 mm 230 mm provided < 300 mm…..( OK ) For distribution steel 5d = 5 x 150 = 750 mm 450 mm 130 mm provided < 450 mm……( OK ) Check for deflection ( IS: 456:2000, Pg-38, Fig 4 ) Allowable l/d = 20 x M.F Pt provided = 100×𝐴𝑠𝑡 𝑏×𝑑 = 100 ×341.43 1000×150 = 0.227%
- 22. 22 M.F = 1.6 Allowable l/d = 20 x 1.6 = 32 Actual l/d = 3350/150 = 22.32 Actual l/d < allowable l/d Check for development length ( IS: 456:2000, Pg-44 ) Ld ≤ 1.3 𝑀1 𝑉 + 𝐿𝑜 d = 150 mm 12∅ = 12 x 10 = 120 mm Taking larger of two value Lo = 150 mm Shear force at support 𝑉 = 𝑤𝑙 2 = 11.82 𝑥 3.20 2 = 18.91 kN
- 23. 23 Ast = 50 % of Ast at mid span = 341.43/2 = 170.71 M1 = 0.87 Fy Ast d (1- 𝐹 𝑦 𝐴𝑠𝑡 𝐹 𝑐𝑘 𝑏 𝑑 ) = 0.87 x 415 x 170.71 x 150 x ( 1- 415×170.71 20×1000×150 ) = 9.026 x 106 1.3 𝑀1 𝑉 + 𝐿0 = 1.3 × 9.026×106 18.91 ×106 + 150 = 770.50 mm For M 20 Fy = 415 N/𝑚𝑚2 10 mm∅ bar in Tension (sp_16, Pg 184, Table -65) 𝐿 𝑑 = 470𝑚𝑚 470𝑚𝑚 < 770.50𝑚𝑚 ….. (OK) OR 𝐿 𝑑 = ∅𝜎 𝑠 4𝜏 𝑏𝑑 where, ∅ = Nominal diameter of the bar 𝜎𝑠 = stress in bar at the Section considered at design load 𝜏 𝑏𝑑 = design bond stress given in 26. 2.1.1
- 24. 24 ∅ = 10𝑚𝑚 𝜎𝑠 = 0.87𝑓𝑦 = 0.87 x 415 = 361.05 N/𝑚𝑚2 𝜏 𝑏𝑑 = 1.2 x 1.6 = 1.92 N/𝑚𝑚2 𝐿 𝑑 = 10×361.05 4×1.92 = 470.11 mm check for shear 𝜏 𝑏𝑑 = 𝑉𝑢 𝑏𝑑 = 1891×103 1000×150 = 0.126 N/𝑚𝑚2 Consider 50% bars are bent up at support to resist shear 𝐴 𝑠𝑡 = 341.43 2 = 170.71𝑚𝑚2 𝑃𝑡 = 100𝐴 𝑠𝑡 𝑏𝑑 = 100×170.71 1000×150 = 0.113 % For Fck = 20 N/𝑚𝑚2 ( IS: 456:2000 Pg.73, Table -19 ) 𝜏 𝑐 = 0. 28 N/𝑚𝑚2 For slab thickness, D = 175 mm K = 1.25 ( IS : 456: 2000, Pg-72, Cl 40.2.1.1 )
- 25. 25 𝜏 𝑐 ′ = 𝑘 𝑥 𝜏 𝑐 = 1.25 x 0.28 = 0.35 𝜏 𝑣 < 𝜏 𝑐′ ….. ( OK )
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- 27. ONE WAY SLAB DESIGN (EXAMPLE) Design a simply supported one way slab for an effective span of 3.0 m to carry total factor load of 9 kN/m2. use M20 concrete and Fe250 steel. Draw sketch with required details. Design Effective span = 3.0 m Total factor load = 9 kN/m2 Fck = 20 N/mm2 Fy = 250 N/mm2 Effective depth ( IS: 456-2000, P-37,38 ) l/d = 20 x M.F Assume 0.4 % steel 27
- 28. 28 Fy = 250 N/mm2 Fs = 145 N/mm2 ( IS : 456-2000, fig- 4 ) M.F = 2.0 3000/d = 20 x 2 d = 75 mm Provided depth, d = 100 mm Assume diameter of bar = 10 mm Clear cover = 20 mm D = 100 + 5 + 20 = 125 mm Bending moment Mu = 𝑊𝑙2 8 = 9 ×32 8 = 10.125 kN.m
