Answer:
[tex]\Delta _{max}=\dfrac{5wL^4}{384EI}[/tex]
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
[tex]\Delta _x=\dfrac{wx}{24EI}(L^3-2Lx^2+x^3)[/tex]
The maximum deflection of beam will at x = L/2 (mid point )
[tex]\Delta _{max}=\dfrac{w\times \dfrac{L}{2}}{24EI}(L^3-2L\times \left (\dfrac{L}{2}\right)^2+\left(\dfrac{L}{2}\right)^3)[/tex]
[tex]\Delta _{max}=\dfrac{5wL^4}{384EI}[/tex]
