Let's work through the questions step-by-step as requested. --- ## **Q 3(a): Axial Load on Steel Column** ### **Step 1: Conceptual Introduction** This problem involves determining the minimum size of a square base plate (“a”) that supports a steel column subjected to an axial load. The column must not exceed allowable stresses: both in the steel column (average normal stress) and in the bearing stress between the base plate and the concrete slab. **Explanation Block:** We’re asked to make sure that the stresses in the steel column and the concrete slab remain within safe limits. The steel column can only carry a certain amount of load per square millimeter (MPa), and the concrete slab can only bear a certain stress before damage occurs. Thus, we need to find the smallest size of the base plate that distributes the load safely. --- ### **Step 2: Formulae** - Average normal stress in the steel column: \[ \sigma_{steel} = \frac{P}{A_{steel}} \] - Bearing stress between base plate and concrete slab: \[ \sigma_{concrete} = \frac{P}{A_{base}} \] Where \( A_{base} = a^2 \) (since the plate is square). **Explanation Block:** These formulas calculate the stress by dividing the applied load by the area over which it is distributed. For the steel, the area is the cross-section of the column; for the concrete, it is the area of the base plate. By setting these stresses to not exceed allowable limits, we can solve for the required dimensions. --- ### **Step 3: Calculation for 1st Question (Steel Column Cross-section)** Given: - \( A_{steel} = 11,400~mm^2 \) - \( \sigma_{steel, max} = 110~MPa \) \[ P = \sigma_{steel, max} \times A_{steel} = 110~MPa \times 11,400~mm^2 \] \[ = 110 \times 11,400~N = 1,254,000~N \] **Explanation Block:** First, we calculate the maximum load the steel column can safely carry using its allowable stress and cross-sectional area. This gives us the maximum allowable axial load, \(P\). --- ### **Step 4: Calculation for 2nd Question (Base Plate Dimension)** Given: - \( \sigma_{concrete, max} = 8~MPa \) - \( P = 1,254,000~N \) - \( \sigma_{concrete} = \frac{P}{a^2} \leq 8~MPa \) \[ a^2 = \frac{P}{\sigma_{concrete, max}} = \frac{1,254,000}{8~N/mm^2} \] \[ a^2 = 156,750~mm^2 \] \[ a = \sqrt{156,750} \approx 396~mm \] **Explanation Block:** Here, we use the maximum allowable bearing stress on the concrete to find the minimum area (and thus minimum side length “a”) for the base plate. We solve for a by rearranging the stress formula and substituting the known values. --- ### **Final Answer** **The minimum required dimension ‘a’ for the base plate is approximately \( \boxed{396~mm} \).** --- *If you need the detailed solution for Q3(b) (the beam section), please confirm so I can proceed step-by-step for that part as well!*