Answer:
![\Delta _(max)=(5wL^4)/(384EI)](https://img.qammunity.org/2020/formulas/engineering/college/pdx4vauf2kbsut6holjaobh3dm5mabc8wt.png)
Step-by-step explanation:
Given that
Load is uniformly distributed load and beam is simply supported.
Ra + Rb= wL
Ra = Rb =wL / 2
Lets x is measured from left side,then the deflection of beam at any distance x is given as
![\Delta _x=(wx)/(24EI)(L^3-2Lx^2+x^3)](https://img.qammunity.org/2020/formulas/engineering/college/pqmdly1lup5odwsinjhp58j1gwbopzd1uc.png)
The maximum deflection of beam will at x = L/2 (mid point )
![\Delta _(max)=(w* (L)/(2))/(24EI)(L^3-2L* \left ((L)/(2)\right)^2+\left((L)/(2)\right)^3)](https://img.qammunity.org/2020/formulas/engineering/college/zqgt0aivjzbqno4rge0hp3dpqddx370a2a.png)
![\Delta _(max)=(5wL^4)/(384EI)](https://img.qammunity.org/2020/formulas/engineering/college/pdx4vauf2kbsut6holjaobh3dm5mabc8wt.png)