Propped cantilever by strain energy method

Method

The beam is statically indeterminate to the first degree. Take the prop reaction $R_B$ as the redundant. Since the prop does not allow vertical movement at B:

$$\delta_B = \frac{\partial U}{\partial R_B} = 0$$

Bending moment

Take $x$ from B toward the fixed end A ($L = 4$ m, $w = 12$ kN/m):

$$M(x) = R_B\,x - \frac{w x^2}{2},\qquad \frac{\partial M}{\partial R_B} = x$$

 A (fixed)                       B
 |#==============================o   <- prop R_B
 |#  v v v v v v v v v v v v v
 |#  w = 12 kN/m,  L = 4 m

Compatibility

$$\begin{aligned} \frac{1}{EI}\int_0^L \left(R_B x - \frac{wx^2}{2}\right)x\,dx &= 0 \\ \frac{R_B L^3}{3} - \frac{wL^4}{8} &= 0 \\ R_B &= \frac{3wL}{8} = \frac{3\times12\times4}{8} = 18.0\ \text{kN} \end{aligned}$$

Reactions at A

$$\begin{aligned} R_A &= wL - R_B = 48 - 18 = 30.0\ \text{kN} \\ M_A &= \frac{wL^2}{2} - R_B L = \frac{12\times16}{2} - 18\times4 = 96 - 72 = 24.0\ \text{kN m (hogging)} \end{aligned}$$

Check: $M_A = wL^2/8 = 24$ kN m. $\checkmark$

Contraflexure

$M(x) = 18x - 6x^2 = 0 \Rightarrow x = 3$ m from B (i.e. 1 m from the fixed end).

Maximum sagging moment

Shear force $= R_B - wx = 0 \Rightarrow x = 18/12 = 1.5$ m from B:

$$M_{max} = 18(1.5) - 6(1.5)^2 = 27 - 13.5 = 13.5\ \text{kN m (sagging)}$$

Bending moment diagram (shape)

 -24 at A                  0 at B
  \                          .
   \____ 0 at x=3 m _____ +13.5 at x=1.5 m

Answer: $R_B = 18$ kN, $R_A = 30$ kN, $M_A = 24$ kN m (hogging), contraflexure 1 m from the fixed end, maximum sagging moment $13.5$ kN m at 1.5 m from the prop.