To solve the problem:

Given data:

1. Length of tendon (L) = 85 m = $85 \times 10^3 \, \text{mm}$
2. Cross-sectional area (A) = $125 \, \text{mm}^2$
3. Extension ($\Delta L$) = $17 \, \text{mm}$
4. Load (F) = $6.8 \, \text{N}$
5. Poisson's ratio ($\nu$) = $0.68$

We need to calculate the shear modulus ($G$).

Step 1: Calculate the Young’s modulus ($E$)

Using the formula for Young’s modulus:

$E = \frac{\text{Stress}}{\text{Strain}}$

• Stress ($\sigma$):

$\sigma = \frac{F}{A} = \frac{6.8}{125} = 0.0544 \, \text{N/mm}^2$

• Strain ($\varepsilon$):

$\varepsilon = \frac{\Delta L}{L} = \frac{17}{85 \times 10^3} = 0.0002$

Thus:

$E = \frac{\sigma}{\varepsilon} = \frac{0.0544}{0.0002} = 272 \, \text{N/mm}^2$

Step 2: Relating $E$, $G$, and $\nu$

The relationship between Young’s modulus ($E$), shear modulus ($G$), and Poisson’s ratio ($\nu$) is:

$G = \frac{E}{2(1 + \nu)}$

Substitute the known values:

$G = \frac{272}{2(1 + 0.68)} = \frac{272}{2 \times 1.68} = \frac{272}{3.36} = 81 \, \text{N/mm}^2$

Final Answer:

The shear modulus ($G$) is:

$G = 81 \, \text{N/mm}^2$

Would you like further clarification or step-by-step verification?

Related Questions:

1. How is Poisson’s ratio related to shear modulus and Young's modulus?
2. Can the formula $G = \frac{E}{2(1 + \nu)}$ be derived?
3. How would the result change if the tendon had a different Poisson's ratio?
4. What units are commonly used for stress, strain, and modulus in engineering?
5. How does material deformation depend on these constants?

Tip:

Always ensure consistent units throughout your calculations to avoid errors