STEM Problem Detail

Problem Statement

Prompt

Give step-wise answer for given example

Solution Rendering...

## Two-Way Slabs

Actual:

$$
\frac{L}{d} = \frac{3000}{86} = 34.88 \approx 34.32 \quad \text{OK}
$$

---

## Slabs

At percentage of tension steel:

$$
\frac{100 A_s}{bd}
= \frac{100 \times 250}{1000 \times 86}
= 0.29\%
$$

Shear strength of concrete:

$$
\tau_c = 0.38~\text{N/mm}^2 \quad \text{(Table 8.2)}
$$

$$
k = 1.3 \quad \text{(Table 14.1)}
$$

$$
\tau'_c = 1.3 \times 0.38 > \tau_v \quad \text{OK}
$$

---

### Check for Development Length

The slab is checked for development length.
Let us take the end support A.

Let us curtail $$50\%$$ bars at $$0.1l$$ from the centre of support A.

Moment:

$$
M_u = 0.87 \sigma_y A_s \left(d - \frac{\sigma_y A_s}{\sigma_{ck} b}\right)
$$

$$
= 0.87 \times 500 \times 200
\left(86 - \frac{500 \times 200}{25 \times 1000}\right)
$$

$$
= 7.13 \times 10^6~\text{Nmm} = 7.13~\text{kNm}
$$

Shear force:

$$
V_u = 0.4 \times 14.625 + 0.45 \times 11.25 = 10.9~\text{kN}
$$

Assuming anchorage length:

$$
L_0 = 200~\text{mm}
$$

Including anchorage value of U-hook:

$$
\phi \le
\frac{4 \tau_{bd}}{0.87 \sigma_y}
\left(\frac{1.3 M_u}{V_u} + L_0\right)
$$

$$
= \frac{4 \times 1.4}{0.87 \times 500}
\left(\frac{1.3 \times 7.13 \times 10^6}{10.9 \times 1000} + 200\right)
= 13.5~\text{mm}
$$

Since the actual bar diameter is less than $$13.5$$ mm, the slab is safe in
development length.

These bars must be embedded into the support by at least $$L_d/3$$ distance and
U-hooks must be provided.

---

### Check for Deflection

The slab is checked for deflection.
Let us consider the end span.

$$
\frac{L}{D} \le \alpha \beta \delta \lambda
$$

$$
\beta = 1 \quad \text{since } L < 10~\text{m}
$$

$$
\alpha = 26 \quad \text{for continuous span}
$$

Percentage tension steel:

$$
p_t
= \frac{100 \times 5 \times 1000 / 250}{1000 \times 86}
= 0.23\%
$$

For service stress of $$290$$ MPa in Fe500 grade steel:

$$
\gamma = 1.32 \quad \text{(Table 10.1)}
$$

$$
\delta = 1 \quad \text{since compression reinforcement is zero}
$$

$$
\lambda = 1
$$

Allowable:

$$
\frac{L}{d} = 26 \times 1.32 = 34.32
$$

---

## One-Way Slabs

Effective depth:

$$
d = 105 - 15 - \frac{8}{2} = 86~\text{mm}
$$

Area of steel:

$$
M_u = 0.87 \sigma_y A_s
\left(d - \frac{\sigma_y A_s}{\sigma_{ck} b}\right)
$$

$$
8.13 \times 10^6
= 0.87 \times 500 \times A_s
\left(86 - \frac{500 A_s}{25 \times 1000}\right)
$$

$$
A_s = 229~\text{mm}^2/\text{m}
$$

Use $$8$$ mm bars @ $$200$$ mm c/c:

$$
A_s = 250~\text{mm}^2/\text{m} > 229~\text{mm}^2/\text{m}
$$

---

At E, area of steel:

$$
7.03 \times 10^6
= 0.87 \times 500 \times A_s
\left(86 - \frac{500 A_s}{25 \times 1000}\right)
$$

