How to Design One-way Slab as per ACI 318-19? | Example Included

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One-way slab is a type of concrete slab in which loads are transferred in one direction to the supporting beams and columns. Therefore, the bending occurs in only one direction. The design of one-way slab is simple and can be carried out easily.

The ACI 318-19 provides a number of requirements regarding slab thickness, concrete cover, and reinforcement ratio which facilitate the design process. For instance, the ACI 318-19 specifies minimum slab thickness that satisfies deflection.

The designer can select smaller slab thickness but they need to check deflection of the slab to make sure that it does not exceed maximum allowable deflection. The procedure of designing a one-way slab is similar to that of a rectangular beam.

Slabs are used to provide flat, useful surfaces. A reinforced concrete slab is a broad, flat plate, usually horizontal, with top and bottom surfaces parallel or nearly so. Slabs may be supported by reinforced concrete beams, masonry or reinforced concrete walls, structural steel members, columns, and continuously by the ground.

How to Differentiate One-way Slab from Two-way Slab?

When a rectangular slab is supported on all four sides, but the ratio of the longer side, L, to the shorter side, S, is 2 or more, L/S ? 2.0, then, the slab will act as a one-way slab, with bending primarily occurring in the short direction. The main reinforcement is placed in the shorter direction which is the span, while shrinkage reinforcement is provided in the longer direction to limit cracking.

When the slab is supported on only two sides, the load will be transferred to these sides regardless of its ratio of longer span to shorter span, and it will be classified as a one-way slab.

Behavior of One-way Slab

The structural action of a one-way slab can be visualized in terms of the deformed shape of the loaded surface. The deflected shape of a rectangular slab, simply supported along its two opposite long edges and free of any support along the two-opposite short edges is shown in Figure-1. The deflected shape is shown by solid lines.

Bending moments are the same in all strips (S) spanning in a short direction between supported edges whereas there is no bending moment in the long strips (l) parallel to the supported edges. The surface is approximately cylindrical.

Unit strip Basis for Flexural Design

For the purpose of analysis and design, a unit of the slab is cut out at right angles to the supporting beams, shown in Figure-2, which may be considered as a rectangular beam of unit width, with depth (h) equal to the slab thickness and the span (l)  equal to the distance between supporting edges.

1. The strip illustrated in Figure-2 can be analyzed by the methods that are used for rectangular beams.
2. The bending moment is computed for the strip of unit width.
3. The load per unit area on the slab becomes the load per unit length on the slab strip.
4. Since all the loads on the slab must be transmitted to the two supporting beams, all the reinforcement should be placed at right angles to these beams, with the exception of any bars that may be placed in the other direction to control shrinkage and temperature cracking.
5. Therefore, a one-way slab consists of a set of rectangular beams side by side.

Reinforcements in Concrete Slabs

Generally, two types of reinforcements are provided in slabs, namely, main reinforcements ( primary reinforcement) and secondary reinforcement (shrinkage and temperature reinforcement). These are discussed below:

1. Main Reinforcement

The main reinforcements are placed perpendicular to the supports of the slab i.e. they are responsible for transferring loads to the supports as shown in Figure-3. The purpose of design calculation is to compute the required quantity of the primary reinforcement.

The primary reinforcement can be computed using the flexural formula of the beam. The process involves estimating loads on the slab and then calculating the applied moment. One can find area of main reinforcement by equating applied moment to resisting moment. This calculation procedure is discussed in the design procedure of the one-way slab below.

Temperature and Shrinkage Reinforcement

Shrinkage and temperature reinforcements are provided to resist shrinkage and temperature stresses in concrete. Slabs are joined rigidly to other parts of the structure and cannot contract freely, this results in tension stresses known as shrinkage stresses.

A decrease in temperature relative to that at which the slab was cast, particularly in outdoor structures such as bridges, may have an effect similar to shrinkage. This means, the slab tends to contract and if restrained from doing so becomes subjected to tensile stresses.

According to ACI 318-19 section 24.4, the minimum temperature and shrinkage reinforcement can be at least equal to or greater than the steel area computed using the following formula:

A_{s, shrinkage and temperature}=0.0018*b*h ------------------------ Equation-1

Where:

b: width of slab strip, 1 m

h: thickness of slab

ACI Provision for Design of One-way Slab

1. Compressive Strength

The compressive strength of concrete is specified based on the following criteria.

