Propped cantilever with UDL by integration

Statically indeterminate beam

 ||A===================B
 ||  w = 20 kN/m       ^ R_B (prop)
 ||<------ 4 m ------>

Let $R_B$ be the prop reaction. Take the origin at A, $x$ towards B, sagging moment positive and $y$ upward.

Let $R_A = wL - R_B$ be the vertical reaction at the fixed end A, and $M_A$ the hogging fixed-end moment. The bending moment at a section $x$ from A gives

$$EI\,y'' = R_Ax - M_A - \frac{wx^2}{2}$$ $$EI\,y' = \frac{R_Ax^2}{2} - M_Ax - \frac{wx^3}{6} \quad (C_1=0\text{ since } y'=0\text{ at }x=0)$$ $$EI\,y = \frac{R_Ax^3}{6} - \frac{M_Ax^2}{2} - \frac{wx^4}{24} \quad (C_2=0\text{ since } y=0\text{ at }x=0)$$

(a) Prop reaction

Conditions: $y = 0$ at $x = L$ (prop), and $M(L) = 0$ at the free end B (no moment is applied there):

$$R_AL - M_A - \frac{wL^2}{2} = 0 \;\Rightarrow\; M_A = R_AL - \frac{wL^2}{2}$$ $$EI\,y_B = \frac{R_AL^3}{6} - \frac{M_AL^2}{2} - \frac{wL^4}{24} = 0$$

Substituting $M_A$: $\dfrac{R_AL^3}{6} - \dfrac{R_AL^3}{2} + \dfrac{wL^4}{4} - \dfrac{wL^4}{24} = 0$, hence $\dfrac{R_AL^3}{3} = \dfrac{5wL^4}{24}$, so $R_A = \dfrac{5wL}{8}$.

$$R_A = \frac{5\times20\times4}{8} = 50\ \text{kN}, \qquad R_B = wL - R_A = 80 - 50 = 30\ \text{kN}$$

(b) Fixed-end reactions and moment

$R_A = 50$ kN (upward), and

$$M_A = R_AL - \frac{wL^2}{2} = 50\times4 - 160 = 40\ \text{kN m (hogging)} \;\left(=\frac{wL^2}{8}\right)$$

Check, moments about A: $30\times4 - 80\times2 + 40 = 0$. ✓

(c) Maximum deflection ($y'=0$)

$$EI\,y' = 25x^2 - 40x - \frac{20x^3}{6} = 0 \;\Rightarrow\; 3.333x^2 - 25x + 40 = 0$$ $$x = \frac{25 - \sqrt{625 - 533.3}}{6.667} = \frac{25 - 9.57}{6.667} = 2.314\ \text{m}\ (=0.5785L\text{ from A, i.e. }0.4215L\text{ from the prop})$$ $$EI\,y = \frac{50(2.314)^3}{6} - \frac{40(2.314)^2}{2} - \frac{20(2.314)^4}{24} = 103.2 - 107.1 - 23.9 = -27.7$$ $$y_{max} = \frac{-27.7}{16000} = -1.73\times10^{-3}\ \text{m} = 1.73\ \text{mm downward}$$

(d) Point of contraflexure

$M(x) = 50x - 40 - 10x^2 = 0 \Rightarrow x^2 - 5x + 4 = 0 \Rightarrow x = 1\ \text{m}$ or $4\ \text{m}$. The root $x = 4$ m is the prop itself, so the point of contraflexure is at $x = 1$ m from A. The maximum sagging moment is where $V = 0$, at $x = 50/20 = 2.5$ m: $M = 50(2.5) - 40 - 10(6.25) = 22.5$ kN m.

Answer: $R_B = 30$ kN, $R_A = 50$ kN, $M_A = 40$ kN m; $y_{max} = 1.73$ mm at 2.31 m from A; contraflexure at 1 m from A.