To determine the thickness $t$ of the plate, we will use the bending stress formula since the force $P$ produces a bending moment in the frame. Here’s the step-by-step solution:

Given Data:

1. Maximum force, $P = 20 \, \text{kN} = 20,000 \, \text{N}$.
2. Tensile yield strength, $\sigma_{\text{yield}} = 380 \, \text{MPa} = 380 \times 10^6 \, \text{Pa}$.
3. Factor of safety, $\text{FoS} = 2.5$.
4. Distances from the diagram:
   • $a = 200 \, \text{mm} = 0.2 \, \text{m}$,
   • $b = 150 \, \text{mm} = 0.15 \, \text{m}$.

The design stress is: $\sigma_{\text{design}} = \frac{\sigma_{\text{yield}}}{\text{FoS}} = \frac{380 \times 10^6}{2.5} = 152 \times 10^6 \, \text{Pa}.$

Step 1: Bending Moment

The bending moment at the critical section of the plate (where the vertical load acts) is: $M = P \cdot a = 20,000 \cdot 0.2 = 4,000 \, \text{Nm}.$

Step 2: Section Modulus

For a rectangular cross-section, the section modulus $Z$ is: $Z = \frac{b \cdot t^2}{6}.$ Here, $b = 0.15 \, \text{m}$, and $t$ is the thickness to be determined.

Step 3: Bending Stress Formula

The maximum bending stress is: $\sigma = \frac{M}{Z}.$ Substitute $Z$ into the equation: $\sigma = \frac{6 \cdot M}{b \cdot t^2}.$ Rearranging for $t$: $t^2 = \frac{6 \cdot M}{b \cdot \sigma_{\text{design}}}.$

Step 4: Substitute Values

$t^2 = \frac{6 \cdot 4,000}{0.15 \cdot 152 \times 10^6}.$ Simplify: [ t^2 = \frac{24,000}{22.8 \times