1. Based on minimum compressive strength specified by ACI 318-19.
2. Based on the strength requirements of a structure under consideration.
3. Based on the durability requirements of the structure. Sometimes, the durability requirements enforce the use of high concrete compressive strength.

2. Minimum Thickness

For solid non-prestressed slabs not supporting or attached to partitions or other construction likely to be damaged by large deflections, overall slab thickness (h) shall not be less than the limits in Table-1, unless the calcu­lated deflection limits of 7.3.2 are satisfied.

Table 1: Minimum Thickness of solid non-prestressed One-way Slab

Support condition	Minimum, h
Simply supported	?/20
One end continuous	?/24
Both ends continuous	?/28
Cantilever	?/10

Notes:

1. If yield strength (f_y) of other than 420 MPa, the values of Table-1 should be multiplied by (0.4+f_y/700).
2. If the slab is constructed with lightweight concrete having (wc) in the range of 1440 to 1840 kg/m³, the values in Table-1 shall be multiplied by the greater of (1.65 – 0.0003wc) and (1.09).
3. The total slab thickness (h) is usually rounded to the next higher 10 mm. Best economy is often achieved when the slab thickness is selected to match the nominal lumber dimension.

3. Concrete Cover

The concrete protection below the reinforcement should follow the requirements of ACI Code 20.5.1.3, calling for 20 mm below the bottom of the steel. In a typical slab, 25 mm below the center of the steel may be assumed.

4. Maximum Reinforcement Ratio

The maximum reinforcement ratio (p_0.005) is computed by using the following expression:

Where:

f_y: yield strength of steel, MPa

fc': concrete compressive strength, MPa

epsilon, cu: concrete compressive strain which is equal to 0.003

B₁ is a constant that can be computed using Equation-3:

5. Minimum Reinforcement Ratio

The minimum primary reinforcement ratio is equal to the shrinkage and temperature reinforcement computed using Equation 1; the usual minimums for flexural steel do not apply.

6. Maximum and Minimum Spacing between Steel bars

1. The lateral maximum spacing of the bars, except those used only to control shrinkage and temperature cracks, should not exceed 3 times the thickness (h) or 450 mm, whichever is smaller.
2. The maximum spacing between shrinkage and temperature reinforcement bars is five times the slab thickness or 450 mm, whichever is smaller.
3. The minimum distance is 25 mm, the diameter of steel bar, or (4/3* maximum aggregate size).

7. Bar Size

The bar size should be selected so that the actual spacing is not less than about 1.5 times the slab thickness, to avoid excessive cost for bar fabrication and handling. Also, to reduce cost, straight bars are usually used for slab reinforcement.

Design Procedure

1. Estimate live load based on the function of the slab. For instance, according to minimum design loads for buildings and other structures, the live load of a slab for office use is 2.4 KN/m².
2. Compute the self-weight of the slab and add it to the superimposed dead load if available. The self-weight is equal to the concrete unit weight times thickness of the slab (h) which is taken from Table-1 based on the span length.
3. Calculate ultimate distributed load on the slab using suitable load combination equation provided by ACI 318-19.
4. Evaluate ultimate moment/applied moment (M_u) using suitable structural analysis methods such as the ACI coefficient method or use equations for cases like cantilever or simply supported slabs.
5. Compute effective depth (d) which is equal to slab thickness (h), minimum (25 mm).
6. Calculate maximum reinforcement ratio using Equation-2.
7. Assume a reinforcement ratio. It is recommended to take 30% of the maximum reinforcement ratio
8. Calculate effective depth from the assumed reinforcement ratio using Equation-4 to check whether the minimum depth computed in Step-2 is adequate or not.
9. Assume a value for rectangular stress block and then compute reinforcement area using Equation 5.
10. After that, calculate rectangular stress block by plugin reinforcement area in Step-9 into Equation 6.
11. Take three trials to reach the correct reinforcement ratio.
12. Calculate shrinkage and temperature reinforcement using Equation-1.
13. Use Table-2 to estimate spacing for both main and secondary reinforcement computed in Step-9 and Step-10, respectively.
14. Check shear strength of the slab.

Where:

d: effective depth measured from the top of slab cross section to the center of steel bars, mm

M_u: Applied or ultimate moment

P: reinforcement ratio

b: width of slab strip which is 1m.

