A) Half
B) One-eighth
C) One-sixteenth
D) Double
Correct Answer: A
Solution :
\[\delta =\frac{W{{L}^{3}}}{3EI}\] \[\delta \propto \frac{W}{I}\] \[{{\delta }_{1}}\propto \frac{12\,W}{b{{h}^{3}}}\] \[{{\delta }_{2}}\propto \frac{12\times \frac{W}{2}}{b\times {{(2h)}^{3}}}=\frac{6W}{8b{{h}^{3}}}=\frac{3}{4}=\frac{W}{b{{h}^{3}}}\] \[\frac{{{\delta }_{2}}}{{{\delta }_{1}}}=\frac{3}{4}\times \frac{1}{12}=\frac{1}{16}\]
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