A UDL of 10 kN/m and 5 m length moves from the left to the right support of a simply supported beam of span 10 m. The maximum bending moment at 4 m from the left support is:

Asked in SSC JE 30/10/2020 (Morning)

Detailed Explanation

For a moving UDL shorter than the span, the bending moment at a given section is greatest when the section divides the load in the same ratio as it divides the span. Equivalently, the average load intensity on each side of the section is the same.

Note on the answer: the textbook's key gives (a) 70 kN·m, which is the moment with the load sitting over the left 5 m only — not the worst position. The correct maximum, accepted by the commission, is 90 kN·m (option c).

Section C at 4 m divides the span 4 : 6. Load length on the left of C:

\[ \frac{x_1}{5} = \frac{4}{10} \Rightarrow x_1 = 2\ \text{m} \]

So the load runs from 2 m to 7 m; its resultant 50 kN acts at 4.5 m from A.

\[ R_A = \frac{50 \times (10 - 4.5)}{10} = 27.5\ \text{kN} \]

\[ M_C = 27.5 \times 4 - 10 \times 2 \times 1 = 110 - 20 = 90\ \text{kN·m} \]

With the load on 0–5 m instead: \( R_A = 37.5,\ M_C = 37.5 \times 4 - 40 \times 2 = 70\ \text{kN·m} \) — smaller.

Why “90 kN·m” Is Correct

The critical position gives 90 kN·m, larger than any other position, so option (c) is correct.

Why the Other Options Are Wrong

• A. 70 kN·m: 70 kN·m is the moment with the load on the left 5 m — a possible position, but not the worst one (this is the book's key).
• B. 30 kN·m: 30 kN·m is far below the moment for any sensible position of the load.
• D. 50 kN·m: 50 kN·m is also below the critical value.

Key Concepts for Students

• For absolute maximum BM, the centre of the span should bisect the distance between the resultant and the nearest load (point loads).
• Influence line for M at a section of a SS beam is a triangle with peak ab/L.
• Max M at C = w × area of ILD under the loaded length.

Practice more: this is question 26 in Bending Moment and Shear Force MCQs – Part 2. Try the whole set and check every answer instantly.