Question
Download Solution PDFA horizontal semi-circular beam of radius ‘R’ is fixed at the ends and carries a uniformly distributed load ‘W’ over the entire length. The bending moment at the fixed support is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
As it is a Semi-Circular Beam Length = 2 R
The semi-circular fixed beam can be depicted as the horizontal beam length of ‘2R’.
The standard result for fixed end moments can be used to obtain the fixed end moments at the supports.
Fixed end moment:
\({{\rm{M}}_{{\rm{AB}}}} = {{\rm{M}}_{{\rm{BA}}}} = \frac{{{\rm{W}}{\ell ^2}}}{{12}} = \frac{{{\rm{W}} \times {{\left( {2{\rm{R}}} \right)}^2}}}{{12}} = \frac{{{\rm{W}} \times {{\rm{R}}^2}}}{3}\)
Important Point:
Fixed end moments developed due to various load combinations is given below:
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Loading |
Fixed End Moment |
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\({{\rm{\bar M}}_{{\rm{ab}}}}\) |
\({{\rm{\bar M}}_{{\rm{ba}}}}\) |
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\( - \frac{{{\rm{WI}}}}{8}\) |
\( + \frac{{{\rm{WI}}}}{8}\) |
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\( - \frac{{{\rm{Wa}}{{\rm{b}}^2}}}{{{{\rm{I}}^2}}}\) |
\( + \frac{{{\rm{Wa}}{{\rm{b}}^2}}}{{{{\rm{I}}^2}}}\) |
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\( - \frac{{{\rm{W}}{{\rm{I}}^2}}}{{12}}\) |
\( + \frac{{{\rm{W}}{{\rm{I}}^2}}}{{12}}\) |
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