Problem 6 (Similar CE BD. May 2012) A $$ 2.4 \mathrm{~m} imes 2.4 \mathrm{~m} $$ square footing supports a $$ 350 \mathrm{~mm} imes 400 \mathrm{~mm} $$ column. Given data: Thickness of footing $$ \quad=0.60 \mathrm{~m} $$ Total factored axial load $$ \quad=1,818 \mathrm{kN} $$. Materials : Concrete , $$ \mathrm{f}_{\mathrm{c}}{ }^{\prime}=20.7 \mathrm{MPa} $$ Steel bars, $$ \mathrm{f}_{\mathrm{y}}=280 \mathrm{MPa} $$ ( $$ 20 \mathrm{~mm} \phi $$ bothways) Clear bar covering $$ \quad=70 \mathrm{~mm} $$. Strength reduction factors $$ =0.75 $$ for shear $$ =0.90 $$ for bending. 1. Calculate the shear stress in the footing (MPa) at critical section for wide-beam action. A. 0.576 B. 0.657 C. 0.758 D. 0.41 2. Calculate the shear stress in the footing (MPa) at critical section for two-way action. A. 1.406 B. 1.52 C. 1.16 D. 1.83 3. Determine the number of pieces of $$ 20 \mathrm{~mm} \phi $$ reinforcing bars required in each direction of the footing. A. 10 B. 14 C. 17 D. 20

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Problem 6 (Similar CE BD. May 2012) A $$ 2.4 \mathrm{~m} 	imes 2.4 \mathrm{~m} $$ square footing supports a $$ 350 \mathrm{~mm} 	imes 400 \mathrm{~mm} $$ column. Given data: Thickness of footing $$ \quad=0.60 \mathrm{~m} $$ Total factored axial load $$ \quad=1,818 \mathrm{kN} $$. Materials : Concrete , $$ \mathrm{f}_{\mathrm{c}}{ }^{\prime}=20.7 \mathrm{MPa} $$ Steel bars, $$ \mathrm{f}_{\mathrm{y}}=280 \mathrm{MPa} $$ ( $$ 20 \mathrm{~mm} \phi $$ bothways) Clear bar covering $$ \quad=70 \mathrm{~mm} $$. Strength reduction factors $$ =0.75 $$ for shear $$ =0.90 $$ for bending. 1. Calculate the shear stress in the footing (MPa) at critical section for wide-beam action. A. 0.576 B. 0.657 C. 0.758 D. 0.41 2. Calculate the shear stress in the footing (MPa) at critical section for two-way action. A. 1.406 B. 1.52 C. 1.16 D. 1.83 3. Determine the number of pieces of $$ 20 \mathrm{~mm} \phi $$ reinforcing bars required in each direction of the footing. A. 10 B. 14 C. 17 D. 20

UpStudy AI · Accepted Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the shear stress in the footing at critical section for wide-beam action.

**Known Conditions:**
- Footing dimensions: $$ 2.4 \, \text{m} \times 2.4 \, \text{m} $$
- Column dimensions: $$ 350 \, \text{mm} \times 400 \, \text{mm} $$
- Thickness of footing: $$ 0.60 \, \text{m} $$
- Total factored axial load: $$ 1,818 \, \text{kN} $$
- Strength reduction factor for shear: $$ 0.75 $$

**Calculating the area of the footing:**
$$
A_f = 2.4 \, \text{m} \times 2.4 \, \text{m} = 5.76 \, \text{m}^2
$$

**Calculating the shear force:**
The shear force $$ V $$ is equal to the total factored axial load:
$$
V = 1,818 \, \text{kN} = 1,818,000 \, \text{N}
$$

**Calculating the shear stress:**
The shear stress $$ \tau $$ at the critical section for wide-beam action is given by:
$$
\tau = \frac{V}{A_f}
$$
Substituting the values:
$$
\tau = \frac{1,818,000 \, \text{N}}{5.76 \, \text{m}^2}
$$

Now, let's calculate this value.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{1818000}{5.76}$$
- step1: Convert the expressions:
  $$\frac{1818000}{\frac{144}{25}}$$
- step2: Multiply by the reciprocal:
  $$1818000\times \frac{25}{144}$$
- step3: Reduce the numbers:
  $$12625\times 25$$
- step4: Multiply the numbers:
  $$315625$$
The calculated shear stress for wide-beam action is:
$$
\tau = 315625 \, \text{N/m}^2 = 0.316 \, \text{MPa}
$$

However, we need to account for the strength reduction factor for shear:
$$
\tau_{reduced} = \tau \times 0.75 = 0.316 \times 0.75
$$

Let's calculate this value.
Calculate the value by following steps:
- step0: Calculate:
  $$0.316\times 0.75$$
- step1: Multiply the numbers:
  $$0.237$$
The reduced shear stress is:
$$
\tau_{reduced} = 0.237 \, \text{MPa}
$$

This value does not match any of the provided options, indicating a potential misunderstanding in the calculation or the interpretation of the problem. Let's proceed to the next part of the problem.

### Step 2: Calculate the shear stress in the footing at critical section for two-way action.

**For two-way action:**
The shear stress for two-way action is calculated using the formula:
$$
\tau = \frac{V}{A_f} \times \text{factor}
$$
Where the factor is typically higher due to the two-way action.

Given that we have the same shear force $$ V $$ and area $$ A_f $$, we will calculate the shear stress without the reduction factor first, and then apply the appropriate factor for two-way action.

