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DESIGN OF STAIRCASE

DESIGN OF STAIRCASE Dr. Izni Syahrizal bin Ibrahim Faculty of Civil Engineering Universiti Teknologi Malaysia Email: T G N Introduction T G R N h Span, L Flight Landing T = Thread R = Riser G = Going h = Waist N = Nosing = Slope Introduction Public building R 180 mm & G 255 mm Private building R 200 mm & 250 mm G 400 mm For comfort: (2 R) + G = 600 mm (UBBL, BS 5395, Reynold et al. 2007) Types of STAIRCASE Straight stair spanning longitudinally Free-standing stair Helical stair Types of STAIRCASE Slabless stair Straight stair spanning horizontally Spiral stair General DESIGN Considerations Loads Permanent action: Weight of steps & finishes. Also consider increased loading on plan (inclination of the waist) Stairs with open well: Two intersecting landings at right-angles to each other, loads on areas common to both spans may be divided equally between spans Bending Moment & Shear Force Stair slab & landing to support unfavourable arrangements of DESIGN load Continuous stairs: Bending moment can be taken as FL/10 (F is the total ultimate load) General DESIGN Considerations Effective Span Stairs between beam or wall: Centreline between the supporting beam or wall Stairs between landing

(UBBL, BS 5395, Reynold et al. 2007) Types of Staircase Straight stair spanning longitudinally Free-standing stair Helical stair . Types of Staircase Slabless stair Straight stair spanning Spiral stair horizontally . ... 10 255 mm = 2550 mm 250 mm 250 mm • Permanent action, g k

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Transcription of DESIGN OF STAIRCASE

1 DESIGN OF STAIRCASE Dr. Izni Syahrizal bin Ibrahim Faculty of Civil Engineering Universiti Teknologi Malaysia Email: T G N Introduction T G R N h Span, L Flight Landing T = Thread R = Riser G = Going h = Waist N = Nosing = Slope Introduction Public building R 180 mm & G 255 mm Private building R 200 mm & 250 mm G 400 mm For comfort: (2 R) + G = 600 mm (UBBL, BS 5395, Reynold et al. 2007) Types of STAIRCASE Straight stair spanning longitudinally Free-standing stair Helical stair Types of STAIRCASE Slabless stair Straight stair spanning horizontally Spiral stair General DESIGN Considerations Loads Permanent action: Weight of steps & finishes. Also consider increased loading on plan (inclination of the waist) Stairs with open well: Two intersecting landings at right-angles to each other, loads on areas common to both spans may be divided equally between spans Bending Moment & Shear Force Stair slab & landing to support unfavourable arrangements of DESIGN load Continuous stairs: Bending moment can be taken as FL/10 (F is the total ultimate load) General DESIGN Considerations Effective Span Stairs between beam or wall: Centreline between the supporting beam or wall Stairs between landing slab.

2 Centreline of the supporting landing slab, or the distance between edges of supporting slab + m (whichever is the smaller) Detailing Ensure that the tension bar may not break through at the kink General DESIGN Considerations Correct detailing General DESIGN Considerations Incorrect detailing DESIGN Procedure Step Task Standard 1 Determine DESIGN life, Exposure class & Fire resistance EN 1990 Table EN 1992-1-1: Table EN 1992-1-2: Sec. 2 Determine material strength BS 8500-1: Table EN 206-1: Table F1 3 Select the waist, h and average thickness, t of STAIRCASE EN 1992-1-1: Table EN 1992-1-2: Table 4 Calculate min. cover for durability, fire and bond requirements EN 1992-1-1: Sec. 5 Estimate actions on STAIRCASE EN 1991-1-1 6 Analyze structure to obtain maximum bending moments and shear forces EN 1992-1-1: Sec.

3 5 7 DESIGN flexural reinforcement EN 1992-1-1: Sec. 8 Check shear EN 1992-1-1: Sec. 9 Check deflection EN 1992-1-1: Sec. 10 Check cracking EN 1992-1-1: Sec. 11 Detailing EN 1992-1-1: & Example 1 STRAIGHT STAIRCASE SPANNING LONGITUDINALLY Example 1: Straight Staicase G = 255 mm R = 175 mm h = 110 mm L = 2800 mm 10 255 mm = 2550 mm 250 mm 250 mm Permanent action, gk = kN/m2 (excluding selfweight) Variable action, qk = = kN/m2 fck = 25 N/mm2 fyk = 500 N/mm2 RC density = 25 kN/m3 Cover, c = 25 mm bar = 8 mm Example 1: Straight STAIRCASE Determine Average Thickness of STAIRCASE = 2+ 2 =1102552+1752255=133 mm Average thickness: = +( + )2=133+(133+175)2= mm G R y t y Example 1: Straight STAIRCASE Action Slab selfweight = = kN/m2 Permanent action (excluding selfweight) = kN/m2 Characteristics permanent action, gk = + = kN/m2 Characteristics variable action, qk = kN/m2 DESIGN action, nd = + = kN/m2 Consider 1 m width, wd = nd 1 m = kN/m/m width Example 1: Straight STAIRCASE Note: F = wd L = m = kN M = FL/10 = kNm M = FL/10 = kNm M = FL/10 = kNm L = m Analysis Example 1: Straight STAIRCASE Main Reinforcement Effective depth, d = 110 25 8/2 = 81 mm = 2= 10625 812= Kbal = Compression reinforcement is NOT required = = = 500 81= mm2/m Example 1.

