--- title: "Muller-Breslau principle; ILD reaction at C" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:mueller-breslau-ild-reaction-c-6" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/mueller-breslau-ild-reaction-c-6 site: BCT Engineering --- # Muller-Breslau principle; ILD reaction at C Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2070 Chaitra (old course), 2+8 marks - Video: [Muller-Breslau Principle for Influence Lines - Intro to Structural Analysis](https://www.youtube.com/watch?v=1VcwsaQeUMY) — Structures with Prof. H - 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-reaction-c-6 - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question Define Muller-Breslau principle. Draw influence line diagram for reaction at C. Find ordinates at 2 m interval. [Figure: beam ABC, A hinged, B and C rollers, AB = 6 m, BC = 8 m.] ## Short answer Muller-Breslau states the ILD of a force quantity is the deflected shape produced by a unit displacement in the direction of the removed restraint. For reaction at C, ordinates are zero at A and B and 1 at C; a small negative loop in AB reaches about -0.06, and BC rises steeply to 1. ## Answer #### Muller-Breslau principle The influence line for any force-type quantity (a reaction, shear or bending moment) in a structure has the same shape as the deflected shape of the structure when the restraint corresponding to that quantity is removed and a unit displacement (or rotation) is given in the direction of the quantity. The ordinate of the ILD at any point is the deflection of the beam at that point. It follows from the principle of virtual work (Betti's law). #### ILD for $R_C$ Beam ABC: A hinged, B and C rollers, $AB = 6$ m, $BC = 8$ m. Remove the support at C and give C a unit upward displacement: BC rotates about B as a curved shape, and AB, held by the supports A and B, bends the other way (it goes below the axis). Ordinates (statics + three-moment equation): the support moment is $M_B = -\dfrac{x(36-x^2)}{168}$ for a load in AB at $x$ from A, and - load in AB: $R_C = \dfrac{M_B}{8}$ (negative, since a load in AB pulls C down), - load in BC at $z$ from B: $R_C = \dfrac{z}{8} + \dfrac{M_B}{8}$ with $M_B = -\dfrac{y(64-y^2)}{224}$, where $y = 8 - z$ is measured from C. Sample: $x = 4$ (in AB): $M_B = -\dfrac{4(36-16)}{168} = -0.476$, $R_C = -0.0595$. $z = 2$: $y = 6$, $M_B = -\dfrac{6(64-36)}{224} = -0.75$, $R_C = \dfrac28 - \dfrac{0.75}{8} = 0.156$. #### Ordinates (2 m interval) | x from A (m) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | |---|---|---|---|---|---|---|---|---| | $R_C$ | 0 | -0.0476 | -0.0595 | 0 | 0.1562 | 0.3929 | 0.683 | 1 | #### ILD shape ```text *** ****** ****** ******* **-----------------********---------------------- **** **** ********* ^ ^ ^ ``` The ordinate at C is 1, and it is 0 at A and B. There is a small negative loop in AB (about $-0.06$ near x = 4 m), and the BC portion is positive and rises steeply to 1 at C. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Muller-Breslau says the ILD for a force-type quantity has the shape of A. The BMD for a unit load B. The shear diagram C. The beam with all supports removed D. The deflected shape when the restraint is removed and a unit displacement is given Answer: D. Ordinate equals deflection of the beam at that point. (Key point: Principle) 2. For a load in AB, R_C is A. M_B/8, negative B. z/8, positive C. Zero D. 1 Answer: A. A load in AB pulls C down, so R_C = M_B/8 < 0. (Key point: Load in AB) 3. The ILD ordinate for R_C at C is A. 0 B. 1 C. 0.156 D. -0.06 Answer: B. A unit load directly at C is carried by C. 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