Staircase design

Design example of the stair,very use for civil engineers.

1. Design Of Staircases Part -1 ~Pritesh Parmar [email protected]

2. Introduction : The staircase is an important component of a building, and often the only means of access between the various floors in the building. It consists of a flight of steps, usually with one or more intermediate landings (horizontal slab platforms) provided between the floor levels.

3. Component of staircase :

4. Component of staircase : Flight : The inclined slab of staircase is called flight. Landing : It is level platform at the top or bottom of a flight between floor. Riser : Vertical projection of the step (i.e., the vertical distance between two neighbouring steps) is called Riser. Tread : The horizontal top portion of a step (where the foot rests) is termed Tread. Nosing : It is the projecting part of the tread beyond the face of riser.

5. Component of staircase : Waist slab : The slab below steps in the stair case is called waist slab. Soffit : Underside of a stair is known as soffit. Going : The horizontal projection (plan) of an inclined flight of steps, between the first and last risers, is termed going. The steps in the flight can be designed in a number of ways: with waist slab, with tread-riser arrangement (without waist slab) or with isolated tread slab.

6. Typical flight in a staircase : Plan Elevation Waist Slab Type

7. Typical flight in a staircase : Tread Riser Type Isolated Tread . Slab Type

8. Type of staircase : A wide variety of staircases are met with in practice. Some of the more common geometrical configurations are : • straight stairs (with or without intermediate landing) • quarter-turn stairs • dog-legged stairs • open well stairs • spiral stairs • helicoidal stairs

9. Type of staircase :

10. Structural Classification : Structurally, staircases may be classified largely into two categories, depending on the predominant direction in which the slab component of the stair undergoes flexure: 1. stair slab spanning transversely (stair widthwise); 2. stair slab spanning longitudinally (along the incline).

11. Stair slab spanning transversely : This category generally includes: 1. slab cantilevered from a spandrel beam or wall ; 2. slab doubly cantilevered from a central spine beam ; 3. slab supported between two stringer beams or walls.

12. Stair slab spanning transversely :

13. Stair slab spanning transversely :

14. Stair slab spanning transversely :

15. Stair slab spanning transversely : When the slab is supported at the two sides by means of ‘stringer beams’ or masonry walls , it may be designed as simply supported, but reinforcement at the top should be provided near the supports to resist the ‘negative’ moments that may arise on account of possible partial fixity. In the case of the cantilevered slabs , it is economical to provide isolated treads (without risers). However, the tread-riser type of arrangement and the waist slab type are also sometimes employed in practice, as cantilevers. The spandrel beam is subjected to torsion (‘equilibrium torsion’), in addition to flexure and shear. The slab supports gravity loads by bending essentially in a transverse vertical plane, with the span along the width of the stair.

16. Stair slab spanning transversely : When the slab is doubly cantilevered from a central (spine) beam, it is essential to ensure, by proper detailing, that the slab does not separate from the beam when loaded on one side only. This can be done by anchoring the slab reinforcement into the beam, so that the same reinforcement acts as a stirrup in the beam. Alternative arrangements are possible; however, it should be ensured that the beam stirrups are ‘closed’, to provide desired torsional resistance.

17. Stair slab spanning transversely : It may be noted that, although the stair slab spans Transversel , the supporting spandrel/spine/stringer beams span Longitudinally along the incline of the stair, framing into supporting columns.

18. Stair slab spanning longitudinally : In this case, the supports to the stair slab are provided parallel to the riser at two or more locations, causing the slab to bend longitudinally between the supports.

19. Stair slab spanning longitudinally :

20. Stair slab spanning longitudinally :

21. Effective span of stair : IS 456 :2000 Cl. 33.1 , Pg. No : 63

22. Effective span of stair : IS 456 :2000 Cl. 33.1 , Pg. No : 63

23. Effective span of stair : IS 456 :2000 Cl. 33.1 , Pg. No : 63

24. Effective span of stair : Another case frequently encountered in residential and office buildings is that of the landings supported on three sides. This case has not been explicitly covered by the IS Code. The ACI Code and BS Code also do not have any special provision as yet for this condition. However, recent studies (based on experiments as well as finite element analysis) reveal that the flight essentially spans between the landing-going junctions, with hogging moments developing at these junctions.

25. Load on stair slab : Stair slabs are usually designed to resist gravity loads, comprising Dead and Live load. In the case of cantilevered tread slabs, the effects of seismic loads should also be investigated. The vertical vibrations induced by earthquakes may induce flexural stresses of considerable magnitude. It is desirable to provide bottom steel in the cantilever slabs (near the support locations) to counter the possibility of reversal of stresses.

