For a cantilever beam carrying uniformly distributed load throughout the span, the standard formula for deflection (δ) and slope (θ) at free end are given below:
\({\rm{\delta }} = {\rm{\;}}\frac{{{\rm{w}}{{\rm{l}}^4}}}{{8{\rm{EI}}}}\) and \({\rm{\theta }} = {\rm{\;}}\frac{{{\rm{w}}{{\rm{l}}^3}}}{{6{\rm{EI}}}}\) (θ is in radian)
Where, w is the uniformly distributed load.
Given that,
θ = 1^{0} = π/180 radian; L = 3 m;
Now,
\(\frac{\delta }{\theta } = \;\frac{{\frac{{{\rm{w}}{{\rm{l}}^4}}}{{8{\rm{EI}}}}}}{{{\rm{\;}}\frac{{{\rm{w}}{{\rm{l}}^3}}}{{6{\rm{EI}}}}{\rm{\;}}}}\)
\(\frac{\delta }{\theta } = \;\frac{3}{4}L\)
\(\frac{\delta }{{\frac{\pi }{{180}}}} = \;\frac{3}{4} \times 3000\)
∴ δ = 39.25 mm
Important point:
Deflection and slope of various beams is given by:
\({y_B} = \frac{{P{L^3}}}{{3EI}}\) 
\({\theta _B} = \frac{{P{L^2}}}{{2EI}}\) 

\({y_B} = \frac{{w{L^4}}}{{8EI}}\) 
\({\theta _B} = \frac{{w{L^3}}}{{6EI}}\) 

\({y_B} = \frac{{M{L^2}}}{{2EI}}\) 
\({\theta _B} = \frac{{ML}}{{EI}}\) 

\({y_B} = \frac{{w{L^4}}}{{30EI}}\) 
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) 

\({y_c} = \frac{{P{L^3}}}{{48EI}}\) 
\({\theta _B} = \frac{{w{L^2}}}{{16EI\;}}\) 


\({y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\) 
\({\theta _B} = \frac{{w{L^3}}}{{24EI}}\) 

\({y_c} = 0\) 
\({\theta _B} = \frac{{ML}}{{24EI}}\) 
\({y_c} = \frac{{M{L^2}}}{{8EI}}\) 
\({\theta _B} = \frac{{ML}}{{2EI}}\) 


\({y_c} = \frac{{P{L^3}}}{{192EI}}\) 
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) 
\({y_c} = \frac{{w{L^4}}}{{384EI}}\) 
\({\theta _A} = {\theta _B} = {\theta _C} = 0\) 
Where, y = Deflection of beam, θ = Slope of beam