$$
A_s = 197~\text{mm}^2/\text{m}
$$

Use $$8$$ mm bars @ $$250$$ mm c/c:

$$
A_s = 200~\text{mm}^2/\text{m} \quad \text{OK}
$$

---

At F, area of steel:

$$
5.55 \times 10^6
= 0.87 \times 500 \times A_s
\left(86 - \frac{500 A_s}{25 \times 1000}\right)
$$

$$
A_s = 153.86~\text{mm}^2/\text{m}
$$

Use $$8$$ mm bars @ $$300$$ mm c/c:

$$
A_s = 167~\text{mm}^2/\text{m}
$$

The Code requires that maximum bar spacing cannot exceed $$3d$$:

$$
3d = 3 \times 86 = 258~\text{mm}
$$

Thus, provide $$8$$ mm bars @ $$250$$ mm c/c.

No bar is curtailed since spacing between bars if any curtailment is carried out
is $$>300$$ mm.

Minimum area of high strength main steel:

$$
= 0.12\% \text{ of total cross-sectional area}
$$

$$
= 0.12 \times \frac{1000 \times 105}{100}
= 126~\text{mm}^2/\text{m}
$$

\[
\therefore \text{The actual area provided is OK.}
\]

Temperature reinforcement:

$$
= 0.15\% \text{ of gross concrete area}
$$

$$
= 0.15 \times 105 \times 1000
= 157.5~\text{mm}^2/\text{m}
$$

Use $$6$$ mm M.S. bars @ $$175$$ mm c/c.

---

### Check for Shear

Maximum shear force occurs at the first interior support B in the span AB:

$$
SF = 0.6 \times 14.625 + 0.6 \times 11.25 = 15.525~\text{kN}
$$

Nominal shear stress:

$$
\tau_v = \frac{SF \times 1000}{bd}
= \frac{15.525 \times 1000}{1000 \times 86}
= 0.18~\text{N/mm}^2
$$

---

## Solution (One Metre Wide Strip)

Consider a $$1$$ m wide strip and assume overall thickness of slab as $$100$$ mm.
Owing to symmetry, only half the slab need be considered.

Dead load of slab (DL):

$$
= 0.1 \times 1.0 \times 25 = 2.5~\text{kN/m}
$$

Dead load of $$3$$ cm thick finish:

$$
= 0.03 \times 1.0 \times 25 = 0.75~\text{kN/m}
$$

Total dead load:

$$
= 3.25~\text{kN/m}
$$

Imposed load (IL):

$$
= 2.5 \times 1.0 = 2.5~\text{kN/m}
$$

Factored dead load:

$$
= 1.5 \times 3.25 = 4.875~\text{kN/m}
$$

Factored imposed load:

$$
= 1.5 \times 2.5 = 3.75~\text{kN/m}
$$

Total factored load:

$$
= 8.625~\text{kN/m}
$$

Maximum factored bending moment:

$$
= 8.625 \times 3 = 25.875~\text{kNm}
$$

Coefficients given in Table 11.1 and are tabulated in Table 14.2.

For a balanced design:

$$
M_u = 0.133 \sigma_{ck} b d^2
$$

$$
d = \sqrt{\frac{8.13 \times 10^6}{0.133 \times 25 \times 1000}}
= 49.45~\text{mm}
$$

This effective depth is too small.

Let us adopt overall depth:

$$
D = 105~\text{mm}
$$

Assume $$8$$ mm bars with $$15$$ mm clear cover.

---

### Table 14.2: Maximum Bending Moments

\[
\begin{array}{|c|c|c|c|c|c|}
\hline
& \text{DL} & \text{IL} & \text{DL} & \text{IL} & \text{DL} + \text{IL} \\
\hline
D & 14.625 \times 3 & 11.25 \times 3 & 14.625 \times 3 & 11.25 \times 3 & \\
E & 12 & 10 & 10 & 9 & \\
\hline
\text{Total} & 7.03 & & 8.13 & & 5.55 \\
\hline
\end{array}
\]

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Problem #61