CHAPTER 9 STAIR CASES 9.1 GENERAL FEATURES Stair cases are provided for connecting successive floors. It is comprised with flights of steps with inter mediate landings which provides rest to the user and support for the flight. A passage is provided at the start of staircase then for the vertical rise a flight is provided with rise and tread. Rise provided in the steps is normally 6 inch which conforms with the comfort of the user. Tread provided is 9.5 inch which can be more if the number of user is more depending on the type of building. The width of the stair can be between 3.5ft to 5 ft depending on the use. Generally public buildings should be provided with larger width. Going is the horizontal projection of the inclined flight between the first and the last riser. A flight is generally consist of two landings with going in between of 10 to steps. Staircases can be designed in many forms as per the requirement of the user and the facility and space available in the construction. Design procedure of few types are discussed in this chapter. 9. TYPES OF STAIR CASES Stair cases can be of varying geometrical shapes and structural behavior. Some of the most common types of staircases are shown is subsequent discussion.
9..1 DOG LEGGED STAIR CASE LANDING PASSAGE Figure 9.1 : Dog legged stair case Most commonly used in buildings. It comprises with two flights and a landing or lobby in between. Normally the landing is provided at mid height. The landing acts as a support of the flight and landing is supported by beams or wall. 31
9.. OPEN WELL STAIR CASE Open well UP UP Figure 9. : Open well stair case 3
Generally adopted in public building where adequate space can be provided for staircases. It ahs quarter landings which provide more comfort to user. Moreover the open well provide adequate ventilation. The flights are consisted of lesser steps in comparison to dog legged staircases. 9..3 TREAD RISER STAIR CASE This type of staircase is normally used for aesthetic beautification. No support for landing is provided. The tread and riser is constructed as folded plates. The construction of this types of staircase is costly as reinforcement required is more. Tread Riser Figure 9.3 : Tread Riser Stair Case 33
9..4 CANTILEVER STAIR CASE Cantilever slab Rise Figure 9.4 : Cantilever Stair Case In this type of staircase cantilever horizontal tread are projected from a wall or an inclined beam. This type of staircase needs complicated formwork and normally used for aesthetic beautification. 34
9.3 DESIGN OF DOGLEGGED STAIR CASE Step 1: General arrangement LANDING PASSAGE Figure 9.5 : Dog legged stair case (general arrangement) 35
The figure above shows the plan of the stair hall. Let the rise be 6 inch and trade be 9.5 inch. The width of each flight is 3.5 inch. 10 Height of each flight 5 ft. 5 No of risers required 10 risers in each flight. 6 No of tread in each flight 10-1 9. Space occupied be trades 9 9.5 7.5 ft. Width of landing 4.5 ft. Width of passage 4.5 ft. Size of stair hall 7 ft 16.5 ft. Step : Design constants For steel f y And for concrete 40,000 psi f c 3000 psi Step 3: Determination of loading The landing slab acts together with the going as a single slab. The bearing of the slab into the wall may be considered 6.5 inch. 6.5 Then the effective span 7.5 + 4.5 +. 17 ft. l Considering one-way slab with both end continuous minimum thickness is. 8 l.17 So, t 5.inches 6 inches. 8 8 6 Self weight of the slab 150 1 75 plf. 1 Tread Riser Tread Self weight of the steps 150 9.5 6 9.5 0.5 150 37.5 plf. 36
Floor finish 0 plf. Total dead load 75+37.5+013.5. Live load 100 plf. So, Design factored load 1.4 13.5 + 1.7 100 355.5 plf. Step 4: Bending Moment Calculation Maximum Moment 1 M max wl 355.5.17 6581.60 lb-ft 78.97 k-in. 8 8 Check for depth ρ 3 87 0.75ρ 0.75 40 87 + 140 max b 0.078 d M max f y φρf yb 1 0. 59ρ f c 78.97 40 0.9 0.078 40 1 0.59 0.078 30 d.9 inch And t.9+13.9 inch (with 1 inch clear cover) t 3.9 inch < 6 inch (Ok) d available 6-1 5 inch 37
Step 5: Reinforcement Calculation Distribution Bar. Minimum reinforcement is provided as temperature and shrinkage reinforcement. Temperature and shrinkage reinforcement, A st 0.00 b t 0.00 6 0.144in / ft # 3 bar can be used. The spacing will be, 0.11 S 0.144 9 inch c/c. Longitudinal Steel. This is selected by trial. Trial No Trial-1 Trial- Assumed a (inch) a1.0 a0.6 M As f y Steel Area, As a a φf y d 0.85 f b c (inch) (inch ) 78.97 0.49 40 A s 0.49 a 0. 64 1 3 0.9 40 5 78.97 0.47 40 A s 0.47 a 0. 61 0.6 3 0.9 40 5 Comments Not OK OK So, A s 0.47inch is provided. It can be furnished by using # 4 bar. 0. Spacing 5.6 6 inch center to center. 0.47 38
