Explanation :
E = 2.5 x 10^{4} MPa
I = 8 x 10^{4} mm^{4}
For P,
$\begin{array}{l}\sum{\mathrm f}_\mathrm v=0\\\Rightarrow\boxed{\mathrm V=80\mathrm N}\end{array}$
$\therefore$ Shear force at support = S_{p} = 80N
⇒ Moment at support = 80 x 8 = 640 Nm.
⇒ M_{p} = 640 Nm
For Q
$\begin{array}{l}\sum{\mathrm f}_\mathrm v=0\\\Rightarrow\boxed{\mathrm V=20\times8=160\mathrm N}\end{array}$
$\therefore$ Shear force at support = S_{Q} = 160N
⇒ Moment at support = M_{Q} = 20 x 8 x $\frac82$
= 640 Nm
For R,
$\sum{\mathrm f}_\mathrm v=0$
V = 0 = S_{R}
Moment at support = M_{R} = 640 Nm
Hence, by above values
$\boxed{{\mathrm S}_\mathrm P<{\mathrm S}_\mathrm Q>{\mathrm S}_\mathrm R}\;\&\;\boxed{{\mathrm M}_\mathrm P={\mathrm M}_\mathrm Q={\mathrm M}_\mathrm R}$