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To solve this design problem, we need to determine the thickness \( t \) of the plate given a few parameters.

### Given:
- Timber column size: \( 6 \, \text{in} \times 6 \, \text{in} \)
- Plate width: \( 6 \, \text{in} \)
- Applied force, \( F = 200 \, \text{kips} = 200 \times 1000 = 200,000 \, \text{lbs} \)
- Timber allowable stress: \( 1100 \, \text{psi} \)
- Steel allowable stress: \( 19 \, \text{ksi} = 19,000 \, \text{psi} \)
- Use same value of \( E \) for timber and steel.

### Approach:
1. **Stress in Timber:**
   The stress in the timber due to the applied load can be calculated as:
   \[
   \sigma_{\text{timber}} = \frac{F}{A_{\text{timber}}}
   \]
   where \( A_{\text{timber}} \) is the cross-sectional area of the timber column. Given the dimensions of \( 6 \, \text{in} \times 6 \, \text{in} \), the area is:
   \[
   A_{\text{timber}} = 6 \times 6 = 36 \, \text{in}^2
   \]
   So,
   \[
   \sigma_{\text{timber}} = \frac{200,000 \, \text{lbs}}{36 \, \text{in}^2} = 5555.56 \, \text{psi}
   \]

2. **Comparing Timber and Steel Stresses:**
   We are given that the allowable timber stress is \( 1100 \, \text{psi} \), which is much lower than the calculated stress, so we need to consider the steel plate for load transfer.

3. **Determining Plate Thickness:**
   The load carried by the steel plate needs to be determined based on the allowable stress for steel, which is \( 19,000 \, \text{psi} \). To prevent failure, the steel plate must carry the same load while keeping the stress under this limit. Thus, the area of the steel plate, \( A_{\text{steel}} \), must satisfy:
   \[
   \sigma_{\text{steel}} = \frac{F}{A_{\text{steel}}} \quad \Rightarrow \quad A_{\text{steel}} = \frac{F}{\sigma_{\text{steel}}}
   \]
   Substituting the values:
   \[
   A_{\text{steel}} = \frac{200,000 \, \text{lbs}}{19,000 \, \text{psi}} = 10.53 \, \text{in}^2
   \]

4. **Plate Dimensions:**
   The width of the steel plate is given as \( 6 \, \text{in} \), so the required thickness \( t \) of the plate is:
   \[
   t = \frac{A_{\text{steel}}}{\text{width}} = \frac{10.53 \, \text{in}^2}{6 \, \text{in}} \approx 1.76 \, \text{in}
   \]

Thus, the required thickness \( t \) of the plate is approximately **1.76 inches**.

---

Would you like further clarification on any step or additional details? 

Here are some questions to expand this topic:
1. How do material properties influence the design of structural components?
2. What are common failure modes in timber and steel under compression?
3. How would a higher or lower steel stress value affect the required plate thickness?
4. How does the modulus of elasticity \( E \) factor into combined timber and steel designs?
5. Can you explain how the safety factor would be applied in this design?

**Tip**: When designing with multiple materials, ensure that the stress distribution across the materials stays within their allowable limits to avoid failure.

If Timber column size is 6 in x 6 in and plate width is 6 in, determine thickness (t) of plate. Given: F=200 kips, timber stress=1100 psi and steel stress=19 ksi, use same value of E. Note: k is kip = 1000 lbs.

Determine the thickness of a steel plate for a timber column with a cross-section of 6 in x 6 in, given an applied force of 200 kips. Includes calculations for stress in timber and steel, using a timber stress of 1100 psi and a steel stress of 19 ksi. Suitable for undergraduate engineering students.

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