- 29. 29 Effective depth for flexure: Fy = 250 N/mm2 Mu = 0.148 Fck b d2 ( Sp-16, Pg-10, Table- C ) 10.125 x 106 = 0.148 x 20 x 1000 x d2 d = 58.48 mm 58.48 mm ( Required ) < 100 mm ( Provided ) Main steel ( Sp-16, Pg-48, Table- 2 ) 𝑀𝑢 𝑏𝑑2 = 10.125 𝑥 106 1000 𝑥 1002 = 1.01 Fck = 20 N/mm2, Fy = 250 N/mm2 Pt = 0.50 % 𝐴 𝑠𝑡 = 𝑃𝑡 100 × 𝑏 × 𝑑 = 0.50 100 𝑥 1000 𝑥 100 = 500 mm2 Consider 10 ∅ bars, ast = 78.53 mm2 Spacing = 𝑎 𝑠𝑡 𝐴 𝑠𝑡 × 1000 = 78.53 500 × 1000 = 157 𝑚𝑚
- 30. 30 Provide spacing 150 mm C/C 150 = 78.53 𝐴 𝑠𝑡 ( 𝑃𝑟𝑜𝑣𝑖𝑑𝑒𝑑 ) × 1000 Ast ( provided ) = 524 mm2 Provided 10 mm ∅ - 150 mm c/c Distribution steel Provided minimum 0.15% of total cs area. ( IS: 456-2000, Pg-48,Cl 26.5.2.1 ) Ast = 0.15 100 × 1000 × 125 = 187.5 mm2 Consider 8 mm ∅ bars, ast = 50.26 mm2 Spacing = 𝑎 𝑠𝑡 𝐴 𝑠𝑡 × 1000 = 268.05 mm ≈ 260 mm c/c Provide spacing 260 mm c/c 260 = 50.26 𝐴 𝑠𝑡 ( 𝑃𝑟𝑜𝑣𝑖𝑑𝑒𝑑 ) × 1000 Ast ( Provided ) = 193.30 mm2 Provided 8 mm ∅ - 260 mm cc
- 31. 31 Check for cracking Main steel i. 3d = 3 x 100 = 300 mm ii. 300 mm 150 mm ( Provided ) < 300 mm ( Required ) ……( OK ) Distribution steel i. 5d = 5 x 100 = 500 mm ii. 450 mm 260 mm ( Provided ) < 450 mm ( Required ) ….. ( OK ) Check for deflection ( IS: 456-2000, Pg-38, Fig-4 ) Allowable l/d = 20 x M.F % Pt provided = 100×𝐴𝑠𝑡 𝑏𝑑 = 100×524 1000×100 = 0.524 % M.F = 1.9 Allowable l/d = 20 x 1.9 = 38 Actual l/d = 3000/100 = 30 Actual l/d < allowable l/d ….. ( OK )
- 32. 32 Check for development length Ld ≤ 1.3 𝑀1 𝑉 + 𝐿0 ( IS: 456;2000, Pg-44 ) i. d = 100 mm ii. 12∅ = 12 x 10 = 120 mm Taking larger of two values Lo = 120 mm S.F at support = 𝑤𝑙 2 = 9×3 2 = 13.5 kN Ast at support = 50 % Ast at mid span = 524/2 = 262 mm2 M1 = 0.87 Fy Ast d ( 1 - 𝐹 𝑦 𝐴𝑠𝑡 𝐹 𝑐𝑘 𝑏 𝑑 ) ( IS: 456:2000, Pg-96 ) = 0.87 x 250 x 262 x 100 x ( 1 - 250×262 20×1000×100 ) = 5.51 x 106 N.mm 1.3 x 𝑀1 𝑉 + 𝐿𝑜 = 1.3 × 5.51 ×106 13.5×103 + 120 = 650.60 mm For M20, Fy-250 10 mm ∅ , tension
- 33. 33 Ld = (0.87𝐹𝑦)∅ 4𝜏 𝑏𝑑 = (0.87×250)∅ 4×1.2 = 45.3∅ = 45.3 x 10 = 453 mm 453 mm < 650.60 mm ….. ( OK ) Check for shear 𝜏 𝑣 = 𝑉 𝑢 𝑏𝑑 = 13.5 ×103 1000×100 = 0.135 N/mm2 Consider 50% bars bent up at support to resist shear. Ast = 524/2 = 262 mm2 Pt = 100 𝐴𝑠𝑡 𝑏𝑑 = 100×262 1000×100 = 0.262 %
- 34. 34 For Fck = 20 N/mm2 ( IS: 456:2000, Pg-73, Table – 19 ) 𝜏 𝑐 = 0.36 𝑁/𝑚𝑚2 For D = 125 mm K = 1.30 𝜏 𝑐 ′ = 𝑘 𝜏 𝑐 = 1.30 x 0.36 = 0.468 N/mm2 𝜏 𝑣 < 𝜏 𝑐′…… ( OK )