4 Straight STAIRCASE Minimum & Maximum Area of Reinforcement , = = As,min = = 1000 81 = 108 mm2/m As,max = = = 1000 110 = 4400 mm2/m Secondary Reinforcement As = 20% of the main reinforcement = 353 = 71 mm2/m Main bar H8-125 (As = 402 mm2/m) Secondary bar H8-350 (As = 144 mm2/m) Example 1: Straight STAIRCASE Shear kN/m kNm/m kNm/m VA VB m M @ B = 0 + ( ) = 0 VA = kN/m VB = kN/m Example 1: Straight STAIRCASE Shear Maximum DESIGN shear force, VEd = kN/m =1+200 =1+20081= Use k = = =402 81= , = 100 1/3 = 251/3 81=45011 N= kN/m = 3/2 = 81=40093 = kN/m VEd ( kN/m) VRd,c ( kN/m) OK Example 1: Straight STAIRCASE Deflection Percentage of required tension reinforcement: = , =353 81= Reference reinforcement ratio: = 10 3=25 10 3= Since o Use Eq.

5 ( ) in EC 2 Cl. Example 1: Straight STAIRCASE Factor or structural system, K = = 11+ + 13/2 (l/d)basic = (11 + + ) = Modification factor for span less than 7 m = Modification for steel area provided = , , =402353= (l/d)allow = = (l/d)actual = 2800/81 = (l/d)allow Deflection OK Example 1: Straight STAIRCASE Cracking h = 110 mm 200 mm Main bar: Smax, slab = 3h (330 mm) 400 mm 330 mm Max bar spacing = 125 mm Smax, slab OK Secondary bar: Smax, slab = (385 mm) 450 mm 385 mm Max bar spacing = 350 mm Smax, slab OK Cracking OK Max bar spacing Example 1: Straight STAIRCASE Detailing 10 175 = 1750 mm 10 255 mm = 2550 mm 250 mm 250 mm 840 mm 840 mm H8-125 H8-350 H8-125 H8-350 Example 2 STAIRCASE WITH LANDING & CONTINUOUS AT ONE END Example 2: STAIRCASE with Landing & Continuous at One End Permanent action, gk = kN/m2 (excluding selfweight) Variable action, qk = = kN/m2 fck = 25 N/mm2 fyk = 500 N/mm2 RC density = 25 kN/m3 Cover, c = 25 mm bar = 10 mm G = 260 mm R = 170 mm h = 160 mm L1 = 2700 mm 10 260 mm = 2600 mm 1500 mm 200 mm L2 = 1600 mm 200 mm L = 4300 mm Example 2: STAIRCASE with Landing & Continuous at One End Determine Average Thickness of STAIRCASE = 2+ 2 =1602602+1702260=191 mm Average thickness: = +( + )2=191+(191+170)2= mm G R y t y Example 2.

6 STAIRCASE with Landing & Continuous at One End Action & Analysis Landing Slab selfweight = = kN/m2 Permanent action (excluding selfweight) = kN/m2 Characteristics permanent action, gk = + = kN/m2 Characteristics variable action, qk = kN/m2 DESIGN action, nd = + = kN/m2 Consider 1 m width, wd, landing = nd 1 m = kN/m/m width Example 2: STAIRCASE with Landing & Continuous at One End Action & Analysis Flight Slab selfweight = = kN/m2 Permanent action (excluding selfweight) = kN/m2 Characteristics permanent action, gk = + = kN/m2 Characteristics variable action, qk = kN/m2 DESIGN action, nd = + = kN/m2 Consider 1 m width, wd, flight = nd 1 m = kN/m/m width Example 2: STAIRCASE with Landing & Continuous at One End Note: F = wd L = ( m) + ( m) = kN M = FL/10 = kNm M = FL/10 = kNm L1 = m Analysis L2 = m kN/m Example 2: STAIRCASE with Landing & Continuous at One End Main Reinforcement Effective depth, d = 160 25 10/2 = 130 mm = 2= 10625 1302= Kbal = Compression reinforcement is NOT required = = = 500 130= mm2/m Example 2: STAIRCASE with Landing & Continuous at One End Minimum & Maximum Area of Reinforcement , = = As,min = = 1000 130 = 173 mm2/m As,max = = = 1000 130 = 6400 mm2/m Secondary Reinforcement As = 20% of the main reinforcement = 485 = 97 mm2/m Main bar H10-150 (As = 524 mm2/m) Secondary bar H10-400 (As = 196 mm2/m) Example 2.

