A 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

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  1. Half
  2. One eight
  3. One sixteenth 
  4. Double

Answer (Detailed Solution Below)

Option 3 : One sixteenth 
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Detailed Solution

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Concept:

Deflection and slope for a cantilever beam carrying point load at the free end is given by

F1 Neel Madhu 10.04.20 D21 \(\delta = \frac{{P{\ell ^3}}}{{3EI}},\;\;\theta = \frac{{P{\ell ^2}}}{{2EI}}\)

 

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 = band 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:

F1 Neel Madhu 10.04.20  D22 \(\delta = \frac{{W{\ell ^4}}}{{8EI}},\;\theta = \frac{{W{\ell ^3}}}{{6EI}}\)
F1 Neel Madhu 10.04.20  D24 \(\delta = \frac{{{\omega _0}{\ell ^4}}}{{30EI}}\;;\theta = \frac{{{\omega _0}{\ell ^3}}}{{24\;EI}}\)
F1 Neel Madhu 10.04.20 D25 \(\delta = \frac{{P{\ell ^3}}}{{48EI}}\;;\theta = \frac{{P{\ell ^2}}}{{16EI}}\)
F1 Neel Madhu 10.04.20 D26 \(\delta = \frac{5}{{384}}\frac{{\omega {\ell ^4}}}{{EI}}\;\;;\;\;\theta = \frac{{\omega {\ell ^3}}}{{24EI}}\)
F1 Neel Madhu 10.04.20 D27 \(\delta = \frac{{{\omega _0}{L^4}}}{{120EI}}\;;\theta = \frac{5}{{192}}\frac{{{\omega _0}{\ell ^3}}}{{EI}}\)
F1 Neel Madhu 10.04.20 D28 \({\theta _1} = \frac{{{M _0}\ell }}{{3EI}}\;;{\theta _2} = \frac{{{M _0}\ell }}{{6EI}}\)
F1 Neel Madhu 10.04.20 D29 \(\theta = \frac{{{M _0}\ell }}{{4EI}}\)
F1 Neel Madhu 10.04.20 D30 \(\delta = \frac{1}{4}\left[ {\frac{{P{\ell ^3}}}{{48EI}}} \right]\)
F1 Neel Madhu 10.04.20 D31 \(\delta = \frac{1}{5}\left[ {\frac{5}{{384}}\frac{{{\omega _0}{\ell ^4}}}{{EI}}} \right]\)
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