Simply supported in both ends - Triangularly distributed load

Table 11 Simply supported in both ends Triangularly distributed load
$M (x)$	$= q_{M A X} \cdot L \cdot x \cdot (\frac{1}{4} - \frac{1}{3} \cdot \frac{x^{2}}{L^{2}})$	for x < L/2
$V (x)$	$= \frac{q_{M A X}}{2 \cdot L} \cdot (L^{2} - 4 \cdot x^{2})$
$u (x)$	$= \frac{q_{M A X} \cdot L \cdot x}{960 \cdot E \cdot I \cdot L^{2}} \cdot {(5 \cdot L^{2} - 4 \cdot x^{2})}^{2}$	for x < L/2
$M_{MAX}$	$= \frac{q_{M A X} \cdot L^{2}}{12}$
$u_{M A X}$	$= \frac{q_{M A X} \cdot L^{4}}{120 \cdot E \cdot I}$
$γ_{A}$	$= \frac{5}{192} \cdot \frac{q \cdot L^{3}}{E \cdot I}$
$γ_{B}$	$= - \frac{5}{192} \cdot \frac{q \cdot L^{3}}{E \cdot I}$ (???????)
$R_{A}$	$= \frac{q \cdot L}{4}$
$R_{B}$	$= \frac{q \cdot L}{4}$

Table 11Simply supported in both ends Triangularly distributed load