7 STAIRCASE with Landing & Continuous at One End Shear kN/m kNm/m VA VB m M @ B = 0 ( ) ( ) = 0 VA = kN/m VB = kN/m kN/m m Example 2: STAIRCASE with Landing & Continuous at One End Shear Maximum DESIGN shear force, VEd = kN/m =1+200 =1+200130= Use k = = =524 130= , = 100 1/3 = 251/3 130=67376 N= kN/m = 3/2 = 130=64347 = kN/m VEd ( kN/m) VRd,c ( kN/m) OK Example 2: STAIRCASE with Landing & Continuous at One End Deflection Percentage of required tension reinforcement: = , =485 130= Reference reinforcement ratio: = 10 3=25 10 3= Since o Use Eq. ( ) in EC 2 Cl. Example 2: STAIRCASE with Landing & Continuous at One End Factor or structural system, K = = 11+ + 13/2 (l/d)basic = (11 + + ) = Modification factor for span less than 7 m = Modification for steel area provided = , , =524485= (l/d)allow = = (l/d)actual = 4300/130 = (l/d)allow Deflection OK Example 2: STAIRCASE with Landing & Continuous at One End Cracking h = 160 mm 200 mm Main bar: Smax, slab = 3h (480 mm) 400 mm 400 mm Max bar spacing = 150 mm Smax, slab OK Secondary bar: Smax, slab = (560 mm) 450 mm 450 mm Max bar spacing = 400 mm Smax, slab OK Cracking OK Max bar spacing Example 2.

8 STAIRCASE with Landing & Continuous at One End Detailing 10 260 mm = 2600 mm 1500 mm 200 mm 10 170 = 170 mm H10-150 H10-400 H10-150 H10-400 H10-400 200 mm = 1290 mm Example 3 STAIRCASE SUPPORTED BY LANDING Example 3: STAIRCASE Supported by Landing Example 3: STAIRCASE Supported by Landing Plan View 200 1500 1500 100 50 50 200 200 1500 200 10 @ 260 = 2600 Example 3: STAIRCASE Supported by Landing G = 260 R = 170 h = 150 h = 150 Permanent action, gk = kN/m2 (excluding selfweight) Variable action, qk = = kN/m2 fck = 25 N/mm2 fyk = 500 N/mm2 RC density = 25 kN/m3 Cover, c = 25 mm bar = 10 mm Section Example 3: STAIRCASE Supported by Landing Determine Average Thickness of STAIRCASE = 2+ 2 =1502602+1702260=179 mm Average thickness: = +( + )2=179+(179+170)2= mm G R y t y Example 3: STAIRCASE Supported by Landing For this type of STAIRCASE , DESIGN for LANDING and FLIGHT should be done SEPARATELY !

9 !! Example 3: STAIRCASE Supported by Landing Action Landing Slab selfweight = = kN/m2 Permanent action (excluding selfweight) = kN/m2 Characteristics permanent action, gk = + = kN/m2 Characteristics variable action, qk = kN/m2 DESIGN action, nd = + = kN/m2 Consider 1 m width, wd, landing = nd 1 m = kN/m/m width Example 3: STAIRCASE Supported by Landing Action Flight Slab selfweight = = kN/m2 Permanent action (excluding selfweight) = kN/m2 Characteristics permanent action, gk = + = kN/m2 Characteristics variable action, qk = kN/m2 DESIGN action, nd = + = kN/m2 Consider 1 m width, wd, flight = nd 1 m = kN/m/m width Example 3: STAIRCASE Supported by Landing Analysis for STAIRCASE Effective span, Le = La + (Lb1 + Lb2) La = Clear distance between supports = 2600 mm Lb1 = The lesser of width support 1 or m = 200 mm Lb2 = The lesser of width support 2 or m = 1500 mm Le = 2600 + (200 + 1500) = 3450 mm Example 3: STAIRCASE Supported by Landing Note: F = wd L = ( m) = kN M = FL/10 = kNm M = FL/10 = kNm L1 = m L2 = m Le = Effective span Support 1 Support 2 Analysis for STAIRCASE Example 3: STAIRCASE Supported by Landing Moment DESIGN Shear Check Deflection Check Cracking Check Detailing Self Study Example 3: STAIRCASE Supported by Landing Analysis for Landing w kN/m L = m w = wlanding + Load from STAIRCASE = ( ) + (reaction from support 2) = kN/m =.

10 = = . Example 3: STAIRCASE Supported by Landing Main Reinforcement Effective depth, d = 150 25 10/2 = 120 mm = 2= 10625 1202= Kbal = Compression reinforcement is NOT required = = = 500 120= mm2/m Example 3: STAIRCASE Supported by Landing Minimum & Maximum Area of Reinforcement , = = As,min = = 1500 120 = 240 mm2/m As,max = = = 1500 120 = 9000 mm2/m Main bar 17H10 (As = 1335 mm2/m) Example 3: STAIRCASE Supported by Landing Shear Maximum DESIGN shear force, VEd = kN/m =1+200 =1+200120= Use k = = =1335 130= , = 100 1/3 = 251/3 120=114350 N= kN/m = 3/2 = 120=89095 = kN/m VEd ( kN/m) VRd,c ( kN/m) OK Example 3: STAIRCASE Supported by Landing Deflection Percentage of required tension reinforcement: = , =854 120= Reference reinforcement ratio: = 10 3=25 10 3= Since o Use Eq.


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