A simply supported beam A carries a point load at its mid span. Another identical beam B carries the same load but uniformly distributed over the entire span. The ratio of the maximum deflections of the beams A and B, will be
A. $$\frac{2}{3}$$
B. $$\frac{3}{2}$$
C. $$\frac{5}{8}$$
D. $$\frac{8}{5}$$
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {\text{Deflection max }}\left( {\text{A}} \right) = \frac{{{\text{P}}{{\text{L}}^3}}}{{48{\text{E}}I}} \cr & {\text{Deflection max }}\left( {\text{B}} \right) = \frac{{5{\text{w}}{{\text{L}}^4}}}{{384{\text{E}}I}};\,\,\,\left( {{\text{w}} = \frac{{\text{P}}}{{\text{L}}}} \right) \cr & = \frac{{5{\text{P}}{{\text{L}}^3}}}{{384{\text{E}}I}} \cr & {\text{Ratio}} = \frac{{{\text{P}}{{\text{L}}^3}}}{{48{\text{E}}I}} \times \frac{{384{\text{E}}I}}{{5{\text{P}}{{\text{L}}^3}}} = \frac{8}{5} \cr} $$This Question Belongs to Civil Engineering >> Theory Of Structures
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Comments ( 5 )
A. $$\frac{2}{3}$$
B. $$\frac{3}{2}$$
C. $$\frac{5}{8}$$
D. $$\frac{8}{5}$$
Principal planes are subjected to
A. Normal stresses only
B. Tangential stresses only
C. Normal stresses as well as tangential stresses
D. None of these
A. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$
B. $$\frac{{\text{I}}}{{\text{M}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$
C. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{F}}}{{\text{Y}}}$$
D. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{Y}}}{{\text{F}}}$$
A. $$\frac{{\text{M}}}{{\text{T}}}$$
B. $$\frac{{\text{T}}}{{\text{M}}}$$
C. $$\frac{{2{\text{M}}}}{{\text{T}}}$$
D. $$\frac{{2{\text{T}}}}{{\text{M}}}$$
Correct answer
IT WILL BE (8:5) OR (1.6:1)
Def.max (A)= PL3/48EI;
Def.max (B)= 5wL4/384EI; w=P/L,
= 5PL3/384EI.
Ratio= (PL3)/48EI * 384EI/(5PL3) = 8/5
Good
Wrong answer
It should be 2/1