26. Dead Load : The components of the dead load to be considered comprise: • self-weight of stair slab (tread/tread-riser slab/waist slab); • self-weight of step (in case of ‘waist slab’ type stairs); • self-weight of tread finish (usually 0.5 – 1.0 kN /m 2 ). The unit weight of reinforced concrete for the slab and step may be taken as 25kN/m 3 as specified in the Code (Cl. 19.2.1).

27. Dead Load : Live loads are generally assumed to act as uniformly distributed loads on the horizontal projection of the flight, i.e., on the ‘going’. The Loading Code [IS 875 : 1987 (Part II)] recommends,

28. Distribution of Load : IS 456 : 2000 Cl. 33.2 , Pg. No 63

29. Distribution of Load : IS 456 : 2000 Cl. 33.2 , Pg. No 63

30. Waist Slab Spanning Longitudinally : Slab thickness t may be taken as approximately l /20 for simply supported end conditions and l /25 for continuous end conditions. The normal load component w n causes flexure in vertical planes containing the span direction (parallel to the longitudinal axis of the slab), and the tangential load component w t causes axial compression (of low order) in the slab. The main bars are placed longitudinally, and designed for the bending moments induced in the vertical planes along the slab span. The distributor bars are provided in the transverse directions. These moments may be conveniently computed by considering the entire vertical load w acting on the projected horizontal span (going), rather than considering the normal load component w n acting on the inclined span s

31. Waist Slab Spanning Longitudinally :

32. Waist Slab Spanning Longitudinally : Example : Design the staircase slab, shown in fig. The stairs are simply supported on beams provided at the first riser and at the edge of the upper landing. Assume a finish load of 0.8 kN /m2 and a live load of 5.0 kN /m2. Use M 20 concrete and Fe 415 steel.

33. Waist Slab Spanning Longitudinally : Example : Riser = 150 mm , Tread = 300 mm Effective span =(Left support width)/2+Going+Landing width-(Right support . width)/2 =150+3000+1500-150 = 4500 mm / 4.5 m Assume waist slab thickness = l/20 = 4500/20 = 225 mm ≈ 250 mm . Initially considering main bar of dia. 12 mm, Effective cover = 20+12/2 = 26 mm. Effective depth , d = D - effective cover = 250 – 26 = 224 mm .

34. Waist Slab Spanning Longitudinally : Example : Load on going on projected plan area : 1) Self weight of waist slab = 25 KN/m 3 X 0.250 m X 335.4/300 = 6.99 KN/m 2 Where, 25 KN/m 3 (Density of concrete) 0.250 m (Waist slab thickness) 335.4 mm(inclined flight distance)=(Riser 2 + Tread 2 )^0.5 )=(150 2 + 300 2 )^0.5 . = 335.4 mm

35. Waist Slab Spanning Longitudinally : Example : Load on going on projected plan area : 2) Self weight of step = 25 KN/m 3 X (0+0.15)/2 m = 1.88 KN/m 2 Where, 25 KN/m 3 (Density of concrete) (0+0.15)/2 m (Avg. thickness of step above waist slab)

36. Waist Slab Spanning Longitudinally : Example : Load on going on projected plan area : 3) Floor finish load = 0.8 KN/m 2 …………………(given) 4) Live load = 5 KN/m 2 …………………(given) …………………………………………………………………………….. Total load on going = 14.67 KN/m 2

37. Waist Slab Spanning Longitudinally : Example : Total load on going = 14.67 KN/m 2 Factored load = 1.5 ( F.o.s ) X 14.67 = 22 KN/m 2 Considering width of stair = 1 m Factored load on going = 22 KN/m

38. Waist Slab Spanning Longitudinally : Example : Load on landing: 1) Self weight of slab = 25 KN/m 3 X 0.250 m = 6.25 KN/m 2 Where, 25 KN/m 3 (Density of concrete) 0.250 m (Landing slab thickness)

39. Waist Slab Spanning Longitudinally : Example : Load on landing : 2) Floor finish load = 0.8 KN/m 2 …………………(given) 3) Live load = 5 KN/m 2 …………………(given) …………………………………………………………………………….. Total load on going = 12.05 KN/m 2

40. Waist Slab Spanning Longitudinally : Example : Total load on landing = 12.05 KN/m 2 Factored load = 1.5 ( F.o.s ) X 12.05 = 18.075 KN/m 2 Considering width of stair = 1 m Factored load on going = 18.075 KN/m

41. Waist Slab Spanning Longitudinally : Example : Loading on staircase can be represent as : 18.075 KN/m 22 KN/m 54.61

42. Waist Slab Spanning Longitudinally : Example : Reaction on left support R 1 = ∑M @ right support =0, R 1 X 4.5 = 18.075 X 1.05 2 / 2 + 22 X 3.45 X (3.45/2 + 1.05) R 1 =49.02 KN Maximum moment will occurs at point of zero shear, Let, at distance x from left support, Shear force = 0 , 49.02 – 22 * x = 0 => x = 2.228 m from left support.

