Dog-leg is that type of stair in which two flight are arranged in such a way that two adjacent flight runs in opposite direction. The flight of stairs ascends to the landing then turns by 180 degree so it consumes less space. The landing is provided to that level at which the direction of flight changes. The design of dog-legged stair is discussed here with an example.
Not 100% sure why it is called dog-legged, but it may be due to its appearance in sectional elevation basically resembling to the dog leg when it is in the position of peeing. If you have noticed it then you know it well.
Table of Contents
Pros of Dog Legged Staircase Design:
High utilization of available space.
Compact layout and better circulation
More economical
Provide more privacy since upper floor is not visible from bottom of stairs
Very common and popular type of staircase used in public and residential buildings.
Design of dog-legged stair:
Design a dog-legged stair for an institution building of floor height 12’. Assume necessary data.
Step1. Design of space required for stair.
Normally, in institution buildings, the tread ranges from 250 – 300 mm whereas the riser ranges from 100 – 150 mm. The minimum width of main-stair in institutional building is 2.0 m.
Since the staircase we are designing is for alternative purpose (not-main), so we can reduce the width of stairs according to space availability.
Take width of stair = 4.5’ = 1.372 m [There is not exact rule to take specific width, it depends upon the expected number of moving people]
Width of landing = width of stairs = 1.372 m
Now, floor height = 12’ = 3.6585 m
Assume, riser = 6” = 0.1524 m [To get number of steps in whole number, slighter greater than 0.15 m is assumed]
Number of riser = 12’ / 6” = 24 nos
Since dog-legged is assumed, take two number of flight.
Number of riser in each flight R = 24/2 = 12 nos
Number of tread = R – 1 = 11 nos
Width of tread = 600 – 2R = 600 – 2 * 152.4 = 295.2 mm
For construction easiness take 1’ = 304.8 mm
So, going = 11 * 1’= 11’ = 3.354 m
Space required for staircase,
Length = entry space + going + landing = 1.372 + 3.354 + 1.372 = 6.098 m
[Entry space is the space required before the beginning of step and it is normally taken equal to landing]
Width = 2 * width of stairs (+ small gap) = 2 * 4.5’ = 9’ = 2.744 m
Step 2: Effective span of stair:
The effective span of stair is calculated as explained under the topic Design of staircase [IS 456:2000, clause 33.1]
Since the stair is supported by beam (width = 250mm) spanning parallel to riser at the start and end of steps in stair but the landing is supported in the masonry wall of thickness 200mm.
Effective span = Going + landing + half of bearing on both side.
= 3.354 + 1.372 + (0.2/2 + 0.25/2)
= 4.95 m
The flight with the greatest span is taken for design of all flights. Since both have same span so take any one.
Let’s take lower flight for designed and analysis which is same for upper flight also.
Step 3: Thickness of waist slab
The thickness of waist slab is taken slightly greater than the normal slab. Since the thickness of slab used in the floor is 130mm so take the thickness of waist slab equal to 150 mm.
Step 4: Load Calculation for stair:
a)Load on going
Self-weight of steps = 25 * (1/2 * R*T) * width of stair * Number of steps
= 25 * ½ * 0.1524 * 0.3048 * 1.372 * 11
= 8.76 KN
Self-weight of waist slab = 25 * √ (R2 + T2) * overall depth * width of stairs * Numbers of steps
= 25 * √ (0.15242 + 0.30482) * 0.15 * 1.372 * 11
= 19.286 KN
Floor finish in going = 1.5 KN/m2 * width of stairs * going
= 1.5 KN/m2 * 1.372 * 3.354 = 6.90 KN
Total DL on going = Self-weight of steps + waist slab + Floor finish
= 8.76 + 19.286 + 6.90
= 34.95 KN
So, UDL due to DL on going = total load / length of going = 34.95/3.354 = 10.42 KN/m
Live load LL = 4.0 KN/m2
So, UDL due to LL on going = 4.0 * width of stairs = 5.488 KN/m
Total UDL on going = 10.42+5.488 = 15.91 KN/m
b)Load on landing
UDL due to DL = (25 * overall depth + floor finish)*width of stairs
= (25 * 0.15 + 1.5) * 1.372
=7.2 KN/m
Total UDL on landing = UDL due to DL + UDL due to LL
= 7.2 + 4 * 1.372
= 12.69 KN/m
Average UDL on the flight = (13.69 * 1.372 + 15.91 * 3.354) / (1.372+3.354) = 15.26 KN/m
Average factored UDL on the flight wu = 1.5 * 15.26 = 22.9 KN/m
Step 5: Calculate bending moment and shear force
Assuming the flight to be simply supported on both ends and is loaded with a UDL, maximum moment is found at mid span as:
Mmax = Wul2/8
= 22.9 * 4.952 / 8
= 70.14 KN-m
Vu,max = Wu l / 2
= 22.9 * 4.95 /2 = 56.68 KN
Step 6: Check for depth
The thickness of slab is derived for calculated value of factored moment.
Mu = Mu,lim = 0.36 fck b xu [d – 0.42 xu]
For Fe 500, xu/d = 0.46 [from IS 456:2000 clause 38.1]
Therefore, Mu = Mu,lim = 0.1336 fck b d²
70.14 * 106 = 0.1336 * 30 * 1372* d²
d = 112.94 mm
Assume, diameter of bar = 10mm and clear cover = 15mm
Overall depth = 112.94 + 5 + 15 = 132.94 < 150 mm ok
Step 7: Calculate reinforcement
The area can be calculated from given formula;
Mu = 0.87 fy As [d – fy As/ fck b]
70.14 * 106 = 0.87 * 500 * As [130 – 500 * As / 30*1372]
As = 1431.9 mm²
Since diameter of bar = 10 mm
Spacing of bar = area of a bar * width of strip / area of reinforcement required
= 78.54 * 1372 / 1431.9 = 75.25 mm
The spacing of main reinforcement bar should not be more than 3 times the effective depth of solid slabs or 300 mm whichever is smaller.
3d = 3* 130 = 390mm
So, provide spacing of 70 mm
Area provided = 78.54 * 1372 / 70 = 1539.38 mm²
And the minimum area of main steel reinforcement is equal to 0.12% of bD.
Min. area required = 0.12* 1372 * 150 / 100 = 246.96 mm² < 1539.38 mm² (area provided), Ok
Provide 10mm dia. bars @ 70mm c/c throughout the width of waist slab
Calculation of distribution bars:
Provide minimum reinforcement as distribution bars:
Ast = Ast,min = 246.96 mm2
Adopting 10mm bars for distribution reinforcement,
Number of bars = 246.96/(π*102/4) = 3.14
Spacing (S) = b/3.14 = 1372/3.14 = 436.94 mm say 300mm < Min(5d,450mm) = 450mm Ok.
Provide 10mm di a. bars @ 300mm c/c throughout the span of waist slab
Step 8: Check for shear:
Shear stress in waist slab τuv = Vu,max/bd = 56.68*1000/(130*1372) = 0.3178N/mm2
Allowable shear stress = k τuc , where k=1.3 for D=150mm
For τuc , percentage reinforcement %pt is required
%pt = 100As/bd = 100*1539.38 /(1372*130) = 0.86%
For M30 grade concrete, from Table 19 of IS456:2000,
τuc = 0.61
Since τuv = 0.3178N/mm2 < k τuc = 1.3*0.61 = 0.793N/mm2, safe in shear
After the completion of design of dog legged staircase, we have to show the arrangement of reinforcement. The sample arrangement for the design of dog-legged stair should be done as shown:
