Conjugate beam: stepped rigidity, central load

Given data

Simply supported beam, span $L$, central load $W$. The middle third ($L/3$) has rigidity $2EI$; the two outer thirds have $EI$. By symmetry the reactions are $W/2$ each and the maximum deflection is at mid-span.

             W
 A --- EI ---|--- 2EI ---|--- EI --- B
 ^    L/3         L/3         L/3     ^

Step 1: Bending moment and $M/EI$ diagram (left half)

$M = \dfrac{Wx}{2}$ for $0 \le x \le \dfrac{L}{2}$. $M/EI$ ordinates:

• At $x = L/3$: in the outer third $M/EI = \dfrac{WL}{6EI}$; in the middle third $M/(2EI) = \dfrac{WL}{12EI}$ (a sudden drop).
• At mid-span: $M/(2EI) = \dfrac{WL}{8EI}$.

Step 2: Conjugate beam

The conjugate beam is simply supported, loaded with the $M/EI$ diagram. Loads (left half, in units of $W L^2/EI$ with centroids from A):

Sum of the half-loads $= \dfrac{WL^2}{EI}\left(\dfrac{8 + 4 + 1}{288}\right) = \dfrac{13\,WL^2}{288\,EI}$.

Step 3: Slope at the supports

The conjugate reaction equals half the total conjugate load (symmetry):

$$\theta_A = \theta_B = R'_A = \frac{13\,WL^2}{288\,EI} = 0.0451\,\frac{WL^2}{EI}$$

Step 4: Maximum deflection (conjugate moment at mid-span)

$$y_{max} = R'_A\frac{L}{2} - \frac{WL^2}{36EI}\left(\frac{L}{2}-\frac{2L}{9}\right) - \frac{WL^2}{72EI}\left(\frac{L}{2}-\frac{5L}{12}\right) - \frac{WL^2}{288EI}\left(\frac{L}{2}-\frac{4L}{9}\right)$$ $$y_{max} = \frac{WL^3}{EI}\left[\frac{13}{576} - \frac{5}{648} - \frac{1}{864} - \frac{1}{5184}\right] = \frac{WL^3}{EI}\cdot\frac{117 - 40 - 6 - 1}{5184} = \frac{70}{5184}\frac{WL^3}{EI}$$ $$y_{max} = \frac{35\,WL^3}{2592\,EI} = 0.0135\,\frac{WL^3}{EI}$$

Answer: slope at each support $= \dfrac{13WL^2}{288EI}$; maximum (mid-span) deflection $= \dfrac{35WL^3}{2592EI}$ downward. For comparison, a uniform beam of rigidity $EI$ has $\dfrac{WL^3}{48EI} = 0.0208\dfrac{WL^3}{EI}$, so the stiffer middle reduces the deflection by about 35%.