This problem involves determining the final unit tensile stress in a steel eye-bar that is accidentally shorter by 0.03 inches than the others in a composite bridge member. The basic approach here involves applying principles of elasticity and mechanics of materials, considering how the load is shared among the bars after assembly.

Given Data:

• The total tension force, $T = 200,000$ lb.
• Number of steel eye-bars = 3.
• Each bar's cross-sectional area: $A = 4 \, \text{in} \times 1 \, \text{in} = 4 \, \text{in}^2$.
• Length of each bar (before assembly): $L = 16 \, \text{ft} = 192 \, \text{in}$.
• Middle bar is shorter by $\delta L = 0.03 \, \text{in}$.
• Modulus of elasticity for steel, $E = 30 \times 10^6 \, \text{psi}$ (a typical value).

Approach:

1. Determine the force shared by the outer two bars (longer bars): Since the middle bar is shorter, it will take on less of the load initially, and the outer bars will take more load. The system will balance such that the total tension in the member is 200,000 lb.

2. Deformation of the bars:

   • The outer two bars will stretch more than the middle bar because the middle bar starts out shorter.
   • The extension in the longer bars can be calculated from the compatibility of deformations in the system.

3. Stress in the middle bar: The tension in each bar can be related to the deformation through Hooke's law: $\sigma = \frac{F}{A}$ where $\sigma$ is the stress, $F$ is the force in the bar, and $A$ is the cross-sectional area.

4. Balance of Forces: The total force in the system is the sum of the forces in the three bars, which equals the total tension of 200,000 lb. You’ll need to account for how much of that force each bar is taking based on the deformation difference due to the initial length discrepancy.

Let's go step-by-step through the calculations. I'll compute the final tensile stress in the middle bar.The forces carried by the bars are:

• The middle bar carries approximately $54,167 \, \text{lb}$.
• Each outer bar carries approximately $72,917 \, \text{lb}$.

Now, we can calculate the unit tensile stress in the middle bar, which is the force divided by the cross-sectional area:

$\sigma_{\text{middle}} = \frac{F_{\text{middle}}}{A}$

Let’s compute the tensile stress in the middle bar.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, based on the previous results, we determined that the middle bar carries a force of approximately 54,167 lb. You can calculate the unit tensile stress using the formula:

$\sigma_{\text{middle}} = \frac{F_{\text{middle}}}{A} = \frac{54,167}{4} \approx 13,542 \, \text{psi}$

This shows that the tensile stress in the middle bar would be approximately 13,542 psi. If you'd like further assistance or have more questions, feel free to ask!

Questions to consider:

1. How would the stress distribution change if the outer bars were shorter instead?
2. What would happen if the discrepancy in length were larger than 0.03 inches?
3. How does the modulus of elasticity influence the stress in this setup?
4. How would the system behave if all bars had slightly different lengths?
5. What if the cross-sectional area of the middle bar were different from the outer bars?

Tip:

Always ensure uniform length and assembly precision in composite members to avoid unequal load sharing and potential failure.