43. Waist Slab Spanning Longitudinally : Example : Now Bending moment @ x = 2.228 m from left support, Mu = 49.02 * 2.228 -22 * 2.228 2 /2 = 54.61 KNm . Main reinforcement design : Mu = 54.61 KNm For Fe415 for balanced case, Mu, lim =0.138* f ck *b*d 2 => d req . = 140.66 mm < d provided = 224 mm Ok. Note : Based on trial and error depth of stair can be reduced considering flexure and deflection criteria.

44. Waist Slab Spanning Longitudinally : Example : Percentage of steel in tension, P t = 50*(1-(1-4.6*Mu/ f ck *b*d 2 )^0.5 )/( f ck / f y ) f ck = 20 Mpa f y = 415 Mpa b = 1000 mm d = 224 mm Mu = 54.61 KNm Ans. P t = 0.33 %

45. Waist Slab Spanning Longitudinally : Example : Area of steel in tension , A st req. = P t X b X d /100 = 0.33*1000*224/100 = 739.2 mm 2 A st min. = 0.12 X b X d /100 = 0.12*1000*250/100 = 300 mm 2 < A st req. Spacing = a st / A st *1000 a st = d 2 /4 = *12 2 /4 =113.097 mm 2

46. Waist Slab Spanning Longitudinally : Example : Spacing = a st / A st *1000 =113.097*1000/ 739.2 =152.99 mm Provide 12 mm dia. Bar @ 150 mm c/c.

47. Waist Slab Spanning Longitudinally : Example : Distribution reinforcement design : A st min. = 0.12 X b X d /100 = 0.12*1000*250/100 = 300 mm 2 Considering 8 mm dia. Bar, Spacing = a st / A st *1000 a st = d 2 /4 = *8 2 /4 =50.26 mm 2

48. Waist Slab Spanning Longitudinally : Example : Spacing = a st / A st *1000 =50.26*1000/ 300 =167.53 mm Provide 8 mm dia. Bar @ 160 mm c/c.

49. Waist Slab Spanning Longitudinally : Example : Check for shear : As reinforcement of stair are confined by compression , shear should be checked at ‘d’ distance from support, ( IS 456 :2000 Cl. 22.6.2.1 ) Shear force at distance ‘d’ from left support : V u = 49.02 – (21.17 × 0.224) = 44.47 kN Shear stress , T v = V u /b*d = 0.197 N/mm 2

50. Waist Slab Spanning Longitudinally : Example : Shear strength of concrete , T c ‘= T c * k k = 1.1 ( IS 456:2000 Cl. 40.2.1.1 ) T c = 0.4 N/mm 2 (For M20 grade and P t = 0.33)( IS 456:2000 , Table -19) T c ‘= 0.4 * 1.1 = 0.44N/mm 2 >>> T v Safe. Nominal reinforcment are provided in order to prevent cracks, shown in detailing .

51. Waist Slab Spanning Longitudinally : Example : Deflection check : Basic l/d ratio = 20 (IS 456 :2000 , Cl. 23.2.1) Modification factor (IS 456:2000 fig 4 , Pg. 38) OR MF = 1/(1+0.625 * log 10 P t ) MF = 1/(1+0.625 * log 10 0.33) (you can check by putting P t = P t,provided , P t = P t,req . is critical ) MF = 1.43 Span / effective depth = 20*1.43 = 28.6 Actual span / effective depth = 4500/224 = 20.089 << 28.6 Deflection is in control.