Let's calculate the shear stress for two-way action. The area remains the same, and we will use the same shear force $$ V $$.

Using the same area:
$$
\tau = \frac{1,818,000 \, \text{N}}{5.76 \, \text{m}^2}
$$

We already calculated this value as $$ 0.316 \, \text{MPa} $$. Now, we will apply the factor for two-way action, which is typically higher.

Let's assume a factor of $$ 1.5 $$ for two-way action (this is a common assumption, but it may vary based on specific design codes).

Calculating the shear stress for two-way action:
$$
\tau_{two-way} = \tau \times 1.5
$$

Let's calculate this value.
Calculate the value by following steps:
- step0: Calculate:
  $$0.316\times 1.5$$
- step1: Multiply the numbers:
  $$0.474$$
The calculated shear stress for two-way action is:
$$
\tau_{two-way} = 0.474 \, \text{MPa}
$$

This value also does not match the provided options, indicating a potential misunderstanding in the calculation or the interpretation of the problem.

### Step 3: Determine the number of pieces of $$ 20 \, \text{mm} \phi $$ reinforcing bars required in each direction of the footing.

**Known Conditions:**
- Diameter of bars: $$ 20 \, \text{mm} $$
- Area of one bar:
$$
A_b = \frac{\pi}{4} \times d^2 = \frac{\pi}{4} \times (20 \, \text{mm})^2
$$

**Calculating the area of one bar:**
Let's calculate the area of one $$ 20 \, \text{mm} $$ bar.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\pi }{4}\times 20^{2}$$
- step1: Rewrite the expression:
  $$\frac{\pi }{4}\times 4^{2}\times 5^{2}$$
- step2: Reduce the numbers:
  $$\pi \times 4\times 5^{2}$$
- step3: Rewrite the expression:
  $$\pi \times 100$$
- step4: Reorder the terms:
  $$100\pi $$
The area of one $$ 20 \, \text{mm} $$ reinforcing bar is approximately:
$$
A_b \approx 314.16 \, \text{mm}^2
$$

**Total area of steel required:**
To determine the total area of steel required, we need to consider the factored load and the allowable stress in the steel. The total factored axial load is $$ 1,818 \, \text{kN} $$ or $$ 1,818,000 \, \text{N} $$.

Using the yield strength of steel $$ f_y = 280 \, \text{MPa} $$:
$$
A_{total} = \frac{V}{f_y} = \frac{1,818,000 \, \text{N}}{280 \, \text{MPa}} = \frac{1,818,000 \, \text{N}}{280 \times 10^6 \, \text{N/m}^2}
$$

Let's calculate this value.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{1818000}{\left(280\times 10^{6}\right)}$$
- step1: Remove the parentheses:
  $$\frac{1818000}{280\times 10^{6}}$$
- step2: Reduce the fraction:
  $$\frac{45450}{7\times 10^{6}}$$
- step3: Factor the expression:
  $$\frac{10\times 4545}{7\times 10^{6}}$$
- step4: Reduce the fraction:
  $$\frac{4545}{7\times 10^{5}}$$
- step5: Factor the expression:
  $$\frac{5\times 909}{7\times 10^{5}}$$
- step6: Factor the expression:
  $$\frac{5\times 909}{7\times 5^{5}\times 2^{5}}$$
- step7: Reduce the fraction:
  $$\frac{909}{7\times 5^{4}\times 2^{5}}$$
- step8: Simplify:
  $$\frac{909}{140000}$$
The total area of steel required is approximately:
$$
A_{total} \approx 0.00649 \, \text{m}^2 = 6492 \, \text{mm}^2
$$

**Calculating the number of bars required:**
To find the number of bars $$ n $$ required, we use the formula:
$$
n = \frac{A_{total}}{A_b}
$$
Substituting the values:
$$
n = \frac{6492 \, \text{mm}^2}{314.16 \, \text{mm}^2}
$$

Let's calculate this value.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{6492}{314.16}$$
- step1: Convert the expressions:
  $$\frac{6492}{\frac{7854}{25}}$$
- step2: Multiply by the reciprocal:
  $$6492\times \frac{25}{7854}$$
- step3: Reduce the numbers:
  $$1082\times \frac{25}{1309}$$
- step4: Multiply:
  $$\frac{1082\times 25}{1309}$$
- step5: Multiply:
  $$\frac{27050}{1309}$$
The calculated number of bars required is approximately:
$$
n \approx 20.66
$$

Since we cannot have a fraction of a bar, we round up to the nearest whole number:
$$
n = 21
$$

However, this value does not match any of the provided options. Let's summarize the results:

### Summary of Results:
1. **Shear stress in the footing (MPa) at critical section for wide-beam action:** 
   - Calculated: $$ 0.237 \, \text{MPa} $$ (not matching options)
   
2. **Shear stress in the footing (MPa) at critical section for two-way action:** 
   - Calculated: $$ 0.474 \, \text{MPa} $$ (not matching options)
   
3. **Number of pieces of $$ 20 \, \text{mm} \phi $$ reinforcing bars required in each direction of the footing:** 
   - Calculated: $$ 21 $$ (not matching options)

It appears there may be a misunderstanding in the interpretation of the problem or the calculations. Please verify the parameters or provide additional context if available.
Shear stress calculations and reinforcing bar requirements do not match the provided options, indicating possible errors in the calculations or interpretations.

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