Minimum Thickness of Two Way Simply Supported Slab

The minimum thickness of a reinforced concrete slab is primarily determined by deflection control requirements as per Indian Standard Code IS 456:2000. For two-way slabs, the span-to-effective depth ratio is governed by the shorter span and modified based on the aspect ratio ($l_y/l_x$), support conditions, and the amount and type of tension reinforcement.

Given Parameters:

• Slab Type: Two-way simply supported
• Shorter Span ($l_x$): 2.5 m = 2500 mm
• Longer Span ($l_y$): 3.5 m = 3500 mm
• Steel Grade: Fe 250
• Live Load: 3 kN/m² (Note: Live load is generally used to calculate required steel area, which in turn affects the modification factor for deflection control, but minimum thickness for deflection is often initially checked based on assumed steel percentage or code-specified limits.)

Calculation Steps based on IS 456:2000

As per IS 456:2000, Clause 23.2 and Annex D, the deflection control for two-way slabs requires that the ratio of the shorter span to the effective depth ($l_x/d$) should not exceed certain modified values.

First, calculate the aspect ratio:

\( \frac{l_y}{l_x} = \frac{3.5}{2.5} = 1.4 \)

According to IS 456:2000, Table 28, for simply supported slabs with an aspect ratio of 1.4, the basic span-to-effective depth ratio (based on the shorter span) needs to be determined by interpolation between the values provided for \(l_y/l_x = 1.25\) and \(l_y/l_x = 1.5\).

Interpolating for \( l_y/l_x = 1.4 \):

\( \text{Ratio} = 28 + (30 - 28) \times \left( \frac{1.4 - 1.25}{1.50 - 1.25} \right) \)

\( \text{Ratio} = 28 + 2 \times \left( \frac{0.15}{0.25} \right) \)

\( \text{Ratio} = 28 + 2 \times 0.6 \)

\( \text{Ratio} = 28 + 1.2 = 29.2 \)

So, the basic ratio of shorter span to effective depth (\( l_x/d \)) for a simply supported two-way slab with \( l_y/l_x = 1.4 \) is 29.2.

This basic ratio is then multiplied by a modification factor \(k_t\) for tension reinforcement (from IS 456:2000, Figure 4). The factor \(k_t\) depends on the percentage of tension reinforcement (\( p_t \)) and the stress in the steel (\( f_s \)). For Fe 250 steel, \( F_y = 250 \) N/mm². The stress \( f_s \) can be taken as 0.58 \( F_y \) for calculation based on required steel area or minimum steel area.

\( f_s = 0.58 \times 250 = 145 \) N/mm²

To find the minimum thickness, we often assume a minimum or reasonable amount of steel. Let's work backward from the provided options to see which one aligns with the code provisions under typical assumptions for cover and bar diameter.

Consider the option of 72 mm overall thickness (\( D \)). Assume a clear cover of 20 mm and a bar diameter of 6 mm (common for slabs). The effective depth (\( d \)) would be:

\( d = D - \text{Clear Cover} - \frac{\text{Bar Diameter}}{2} \)

\( d = 72 - 20 - \frac{6}{2} = 72 - 20 - 3 = 49 \) mm

Now, calculate the actual span-to-effective depth ratio for this assumed thickness:

\( \frac{l_x}{d} = \frac{2500}{49} \approx 51.02 \)

According to IS 456, the actual ratio (\( 51.02 \)) must be less than or equal to the modified basic ratio (\( 29.2 \times k_t \)).

\( 51.02 \le 29.2 \times k_t \)

This implies that the required modification factor \( k_t \) must be at least:

\( k_t \ge \frac{51.02}{29.2} \approx 1.747 \)

Referring to Figure 4 of IS 456:2000, for \( f_s = 145 \) N/mm², a \( k_t \) value of approximately 1.75 corresponds to a percentage of tension reinforcement (\( p_t \)) between 0.18% and 0.20%. The minimum percentage of tension reinforcement for Fe 250 steel in slabs is 0.12% of the gross cross-sectional area (IS 456:2000, Clause 26.5.2). Providing 0.18% to 0.20% steel is a reasonable amount for a slab.

Therefore, an effective depth of 49 mm (corresponding to an overall depth of 72 mm with 20 mm cover and 6 mm bars) is permissible for deflection control if approximately 0.18% to 0.2% tension steel is provided. This shows that 72 mm is a plausible minimum thickness considering typical detailing practices and code requirements for deflection.

Calculations with other options:

• If D = 62 mm, $d \approx 62 - 20 - 3 = 39$ mm. $l_x/d = 2500/39 \approx 64.1$. Required $k_t = 64.1/29.2 \approx 2.19$. Figure 4 shows maximum $k_t$ is around 2 for very low steel, so 62mm might be too thin.
• If D = 90 mm, $d \approx 90 - 20 - 3 = 67$ mm. $l_x/d = 2500/67 \approx 37.3$. Required $k_t = 37.3/29.2 \approx 1.27$. This $k_t$ corresponds to a higher steel percentage (~0.5-0.6%), meaning 90mm is likely thicker than the minimum required for deflection, especially if less steel is used.
• If D = 100 mm, $d \approx 100 - 20 - 3 = 77$ mm. $l_x/d = 2500/77 \approx 32.5$. Required $k_t = 32.5/29.2 \approx 1.11$. This corresponds to even higher steel percentage (~0.8%), indicating it is much thicker than the minimum.

Based on the analysis, 72 mm is the smallest thickness among the options that satisfies the deflection control criteria with a reasonable amount of tension reinforcement above the minimum required by the code.

Conclusion on Minimum Thickness

Considering the spans, support conditions, steel grade, and the requirements of IS 456 for deflection control, a minimum thickness of 72 mm is found to be adequate based on typical assumptions for cover and bar size and a reasonable percentage of tension reinforcement.