Act like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:The couple A= 18 kN"m acts in a vertical plane with h=70mm and b=25mm. Determine the stress at (a)Aand (b) B - 20 40 20 - b M A 1 = - PF Am ’ i Dimensions in mm
Question:
Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:
The couple A= 18 kN"m acts in a vertical plane with h=70mm and b=25mm. Determine the stress at
(a)Aand (b) B
- 20 40 20 -
b M
A 1 = -
PF Am
’ i
Dimensions in mm
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Asked by: manish kumar
Created at: 2025-10-06 20:32:08
Asked by: manish kumar
Created at: 2025-10-06 20:32:08
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Let's break this problem down step by step:
---
# Step 1: **Identify the Problem**
Given:
- Couple \( M = 18 \) kN·m = **18,000 N·m**
- Rectangle section with:
- \( h = 70 \) mm = **.070 m**
- \( b = 25 \) mm = **.025 m**
We are to find **the stress at points A and B**.
---
# Step 2: **Find the Moment of Inertia (I)**
For a rectangle about its centroid (neutral axis):
\[
I = \frac{1}{12} b h^3
\]
Plugging in the values (convert mm to m):
\[
I = \frac{1}{12} \times .025 \times (.070)^3
\]
\[
I = \frac{1}{12} \times .025 \times .000343
\]
\[
I = \frac{1}{12} \times .000008575
\]
\[
I = .0000007146 \ \text{m}^4
\]
---
# Step 3: **Bending Stress Formula**
Bending stress at a point is:
\[
\sigma = \frac{M y}{I}
\]
Where:
- \( M \): Moment (in N·m)
- \( y \): Distance from the neutral axis to the point (in meters)
- \( I \): Moment of inertia (in m\(^4\))
---
# Step 4: **Find Distance \( y \) for Points A and B**
- **Neutral axis** is at the centroid: at \( \frac{h}{2} \) from either edge.
- \( y \) is measured from the centroid:
- For **point A**:
- From centroid to A: \( y_A = \frac{h}{2} - .020 = .035 - .020 = .015 \) m (since centroid is 35 mm from bottom, A is 20 mm above centroid).
- For **point B**:
- From centroid to B: \( y_B = -\frac{h}{2} = -.035 \) m (since B is at the very bottom, so it's -35 mm from centroid).
---
# Step 5: **Calculate Stress at Each Point**
## At Point A
\[
\sigma_A = \frac{M y_A}{I}
= \frac{18000 \times .015}{.0000007146}
\]
\[
\sigma_A = \frac{270}{.0000007146}
= 3.78 \times 10^8 \ \text{N/m}^2
= 378 \ \text{MPa}
\]
## At Point B
\[
\sigma_B = \frac{M y_B}{I}
= \frac{18000 \times (-.035)}{.0000007146}
\]
\[
\sigma_B = \frac{-630}{.0000007146}
= -8.82 \times 10^8 \ \text{N/m}^2
= -882 \ \text{MPa}
\]
---
# Step 6: **Interpretation**
- **Positive stress**: tension
- **Negative stress**: compression
---
## **Final Answers**
**(a) At A:**
\[
\boxed{\sigma_A = 378 \ \text{MPa} \ \text{(tension)}}
\]
**(b) At B:**
\[
\boxed{\sigma_B = -882 \ \text{MPa} \ \text{(compression)}}
\]
---
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