The ratio of maximum shear stress to average shear stress of a circular beam, is
A. $$\frac{2}{3}$$
B. $$\frac{3}{2}$$
C. $$\frac{3}{4}$$
D. $$\frac{4}{3}$$
Answer: Option D
Solution (By Examveda Team)
Shear stress is the stress induced when a force is applied parallel or tangential to the surface of a material.Average shear stress is calculated as the total shear force divided by the cross-sectional area.
Maximum shear stress is the highest shear stress value in a beam's cross-section, and in the case of a circular cross-section, it occurs at the neutral axis.
For a circular beam, the relationship between maximum shear stress and average shear stress is given by:
τmax = (4/3) × τavg
Therefore, the ratio of maximum shear stress to average shear stress is:
τmax / τavg = 4⁄3
This means the maximum shear stress is 1.33 times the average shear stress in a circular beam
Hence, the correct answer is Option D: 4⁄3
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. $$\frac{{2{\text{T}}}}{{\text{M}}}$$
Is it correct for rectangular , it is 3/2 ?
@Muhammad Saqib.
Can u plz Explain!!!
While in rectangular it is 3/2