STEM Problem Detail

Problem Statement

Prompt

Give step-wise answer for given example 14.5

Solution Rendering...

## Circular Slabs

### Table 14.8: Net values of moment and shear force

\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
r & 0 & 2 & 3 & 4.5 & 5 & 6 & M \\
\hline
M_r & 161.8 & 85.3 & -10.35 & -225.3 & -107.97 & 0 & \text{kNm/m} \\
M_\theta & 161.8 & 136.3 & 104.40 & 32.70 & 13.18 & 4 & \text{kNm/m} \\
V & 0 & 102 & 153 & -232.75 & -161 & -40.7 & \text{kN/m} \\
\hline
\end{array}
\]

Maximum bending moment:

$$
M_{\max} = 225.5~\text{kNm}
$$

Effective depth of base slab is given by:

$$
BM = 0.138 \sigma_{ck} b d^2
$$

$$
d = \sqrt{\frac{225.5 \times 10^6}{0.138 \times 25 \times 1000}}
= 255~\text{mm}
$$

Adopt:

$$
d = 26~\text{cm}, \quad D = 30~\text{cm}
$$

---

### Steel for Maximum Radial Hogging Moment

$$
0.87 \sigma_y A_s = 0.36 \sigma_{ck} b x_m
$$

$$
A_s
= \frac{0.36 \times 25 \times 1000 \times 0.48 \times 260}
{0.87 \times 415}
= 3110~\text{mm}^2
$$

---

### Steel for Maximum Sagging Circumferential Moment

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

$$
161.8 \times 10^6
= 0.87 \times 415 \times A_s
\left(260 - \frac{415 A_s}{25 \times 1000}\right)
$$

$$
A_s = 1970~\text{mm}^2
$$

Area of steel for maximum sagging radial moment:

$$
= 161.8 \times 10^6~\text{Nmm} = 1970~\text{mm}^2
$$

---

### Reinforcement Details

Provide $$16$$ mm bars @ $$100$$ mm c/c:

$$
A_s = 2010~\text{mm}^2 > 1970~\text{mm}^2
$$

These bars are placed circumferentially at the bottom of the slab.
This reinforcement is required in the central portion of the slab up to a radius
of $$3r$$ beyond which the moment is hogging.

Provide $$12$$ mm bars @ $$100$$ mm c/c in the form of rings beyond $$2.5$$ m radius up to $$4.0$$ m.

Provide $$8$$ mm bars @ $$100$$ mm c/c in the form of rings for circumferential sagging moment beyond $$4.0$$ m.

Provide $$20$$ mm bars @ $$100$$ mm c/c:

$$
A_s = 3140~\text{mm}^2 > 3110~\text{mm}^2
$$

These bars are placed at the top of the slab over the centre line of the support to
resist hogging moment.

The slab should be checked for shear, development length, and deflection.
The reinforcement details are shown in Fig. 14.20.

---

## Moments and Shear Due to UDL

### Table 14.6: Moment and shear force due to UDL

\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
r & 0 & 2 & 3 & 4.5 & 5 & 6 \\
\hline
M_r & 685.5 & 612 & 516.37 & 301.22 & 210.37 & 0 \\
M_\theta & 688.5 & 663 & 631.12 & 559.40 & 529.12 & 459 \\
V & 0 & 102 & 153 & 229.5 & 255 & 306 \\
\hline
\end{array}
\]

---

### Forces Due to 1370 kN Acting Upward

Let:

$$
a = 6~\text{m}, \quad b = 4.5~\text{m}
$$

For $$r < b$$, radial moment, circumferential moment, and shear force:

$$
M_r = M_\theta
= -\frac{P}{8}
\left[2 \log_e\left(\frac{a}{b}\right)
+ 1 - \left(\frac{b}{a}\right)^2\right]
$$

$$
= -\frac{1370}{8}
\left[2 \log_e\left(\frac{6}{4.5}\right)
+ 1 - \left(\frac{4.5}{6}\right)^2\right]
$$

$$
M_r = M_\theta = -527~\text{kNm/m}, \quad V = 0
$$

For $$r > b$$:

$$
M_r = -\frac{P}{8}
\left[2 \log_e\left(\frac{a}{r}\right)
+ \left(\frac{b}{r}\right)^2 - \left(\frac{b}{a}\right)^2\right]
$$

