Question
Download Solution PDFThe ratio of the deflections of the free end of a cantilever due to an isolated load at 1/3rd and 2/3rd of the span is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFIst condition:
For a cantilever beam subjected to load W at distance of L/3 from free end, the deflection is given by:
\({\rm{y}}{{\rm{c}}_1} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{2{\rm{L}}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{2{\rm{L}}}}{3}} \right)^2} \times \frac{{\rm{L}}}{3} = {\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}^3}\)
IInd condition:
For a cantilever beam subjected to load W at distance of 2L/3 from free end:
\({\rm{y}}{{\rm{c}}_2} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{\rm{L}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{\rm{L}}}{3}} \right)^2} \times \frac{{2{\rm{L}}}}{3} = \frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}\)
\({\rm{Ratio\;}}\left( {\rm{r}} \right) = \frac{{{\rm{y}}{{\rm{c}}_1}}}{{{\rm{y}}{{\rm{c}}_2}}} = \frac{{{{\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}}^3}}}{{\frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}}} = \frac{{28}}{8} = \frac{7}{2}\)
\(\therefore \frac{{{\rm{y}}{{\rm{c}}_2}}}{{{\rm{y}}{{\rm{c}}_1}}} = \frac{2}{7}\)
Last updated on Jul 1, 2025
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