--- title: "ILD: shear at section of propped cantilever" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:ild-shear-propped-10m-sec4" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-shear-propped-10m-sec4 site: BCT Engineering --- # ILD: shear at section of propped cantilever Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2079 Baishakh, 6 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-propped-10m-sec4 - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question Draw the ILD for the shear at section 1-1 of the propped cantilever beam given in the figure. [Figure: beam AB, A fixed, B roller, span 10 m, EI constant, section 1-1 at 4 m from A.] ## Short answer Use Muller-Breslau by releasing B; shear ILD shape is -R_B left of section, 1-R_B right. At section 4 m from A, ordinates are -0.208 just left and +0.792 just right; zero at A and B. ## 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 (Muller-Breslau / equilibrium) Beam AB: A fixed, B roller, $L = 10$ m, section 1-1 at 4 m from A. The beam is singly indeterminate; take $R_B$ as the redundant. With support B removed, the cantilever deflection at B due to a unit load at $x$ from A is $\delta_{Bx} = \dfrac{x^2(3L-x)}{6EI}$, and due to a unit load at B it is $f_{BB} = \dfrac{L^3}{3EI}$. Compatibility (zero deflection at B) gives $$ R_B = \frac{\delta_{Bx}}{f_{BB}} = \frac{x^2(3L-x)}{2L^3} $$ Shear at the section (left part): - load to the left of the section ($x < 4$): $V = R_A - 1 = -R_B$ - load to the right of the section ($x > 4$): $V = R_A = 1 - R_B$ So the ILD is the $R_B$ curve turned upside down ($-R_B$) to the left of the section and $1 - R_B$ to the right of it, with a unit jump at the section. Sample: $x = 6$: $R_B = \dfrac{36(30-6)}{2000} = 0.432$, so $V = 1 - 0.432 = 0.568$. At the section: $R_B(4) = \dfrac{16\times26}{2000} = 0.208$, so $V = -0.208$ just left and $+0.792$ just right. #### Ordinates | x from A (m) | 0 | 2 | 4 | 6 | 8 | 10 | |---|---|---|---|---|---|---| | $V_{1-1}$ | 0 | -0.056 | -0.208 / 0.792 | 0.568 | 0.296 | 0 | #### ILD shape ```text |**** | ******** | ******* | ****** *************------|-------------------------**** ******| ^ ^ ``` The maximum negative ordinate is $-0.208$ and the maximum positive ordinate is $+0.792$, both at the section (a jump of 1 unit). The ordinates are 0 at A and at B. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is R_B found for a unit load at x from A on the propped cantilever? A. By compatibility: R_B = delta_Bx / f_BB (zero deflection at B) B. By taking moments about B only C. R_B = 1/2 D. R_B = x/L Answer: A. The cantilever deflection at B due to the load is cancelled by R_B times the flexibility f_BB = L^3/3EI. (Key point: Redundant reaction) 2. For a unit load to the left of section 1-1, the shear at the section is... A. 1 - R_B B. R_A C. +R_B D. -R_B (the load is on the left part, so V = R_A - 1 = -R_B) Answer: D. The left part carries R_A and the unit load, so V = R_A - 1 = -R_B. (Key point: Shear for load left of section) 3. What happens to the ILD at the section? A. A jump of 1 unit, from -0.208 to +0.792 B. It is continuous C. A jump of 0.5 D. It becomes zero Answer: A. When the unit load crosses the section, the shear changes by 1: -R_B becomes 1 - R_B, with R_B(4) = 0.208. (Key point: Jump at section) 4. For a load at x = 6 m (L = 10), what is V at the section? A. 0.432 B. 0.568 C. -0.432 D. 0.792 Answer: B. R_B = 36(30-6)/2000 = 0.432, and the load is to the right of the section, so V = 1 - 0.432 = 0.568. 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