--- title: "ILD: shear at A, propped cantilever 7 m" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:ild-shear-a-propped-7m-1m" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-shear-a-propped-7m-1m site: BCT Engineering --- # ILD: shear at A, propped cantilever 7 m Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2066 Bhadra, 10 marks - Video: [Influence Line Diagrams for Propped Cantilever Beams - Problem No 2 (Shear Force at C)](https://www.youtube.com/watch?v=p24SaLdKj3k) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-shear-a-propped-7m-1m - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question Draw Influence Line Diagram for Shear Force at A of the propped cantilever beam shown below. Calculate ordinates at 1.0 m interval. [Figure: beam, left end fixed, right end roller, span 7.0 m, section A at 1.0 m from the right end.] ## Short answer For a propped cantilever of span 7m, section at6m from fixed end, roller reaction R_B=x²(21-x)/686. Shear V_A equals -R_B for x<6, 1-R_B for x>6. Ordinates (m from fixed end): 0, -0.029, -0.111, -0.236, -0.397, -0.583, -0.787/0.213, 0. Maximum negative -0.787 just left of section. ## Answer Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam. #### Method Propped cantilever, span $L = 7$ m, fixed at the left end and on a roller at the right end B. Section A is 1 m from the right end, i.e. 6 m from the fixed end. Let $x$ be the position of the unit load measured from the fixed end. Redundant: $R_B$. $$ R_B = \frac{x^2(3L-x)}{2L^3} = \frac{x^2(21-x)}{686} $$ Shear at A (left part): load to the left of A ($x < 6$): $V_A = R_{fixed} - 1 = -R_B$; load to the right of A ($x > 6$): $V_A = R_{fixed} = 1 - R_B$. Sample: $x = 3$: $R_B = \dfrac{9\times18}{686} = 0.236$, so $V_A = -0.236$. At the section ($x = 6$): $R_B = \dfrac{36\times15}{686} = 0.787$, so $V_A = -0.787$ (just left) and $+0.213$ (just right). #### Ordinates (1 m interval) | x from the fixed end (m) | 0 | 1 | 2 | 3 | 4 | 5 | 6 (section A) | 7 | |---|---|---|---|---|---|---|---|---| | $V_A$ | 0 | -0.0292 | -0.1108 | -0.2362 | -0.3965 | -0.5831 | -0.7872 / 0.2128 | 0 | #### ILD shape ```text |*** *************----------------------------|---**** *********** | ******** | ****** | ***| ^ ^ ``` The ILD is negative from the fixed end up to the section, with a maximum of $-0.787$ just left of A. It jumps by +1 at the section to $+0.213$ and falls to zero at the roller. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a unit load to the left of section A, what is the shear at A (left part)? A. V_A = 1 - R_B B. V_A = -R_B C. V_A = R_B D. V_A = +1 Answer: B. The fixed-end reaction is 1 - R_B; with the unit load on the left part, V_A = (1 - R_B) - 1 = -R_B. (Key point: Shear left of A) 2. What happens to the ILD as the unit load crosses the section A? A. It jumps by +1, from -0.787 to +0.213 B. It jumps by -1 C. It is continuous with no jump D. It jumps to zero Answer: A. Load to the right gives V_A = 1 - R_B, so the ordinate jumps by 1 at the section. (Key point: Jump at section) 3. Which formula is used for the redundant R_B at the roller with L = 7 m? A. R_B = x(3L-x)/2L^2 B. R_B = x^2(3L-x)/2L^3 C. R_B = x^3/L^3 D. R_B = x^2(L-x)/L^3 Answer: B. Compatibility of deflection at the roller gives R_B = x^2(3L-x)/(2L^3) = x^2(21-x)/686. (Key point: Redundant reaction) 4. What is V_A for a load at x = 3 m, and what is the value at the roller end? A. -0.236; zero B. +0.236; zero C. -0.236; -1 D. -0.764; +1 Answer: A. R_B = 9 x 18/686 = 0.236 so V_A = -0.236; with the load at the roller, the shear at A is zero. (Key point: Ordinates) ## Close variants set in other exams - [ILD: shear at section 2-2, propped cantilever](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-shear-section-2-2-propped) — Theory of Structures II, chapter 4, asked 3× ## Related in Theory of Structures II - [ILD: shear at section of propped cantilever](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-shear-propped-10m-sec4) — Theory of Structures II, chapter 4 - [ILD: fixed-end reaction, propped cantilever 5 m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-fixed-end-reaction-propped-5m) — Theory of Structures II, chapter 4, asked 2× - [ILD: shear at section x-x, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-shear-section-xx-two-span) — Theory of Structures II, chapter 4 - [ILD: fixed-end moment of propped cantilever](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-reaction-moment-fixed-propped-10m) — Theory of Structures II, chapter 4, asked 2× - [ILD: moment at support B, fixed-propped beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-b-propped-2l-l) — Theory of Structures II, chapter 4 - [ILD: shear at C, propped cantilever 14 m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-shear-c-propped-14m) — Theory of Structures II, chapter 4 ## Related in other subjects - [Overhang beam ILD for shear force at C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aoverhang-ild-sf-c-2071o) — Theory of Structures I, chapter 5 - [Propped cantilever with UDL, two elements](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Apropped-cantilever-two-elements) — Computational Techniques in Civil Engineering, chapter 4 - [Arch: reactions, BM, thrust and shear at D by ILD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-50m-udl-50kn-ild-d-15m) — Theory of Structures I, chapter 6 - [ILD at mid-span: BM and SF under given loads](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-mid-span-bm-sf-50-10-30) — Theory of Structures I, chapter 5