--- title: "BM at section under moving load train" subject: "Theory of Structures I" chapter: 5 key: "semester_4/theory_of_structures_i:load-train-bm-section-method" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/load-train-bm-section-method site: BCT Engineering --- # BM at section under moving load train Theory of Structures I · Chapter 5: Influence Lines for Simple Structures - Asked in: 2078 Chaitra, 6 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/load-train-bm-section-method - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5 ## Question How will you determine the bending moment at a section when a load train $P_1, P_2, P_3, P_4$ and $P_5$ move in a simply supported beam. Assume all missing data suitably. ## Short answer Place the train so that a load stands over the section satisfying (W_L+P_k)/a ≥ W/L ≥ W_L/a, where W_L is total load left of section before placing P_k. Then compute M_C = ∑ P_i y_i using the triangular ILD; evaluate for each critical load and take the maximum. ## Answer For a train of loads $P_1,\dots,P_5$ at fixed spacings crossing a simply supported beam of span $L$, the bending moment at a section $C$ is found from the ILD of $M_C$ by trying the critical position of the train. #### ILD for $M_C$ Let $C$ be at $a$ from the left support and $b = L-a$. The ILD is a triangle with peak $ab/L$ under $C$. Left slope $= b/L$, right slope $= -a/L$. ```text A o-------------C-------------o B ^ ab/L P1 P2 P3 P4 P5 -> moving ``` $$ M_C = \sum P_i\,y_i $$ #### Condition for maximum Let $W$ be the total load on the span and $W_L$ the load on the left of $C$. Shifting the train to the right by $\delta$: $$ \Delta M_C = \frac{\delta}{L}\left(W_L\,b - W_R\,a\right) = \delta\left(W_L - \frac{W\,a}{L}\right) $$ $M_C$ is maximum when $\Delta M_C$ changes sign from positive to negative as a load $P_k$ crosses $C$: $$ \frac{W_L + P_k}{a} \ge \frac{W}{L} \ge \frac{W_L}{a} $$ where $W_L$ excludes $P_k$ (load just right of $C$) and $W_L + P_k$ includes it (just left of $C$). At such a position the average load on the left of $C$ equals the average load on the whole span, and **$P_k$ stands over the section**. #### Steps 1. Place the heaviest load near $C$ and test the condition; move to the next load if it fails. Remember that loads leaving or entering the span change $W$. 2. At the critical position, read the ordinates under every load of the train from the ILD (similar triangles). 3. Compute $M_{C,max} = P_1y_1 + P_2y_2 + P_3y_3 + P_4y_4 + P_5y_5$. 4. If more than one load satisfies the test, compute for each and take the largest. 5. For the absolute maximum moment anywhere, put a load and the resultant of the loads on the span symmetric about mid-span. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the bending moment at section C under a load train computed? A. M_C = sum of P_i y_i from the ILD of M_C B. M_C = total load x peak ordinate C. M_C = W x L / 4 D. M_C = R_A x a only Answer: A. Each wheel load is multiplied by the ILD ordinate beneath it. (Key point: Basis) 2. Which condition shows a load P_k at C gives the maximum M_C? A. W_L = W_R B. (W_L + P_k)/a >= W/L >= W_L/a C. W_L/a = W_R/b exactly at mid-span D. P_k is the lightest load Answer: B. The change in M_C changes sign from positive to negative as P_k crosses C, giving this inequality. (Key point: Critical condition) 3. At the critical position, the average load on the left of C equals A. the average load on the right of C only B. half the total load C. the average load on the whole span D. the heaviest load Answer: C. W_L/a is approximately W/L at the critical position. (Key point: Meaning) 4. If more than one load satisfies the test, what is done? A. Take the first one only B. Take the average C. Take the lightest load position D. Compute M_C for each and take the largest Answer: D. Several positions may satisfy the condition, so the largest computed M_C governs. 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