The ratio of the length and diameter of a simply supported uniform circular beam which experiences maximum bending stress equal to tensile stress due to same load at its mid span, is
A. $$\frac{1}{8}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{1}{3}$$
Answer: Option C
This Question Belongs to Civil Engineering >> Theory Of Structures
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Comments ( 2 )
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. $$\frac{{2{\text{T}}}}{{\text{M}}}$$
Bending Stress = My/I
Tensile stress = P/A
FROM QUESTION,
Bending stress = Tensile stress
i.e. My/I= P/A
Where,
M = PL/4
y=D/2
I=π D^4/64
A =π D^2/4
after putting these values
L/D = 1/2
Bending Stress = M/Z
Tensile stress = P/A
FROM QUESTION,
Bending stress = Tensile stress
i.e. M/Z = P/A
M = PL/4
Z = (Pie * D^3)/32
A = (Pie*D^2)/4
after putting these values
L/D = 1/2