Mueller-Breslau principle for influence lines

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: $x = 3$: $M_B = -\dfrac{3(36-9)}{144} = -0.5625$.

**----------------------*----------------------**
  **                   * *                   **
    ***              **   **              ***
       ***        ***       ***        ***
          ********             ********
^                       ^                       ^

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):

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) $= 7.5w$.