Responder
To support an axial pull of 167.7 kN with a factor of safety of 6, the tie-bar should have a base width of approximately 50 mm, a height of about 116.4 mm, and a mean thickness of 12.7 mm.
Solución
To find suitable dimensions for the tie-bar of angle steel, we can follow these steps:
1. **Extract Known Conditions:**
- Breaking strength of mild steel, \( \sigma_b = 389 \, \text{MPa} \)
- Axial pull, \( P = 167.7 \, \text{kN} = 167700 \, \text{N} \)
- Mean thickness of metal, \( t = 12.7 \, \text{mm} = 0.0127 \, \text{m} \)
- Factor of safety, \( n = 6 \)
2. **Calculate Allowable Stress:**
The allowable stress \( \sigma_a \) can be calculated using the breaking strength and the factor of safety:
\[
\sigma_a = \frac{\sigma_b}{n}
\]
3. **Calculate Required Cross-Sectional Area:**
The required cross-sectional area \( A \) can be calculated using the formula:
\[
A = \frac{P}{\sigma_a}
\]
4. **Determine Dimensions:**
For an angle steel section, we can assume a standard shape and calculate the dimensions based on the required area and the mean thickness.
Let's perform the calculations step by step.
### Step 1: Calculate Allowable Stress
\[
\sigma_a = \frac{389 \, \text{MPa}}{6}
\]
### Step 2: Calculate Required Cross-Sectional Area
\[
A = \frac{167700 \, \text{N}}{\sigma_a}
\]
Now, let's calculate these values.
Calculate the value by following steps:
- step0: Calculate:
\(\frac{389}{6}\)
Calculate or simplify the expression \( 167700/(389/6) \).
Calculate the value by following steps:
- step0: Calculate:
\(\frac{167700}{\frac{389}{6}}\)
- step1: Multiply by the reciprocal:
\(167700\times \frac{6}{389}\)
- step2: Multiply:
\(\frac{167700\times 6}{389}\)
- step3: Multiply:
\(\frac{1006200}{389}\)
### Step 1: Calculate Allowable Stress
The allowable stress \( \sigma_a \) is calculated as follows:
\[
\sigma_a = \frac{389 \, \text{MPa}}{6} \approx 64.83 \, \text{MPa}
\]
### Step 2: Calculate Required Cross-Sectional Area
Now, we can calculate the required cross-sectional area \( A \):
\[
A = \frac{167700 \, \text{N}}{64.83 \, \text{MPa}} \approx 2586.63 \, \text{mm}^2
\]
### Step 3: Determine Dimensions
For an angle steel section, the cross-sectional area can be approximated based on the dimensions of the angle. The area \( A \) of an angle section can be calculated using the formula:
\[
A = 2 \left( b \cdot t + \frac{(h - t) \cdot t}{2} \right)
\]
where:
- \( b \) is the base width,
- \( h \) is the height of the angle,
- \( t \) is the thickness.
Assuming a standard angle section, we can rearrange the formula to find suitable dimensions \( b \) and \( h \) that satisfy the area requirement of \( 2586.63 \, \text{mm}^2 \).
Let's assume a base width \( b \) and calculate the height \( h \) based on the required area. We can set \( b = 50 \, \text{mm} \) as a starting point and solve for \( h \).
The equation becomes:
\[
2586.63 = 2 \left( 50 \cdot 12.7 + \frac{(h - 12.7) \cdot 12.7}{2} \right)
\]
Now, let's solve for \( h \).
Solve the equation by following steps:
- step0: Solve for \(h\):
\(2586.63=2\left(50\times 12.7+\frac{\left(h-12.7\right)\times 12.7}{2}\right)\)
- step1: Simplify:
\(2586.63=\frac{110871}{100}+\frac{127}{10}h\)
- step2: Swap the sides:
\(\frac{110871}{100}+\frac{127}{10}h=2586.63\)
- step3: Move the constant to the right side:
\(\frac{127}{10}h=2586.63-\frac{110871}{100}\)
- step4: Subtract the numbers:
\(\frac{127}{10}h=\frac{36948}{25}\)
- step5: Multiply by the reciprocal:
\(\frac{127}{10}h\times \frac{10}{127}=\frac{36948}{25}\times \frac{10}{127}\)
- step6: Multiply:
\(h=\frac{73896}{635}\)
Calculate or simplify the expression \( 73896/635 \).
Calculate the value by following steps:
- step0: Calculate:
\(\frac{73896}{635}\)
### Step 3: Solve for Height \( h \)
From the equation, we found:
\[
h = \frac{73896}{635} \approx 116.37 \, \text{mm}
\]
### Summary of Dimensions
Based on our calculations, we can propose the following dimensions for the angle steel tie-bar:
- Base width \( b \approx 50 \, \text{mm} \)
- Height \( h \approx 116.37 \, \text{mm} \)
- Mean thickness \( t = 12.7 \, \text{mm} \)
These dimensions should provide a suitable cross-sectional area to safely support the axial pull of \( 167.7 \, \text{kN} \) with a factor of safety of 6.
Respondido por UpStudy AI y revisado por un tutor profesional
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