Question
Download Solution PDFA cantilever beam of rectangular cross section is subjected to a load 'W' at its free end .If the depth of the beam is doubled and the load is halved, the deflection of free end as compared to original deflection will be
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Deflection and slope for a cantilever beam carrying point load at the free end is given by
\(\delta = \frac{{P{\ell ^3}}}{{3EI}},\;\;\theta = \frac{{P{\ell ^2}}}{{2EI}}\) |
Calculation:
Given,
bA and bB is width of beam A and B respectively
dA and dB is depth of beam A and B respectively
dB = 2dA , bA = bB and PB = 0.5PA
Deflection for a cantilever beam (A) carrying load at free end is given by
\({\rm{\delta_A = }}\frac{{{\rm{P_A}}{{\rm{l}}^{\rm{3}}}}}{{{\rm{3EI_A}}}}\)
where IA is area of moment of inertia of beam A
\({{\rm{I}}_{\rm{A}}} = \frac{{{{\rm{b}}_{\rm{A}}}{\rm{d}}_{\rm{A}}^3}}{{12}}\)
Deflection for a cantilever beam (B) carrying load at free end is given by
\({\rm{\delta_B = }}\frac{{{\rm{P_B}}{{\rm{l}}^{\rm{3}}}}}{{{\rm{3EI_B}}}}\)
where IB is area of moment of inertia of beam A
\({{\rm{I}}_{\rm{B}}} = \frac{{{{\rm{b}}_{\rm{B}}}{\rm{d}}_{\rm{B}}^3}}{{12}}\)
Taking ratio of deflection of beam B to deflection of beam A
\(\frac{{{{\rm{\delta }}_{\rm{B}}}}}{{{{\rm{\delta }}_{\rm{A}}}}} = \frac{{{\rm{P_B}}{{\rm{l}}^3}}}{{3{\rm{E}}{{\rm{I}}_{\rm{B}}}}} \times \frac{{3{\rm{E}}{{\rm{I}}_{\rm{A}}}}}{{{\rm{P_A}}{{\rm{l}}^3}}}\)
\(\frac{{{{\rm{\delta }}_{\rm{B}}}}}{{{{\rm{\delta }}_{\rm{A}}}}} = \frac{\rm{P_B}I_A}{\rm{P_A}I_B}\)
\(\frac{{{{\rm{\delta }}_{\rm{B}}}}}{{{{\rm{\delta }}_{\rm{A}}}}} = \frac{\rm{P_B}b_A{d_A}^3}{\rm{P_A}b_B{d_B}^3}\)
\(\frac{{{{\rm{\delta }}_{\rm{B}}}}}{{{{\rm{\delta }}_{\rm{A}}}}} = \frac{1}{{16}} \)
Hence the deflection of beam B = 6.25 % of deflection of beam A
Imoprtant point:
Deflection calculations using standard results:
\(\delta = \frac{{W{\ell ^4}}}{{8EI}},\;\theta = \frac{{W{\ell ^3}}}{{6EI}}\) | |
\(\delta = \frac{{{\omega _0}{\ell ^4}}}{{30EI}}\;;\theta = \frac{{{\omega _0}{\ell ^3}}}{{24\;EI}}\) | |
\(\delta = \frac{{P{\ell ^3}}}{{48EI}}\;;\theta = \frac{{P{\ell ^2}}}{{16EI}}\) | |
\(\delta = \frac{5}{{384}}\frac{{\omega {\ell ^4}}}{{EI}}\;\;;\;\;\theta = \frac{{\omega {\ell ^3}}}{{24EI}}\) | |
\(\delta = \frac{{{\omega _0}{L^4}}}{{120EI}}\;;\theta = \frac{5}{{192}}\frac{{{\omega _0}{\ell ^3}}}{{EI}}\) | |
\({\theta _1} = \frac{{{M _0}\ell }}{{3EI}}\;;{\theta _2} = \frac{{{M _0}\ell }}{{6EI}}\) | |
\(\theta = \frac{{{M _0}\ell }}{{4EI}}\) | |
\(\delta = \frac{1}{4}\left[ {\frac{{P{\ell ^3}}}{{48EI}}} \right]\) | |
\(\delta = \frac{1}{5}\left[ {\frac{5}{{384}}\frac{{{\omega _0}{\ell ^4}}}{{EI}}} \right]\) |
Last updated on May 30, 2025
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