A simple girder of 20 m span is traversed by a moving udl of 6 m length with an intensity of 20 kN/m from left to right. The maximum bending moment and maximum positive and negative shear forces at section 4 m from left support are ___, ___ and ___ respectively.
- A326.4 kN-m, 78 kN and 8 kNCorrect
- B362.4 kN-m, 87 kN and 6 kN
- C342.6 kN-m, 77 kN and 9 kN
- D324.6 kN-m, 75 kN and 9 kN
Solution
Concept
To find maximum bending moment and shear forces at a section due to a moving uniformly distributed load (udl), we use Influence Line Diagrams (ILD). For maximum bending moment, the section divides the load in the same ratio as it divides the span.
Formula
Peak ILD ordinate for BM at section = x(L - x) / L Peak ILD ordinate for positive SF = (L - x) / L Peak ILD ordinate for negative SF = -x / L
Step-by-step solution
1. Span of beam, L = 20 m. Length of moving load, d = 6 m. Load intensity, w = 20 kN/m. Section distance from left, x = 4 m. So, L - x = 16 m. 2. For Maximum Bending Moment at x = 4 m, the load must be divided by the section in the ratio x : (L - x) = 4 : 16 = 1 : 4. Load to the left of section = 6 × (4 / 20) = 1.2 m. Load to the right of section = 6 × (16 / 20) = 4.8 m. Peak ordinate of ILD for BM at section = x(L - x) / L = 4 × 16 / 20 = 3.2 m. Ordinate at left end of load = 3.2 × (4 - 1.2) / 4 = 3.2 × 2.8 / 4 = 2.24 m. Ordinate at right end of load = 3.2 × (16 - 4.8) / 16 = 3.2 × 11.2 / 16 = 2.24 m. Area of ILD under load = (1/2) × (2.24 + 3.2) × 1.2 + (1/2) × (3.2 + 2.24) × 4.8 = (1/2) × (2.24 + 3.2) × 6 = 16.32 m². Maximum Bending Moment = w × Area = 20 × 16.32 = 326.4 kN-m. 3. For Maximum Positive Shear Force, place the load to the right of the section with its head just at the section. Peak ordinate of positive SF ILD = (L - x) / L = 16 / 20 = +0.8. Ordinate at tail of load (6 m from section) = (16 - 6) / 20 = 10 / 20 = +0.5. Area of ILD under load = (1/2) × (0.8 + 0.5) × 6 = 3.9 m. Maximum Positive Shear Force = w × Area = 20 × 3.9 = 78 kN. 4. For Maximum Negative Shear Force, place the load to the left of the section with its head just at the section. Since load length is 6 m and distance to left support is only 4 m, only 4 m of load can be on the beam. Peak ordinate of negative SF ILD = -x / L = -4 / 20 = -0.2. Ordinate at support (left end) = 0. Area of ILD triangle = (1/2) × 4 × 0.2 = 0.4 m. Maximum Negative Shear Force (magnitude) = w × Area = 20 × 0.4 = 8 kN. Answer: (A) 326.4 kN-m, 78 kN and 8 kN
Explanation
Maximum bending moment occurs when the section divides the moving udl in the same ratio as the span, giving 326.4 kN-m. Maximum positive shear force occurs when the udl covers the right part starting from the section, giving 78 kN. Maximum negative shear force occurs when the udl covers the span to the left of the section (only 4 m can fit), giving 8 kN.
Why the other options are wrong
B: Gives 362.4 kN-m, 87 kN and 6 kN due to incorrect placement of load. C: Gives 342.6 kN-m, 77 kN and 9 kN due to calculation errors. D: Gives 324.6 kN-m, 75 kN and 9 kN due to incorrect ILD areas.
Shortcut
Check negative SF first: triangle area = 0.5 × 4 × (4/20) = 0.4, Force = 20 × 0.4 = 8 kN. Only option (A) has 8 kN.
Common mistake
Using the full 6 m length of load for negative shear force even though only 4 m span is available to the left of the section.
Correct answer: A) 326.4 kN-m, 78 kN and 8 kN