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Let's break down the problem step by step to solve it.

### Problem Summary:
- The tension member is held together by two bolts, each with a diameter of 20 mm.
- The maximum allowable shear stress \( \tau \) for the bolts is 83 MPa.
- The maximum allowable average normal stress \( \sigma \) for the tension member is 135 MPa.
- We need to determine the maximum load \( P \) that can be applied to the member.

### Step 1: Calculate the shear force on the bolts
The shear force in each bolt is related to the applied force \( P \). Since there are two bolts, the shear force \( V \) in each bolt is:
\[
V = \frac{P}{2}
\]
The shear stress \( \tau \) is given by:
\[
\tau = \frac{V}{A_{\text{bolt}}}
\]
where \( A_{\text{bolt}} \) is the cross-sectional area of the bolt.

The area \( A_{\text{bolt}} \) of a bolt with a diameter \( d \) is:
\[
A_{\text{bolt}} = \frac{\pi d^2}{4}
\]
Substitute \( d = 20 \, \text{mm} = 0.02 \, \text{m} \):
\[
A_{\text{bolt}} = \frac{\pi (0.02)^2}{4} = 3.1416 \times 10^{-4} \, \text{m}^2
\]

Now, solve for the force \( V \) in each bolt using the allowable shear stress:
\[
V = \tau \times A_{\text{bolt}} = 83 \times 10^6 \, \text{Pa} \times 3.1416 \times 10^{-4} \, \text{m}^2
\]
\[
V = 26,095.9 \, \text{N}
\]
Since \( V = \frac{P}{2} \), the total applied force \( P \) due to shear is:
\[
P = 2V = 2 \times 26,095.9 = 52,191.8 \, \text{N}
\]

### Step 2: Calculate the normal stress in the tension member
The normal stress \( \sigma \) in the tension member is related to the applied force \( P \) by:
\[
\sigma = \frac{P}{A_{\text{member}}}
\]
where \( A_{\text{member}} \) is the cross-sectional area of the tension member.

From the problem, the allowable average normal stress \( \sigma = 135 \, \text{MPa} = 135 \times 10^6 \, \text{Pa} \).

Rearranging for \( P \):
\[
P = \sigma \times A_{\text{member}}
\]
However, since the cross-sectional area of the member is not directly provided in the problem, we are assuming that the critical condition will be controlled by the shear in the bolts. Therefore, we only use the shear stress to determine \( P \).

### Step 3: Conclusion
Thus, the maximum load \( P \) that can be applied to the member is **52.19 kN** based on the shear stress in the bolts.

Would you like more details on any of these calculations or explanations?

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Here are 5 related questions for further understanding:
1. How would the maximum load change if the bolt diameter increased to 25 mm?
2. How can you determine if the tension member's normal stress controls the design instead of shear in the bolts?
3. What is the impact of doubling the number of bolts on the maximum allowable load?
4. How is shear stress distributed in a bolt group like this one?
5. How can one calculate the factor of safety for this design?

**Tip:** Always check both shear and normal stress conditions when analyzing connections to ensure you find the governing limit.

The tension member is fastened together using two bolts, one on each side of the member. Each bolt has a diameter of 20 mm. Determine the maximum load P that can be applied to the member if the allowable shear stress for the bolts is τ = 83 MPa and the allowable average normal stress is σ = 135 MPa.

Determine the maximum load P applied to a tension member using two bolts, considering an allowable shear stress of 83 MPa and normal stress of 135 MPa. Detailed solution includes calculating shear force, bolt cross-sectional area, and applying normal stress principles.

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