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- 36. ONE WAY SLAB DESIGN ( EXAMPLE ) for practice i. Design a simply supported slab on 350 mm wide brick masonry for a clear room size 4 m x 10 m. use material grades M-20, and Fe250. take live load as 3.5 kN/m2 and floor finish as 1 kN/m2. check your design for any two as shear, development length, deflection control and cracking. ANS : main steel ( Ast ) = 952.5 mm2, Distribution steel ( Ast ) = 262.5 mm2 i. Design a simply supported slab on 300 mm wide brick masonry walls for a clear room dimension 3 m x 8 m. assume floor finish 0.75 kN/mm2 and live load 3 kN/mm2. check for limit state of serviceability. Take M-20 concrete and Fe415 steel. ANS : Main steel ( Ast ) = 292.5 mm2, Distribution steel ( Ast ) = 262.5 mm2 36
- 37. ONE WAY SLAB DESIGN STAIR CASE DESIGN o Term related to stair case • Flight : the incline slab of stair case is called flight. • Landing : it is the level platform at the top or bottom or a flight between the floors. • Rise : it is the vertical distance between two successive tread faces. • Tread : it is the horizontal distance between two successive riser faces. • Nosing : it is the projecting part of the trend beyond the face of riser. • Waist slab : the slab below steps in the stair case is called waist slab. • Soffit : it is underside of a stair. 37
- 38. ONE WAY SLABS DESIGN Requirement for staircase one flight design o Live load = as per IS:875, • 3 kN/m2 ….. If there is no possibility of overcrowding • 5 kN/m2 ….. If there is possibility of overcrowding o Effective span • As per IS : 456-2000, Pg-63, Cl 33.1 (a) • Effective span : center to center of beam 38
- 39. ONE WAY SLABS DESIGN o Pitch • The slop of staircase flight should not exceed 38’ • For steps, riser ( R ) should not be more than 200 mm and tread ( T ) should not be less than 240 mm • Normally, R = 175 to 200 mm T = 250 to 280 mm 39
- 40. ONE WAY SLAB DESIGN ( EXAMPLE ) STAIRCASE A one meter wide single flight R.C.C stair is to be provided for a height 3.20 m in a residential building. Stair is supported at top and bottom risers by beams 300 mm wide. Waist slab is 150 mm thick. Riser is 160 mm and tread is 300 mm Evaluate: a) Effective span b) Design load c) Reinforcement in waist slab. Prepared a sketch. Use M-20 and Fe-250 grade steel 40
- 41. 41 Effective span of staircase: ( IS: 456:2000, Pg-63, Cl 33.1 (a) ) R = 160 mm ( riser ) T = 300 mm ( tread ) No. of riser = 𝑓𝑙𝑜𝑜𝑟 ℎ𝑒𝑖𝑔ℎ𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑟𝑖𝑠𝑒𝑟 = 3200/160 = 20 Nos Total number of treads = Number of riser – 1 = 20-1 = 19 Nos Horizontal length =Number of treads x length of tread = 19 x 300 = 5700 mm Effective span = centre to centre of beam = 5700 + 300/2 + 300/2 = 6000 mm Load calculations: Inclined length of one step = 3002 + 1602 = 340 mm Area of step = ½ x 0.3 x 0.16 = 0.024 m2 Area of inclined slab = 0.34 x 0.15 = 0.051 m2
