--- title: "Four wheel loads on 20 m beam: absolute max BM" subject: "Theory of Structures I" chapter: 5 key: "semester_4/theory_of_structures_i:four-wheel-30-80-60-30-20m" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/four-wheel-30-80-60-30-20m site: BCT Engineering --- # Four wheel loads on 20 m beam: absolute max BM Theory of Structures I · Chapter 5: Influence Lines for Simple Structures - Asked in: 2078 Chaitra, 7 marks - Video: [Influence Line Diagrams for Simply Supported Beams - Problem No 6 (with 4 wheel loads)](https://www.youtube.com/watch?v=bbsIf1bZvVg) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/four-wheel-30-80-60-30-20m - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5 ## Question Four-wheel loads of 30 kN, 80 kN, 60 kN and 30 kN as shown in figure move from right to left on a simply supported beam of span 20 m. Calculate absolute maximum bending moment that can be expected in the beam. [Figure: loads 30, 80, 60, 30 kN with spacings 3 m, 4 m, 3 m.] ## Short answer Place the 80 kN load 0.90 m from mid-span with the resultant opposite. Absolute maximum bending moment is 738.1 kN m under the 80 kN load. ## Answer **Data:** span $L = 20$ m; loads 30, 80, 60, 30 kN with spacings 3 m, 4 m, 3 m (taken from the 30 kN at one end). Total $W = 200$ kN. The absolute maximum moment does not depend on the direction of travel. #### Position of the resultant Measured from the first 30 kN load: $$ \bar{x} = \frac{30(0) + 80(3) + 60(7) + 30(10)}{200} = \frac{960}{200} = 4.8\ \text{m} $$ The 80 kN load (3 m from the first load) is the nearest to the resultant, at 1.8 m from it. #### Absolute maximum BM It occurs under a load when that load and the resultant are equidistant from the centre of the span. Take the 80 kN load as the critical one: it is placed 0.90 m from mid-span, and the resultant is on the other side of mid-span. ```text 30 80 60 30 (kN) | 3 | 4 | 3 | x=6.10 9.10 13.10 16.10 m from left support 0.....................20 ``` Positions from the left support: 30 kN at 6.10 m, 80 kN at 9.10 m, 60 kN at 13.10 m, 30 kN at 16.10 m; resultant at 10.90 m. All four loads are on the span. $$ R_A = 200\times\frac{20-10.90}{20} = 91.0\ \text{kN} $$ $$ M_{max} = R_A\times 9.10 - 30\times 3 = 91.0\times 9.10 - 90 = 738.1\ \text{kN m} $$ Check: the other loads give a smaller moment (under the 60 kN load, $M = 702.1$ kN m). **Answer: absolute maximum bending moment = 738.1 kN m, under the 80 kN load.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the resultant of the loads and where is it from the first 30 kN load? A. 200 kN at 4.8 m B. 200 kN at 5 m C. 200 kN at 3 m D. 160 kN at 4.8 m Answer: A. x = 960/200 = 4.8 m. (Key point: Resultant) 2. Which load is chosen as critical? A. The 60 kN load B. The 80 kN load, nearest the resultant C. The 30 kN load D. The heaviest and farthest load Answer: B. The 80 kN load is 1.8 m from the resultant, the nearest. (Key point: Critical load) 3. How is the critical load placed? A. The load at mid-span B. The resultant at mid-span C. Load and resultant equidistant from mid-span (0.90 m each side) D. The load at the support Answer: C. The mid-point lies halfway between the load and the resultant, so each is 0.90 m from it. (Key point: Placement) 4. What is the absolute maximum bending moment? A. 702.1 kN m B. 800 kN m C. 600 kN m D. 738.1 kN m Answer: D. M = 91 x 9.10 - 30 x 3 = 738.1 kN m. 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