Combined footing design — verify geometry and area from load data and SBC Two reinforced concrete columns of sizes 50 cm × 50 cm and 60 cm × 60 cm carry 80 t and 120 t respectively. The centre-to-centre spacing between the columns is 5.00 m. The allowable (permissible) bearing capacity of soil is 20 t/m². If the footing must not project more than 0.25 m beyond the outside face of the smaller column, which set of design parameters is correct?

Difficulty: Medium

Correct Answer: All the above

Explanation:


Introduction / Context:
Combined footings are used when two columns are close and individual footings would overlap or when eccentric loading must be balanced. The footing is proportioned so that the resultant of column loads passes through the centroid of the footing area, keeping soil pressure uniform (or linearly distributed without tension).


Given Data / Assumptions:

  • Column loads: 80 t and 120 t; spacing = 5.00 m.
  • Soil bearing capacity (allowable) q_allow = 20 t/m².
  • Projection limit beyond the smaller column = 0.25 m.
  • Neglect footing self-weight, or include a small allowance folded into rounding.


Concept / Approach:

The total area A needed is approximately total load divided by allowable pressure. The centroid of the two loads from the smaller column equals the weighted average of positions. The footing length is chosen so that the centroid of area coincides with the resultant load location; width follows from A = L * B within projection constraints.


Step-by-Step Solution:

Compute total load W = 80 + 120 = 200 t.Area A ≈ W / q_allow = 200 / 20 = 10 m². Allowing ≈10% for self-weight and rounding → about 11 m².Centroid of loads from the smaller column: x̄ = (080 + 5120) / 200 = 600 / 200 = 3.00 m → matches (a).Limit the near-side projection to 0.25 m; choose an overall length L so that the load resultant coincides with the footing area centroid. A practical solution consistent with the options is L = 7.00 m.Width B = A / L ≈ 11 / 7 ≈ 1.57 m → matches (d).


Verification / Alternative check:

With L = 7 m and B ≈ 1.57 m, the centroid of the rectangular area is at 3.5 m from either end. By arranging the smaller-column face 0.25 m from the near end, the smaller-column centre lies at 0.25 + 0.25 (half column) = 0.50 m from the end; hence the centroid of loads at 3.00 m from that centre puts the resultant at 3.50 m from the same end, aligning with the area centroid.


Why Other Options Are Wrong:

  • Each individual statement (a)–(d) is correct; together they form a consistent design set, so the best choice is “All the above”.


Common Pitfalls:

  • Forgetting to include projection and column sizes when positioning the footing.
  • Not checking for shear and bending after sizing based on area alone.


Final Answer:

All the above.

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