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To solve the problem of determining the downward deflection \( \delta \) due to the applied load \( P \), we can use principles from mechanics of materials. This specific problem is related to the shear deformation of rubber blocks, given their shear modulus \( G \) and the dimensions provided.

### Parameters:
- Applied load: \( P = 890 \, \text{N} \)
- Shear modulus: \( G = 330 \, \text{kPa} = 330 \times 10^3 \, \text{Pa} \)
- Dimensions:
  - \( a = 20 \, \text{mm} = 0.02 \, \text{m} \)
  - \( b = 40 \, \text{mm} = 0.04 \, \text{m} \)
  - \( c = 75 \, \text{mm} = 0.075 \, \text{m} \)

### Formula for Shear Deformation:
The deflection \( \delta \) caused by shear deformation is given by:

\[
\delta = \frac{P \cdot a}{G \cdot A}
\]

Where:
- \( P \) is the applied load,
- \( a \) is the thickness of the rubber block,
- \( G \) is the shear modulus,
- \( A \) is the area of the block through which the load is applied.

The area \( A \) is the cross-sectional area of one block, which can be calculated as:

\[
A = b \times c = 0.04 \, \text{m} \times 0.075 \, \text{m} = 0.003 \, \text{m}^2
\]

### Substituting into the Formula:
Now we can substitute the known values into the equation for \( \delta \):

\[
\delta = \frac{890 \times 0.02}{330 \times 10^3 \times 0.003}
\]

First, calculate the numerator:

\[
890 \times 0.02 = 17.8 \, \text{N} \cdot \text{m}
\]

Now calculate the denominator:

\[
330 \times 10^3 \times 0.003 = 990 \, \text{Pa} \cdot \text{m}^2
\]

Finally, the deflection \( \delta \) is:

\[
\delta = \frac{17.8}{990} \approx 0.018 \, \text{m} = 18 \, \text{mm}
\]

### Conclusion:
The downward deflection \( \delta \) is approximately \( 18 \, \text{mm} \).

---

Do you want further clarification or more details? Here are 5 related questions:

1. How does shear modulus \( G \) affect the deflection of materials?
2. What would happen to the deflection if the load \( P \) increased?
3. How would changing the block thickness \( a \) influence the deflection?
4. How do the area dimensions \( b \) and \( c \) affect the performance of the anti-vibration mount?
5. What are common applications of rubber blocks in vibration damping systems?

**Tip:** Always double-check units when working with different measurements to ensure consistency in calculations.

Two hard rubber blocks (G = 330 kPa) are used in an anti-vibration mount to support a small machine as shown in the figure. Determine the downward deflection δ that will occur for an applied load of P = 890 N. Assume a = 20 mm, b = 40 mm, and c = 75 mm.

Learn how to calculate the downward deflection of a rubber block in an anti-vibration mount with an applied load of 890 N. The problem involves shear deformation, using the shear modulus (G = 330 kPa) and block dimensions of 20 mm, 40 mm, and 75 mm. This is a typical mechanics of materials problem suitable for mechanical engineering students.

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