Muller-Breslau principle; ILD reaction at C

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)

ILD shape

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

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.