52. Waist Slab Spanning Longitudinally : Example : Check for development length : M1/ V+Lo <= Ld (IS 456 :2000 , Cl. 26.2.3.3 (C)) As reinforcements are confined by compressive reaction M1/V increased by 30 %. M1=moment capacity for section V = Shear force at support Pt provided = Ast provided*100/b*d Ast provided = a st / Spacing *1000 =113.097*1000/ 150 =753.98 mm Pt provided = 0.34 %

53. Waist Slab Spanning Longitudinally : Example : Check for development length : P t = 50*(1-(1-4.6*Mu/ f ck *b*d 2 )^0.5 )/( f ck / f y )  Mu =57.22 KNm Lo = 8* dia of bar ( Anchorage value for 90degree bent ) Ld = 47 dia. For fe415 and M20 (IS 456 : 2000, Cl. 26.2.2.1) M1/V + Lo = 57.22*1000/49.02+8*12=1263.27 >> 47*12=564 mm OK

54. Waist Slab Spanning Longitudinally : Example : Detailing : 150 160 150 250 250

55. Waist Slab Spanning Longitudinally : Example : Design a (‘waist slab’ type) dog-legged staircase for an office building, given the following data: • height between floor = 3.2 m; • riser = 160 mm, tread = 270 mm; • width of flight = landing width = 1.25 m • live load = 5.0 kN /m 2 • finishes load = 0.6 kN /m 2 Assume the stairs to be supported on 230 mm thick masonry walls at the outer edges of the landing, parallel to the risers. Use M 20 concrete and Fe 415 steel.

56. Waist Slab Spanning Longitudinally : Example : Riser = 160 mm , Tread = 270 mm Effective span = C/C distance between supports =230+2*1250+270*9 = 5160 mm / 5.16 m Assume waist slab thickness = l/20 = 5160/20 = 258 mm ≈ 280 mm . Initially considering main bar of dia. 12 mm, Effective cover = 20+12/2 = 26 mm. Effective depth , d = D - effective cover = 280 – 26 = 254 mm . For economy in design landing slab is taken as 200mm as landing are subjected to low shear force and bending moment compared to going.

57. Waist Slab Spanning Longitudinally : Example : Load on going on projected plan area : 1) Self weight of waist slab = 25 KN/m 3 X 0.280 m X 314/270 = 8.14 KN/m 2 Where, 25 KN/m 3 (Density of concrete) 0.280 m (Waist slab thickness) 314 mm(inclined flight distance)=(Riser 2 + Tread 2 )^0.5 )=(160 2 + 270 2 )^0.5 . = 314 mm

58. Waist Slab Spanning Longitudinally : Example : Load on going on projected plan area : 2) Self weight of step = 25 KN/m 3 X (0+0.16)/2 m = 2.0 KN/m 2 Where, 25 KN/m 3 (Density of concrete) (0+0.16)/2 m (Avg. thickness of step above waist slab)

59. Waist Slab Spanning Longitudinally : Example : Load on going on projected plan area : 2) Floor finish load = 0.6 KN/m 2 …………………(given) 3) Live load = 5 KN/m 2 …………………(given) …………………………………………………………………………….. Total load on going = 15.74 KN/m 2

60. Waist Slab Spanning Longitudinally : Example : Total load on going = 15.74 KN/m 2 Factored load = 1.5 ( F.o.s ) X 15.74 = 23.61 KN/m 2 Considering width of stair = 1 m Factored load on going = 23.61 KN/m

61. Waist Slab Spanning Longitudinally : Example : Load on landing: 1) Self weight of slab = 25 KN/m 3 X 0.20 m = 5.0 KN/m 2 Where, 25 KN/m 3 (Density of concrete) 0.20 m (Landing slab thickness)

62. Waist Slab Spanning Longitudinally : Example : Load on landing : 2) Floor finish load = 0.6 KN/m 2 …………………(given) 2) Live load = 5 KN/m 2 …………………(given) …………………………………………………………………………….. Total load on going = 10.60 KN/m 2

63. Waist Slab Spanning Longitudinally : Example : Total load on landing = 10.60 KN/m 2 Factored load = 1.5 ( F.o.s ) X 10.60 = 15.90 KN/m 2 Considering width of stair = 1 m Factored load on going = 15.90 KN/m

64. Waist Slab Spanning Longitudinally : Example : Loading on staircase can be represent as : 23.61 c 71.40

65. Waist Slab Spanning Longitudinally : Example : Reaction on left support R 1 = As loading is symmetric,half -half load transferred to supports, R 1 = (2*15.90*1.365+23.61*2.43)/2 R 1 =50.38 KN Maximum moment will occurs at point of zero shear( i.e at mid span),

66. Waist Slab Spanning Longitudinally : Example : Now Bending moment @ mid span , Mu = 50.39*2.58 -15.90*1.365*(2.58-1.365/2)-23.61*(2.58-1.365)^2/2 = 71.40 KNm . Main reinforcement design : Mu = 71.40 KNm For Fe415 for balanced case, Mu, lim =0.138* f ck *b*d 2 => d req . = 160.84 mm < d provided = 254 mm Ok. Note : Based on trial and error depth of stair can be reduced considering flexure and deflection criteria.