$$
M_\theta = -\frac{P}{8}
\left[2 \log_e\left(\frac{a}{r}\right)
- \left(\frac{b}{r}\right)^2 + 2 - \left(\frac{b}{a}\right)^2\right]
$$

$$
V = -\frac{P}{2\pi r}
$$

---

### Table 14.7: Moment and shear force due to concentric load

\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
r & 0 & 2 & 3 & 4.5 & 5 & 6 \\
\hline
M_r & -526.72 & -526.72 & -526.72 & -526.72 & -318.34 & 0 \\
M_\theta & -526.72 & -526.72 & -526.72 & -526.72 & -515.95 & -455 \\
V & 0 & 0 & 0 & 462.23 & -416.0 & -346.70 \\
\hline
\end{array}
\]

The net moments and shear force are obtained by algebraically adding the moments
due to each type of load shown in Table 14.8.

---

## Loading on Circular Base Slab

Factored load from roof:

$$
= 1.5 \times 15.9 = 23.9~\text{kN/m}
$$

Factored load from wall:

$$
= 1.5 \times 25 \times \pi \times 11.90 \times 0.20 \times 3.85 = 1080~\text{kN}
$$

Factored load per unit length from wall:

$$
= \frac{1080}{12 - 0.12} = 28.90~\text{kN/m}
$$

Total factored load per unit length:

$$
= 23.9 + 28.9 = 52.8~\text{kN/m}
$$

Downward factored load:

$$
= 1.5 \times 57.75 \times \frac{\pi}{4}(12 - 0.24)^2
+ 1.5 \times 10 \times \frac{\pi}{4} \times 12^2
+ 1.5 \times 593.6
= 13700~\text{kN}
$$

Total vertical upward load:

$$
= 9\pi w = 13700
$$

$$
w = \frac{13700}{9\pi} = 462.25 \approx 465~\text{kN/m}
$$

For analysis of the base slab, it may be assumed to be supported on elastic
supports at its outer periphery and loaded with:

- Uniformly distributed factored load of $$102$$ kN/m$$^2$$ acting downward,
- Upward load of $$465$$ kN/m acting on the periphery of a concentric circle of
$$8$$ m diameter,
- Load of $$52.8$$ kN/m acting downward at the outer periphery.

The edge moment due to monolithic connection with wall is generally very small
compared with the slab moments under loads; therefore, the base slab is treated
as simply supported at its periphery.

---

### For Uniformly Distributed Load of 102 kN/m$$^2$$
Radial moment:
$$
M_r = \frac{3w}{16}(a^2 - r^2)
$$
If Poisson's ratio $$\mu = 0$$.
Circumferential moment:
$$
M_\theta = \frac{w}{16}
\left[(3+\mu)a^2 - r^2(1+3\mu)\right]
$$
If $$\mu = 0$$:
$$
M_\theta = \frac{w}{16}(3a^2 - r^2)
$$
Shear force:
$$
V = 0.5 w r
$$
Radius of slab:
$$
a = 6~\text{m}
$$
---
## Circular Tank Roof Slab (Elevation)
Clear height of tank:
$$
= 4 - 0.15 = 3.85~\text{m}
$$
Load due to filling material:
$$
= 3.85 \times 15 = 57.75~\text{kN/m}^2
$$
Density of material to be stored:
$$
= 15~\text{kN/m}^3
$$
Adopt overall depth of slab:
$$
D = 40~\text{cm}
$$
Dead load due to base slab:
$$
= 0.4 \times 25 = 10~\text{kN/m}^2
$$
Total load:
$$
= 57.75 + 10 = 67.75~\text{kN/m}^2
$$
With load factor of $$1.5$$, factored load on base slab:
$$
= 1.5 \times 67.75 = 102~\text{kN/m}^2
$$
Dead load of roof slab:
$$
= \frac{\pi}{4} \times 12^2 \times 0.15 \times 25 = 424~\text{kN}
$$
Live load on roof slab:
$$
= 1.5~\text{kN/m}^2
$$
Total live load:
$$
= 1.5 \times \frac{\pi}{4} \times 12^2 = 169.6~\text{kN}
$$
Total load from roof:
$$
= 424 + 169.6 = 593.6~\text{kN}
$$
Load per unit length from roof:
$$
= \frac{593.6}{\pi(12 - 0.12)} = 15.9~\text{kN/m}
$$

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Problem #59