Step 6: Detailing The following points are to be remembered in detailing: The main reinforcement should be bent to follow the bottom profile of the stair. Near the landing the reinforcement should be taken straight up and then bent in the compression zone of landing. For tensile stress in the landing zone separate set of bars should be used as shown in the detailing. The length of each type of bar on either side of the crossing should be at least equal to ft inches. All the bars of the tensile reinforcement should be taken into the supports and anchorage and development length requirement must be fulfilled. Distribution bars should be used parallel to the width of the steps. LANDING # 3 bar @ 9 inch c/c # 4 bar @ 6 inch c/c PASSAGE # 3 bar @ 9 inch c/c # 4 bar @ 6 inch c/c Figure 9.6: Detailing of Stair Case 39
PASSAGE LANDING Figure 9.6.: Detailing of Stair Case (continued) 330
9.4 DESIGN OF OPEN WELL STAIR CASE 4.5 ft 7.67 ft 4.79 ft 4.5 ft 4.5 ft 4 ft 13.79 ft Figure 9.7 : Plan View of Open Well Stair Case 331
Step 1: General Arrangements Width of steps 4.5 ft Height of first flight 4.5 ft Height of nd flight 3 ft Total height between the floors 15 ft First landing 4.5 ft nd landing 4.5 ft Riser raised in first flight 9 Riser 6 inch Tread 11.5 inch The size of stair hall 13.79 ft 16.17 ft Step : Design Constants Let for steel f y And for concrete 40000 f c psi 3000 psi Step 3: Design of First Flight The bearing of the landing slab into the wall is 6.5 inch. 6.5 Therefore the effective span 7.67 + 4.5 +.71ft l Considering one-way slab with both end continuous minimum thickness is 8 So, l.71 t 5.45inches 5. 5inches 8 8 5.5 Self weight of the slab 150 1 68.75plf 1 tread riser tread Self weigh of the steps 150 11.5 6 11.5 0.5 150 37.5plf Floor finish0 plf Live load 100 plf [can be determined by table 1.1 of ACI code] 33
Total Dead load 68.75+37.5+06.5 plf Design factored load 1.4 6.5+1.7 100346.75 plf Bending Moment Calculation: 1 M max wl 346.75.71 7001. 93 lb-ft 84.0 k-in 8 8 Check for the depth 3 87 ρ max 0.75ρ b 0.75 0.078 4 87 + 40 M max 84.0 d f y φρf 1 0.59 0.9 0.078 40 1 0.59 0.078 yb ρ f c d.99 inch 3 inch Provide 1inch clear cover t3+14 inch<6 inch or 5.5 inch So, Design is OK. Available d 5.5-14.5 inch 3 40 Reinforcement Calculation: Distribution Bar Only minimum reinforcement is provided as temperature and shrinkage reinforcement. Temperature and shrinkage reinforcement A st 0.00 bt 0.00 5.5 0.13in / ft So # 3 Bar can be used. 0.11 Spacing 10 inch c/c. 0.13 333
Longitudinal bar This is selected through trials. Trial No Trial-1 Trial- Assumed a (inch) a1.0 a0.7 M As f y Steel Area, As a a φf y d 0.85 f b c (inch) (inch ) 84.0 0.58 40 A s 0.58 a 0. 76 1 3 0.9 40 4.5 84.0 0.56 40 As 0.56 a 0. 73 0.7 3 0.9 40 4.5 Comments Not OK OK So, A s 0.56in can be provided. It can be furnished by using # 4 bar. 0. Spacing 4.71 5 inch c/c. 0.56 Step 4: Design of Second Flight Let the bearing of the landing slab into the wall is 6.5 inch. 6.5 6.5 The effective span + 4.5 + 4.79 + 4.5 + 14.87 ft l Considering one way slab with both end continuous minimum thickness is 8 l So, t 8 14.87 8 6.5inch 6.5 Self weight of the slab 150 1 81. 5 plf 1 tread riser tread Self weight of the steps 150 11.5 6 11.5 0.5 150 37.5 plf 334
Floor finish0 plf Total dead load 81.5+37.5+0138.75 Live load 100 plf So, designed factored load1.4 138.75+1.7 100364.5 plf Bending Moment Calculation: 1 M max wl 364.5 14.87 10067. 73 lb-ft 0.81 k-in. 8 8 Check for depth: 3 87 ρ max 0.75ρ b 0.75 0.078 4 87 + 40 M max 0.81 d f y φρf 1 0.59 0.9 0.078 40 1 0.59 0.078 yb ρ f c d3.59 4 inch Provide 1-inch clear cover t4+15 inch <6.5 inch So, design is Ok d available 6.5 1 5. 5inch 3 40 Reinforcement Calculation: Distribution Bar Only minimum reinforcement is provided as temperature and shrinkage reinforcement. Temperature and shrinkage reinforcement A st 0.00 6.5 0.156in / ft If # 3 bar is used as distribution reinforcement 0.11 Spacing 8.46 8 inch c/c. 0.156 335
Longitudinal bar This is selected through trials. Trial No Trial-1 Trial- Assumed a (inch) a1.0 a0.85 M As f y Steel Area, As a a φf y d 0.85 f b c (inch) (inch ) 0.81 0.67 40 A s 0.67 a 0. 88 1 3 0.9 40 5.5 0.81 0.66 40 As 0.66 a 0. 86 0.85 3 0.9 40 5.5 Comments Not OK OK So, A s 0.66in can be provided. It can be furnished by using # 4 bar. 0. Required spacing 4 inch c/c. 0.66 336
Step 5: Detailing # 3 Bar @10 inch c/c # 4 bar @5 inch c/c # 4 bar @5 inch c\c 11.5 inch 6inch 4.5 ft 7.67 ft 4ft Figure 9.8: First Flight 337
# 4 bar @ 4 inch c\c 6inch 11.5 inch # 4 bar @ 4 inch c/c # 3 Bar @ 8 inch c/c 4.5 ft 7.67 ft 4.5 ft # 3 Bar @ 8 inch c/c Figure 9.9: Second Flight 338