67. Waist Slab Spanning Longitudinally : Example : Percentage of steel in tension, P t = 50*(1-(1-4.6*Mu/ f ck *b*d 2 )^0.5 )/( f ck / f y ) f ck = 20 Mpa f y = 415 Mpa b = 1000 mm d = 254 mm Mu = 71.40 KNm Ans. P t = 0.33 %

68. Waist Slab Spanning Longitudinally : Example : Area of steel in tension , A st req. = P t X b X d /100 = 0.33*1000*254/100 = 838.2 mm 2 A st min. = 0.12 X b X d /100 = 0.12*1000*280/100 = 336 mm 2 < A st req. Spacing = a st / A st *1000 a st = d 2 /4 = *12 2 /4 =113.097 mm 2

69. Waist Slab Spanning Longitudinally : Example : Spacing = a st / A st *1000 =113.097*1000/ 838.2 =134.92 mm Provide 12 mm dia. Bar @ 130 mm c/c.

70. Waist Slab Spanning Longitudinally : Example : Distribution reinforcement design : A st min. = 0.12 X b X d /100 = 0.12*1000*280/100 = 336 mm 2 Considering 8 mm dia. Bar, Spacing = a st / A st *1000 a st = d 2 /4 = *8 2 /4 =50.26 mm 2

71. Waist Slab Spanning Longitudinally : Example : Spacing = a st / A st *1000 =50.26*1000/ 336 =149.588 mm Provide 8 mm dia. Bar @ 140 mm c/c.

72. Waist Slab Spanning Longitudinally : Example : Check for shear : As reinforcement of stair are confined by compression , shear should be checked at ‘d’ distance from support, ( IS 456 :2000 Cl. 22.6.2.1 ) Shear force at distance ‘d’ from left support : V u = 49.33 – (15.90 × 0.254) = 45.29 kN Shear stress , T v = V u /b*d = 0.178 N/mm 2

73. Waist Slab Spanning Longitudinally : Example : Shear strength of concrete , T c ‘= T c * k k = 1.0 ( IS 456:2000 Cl. 40.2.1.1 ) T c = 0.4 N/mm 2 (For M20 grade and P t = 0.38)( IS 456:2000 , Table -19) T c ‘= 0.4 * 1.0 = 0.4 N/mm 2 >>> T v Safe. Nominal reinforcment are provided in order to prevent cracks.

74. Waist Slab Spanning Longitudinally : Example : Deflection check : Basic l/d ratio = 20 (IS 456 :2000 , Cl. 23.2.1) Modification factor (IS 456:2000 fig 4 , Pg. 38) OR MF = 1/(1+0.625 * log 10 P t ) MF = 1/(1+0.625 * log 10 0.33) (you can check by putting P t = P t,provided , P t = P t,req . is critical ) MF = 1.43 Span / effective depth = 20*1.43 = 28.6 Actual span / effective depth = 5160/254 = 20.31 << 28.6 Deflection is in control.

75. Waist Slab Spanning Longitudinally : Example : Check for development length : M1/ V+Lo <= Ld (IS 456 :2000 , Cl. 26.2.3.3 (C)) As reinforcements are confined by compressive reaction M1/V increased by 30 %. M1=moment capacity for section V = Shear force at support Pt provided = Ast provided*100/b*d Ast provided = a st / Spacing *1000 =113.097*1000/ 130 =869.97 mm Pt provided = 0.34 %

76. Waist Slab Spanning Longitudinally : Example : Check for development length : P t = 50*(1-(1-4.6*Mu/ f ck *b*d 2 )^0.5 )/( f ck / f y )  Mu =73.57 KNm Lo = 8* dia of bar ( Anchorage value for 90degree bent ) Ld = 47 dia. For fe415 and M20 (IS 456 : 2000, Cl. 26.2.2.1) M1/V + Lo = 73.57*1000/50.38+8*12=1541.66 >> 47*12=564 mm OK

77. Waist Slab Spanning Longitudinally : Example : Detailing : 12mm @ 130 c/c 12mm @ 130 c/c 8mm @ 140 c/c 280

78. Waist Slab Spanning Longitudinally : Example : Detailing : Some nominal main reinforcement ( 10 mm dia. @ 220mm c/c) is provided in the landing slab near the support at the top to resist possible ‘ negative ‘ moments on account of partial fixity. 8 mm dia. @ 250mm c/c is also provided as distribution reinforcement. From crossing of bar , the bars must be extent up to Ld (Development length, IS 456 : 2000, Cl. 26.2.1)

