--- title: "Muller-Breslau: ILD of shear force at B" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:mueller-breslau-ild-shear-b-5m" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/mueller-breslau-ild-shear-b-5m site: BCT Engineering --- # Muller-Breslau: ILD of shear force at B Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2065 Shrawan, 10 marks - Video: [Influence Lines with Muller-Breslau Principle (part 1) - Structural Analysis](https://www.youtube.com/watch?v=Fj5SVD0Rmy0) — structurefree - 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/mueller-breslau-ild-shear-b-5m - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question Use Muller Breslau Principle and draw influence line diagram with ordinates at 1 m interval for the shear force at B, of the beam shown in the figure. EI = constant. [Figure: beam A-B-C, A fixed, span 5 m, support B near the right end, C a 1 m overhang/roller as drawn.] ## Short answer By Müller-Breslau, cut just left of B, impose unit relative vertical displacement. For load on AB, V_left=-x²(15-x)/250; on overhang, -0.3z. Ordinates at 1m intervals from A: 0, -0.056, -0.208, -0.432, -0.704, -1.000, 0, -0.300. ## 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. #### Reading of the figure Beam: A fixed, B a roller at 5 m from A, and a free overhang BC of 1 m. The shear force just to the left of the support B (the section where the support reaction acts) is required. $EI$ is constant. #### Muller-Breslau principle Cut the beam just to the left of B and give the two faces a unit relative vertical displacement (the left face moves down by 1 relative to the right face, the faces stay parallel so the slope does not change). The resulting deflected shape is the ILD for the shear just left of B. - Right of the cut: the support B holds the point at B, so the ordinate at B (right) is 0. The overhang BC continues with the slope of the left part: a straight line. - Left of the cut: A is fixed, B moved down by 1: the shape is that of a propped-cantilever deflected curve, $y(x) = -\dfrac{x^2(3L-x)}{2L^3}$ with $L = 5$. #### Ordinates For a unit load at $x < 5$: $V_B^{left} = -R_B = -\dfrac{x^2(15-x)}{250}$. For a load on the overhang at $z = x - 5$ from B: AB acts as a propped cantilever with an applied moment $z$ at B, so $V_B^{left} = R_A = -\dfrac{3z}{2L} = -0.3z$. Sample: $x = 3$: $-\dfrac{9\times12}{250} = -0.432$. Overhang end ($z = 1$): $-0.3$. | x from A (m) | 0 | 1 | 2 | 3 | 4 | 5 (just left of B) | 5 (just right of B) | 6 (C) | |---|---|---|---|---|---|---|---|---| | $V_B^{left}$ | 0 | -0.056 | -0.208 | -0.432 | -0.704 | -1.000 | 0 | -0.300 | #### ILD shape ```text *************---------------------------|***----- ********** | ***** ******* | ******* | ***| ^ ^ ``` (The ordinate falls from 0 at A to -1 at B on the left, jumps up by 1 at B to 0, and then falls linearly to -0.3 at the free end C.) The shear just to the right of B has the simple ILD $V_B^{right} = 0$ for a load on AB and $+1$ for a load on the overhang. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the ILD of shear just left of B obtained by Muller-Breslau? A. Insert a hinge at B and rotate it. B. Remove the support at B and displace it by 1. C. Cut just left of B and give a unit relative vertical displacement with the faces kept parallel. D. Apply a unit horizontal displacement at B. Answer: C. For shear, the section is cut and the faces displaced vertically by unit amount without relative rotation. (Key point: Muller-Breslau for shear) 2. What is the ILD on the overhang for the shear just left of B? A. A straight line from 0 at B to -0.3 at C B. A parabola C. Zero everywhere D. A straight line from -1 at B to 0 at C Answer: A. The support holds B, so ordinate there is 0, and the overhang continues as a straight line; the ordinate at C is -0.3. (Key point: Right of the cut) 3. For a unit load at x < 5 m from A, what is V_B just left of B? A. -x^2(15-x)/250 B. +x^2(15-x)/250 C. -x(5-x)/250 D. -x^2/25 Answer: A. V_B(left) = -R_B and R_B = x^2(3L-x)/2L^3 with L = 5; at x = 3 this gives -0.432. (Key point: Ordinates on AB) 4. What happens to the ordinate as the load crosses B? A. Falls from 0 to -1 B. Jumps up from -1 to 0 C. Stays at -1 D. Jumps from -0.3 to 0 Answer: B. The ordinate is -1 just left of B and jumps by +1 to 0 just right of B. 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