Related Questions

In a triaxial compression test when drainage is allowed during the first stage (i. e. application of cell pressure) only and not during the second stage (i.e. application of deviator stress at constant cell pressure), the test is known as
A consolidated drained test
B consolidated undrained test
C unconsolidated drained test
D unconsolidated undrained test
When an open coiled helical compression spring is subjected to an axial compressive load, the maximum shear stress induced in the wire is (where D = Mean diameter of the spring coil, d = Diameter of the spring wire, K = Wahl's stress factor and W = Axial compressive load on the spring)
A $$\frac{{{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
B $$\frac{{2{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
C $$\frac{{4{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
D $$\frac{{8{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$
A closely coiled helical spring is acted upon by an axial force. The maximum shear stress developed in the spring is $$\tau $$. The half of the length of the spring if cut off and the remaining spring is acted upon by the same axial force. The maximum shear stress in the spring in new condition will be
A $$\frac{\tau }{2}$$
B $$\tau $$
C $$2\tau $$
D $$4\tau $$
A tri-axial shear test is preferred to direct shear test, because
A It can be performed under all three drainage conditions with complete control
B Precise measurement of pore pressure and change in volume during test, is not possible
C Stress distribution on the failure plane, is non uniform
D None of these
K produces 100 parts twice as fast as G does. G produces 100 parts in 40 minutes. If each produces parts at a constant rate, how many parts will K produces in 6 minutes.
A 30
B 25
C 20
D 15
The strain energy stored in a body due to shear stress, is (where $$\tau $$ = Shear stress, C = Shear modulus and V = Volume of the body)
A $$\frac{{\tau {\text{V}}}}{{2{\text{C}}}}$$
B $$\frac{{2{\text{C}}}}{{\tau {\text{V}}}}$$
C $$\frac{{{\tau ^2}{\text{V}}}}{{2{\text{C}}}}$$
D $$\frac{{2{\text{C}}}}{{{\tau ^2}{\text{V}}}}$$
When a body is subjected to biaxial stress i.e. direct stresses ($${\sigma _{\text{x}}}$$) and ($${\sigma _{\text{y}}}$$) in two mutually perpendicular planes accompanied by a simple shear stress ($${\tau _{{\text{xy}}}}$$ ), then maximum shear stress is
A $$\frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
B $$\frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
C $$\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + \tau _{{\text{xy}}}^2} $$
D $$\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + \tau _{{\text{xy}}}^2} $$
When a body is subjected to a direct tensile stress ($${\sigma _{\text{x}}}$$) in one plane accompanied by a simple shear stress ($${\tau _{{\text{xy}}}}$$ ), the maximum shear stress is
A $$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
B $$\frac{{{\sigma _{\text{x}}}}}{2} - \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
C $$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 - 4\tau _{{\text{xy}}}^2} $$
D $$\frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
According to maximum shear stress failure criterion, yielding in material occurs, when the maximum shear stress is equal to __________ times the yield stress.
A 0.5
B 2
C $$\sqrt 2 $$
D $$\sqrt {\frac{3}{2}} $$
The ratio of the largest load in a test to the original cross-sectional area of the test specimen is called the __________ stress.
A Yield point
B Breaking
C Ultimate
D None of these

Next steps