A_s: area of reinforcement, mm²

a: depth of rectangular stress block, mm

Table-2: Areas of Bars in Slabs mm/m²

Bar No.	10	13	16	19	22	25	29	32	36
Spacing, mm	-	-	-	-	-	-	-	-	-
75	947	1720	2653	3787	5160	6800	8600	10920	11413
80	888	1613	2488	3550	4838	6375	8063	10238	12575
90	789	1433	2211	3156	4300	5667	7167	9100	11178
100	710	1290	1990	2840	3870	5100	6450	8190	10060
110	645	1173	1809	2582	3518	4636	5864	7445	9145
120	592	1075	1658	2367	3225	4250	5375	6825	8383
130	546	992	1531	2185	2977	3923	4962	6300	7738
140	507	921	1421	2029	2764	3643	4607	5850	7186
150	473	860	1327	1893	2580	3400	4300	5460	6707
160	444	806	1244	1775	2419	3188	4031	5119	6288
170	418	759	1171	1671	2276	3000	3794	4818	5918
180	394	717	1106	1578	2150	2833	3583	4550	5589
190	374	679	1047	1495	2037	2684	3395	4311	5295
200	355	645	995	1420	1935	2550	3225	4095	5030
225	316	573	884	1262	1720	2267	2867	3640	4471
250	284	516	796	1136	1548	2040	2580	3276	4024
300	237	430	663	947	1290	1700	2150	2730	3353

Reinforcement Details

1. Straight-bar System

Straight bars are used for the top and bottom reinforcement in all spans. The time and cost to produce straight bars are less than that required to produce bent bars; thus, the straight-bar system is widely used in construction.

2. Bent-bar

Straight and bent bars are placed alternately in the floor slab. The location of bent points should be checked for flexural, shear, and development length requirements. For normal loading in buildings, the bar details at the end and interior spans of one-way solid slabs may be adopted.

Example:

Design a simply supported one-way solid slab, a span of 4 m, required to support service live load  LL=3 KN/m² and self-weight only. fc?=28 MPa and f_y=420 Mpa.

Solution

Calculate slab thickness based on Table-1:

For simply supported slab, h= L/24= 4000/24= 166.6 mm= 180 m

Estimate self-weight of the slab:

Self-weight of the slab=0.18*24= 4.32 KN/m²

Calculate ultimate distributed load on the slab:

w_u=1.2*(4.32)+1.6*(3)= 9.984 KN/m²

Compute Applied moment/ Ultimate moment on the slab (M_u):

For simply supported slab, M_u=wl²/8= (9.984*(4)²)/8= 19.968 KN.m/m

Compute effective depth (d):

d=h-25= 180-25= 155 mm

Calculate maximum reinforcement ratio using Equation-2, and then assume a reinforcement ratio which 30% of (p_0.005), and then check whether it would be adequate or not:

Take strength reduction factor as 0.9

since fc'=28 MPa, so B₁=0.85

p_0.005=0.85*0.85*(28/420)*(0.003/(0.003+0.005))= 0.001806

Assume reinforcement ratio is 0.3*0.001806= 0.005418

d=(19.968*10⁶/(0.9*0.001806*420*1000(1-((0.59*0.005418*420)/28))))^(0.5)= 101.19 mm

The effective depth found from the applied moment (101.19 mm) is smaller than that found from Code restriction (155 mm), the latter will be adopted.

Assume a rectangular stress block (a), then calculate reinforcement area (As) using Equation-5. After that, compute the rectangular stress block from Equation 6. Repeat this process three times to reach the correct reinforcement area:

Assume, a=20 mm

As= (19.968*10⁶)/(0.9*420*(155-20/2)= 364.31 mm²

a=(364.31*420)/(0.85*28*1000)= 6.429 mm

Second trial:

A_s= (19.968*10⁶)/(0.9*420*(155-6.429/2)= 348.026 mm²

a=(348.026*420)/(0.85*28*1000)= 6.141 mm

Since the difference between rectangular stress block reduced considerably in the second trial, so there is no need to conduct third trial.

Take A_s= 348.026 mm²

Compute shrinkage and temperature reinforcement using equation 1:

A_{s, shrinkage and temperature}= 0.0018*1000*180= 324 mm²

Estimate lateral spacing between steel rebars for both main and secondary reinforcement using Table-2:

Select size of steel, consider the use of NO. 13 steel bar

From Table-2, take column 3, select area based on the calculated steel area which is 348.026 mm² for primary reinforcement and 324 mm² for secondary reinforcement.

One can select a steel area of 420 mm² for which lateral spacing is 300 mm. This spacing can be used for both primary and secondary reinforcements since the calculated reinforcement area for both types of reinforcement area close to 420 mm².

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