A post is made of two timber parts glued together as shown in the figure below. The joint allowable compressive stress is 5.0 MPa and allowable shearing stress is 3.6 MPa. Load P = 40 kN and angle a = 30°. Determine the minimum width of the post (in mm) considering compressive stress alone. Determine the minimum width of the post (in mm) considering shearing stress alone. Which value of the angle, a, (in degrees) gives the maximum shearing stress in the joint?

To solve this problem, we need to analyze both the compressive stress and the shearing stress acting on the glued joint. Let's break it down into steps.

### 1. **Determine Minimum Width Considering Compressive Stress**

The compressive force \( P_c \) can be determined from the load \( P \) acting at an angle \( \alpha \):

\[
P_c = P \cdot \cos(\alpha)
\]

Given:
- \( P = 40 \, \text{kN} = 40,000 \, \text{N} \)
- \( \alpha = 30° \)

Calculating \( P_c \):

\[
P_c = 40,000 \cdot \cos(30°) = 40,000 \cdot \frac{\sqrt{3}}{2} \approx 34,641.0 \, \text{N}
\]

The allowable compressive stress \( \sigma_c \) is given as 5.0 MPa:

\[
\sigma_c = \frac{P_c}{A}
\]
where \( A \) is the cross-sectional area of the post.

To find the area, we need the width \( b \) and height \( h \):

\[
A = b \cdot h
\]

Rearranging for width:

\[
b = \frac{P_c}{\sigma_c \cdot h}
\]

Substituting known values:

\[
b = \frac{34,641.0 \, \text{N}}{5.0 \times 10^6 \, \text{Pa} \cdot h}
\]
\[
b \approx \frac{34,641.0}{5,000,000 \cdot h} = \frac{0.0069282}{h}
\]

### 2. **Determine Minimum Width Considering Shearing Stress**

The shearing force \( P_s \) can also be determined:

\[
P_s = P \cdot \sin(\alpha)
\]

Calculating \( P_s \):

\[
P_s = 40,000 \cdot \sin(30°) = 40,000 \cdot 0.5 = 20,000 \, \text{N}
\]

The allowable shearing stress \( \tau \) is given as 3.6 MPa:

\[
\tau = \frac{P_s}{A_s}
\]

Assuming the shear area is similar to the area used for compressive stress (considering the glued joint), we can use:

\[
A_s = b \cdot h
\]

Rearranging for width again gives:

\[
b = \frac{P_s}{\tau \cdot h}
\]

Substituting known values:

\[
b = \frac{20,000 \, \text{N}}{3.6 \times 10^6 \, \text{Pa} \cdot h}
\]
\[
b \approx \frac{20,000}{3,600,000 \cdot h} \approx \frac{0.0055556}{h}
\]

### 3. **Determine Angle for Maximum Shearing Stress**

The maximum shearing stress occurs when the angle \( a \) maximizes \( P_s \):

\[
P_s = P \cdot \sin(a)
\]

The maximum shearing stress is dependent on the angle, but generally, \( \sin(a) \) achieves its maximum value at \( a = 90° \). However, in practical applications for joints, this often varies based on the configuration.

### Summary of Results

1. **Minimum width considering compressive stress**:
   \[
   b_c = \frac{0.0069282}{h}
   \]

2. **Minimum width considering shearing stress**:
   \[
   b_s = \frac{0.0055556}{h}
   \]

3. **Maximum shearing stress occurs theoretically at \( a = 90° \)**, but practical design considerations must be taken into account.

### Follow-Up Questions

1. What is the height \( h \) of the post?
2. How do the values of compressive and shearing stress interact under load?
3. What other factors might affect the joint's performance?
4. How does changing the angle \( a \) affect the overall strength of the joint?
5. What are the implications of the calculated widths on material selection and joint design?

### Tip

Always consider the combined effects of both compressive and shearing stresses in joint design, as they can interact and lead to failure if not properly accounted for.

Explore the calculation of minimum width for a timber post subjected to compressive and shearing stresses with an allowable compressive stress of 5.0 MPa and shearing stress of 3.6 MPa. Learn how to determine the impact of load and angle on the structural integrity. Suitable for advanced high school students.

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