--- title: "Mueller-Breslau principle for influence lines" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:mueller-breslau-principle" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/mueller-breslau-principle site: BCT Engineering --- # Mueller-Breslau principle for influence lines Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2074 Asoj, 5 marks; 2069 Asar, 2 marks; 2068 Chaitra, 8 marks; 2067 Asar, 5 marks (asked 4 times) - 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-principle - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question Explain the Mueller-Breslau principle for influence line diagrams and show with an example (such as a continuous beam) how it is applied to obtain the shape of an influence line. ## Short answer The Müller-Breslau principle states that the influence line for a force response is the deflected shape when the corresponding restraint is removed and a unit displacement is imposed. For a continuous beam, removing the moment restraint at a support and applying a unit rotation yields a curved shape; ordinates give influence values. ## Answer #### Statement **Muller-Breslau principle:** the influence line for a force-type response (a support reaction, shear force or bending moment) at a point in a structure has the same shape as the deflected shape of the structure obtained when the restraint that carries that response is removed and a unit displacement is imposed in its direction at that point. The ordinate of the ILD at any point is the displacement of the structure at that point, measured in the direction of the unit load. **Basis (virtual work):** after removing the restraint, the virtual work of the real forces Q (the response) and the unit load P = 1 moving through the displaced shape must be zero, so $Q\cdot\Delta_Q = 1\cdot y(x)$. With $\Delta_Q = 1$ the displacement $y(x)$ equals the influence ordinate. #### Procedure 1. Remove the restraint related to the response and apply the corresponding unit displacement. 2. Reaction: remove the support and displace it by 1 vertically. Shear: cut the beam at the section and give a unit relative vertical displacement (the two parts stay parallel). Bending moment: insert a hinge at the section and give a unit relative rotation. 3. Draw the deflected shape; the ordinate at any point is the influence value. #### Determinate and indeterminate structures - In a statically determinate beam the displaced shape consists of straight lines, so the ordinates are exact. - In a continuous (indeterminate) beam the rest of the beam remains supported, so the shape is curved and is found from the elastic curve (moment-area/compatibility). The principle gives the correct qualitative shape immediately, so the critical loading can be decided (for example, load alternate spans for maximum positive moment). #### Example: two equal spans of 6 m (A hinged, B and C rollers), ILD for the support moment $M_B$ Insert a hinge at B (so the beam cannot carry moment there) and give the two sides a unit relative rotation in the hogging sense (the beam kinks into an inverted V at B). A, B and C stay on the axis, so each span bends into a curve below the axis, with a total kink of 1 radian at B. The deflected shape is the ILD: both spans lie below the axis, i.e. a load on either span produces a hogging moment at B (negative in the sagging-positive convention). Ordinates from the elastic curve (equal spans, constant $EI$): $M_B = -\dfrac{x(L^2 - x^2)}{4L^2} = -\dfrac{x(36 - x^2)}{144}$ for a load at $x$ from A in span AB (and symmetric in BC). Sample: $x = 3$: $M_B = -\dfrac{3(36-9)}{144} = -0.5625$. | x from A (m) | 0 | 1.5 | 3 | 4.5 | 6 | 7.5 | 9 | 10.5 | 12 | |---|---|---|---|---|---|---|---|---|---| | $M_B$ (kN·m per kN) | 0 | -0.3516 | -0.5625 | -0.4922 | 0 | -0.4922 | -0.5625 | -0.3516 | 0 | ```text **----------------------*----------------------** ** * * ** *** ** ** *** *** *** *** *** ******** ******** ^ ^ ^ ``` Thus a load on either span produces hogging at B; for the maximum hogging moment at B, load both spans. Similarly, the ILD of the reaction $R_B$ is the deflected shape when the support B is removed and B is lifted by 1 (positive in both spans, equal to 1 at B): | x from A (m) | 0 | 1.5 | 3 | 4.5 | 6 | 7.5 | 9 | 10.5 | 12 | |---|---|---|---|---|---|---|---|---|---| | $R_B$ | 0 | 0.3672 | 0.6875 | 0.9141 | 1 | 0.9141 | 0.6875 | 0.3672 | 0 | The largest hogging ordinate of $M_B$ is about 0.577 (kN·m per kN) at $x = L/\sqrt3 = 3.46$ m from A (and at the mirror point in BC), slightly more than the mid-span value 0.5625; the $R_B$ ordinate is 1.0 under B. A UDL $w$ over both spans gives $R_B = w\times$ (area of the ILD) $= 7.5w$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. According to the Mueller-Breslau principle, the ILD for a force-type response has the shape of... A. The bending moment diagram under a unit load B. The deflected shape after the corresponding restraint is removed and a unit displacement is imposed C. The shear force diagram of the beam D. The structure's mode shape in vibration Answer: B. The ordinate at any point equals the displacement there when the restraint is released and displaced by unity. (Key point: Statement) 2. To get the ILD of shear at a section, you... A. Insert a hinge and rotate it B. Remove a support and lift it C. Fix the section D. Cut the beam and give a unit relative vertical displacement, keeping the two parts parallel Answer: D. The cut releases shear only, so the two parts displace relatively and remain parallel. (Key point: Shear ILD) 3. To get the ILD of bending moment at a section, you... A. Insert a hinge and give a unit relative rotation B. Give a unit settlement C. Cut and shift vertically D. Remove the section Answer: A. A hinge removes the moment restraint; the kink of one radian shows the ILD. (Key point: Moment ILD) 4. How does the ILD shape differ for a determinate and a continuous beam? A. Both are curved B. Both are straight C. Determinate: straight lines (exact); continuous: curved, found from the elastic curve D. Determinate: curved; continuous: straight Answer: C. Released determinate beams move as rigid straight parts; indeterminate beams bend since the rest remains supported. 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