DESIGN OF STAIR CASE IN THIS SCENARIO THE LONGITUDINAL REINFORCEMENT SHALL BE USED AS MAIN REINFORCEMENT DIMENSION OF STAIRCASE Total LENGTH of stair case Total HEIGHT of stair case Total WIDTH of stair case Height of riser Length of run Minimum Thickness of slab
ft 10 7 12
No. of Riser =(Total Height/Height of Riser)
inch 0 6 0 7.5 10 8
input
12
No.of Run =(Total Length/ Length of run) 12 Reinforcement yield strength 60000 psi Concrete compressive strength 4000 psi CALCULATION OF DEAD WEIGHT Volume of each step = area of shaded regionxwidth of stair case
shaded area
Ф = Angle of incline of staircase Tan-1 = Height of staircase/Horizontal length of staircase tan-1 Ф =
0.75 36.87
Sum of Vertical side = Riseof step+[2x(thickness of slab/cosФ)] = 27.5000 in² Area of shaded region = 0.5x (sum of vertical sides)xrun =
137.5 in²
Dead Weight = Area of shaded regionx unit weight of concrete = =
143.229 lb/step 171.875 lb/ft² Unit Weight of Concrete =
CALCULATION FOR MAXIMUM REINFORCEMENT 0.85fc'ba = Asfy Where a =0.85C For tension controlled system c/d <= 0.375 C= Therefore a =
2.367 2.012 in
As = 0.85fc'ba/fy Asmax =
1.368 in²
CALCULATION FOR MINIMUM REINFORCEMENT 1/2 Asmin = 3(fc') bw d/fy
Asmin =
0.240 in²
CALCULATING FOR MAXIMUM REINFORCEMENT SPACING CONTROL OF CRACK BY DISTRIBUTION OF REINFORCEMENT. THE SPACING "S" OF STEEL REINFORCEMENT CLOSEST TO THE TENSION FACE SHALL NOT EXCEED S= 15(40,000/fs) -2.5Cc where Cc = Concrete cover fs = 2/3xfy S=
11.40 in
CLACULATION FOR REINFORCEMENT REQUIREMENT The factored nominal flexural strength is given by ФMn =
Ф Asfy [d-(a/2)]
Where Ф =
0.9
Where "a" is given by the following equation 0.85fc'ba = Asfy Therefore a=
Asfy/0.85fc'b
a=
1.471As
a/2 =
0.735As
ФMn =
Ф Asfy [d-(a/2)]
The above equation gives the following for the bottom of slab
85839.844 =
54000As x [6.3125-
0.74As
85839.844 =
340875As -39705.8823529412As²
By Hit & Trail mathod 85839.844 =
340875As -39705.8823529412As²
Let As =
0.26 in²
85839.844 =
put value here for trail & error
85943.38
Use Asreq =
-103.539
0.26 in²/ft width
CHECK REINFORCEMENT Assume reinforcement with the following specification Bar Dia = Area = Spacing
0.375 in 0.110447 in² 8 in
Therefore As/ft width is given by
As =
0.17 in²/ft width
Hence Provided Reinforcement in slab =
Provide
0.24 in²/ft width
D0.375 @ 8 c/c
DESIGN FOR SHEAR Required shear strength Vu = wl/2 Vu =
2.29 kips
Vu = Where
Vc+Vs
Vc = 2 (fc')
1/2
bw d
ФVc/2 =
4.791
Vu<ФVc/2
OK
9581.701 lbs kips
Transverse reinforcement shall be used to account for shrinkage. Vs =
0.2538 in²/ft
Assume reinforcement with the following specification Bar Dia = Area = Spacing