79. Waist Slab Spanning Longitudinally : Example : Design a (‘waist slab’ type) dog-legged staircase for an office building, given the following data: • height between floor = 3.2 m; • riser = 160 mm, tread = 270 mm; • width of flight = landing width = 1.25 m • live load = 5.0 kN /m 2 • finishes load = 0.6 kN /m 2 Assume the landing to be supported On two edges perpendicular to the risers. Use M 20 concrete and Fe 415 steel.

80. Waist Slab Spanning Longitudinally : Example : Riser = 160 mm , Tread = 270 mm Effective span = C/C distance between landing =2*625+2430 = 3680 mm / 3.68 m Assume waist slab thickness = l/20 = 3680/20 = 184 mm ≈ 185 mm . Initially considering main bar of dia. 12 mm, Effective cover = 20+12/2 = 26 mm. Effective depth , d = D - effective cover = 185 – 26 = 159 mm .

81. Waist Slab Spanning Longitudinally : Example : Load on going on projected plan area : 1) Self weight of waist slab = 25 KN/m 3 X 0.185 m X 314/270 = 5.38 KN/m 2 Where, 25 KN/m 3 (Density of concrete) 0.185 m (Waist slab thickness) 314 mm(inclined flight distance)=(Riser 2 + Tread 2 )^0.5 )=(160 2 + 270 2 )^0.5 . = 314 mm

82. Waist Slab Spanning Longitudinally : Example : Load on going on projected plan area : 2) Self weight of step = 25 KN/m 3 X (0+0.16)/2 m = 2.0 KN/m 2 Where, 25 KN/m 3 (Density of concrete) (0+0.16)/2 m (Avg. thickness of step above waist slab)

83. Waist Slab Spanning Longitudinally : Example : Load on going on projected plan area : 2) Floor finish load = 0.6 KN/m 2 …………………(given) 3) Live load = 5 KN/m 2 …………………(given) …………………………………………………………………………….. Total load on going = 12.98 KN/m 2

84. Waist Slab Spanning Longitudinally : Example : Total load on going = 12.98 KN/m 2 Factored load = 1.5 ( F.o.s ) X 12.98 = 19.47 KN/m 2 Considering width of stair = 1 m Factored load on going = 19.47 KN/m

85. Waist Slab Spanning Longitudinally : Example : Load on landing: 1) Self weight of slab = 25 KN/m 3 X 0.185 m = 4.625 KN/m 2 Where, 25 KN/m 3 (Density of concrete) 0.185 m (Landing slab thickness)

86. Waist Slab Spanning Longitudinally : Example : Load on landing : 2) Floor finish load = 0.6 KN/m 2 …………………(given) 2) Live load = 5 KN/m 2 …………………(given) …………………………………………………………………………….. Total load on going = 10.23 KN/m 2

87. Waist Slab Spanning Longitudinally : Example : Total load on landing = 10.23 KN/m 2 Factored load = 1.5 ( F.o.s ) X 10.23 = 15.35 KN/m 2 Considering width of stair = 1 m Factored load on going = 15.35 KN/m Only 50% load is acts longitudionally , (as landing slab is one way) i.e 15.35/2 = 7.68 KN/m

88. Waist Slab Spanning Longitudinally : Example : Loading on staircase can be represent as : c

89. Waist Slab Spanning Longitudinally : Example : Design of waist slab : Reaction on left support R 1 = As loading is symmetric,half -half load transferred to supports, R 1 = (7.68*0.625+19.47*2.43)/2 R 1 =28.46 KN Maximum moment will occurs at point of zero shear( i.e at mid span),

90. Waist Slab Spanning Longitudinally : Example : Now Bending moment @ mid span , Mu = 28.46*3.68/2 -7.68*0625*(1.84-0.625/2)-19.47*1.215^2/2 = 30.69 KNm . Main reinforcement design : Mu = 30.69 KNm For Fe415 for balanced case, Mu, lim =0.138* f ck *b*d 2 => d req . = 105.44 mm < d provided = 159 mm Ok. Note : Based on trial and error depth of stair can be reduced considering flexure and deflection criteria.

91. Waist Slab Spanning Longitudinally : Example : Percentage of steel in tension, P t = 50*(1-(1-4.6*Mu/ f ck *b*d 2 )^0.5 )/( f ck / f y ) f ck = 20 Mpa f y = 415 Mpa b = 1000 mm d = 159 mm Mu = 30.69 KNm Ans. P t = 0.37 %

92. Waist Slab Spanning Longitudinally : Example : Area of steel in tension , A st req. = P t X b X d /100 = 0.37*1000*159/100 = 588.3 mm 2 A st min. = 0.12 X b X d /100 = 0.12*1000*185/100 = 222 mm 2 < A st req. Spacing = a st / A st *1000 a st = d 2 /4 = *12 2 /4 =113.097 mm 2

93. Waist Slab Spanning Longitudinally : Example : Spacing = a st / A st *1000 =113.097*1000/ 588.3 =192.25 mm Provide 12 mm dia. Bar @ 190 mm c/c.

94. Waist Slab Spanning Longitudinally : Example : Distribution reinforcement design : A st min. = 0.12 X b X d /100 = 0.12*1000*185/100 = 222 mm 2 Considering 8 mm dia. Bar, Spacing = a st / A st *1000 a st = d 2 /4 = *8 2 /4 =50.26 mm 2

95. Waist Slab Spanning Longitudinally : Example : Spacing = a st / A st *1000 =50.26*1000/ 222 =226.39 mm Provide 8 mm dia. Bar @ 220 mm c/c.

96. Waist Slab Spanning Longitudinally : Example : Check for shear : As reinforcement of stair are confined by compression , shear should be checked at ‘d’ distance from support, ( IS 456 :2000 Cl. 22.6.2.1 ) Shear force at distance ‘d’ from left support : V u = 28.46 – (7.68 × 0.159) = 27.23 kN Shear stress , T v = V u /b*d = 0.172 N/mm 2

97. Waist Slab Spanning Longitudinally : Example : Shear strength of concrete , T c ‘= T c * k k = 1.23 ( IS 456:2000 Cl. 40.2.1.1 ) T c = 0.416 N/mm 2 (For M20 grade and P t = 0.3 7 )( IS 456:2000 , Table -19) T c ‘= 0.416 * 1.23 = 0.511 N/mm 2 >>> T v Safe. Nominal reinforcment are provided in order to prevent cracks.

98. Waist Slab Spanning Longitudinally : Example : Deflection check : Basic l/d ratio = 20 (IS 456 :2000 , Cl. 23.2.1) Modification factor (IS 456:2000 fig 4 , Pg. 38) OR MF = 1/(1+0.625 * log 10 P t ) MF = 1/(1+0.625 * log 10 0.37) (you can check by putting P t = P t,provided , P t = P t,req . is critical ) MF = 1.36 Span / effective depth = 20*1.36 = 27.2 Actual span / effective depth = 3680/159 = 23.144 << 27.2 Deflection is in control.

99. Waist Slab Spanning Longitudinally : Example : Check for development length : M1/ V+Lo <= Ld (IS 456 :2000 , Cl. 26.2.3.3 (C)) As reinforcements are confined by compressive reaction M1/V increased by 30 %. M1=moment capacity for section V = Shear force at support Pt provided = Ast provided*100/b*d Ast provided = a st / Spacing *1000 =113.097*1000/ 190 =595.54 mm Pt provided = 0.374 %

100. Waist Slab Spanning Longitudinally : Example : Check for development length : P t = 50*(1-(1-4.6*Mu/ f ck *b*d 2 )^0.5 )/( f ck / f y )  Mu =31.47 KN/m Lo = 8* dia of bar ( Anchorage value for 90degree bent ) Ld = 47 dia. For fe415 and M20 (IS 456 : 2000, Cl. 26.2.2.1) M1/V + Lo = 31.47*1000/28.46+8*12=1201.76 >> 47*12=564 mm OK

101. Waist Slab Spanning Longitudinally : Example : Design of landing slab : Load : Direct loaing on landing : 15.35*1.25(width of landing) =19.19KN/m Load from going : (19.47*2.43)*0.5 (Half load transferd to lower and half load to upper landing) ……………………………………………………………………………………………... Total load = 42.85KN/m

102. Waist Slab Spanning Longitudinally : Example : Design of landing slab : Reaction on left support R 1 = As loading is symmetric,half -half load transferred to supports, R 1 = 42.85*2.60/2 R 1 =55.70 KN Maximum moment will occurs at point of zero shear( i.e at mid span),

103. Waist Slab Spanning Longitudinally : Example : Now Bending moment @ mid span , Mu = 42.85*2.60^2/8 = 36.20 KNm . Main reinforcement design : Mu = 36.20 KNm For Fe415 , b=1.25m for balanced case, Mu, lim =0.138* f ck *b*d 2 => d req . = 102.43 mm < d provided = 159 mm Ok. Note : Based on trial and error depth of landing can be reduced considering flexure and deflection criteria.

