Question:
A simply-supported rectangular reinforced concrete beam has a width 250 mm and an overall depth 600 mm. The effective span of the beam is 6.23 m. The beam carries a live load of 5 kN/m and super imposed dead load of 5 kN/m, in addition to its own weight. The unit weight of reinforced concrete is 25 kN/m3. Consider the load factor of 1.5 for all stated loads.
The design bending moment (in kN-m) for the limit state of collapse is (rounded off to two decimal places).
A simply-supported rectangular reinforced concrete beam has a width 250 mm and an overall depth 600 mm. The effective span of the beam is 6.23 m. The beam carries a live load of 5 kN/m and super imposed dead load of 5 kN/m, in addition to its own weight. The unit weight of reinforced concrete is 25 kN/m3. Consider the load factor of 1.5 for all stated loads.
The design bending moment (in kN-m) for the limit state of collapse is (rounded off to two decimal places).
The design bending moment (in kN-m) for the limit state of collapse is (rounded off to two decimal places).
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Add the self weight, superimposed dead load and live load first, apply the single load factor of 1.5 to the total, then use $w_uL^2/8$ for the simply supported span.
Updated On: Jul 22, 2026
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Correct Answer: 100.06
Solution and Explanation
Step 1: Understanding the Question.
We need the factored (design) bending moment on a simply supported beam under a uniformly distributed load. The load has three parts: the beam's own weight, the superimposed dead load, and the live load, and every one of them is scaled by the same load factor of 1.5 for the limit state of collapse (ultimate limit state).
Step 2: Compute the self weight.
Self weight per metre length = unit weight $\times$ cross-section area:
\[ w_{self} = 25 \times (0.25 \times 0.6) = 25 \times 0.15 = 3.75 \text{ kN/m} \]
Step 3: Total service (unfactored) load.
\[ w = w_{self} + w_{SDL} + w_{LL} = 3.75 + 5 + 5 = 13.75 \text{ kN/m} \]
Step 4: Apply the load factor.
The problem states the same load factor of 1.5 applies to all the stated loads, so we can factor the total directly:
\[ w_u = 1.5 \times 13.75 = 20.625 \text{ kN/m} \]
Step 5: Compute the design bending moment.
For a simply supported beam under a uniform load, the maximum bending moment occurs at midspan and equals $w_u L^2/8$, where $L = 6.23$ m is the effective span:
\[ M_u = \frac{w_u L^2}{8} = \frac{20.625 \times (6.23)^2}{8} = \frac{20.625 \times 38.8129}{8} = \frac{800.516}{8} \]
\[ M_u = 100.06 \text{ kN-m} \]
Final Answer:
The design bending moment for the limit state of collapse is 100.06 kN-m.
\[ \boxed{M_u = 100.06 \text{ kN-m}} \]
We need the factored (design) bending moment on a simply supported beam under a uniformly distributed load. The load has three parts: the beam's own weight, the superimposed dead load, and the live load, and every one of them is scaled by the same load factor of 1.5 for the limit state of collapse (ultimate limit state).
Step 2: Compute the self weight.
Self weight per metre length = unit weight $\times$ cross-section area:
\[ w_{self} = 25 \times (0.25 \times 0.6) = 25 \times 0.15 = 3.75 \text{ kN/m} \]
Step 3: Total service (unfactored) load.
\[ w = w_{self} + w_{SDL} + w_{LL} = 3.75 + 5 + 5 = 13.75 \text{ kN/m} \]
Step 4: Apply the load factor.
The problem states the same load factor of 1.5 applies to all the stated loads, so we can factor the total directly:
\[ w_u = 1.5 \times 13.75 = 20.625 \text{ kN/m} \]
Step 5: Compute the design bending moment.
For a simply supported beam under a uniform load, the maximum bending moment occurs at midspan and equals $w_u L^2/8$, where $L = 6.23$ m is the effective span:
\[ M_u = \frac{w_u L^2}{8} = \frac{20.625 \times (6.23)^2}{8} = \frac{20.625 \times 38.8129}{8} = \frac{800.516}{8} \]
\[ M_u = 100.06 \text{ kN-m} \]
Final Answer:
The design bending moment for the limit state of collapse is 100.06 kN-m.
\[ \boxed{M_u = 100.06 \text{ kN-m}} \]
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