Design of One-Way Slabs 1 Saurav Shrestha (Kantipur Engineering College)
Design Steps of ONE WAY SLAB! INTRODUCTION: ONE WAY SLAB: One way slabs are those supported continuously on the two oppostite sides so that the loads are carried along one direction only. The direction in which the load is carried in one way slab is called the SPAN. One way slabs are usually made to span in shorter direction, since the corresponding bending moments and shear forces are the least. Distribution steels are provided to distribute any unevenness that may occure in loading and for temperature and shrinkage effects in that direction.
NOTE: 1. When the slab is supported only on two opposite sides, the slab bends in one direction only; hence, it is called a one way slab. 2. When slab is supported on all four sides and the plan is a long rectangle,(i.e. length is greater than twice the width), the bending along longitudinal direction is negligible resulting one way slab.
Figure 1. (a) One way slab with supports on two opposite sides and (b) One way slab with supports on all sides but length of longer direction is greater than twice the length of shorter side
1
Design of One-Way Slabs 2 Saurav Shrestha (Kantipur Engineering College)
DESIGN PROCEDURES OF ONE WAY SLAB 1. Calculation of thickness. Thickness of slab is basically decided from span to depth ratio. The effective span to depth ratio is as explained in [Clause 23.2.1]… = basic value x modification factors Where, l = effective span, which is calculated as given in Clause 22.2 of IS CODE.
d = effective depth
Basic values are given by IS CODE Clause 23.2.1 (a), page 37. 2
Design of One-Way Slabs 3 Saurav Shrestha (Kantipur Engineering College)
Here, modification factors F2 and F3 are taken as unity (1) since compression steel is usually avoided and the section are to be designed as rectangular. Modification factor, F1 for tension reinforcement is usually taken 1.25.
Hence effective depth can be calculated as
d=
2. Calculate ultimate moment and shear in shorter direction of slab, in 1000mm width of slab.
For simply supported section,
Ultimate Moment (Mu) =
Ultimate shear force (Vu) =
For continuous support, bending moment is calculated from Table 12 of IS Code.
3
Design of One-Way Slabs 4 Saurav Shrestha (Kantipur Engineering College)
3. CHECK FOR DEPTH The calculated depth of slab can be checked through following equation given in CODE [G-1.1(c)]
The values of Xu,max/d for different grades of steel are as follows. (Page 70 of IS CODE)
If Mu
If Mu>Mu,lim , increase the depth and design the singly reinforced section.
4
Design of One-Way Slabs 5 Saurav Shrestha (Kantipur Engineering College) 4. DESIGN OF REINFORCEMENT Design of reinforcement includes calculation of area of reinforcement, no. of bars and its spacing(S). The area of reinforcement can be calculated easily using following equation given in IS CODE ANNEX G [G1.1(b)]
The required area of steel, Ast is to be provided in 1000mm width. Instead of finding number of bars for 1000mm width, find, Spacing(S) of the bars using following formula,
S=
x 1000
5. CHECK i.
Minimum reinforcement, Ast
ii.
Maximum Diameter As per Clause 26.5.2.2 of IS CODE,
iii.
Spacing As per Clause 26.3.3 of IS CODE.
6. CHECK FOR SHEAR in 1000mm strip. Find
5
Design of One-Way Slabs 6 Saurav Shrestha (Kantipur Engineering College)
The permissible shear stress in concrete in beams without shear reinforcements Is given in TABLE 23 of CODE (page no. 84)
7. CHECK FOR DEFLECTION CONTROL As mentioned earlier, F2 and F3 equal unity. We have now, , max = basic value x F1
F1 = modification factor for tension reinforcement is calculated from fig.4 of IS CODE.
6
Design of One-Way Slabs 7 Saurav Shrestha (Kantipur Engineering College)
Basic value is given as mentioned earlier. Check: , provided <
, max
8. PROVIDE DISTRIBUTION REINFORCEMENT For distribution reinforcement, the bars of same diameter as of main bars or smaller may be used. According to Clause 26.5.2.1 for distribution reinforcement we provide reinforcement 12% of the total cross sectional area. i.e Ast = 0.12% of bD and spacing is calculated as S =
x 1000
9. PROVIDE DETAILING Detailing is done as per standard practice mentioned in SP 34. 7