104. Waist Slab Spanning Longitudinally : Example : Percentage of steel in tension, P t = 50*(1-(1-4.6*Mu/ f ck *b*d 2 )^0.5 )/( f ck / f y ) f ck = 20 Mpa f y = 415 Mpa b = 1250 mm d = 159 mm Mu = 36.20 KNm Ans. P t = 0.342 %

105. Waist Slab Spanning Longitudinally : Example : Area of steel in tension , A st req. = P t X b X d /100 = 0.342*1000*159/100 = 544 mm 2 A st min. = 0.12 X b X d /100 = 0.12*1000*185/100 = 222 mm 2 < A st req. Spacing = a st / A st *1000 a st = d 2 /4 = *12 2 /4 =113.097 mm 2

106. Waist Slab Spanning Longitudinally : Example : Spacing = a st / A st *1000 =113.097*1000/ 544 =207.89 mm Provide 12 mm dia. Bar @ 200 mm c/c.

107. Waist Slab Spanning Longitudinally : Example : Distribution reinforcement design : A st min. = 0.12 X b X d /100 = 0.12*1000*185/100 = 222 mm 2 Considering 8 mm dia. Bar, Spacing = a st / A st *1000 a st = d 2 /4 = *8 2 /4 =50.265 mm 2

108. Waist Slab Spanning Longitudinally : Example : Spacing = a st / A st *1000 =50.265*1000/ 222 =226.39 mm Provide 8 mm dia. Bar @ 220 mm c/c.

109. Waist Slab Spanning Longitudinally : Example : Shear strength of concrete , T c ‘= T c * k k = 1.23 ( IS 456:2000 Cl. 40.2.1.1 ) T c = 0.4032 N/mm 2 (For M20 grade and P t = 0.34)( IS 456:2000 , Table -19) T c ‘= 0.4032 * 1.23 = 0.495 N/mm 2 >>> T v Safe. Nominal reinforcment are provided in order to prevent cracks.

110. Waist Slab Spanning Longitudinally : Example : Deflection check : Basic l/d ratio = 20 (IS 456 :2000 , Cl. 23.2.1) Modification factor (IS 456:2000 fig 4 , Pg. 38) OR MF = 1/(1+0.625 * log 10 P t ) MF = 1/(1+0.625 * log 10 0.342) (you can check by putting P t = P t,provided , P t = P t,req . is critical ) MF = 1.414 Span / effective depth = 20*1.414 = 28.28 Actual span / effective depth = 2600/159 = 16.35 << 27.2 Deflection is in control.

111. Waist Slab Spanning Longitudinally : Example : Check for development length : M1/ V+Lo <= Ld (IS 456 :2000 , Cl. 26.2.3.3 (C)) As reinforcements are confined by compressive reaction M1/V increased by 30 %. M1=moment capacity for section V = Shear force at support Pt provided = Ast provided*100/b*d Ast provided = a st / Spacing *1000 =113.097*1000/ 200 =565.485 mm Pt provided = 0.355 %

112. Waist Slab Spanning Longitudinally : Example : Check for development length : P t = 50*(1-(1-4.6*Mu/ f ck *b*d 2 )^0.5 )/( f ck / f y )  Mu =45 KNm Lo = 8* dia of bar ( Anchorage value for 90degree bent ) Ld = 47 dia. For fe415 and M20 (IS 456 : 2000, Cl. 26.2.2.1) M1/V + Lo = 45*1000/55.70+8*12=903.899 >> 47*12=564 mm OK

113. Waist Slab Spanning Longitudinally : Example : Detailing : 8mm @ 250 c/c 220 c/c 300 8mm @ 220 c/c 8mm @ 220 c/c

114. Waist Slab Spanning Longitudinally : Example : Detailing : Some nominal main reinforcement ( 10 mm dia. @ 220mm c/c) is provided in the landing slab near the support at the top to resist possible ‘ negative ‘ moments on account of partial fixity. 8 mm dia. @ 250mm c/c is also provided as distribution reinforcement. From crossing of bar , the bars must be extent up to Ld (Development length, IS 456 : 2000, Cl. 26.2.1)

115. Thank you Presentation is limited to design of waist slab type stair case , different type of staircase design will cover in next presentation.