The ratio of the length and depth of a simply supported rectangular beam which experiences maximum bending stress equal to tensile stress, due to same load at its mid span, is
A. $$\frac{1}{2}$$
B. $$\frac{2}{3}$$
C. $$\frac{1}{4}$$
D. $$\frac{1}{3}$$
Answer: Option B
This Question Belongs to Civil Engineering >> Theory Of Structures
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Comments ( 4 )
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. $$\frac{{2{\text{T}}}}{{\text{M}}}$$
Please any one will explain it
Bending stress(sigma) to tensile(P) ratio=?
M/I=Sigma/y
Or Sigma=(M/I)×y=M/Z
Or sigma=(wl/4)/(bd^2/6)=(3wl)/(2bd^2)
Tensile stress , P =force /area=w/bd
So..... ratio of bending stress to tensile stress ,
Sigma=P
Or (3wl)/(2bd^2)= w/bd
Or l/d =2/3
From these relations
M/I=F/Y
TENSILE STRESS= P/A
GIVEN F=TENSILE STRESS
SO L/D= 2/3
Bending Stress=M(max)/Z
=(PL/4)/(I/y)
=(PL/4)/[(bd^3/12)/(d/2)]
=(PL/4)/[bd^2/6]
=3PL/2bd^2
Now,
Tensile Stress=P/A
=P/bd.
Since, according to question,
Bending Stress=Tensile Stress
PL/2bd^2=P/bd.
====> L/d=2/3