The ratio of maximum and average shear stresses on a rectangular section, is
A. 1
B. 1.25
C. 1.5
D. 2.5
Answer: Option C
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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. $$\frac{{2{\text{T}}}}{{\text{M}}}$$
max shear stress = 1.5* average shear shress
tmax/ tavg = 1.5
The resistance offered by the internal stress to shear is known as shearing stress. Shearing stress is zero at extreme fibres of the beam. The bending stresses are maximum at extreme fibres of the beam cross section. Maximum shear stress is 1.5 times that of average shear stress.