- 42. 42 Total area = 0.075 m2 D.L of step section for 1 m width and 300 mm plan length = 0.075 x 25 = 1.875 kN/m D.L per m2 on plan area = 1.875 x 1000 300 = 6.25 kN/m2 Assume L.L per m2 on plan area = 3.0 kN/m2 Total load = 6.25 + 3.0 = 9.25 kN/m2 Factor load = 1.5 x 9.25 = 13.875 kN/m2 Width of stair case = 1m Factor load = 13.875 x 1.0 = 13.875 kN/m Reinforcement ( SP-16, Pg-10, Table C ) Mu = 𝑤𝑙2 8 = 13.875 ×106 8 = 62.43 kN.m Mu = 0.148 Fck b d2 62.43 x 106 = 0.148 x 20 x 1000 x d2 d = 145.22 mm Provided d = 150 mm Assume 10 mm ∅ bars
- 43. 43 D = 150 + 5 + 20 = 175 mm 𝑀𝑢 𝑏𝑑2 = 62.43×106 1000×1502 = 2.77 ( SP-16, Pg-48, Table -2 ) Pt = 1.593 % Ast = 𝑃𝑡 100 × 𝑏 × 𝑑 = 1.593 100 × 1000 × 150 = 2389 mm2 Consider 12 mm ∅ bars, ast = 113.09 mm2 Number of bars = 𝐴𝑠𝑡 𝑎𝑠𝑡 = 2389 113.09 = 21.12 ≈ 22 Nos
- 44. 44 Ast ( Provided ) = 22 x 113.09 = 2487.98 mm2 Provided 22 Nos- 12 mm ∅ bars Distribution steel: ( IS: 456:2000, Pg-48, Cl 26.5.2 ) Provide, minimum 0.15% steel of total cross section area Ast = 0.15 100 × 1000 × 175 = 262.5 mm2 For 8 mm ∅ bars, Spacing = 50.26 262.5 × 1000 = 191.46 mm Provide spacing = 190 mm 190 = 50.26 𝐴 𝑠𝑡 × 1000 Ast = 265 mm2 Provide, 8 mm ∅ - 190 mm c/c Provide 1 Number -10 mm ∅ bar as temperature reinforcement in each riser.
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- 46. ONE WAY SLAB ( ANALYSIS ) A reinforced concrete slab of 125 mm thick is reinforced with 10 mm bars @ 150 mm c/c. the reinforced is located at an effective depth of 100 mm form top. Calculate the moment of resistance of the section and safe load. Use M-20 concrete and Fe415 steel. 46
- 47. 47 Assume 1 m wide slab M-20 Fe415 For 10 mm ∅ -150 mm c/c Ast = 𝑎 𝑠𝑡 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 × 1000 = 78.53 150 × 1000 = 523.53 mm2 d = 100 mm Xu = 0.87×𝐹𝑦×𝐴𝑠𝑡 0.36×𝐹𝑐 𝑘 ×𝑏 = 0.87 ×415×523.53 0.36×20×1000 = 26.25 mm Xu max = 0.48 d = 0.48 x 100 = 48 mm Xu < Xu max
- 48. 48 Under reinforced section Mu = 0.87 Fy Ast d (1- 𝐹 𝑦 ×𝐴𝑠𝑡 𝐹 𝑐𝑘 ×𝑏×𝑑 ) = 0.87 x 415 x 523.53 x 100 (1- 415×523.53 20×1000×100 ) = 16.81 x 106 N.mm Mu = 𝑤𝑙2 8 16.81 x 106 = 𝑤×10002 8 W = 134.48 N/mm = 134.48 kN/m Safe U.D.L = 134.48/2 = 89.65 kN/m
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