Design of a one way RC slab

Keep in mind that this is (RCD) reinforced concrete design. Prior to initiating the design calculation, let’s examine what is one way slab.

One way slab bends in only one direction and is of course supported on two edges. However, keep in mind that bending in only one direction and the slab supports cannot simply define one way slab or two way slab.

Even when a rectangular slab is supported on all the 4 edges, the slab can be considered as a one-way slab in case the length-to-breadth ratio (L/B) ratio of the slab is equal to 2 or more than 2.

Let’s suppose a rectangular slab with length equal to 20 meters and the breadth is equal to 10 meters. It is a one way slab because 20 divided by 10 is equal to 2.

A one way slab is designed for the spanning direction alone. The main tension reinforcing bars of such slabs run just parallel to the span. For the transverse direction, a minimum amount of shrinkage reinforcement is provided.

Design a one way slab in case

a. Concrete strength = fc’ = 4 Ksi

b. Steel grade = fy = 60 Ksi

c. Live load = L.L = 220 psf

d. Dead load = D.L = slab self weight

e. Span of slab = 10 feet

Let’s suppose there are 2 supports and a slab and the span is just 10 feet. Keep in mind that the design of one way slab is just the same of a rectangular beam. So, we will design a slab only for 1 feet strap. So, we will design the slab only for 12 inches and then we will apply the 12 inches design to overall slab.

Design solution –

Let us assume that effective depth, d = H – 1 inches

H = L/20 (simply supported) = 10/20 X 12 = 6 inches

Then, d = 6 -1 = 5 inches

Dead load = density of concrete X volume of the slab

But the volume of the slab will be the volume of the 12 inches strap.

Dead load = 150 X bH (cross-section area of the strap) = 0.15 X 1 X 0.5 = 0.075 k/feet

Live Load = 0.22 K/feet

Design Load = 1.2 D.L + 1.6 L.L

Design Load = 1.2(0.075) + 1.6(0.22)

W_u = 0.442 k/feet

Moment = Mu = W_ul²/8

Mu = 0.442 x (10)²/8 = 5.525 k.feet

Mu = 5.525 k.feet X 12 = 66.30 k.inches

∫_req = 0.00425

Main steel = ∫_req bd

Main steel = 0.00425 x 12 X 5 = 0.255 inches²

Temperature = 0.0018 x bH = 0.0018 x 12 x 6 = 0.1296 in²

Reinforcement pattern

Main steel = S_max = Mini (3H, 18”)

S_max = Mini(3 x 6’’,18’’) = Mini (18’’, 18’’)

Temp steel = S_max = Mini(5H, 18’’) = Mini(5 x 6’’, 18’’)

S_max = Mini(30’’, 18’’)

S_max = 18’’

The main bar will number 4 bar having spacing equal to 9 inches centre to centre.

For temperature bars, as we know that temperature bar area = 0.1296 in² = 0.13 in²

0.13 is much least with respect to main steel area.

So, we will pick the number 3 bar, the spacing will be 10 centre to centre.