STEM Problem Detail

Problem Statement

Prompt

Give step-wise answer for given example 14.1

Solution Rendering...

## Distribution of Torsion Reinforcement

Required tension reinforcement:

$$
A_s = 98.4~\text{mm}^2 \quad \text{OK}
$$

Minimum tension reinforcement:

$$
A_{s,\min} = \frac{0.85 b d}{\sigma_y}
= \frac{0.85 \times 500 \times 660}{415}
= 680~\text{mm}^2 < A_s
$$

Provide $$5$$–$$16$$ mm bars:

$$
A_s = 1005~\text{mm}^2 > 981~\text{mm}^2 \quad \text{OK}
$$

---

### Transverse Reinforcement

$$
0.87 \sigma_y A_{sv}
= \frac{T_u x}{b_1 d_1} + \frac{V_u x}{2.5 d}
$$

where,

$b_1 =$ centre-to-centre distance between bars in direction of width

$$
= 500 - 25 - 25 - 8 - 8 - \frac{16}{2} - \frac{16}{2}
= 418~\text{mm}
$$

$d_1 =$ centre-to-centre distance between bars in direction of depth

$$
= d - 25 - 8 - \frac{12}{2}
= 660 - 25 - 8 - 6
= 621~\text{mm}
$$

A clear cover of $$25$$ mm is assumed all around the shear stirrups.

Use $$8$$ mm two-legged vertical stirrups:

$$
A_{sv} = 100.5~\text{mm}^2
$$

Spacing of shear reinforcement:

$$
0.87 \times 415 \times 100.5
= \left( \frac{15 \times 10^6}{418 \times 621}
+ \frac{150 \times 1000}{2.5 \times 621} \right) x
$$

$$
x = 237~\text{mm}
$$

$$
x_1 = 500 - 25 - 25 - 8 = 442~\text{mm}
$$

$$
y_1 = 700 - 25 - 25 - 8 = 642~\text{mm}
$$

$$
x < \frac{x_1 + y_1}{4}
= \frac{442 + 642}{4}
= 271~\text{mm}
$$

$$
x = 237~\text{mm} < 271~\text{mm}
$$

Adopt a spacing of:

$$
x = 200~\text{mm c/c}
$$

---

### Minimum Shear Reinforcement

$$
A_{sv}
= \frac{(\tau_c - \tau_v) b x}{0.87 \sigma_y}
$$

Percentage tension steel:

$$
p = \frac{100 A_s}{bd}
= \frac{100 \times 1005}{500 \times 660}
= 0.30\%
$$

Shear strength of concrete:

$$
\tau_c = 0.38~\text{N/mm}^2
$$

$$
A_{sv}
= \frac{(0.60 - 0.38) \times 500 \times 200}{0.87 \times 415}
= 61~\text{mm}^2
$$

$$
A_{sv} = 100.5~\text{mm}^2 > 61~\text{mm}^2 \quad \text{OK}
$$

Use $$8$$ mm two-legged vertical stirrups @ $$200$$ mm c/c.

Since the depth exceeds $$450$$ mm, provide $$0.1\%$$ steel along the vertical sides,
that is, $$4$$–$$12$$ mm bars.

The reinforcement details are shown in Fig. 12.5.

---

## Solution

Given:

$$
V_u = 150~\text{kN}, \quad T_u = 15~\text{kNm}, \quad M_u = 200~\text{kNm}
$$

Equivalent shear force:

$$
V_e = V_u + 1.6 \frac{T_u}{b}
$$

$$
= 150 + 1.6 \times \frac{15}{0.5}
= 198~\text{kN}
$$

Assume effective cover:

$$
= 40~\text{mm}
$$

Equivalent nominal shear stress:

$$
\tau_e = \frac{V_e}{bd}
= \frac{198 \times 1000}{500 \times 665}
= 0.60~\text{N/mm}^2
$$

Maximum shear stress in M20 concrete:

$$
\tau_{c,\max} = 2.8~\text{N/mm}^2 > 0.60~\text{N/mm}^2 \quad \text{OK}
$$

Assume tension steel:

$$
p = 0.25\%
$$

Shear strength of concrete (Table 8.2):

$$
\tau_c = 0.36~\text{N/mm}^2
$$

$$
\tau_e > \tau_c
$$

\[
\therefore \text{Torsion reinforcement is required in the form of longitudinal
and transverse steel.}
\]

---

### Longitudinal Reinforcement

Equivalent bending moment:

$$
M_{e1} = M_u + M_t
$$

$$
= M_u + \frac{T_u (1 + D/b)}{1.7}
$$

$$
= 200 + \frac{15 \times (1 + 70/50)}{1.7}
= 221.1~\text{kNm}
$$

Since $$M_{e1} > M_u$$, there is no need of compression reinforcement due to twisting moment.

Maximum depth of neutral axis:

$$
x_m = 0.48 d = 0.48 \times 665 = 319~\text{mm}
$$

Factored equivalent bending moment:

$$
M_{e1} = \text{force of tension} \times z
$$

$$
221.1 \times 10^6
= 0.87 \sigma_y A_s
\left( d - \frac{0.87 \sigma_y A_s}{0.36 \sigma_{ck} b} \right)
$$

$$
= 0.87 \times 415 \times A_s
\left( 660 - \frac{415 A_s}{20 \times 500} \right)
$$

or,

$$
A_s = 981~\text{mm}^2
$$

Depth of neutral axis:

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

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Problem #65