Flexibility method: beam reactions at B and C

Assumptions (figure values read as given): A is fixed at $x=0$; 200 kN acts at $x=5$ m; B is a roller support at $x=10$ m; 100 kN acts at $x=15$ m; C is a roller support at $x=20$ m. EI is constant. The beam is indeterminate to the second degree. Choose the reactions $R_B$ (coordinate 1) and $R_C$ (coordinate 2) as redundants, acting upward. The released structure is the cantilever fixed at A.

Step 1: Flexibility matrix of the cantilever (L = 20 m)

For a unit load at distance $a$ from the fixed end, the deflection at $x$ is $x^2(3a-x)/6EI$ for $x\le a$ and $a^2(3x-a)/6EI$ for $x\ge a$.

$$f_{11}=\frac{10^3}{3EI}=\frac{333.33}{EI},\quad f_{22}=\frac{20^3}{3EI}=\frac{2666.67}{EI},\quad f_{12}=f_{21}=\frac{10^2(3\times20-10)}{6EI}=\frac{833.33}{EI}$$

Step 2: Deflections due to the loads (downward) on the released beam

At B (x = 10): the 200 kN load gives $\dfrac{200\times5^2(3\times10-5)}{6EI}=\dfrac{20833.3}{EI}$ and the 100 kN load (at 15 m, beyond B) gives $\dfrac{100\times10^2(3\times15-10)}{6EI}=\dfrac{58333.3}{EI}$. At C (x = 20): $\dfrac{200\times5^2(60-5)}{6EI}=\dfrac{45833.3}{EI}$ and $\dfrac{100\times15^2(60-15)}{6EI}=\dfrac{168750}{EI}$.

$$\Delta_1=\frac{79166.7}{EI},\qquad \Delta_2=\frac{214583.3}{EI}$$

Step 3: Compatibility (deflection at B and C must be zero)

$$\begin{bmatrix}333.33&833.33\\833.33&2666.67\end{bmatrix}\begin{Bmatrix}R_B\\R_C\end{Bmatrix}=\begin{Bmatrix}79166.7\\214583.3\end{Bmatrix}$$ $$R_B=166.07\ \text{kN}\uparrow,\qquad R_C=28.57\ \text{kN}\uparrow$$

Step 4: Remaining reactions

$$R_A=200+100-166.07-28.57=105.36\ \text{kN}\uparrow,\qquad M_A=-200(5)-100(15)+166.07(10)+28.57(20)\;\Rightarrow\; M_A=-267.86\ \text{kNm (hogging)}$$

Check: moments of the loads and support reactions about A: $166.07\times10+28.57\times20-200\times5-100\times15=-267.86$ kNm, which is balanced by the fixed-end reaction moment of 267.86 kNm (anticlockwise) ✓.

Shear force (kN)

Bending moment (kNm, sagging +)

The SF changes sign under the 200 kN load (maximum sagging BM +258.93 kNm), at B (hogging peak of -214.29 kNm, where the SF jumps) and under the 100 kN load. The BM is zero at C.

Answer: $R_B=166.07$ kN, $R_C=28.57$ kN (both upward); also $R_A=105.36$ kN, $M_A=267.86$ kNm (hogging).