Related Questions

The tensile strength of the welded joint for double fillet is (where s = Leg or size of the weld, $$l$$ = Length of weld and $${\sigma _{\text{t}}}$$ = Allowable tensile stress for weld metal)
A $$0.5\,{\text{s}}l{\sigma _{\text{t}}}$$
B $${\text{s}}l{\sigma _{\text{t}}}$$
C $$\sqrt 2 \,{\text{s}}l{\sigma _{\text{t}}}$$
D $$2\,{\text{s}}l{\sigma _{\text{t}}}$$
Shear strength of the welded joint for double parallel fillet is (where $$\tau $$ = Allowable shear stress for weld metal)
A $$0.5\,{\text{s}}l\tau $$
B $${\text{s}}l\tau $$
C $$\sqrt 2 \,{\text{s}}l\tau $$
D $$2\,{\text{s}}l\tau $$
In a transverse fillet welded joint, the size of weld is __________ the throat of weld.
A 0.5 times
B Equal to
C 2 times
D Double
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 $$\frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
In a butt welded joint, the size of weld is __________ the throat of weld.
A 0.5 times
B Equal to
C 2 times
D Double
The leg of a barometric leg pump is longer than the Toricellian leg, which is about __________ metres.
A 1
B 5
C 34
D 10
When a body is subjected to a direct tensile stress (σ) in one plane, the maximum shear stress is __________ the maximum normal stress.
A Equal to
B One-half
C Two-third
D Twice
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 $$\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + \tau _{{\text{xy}}}^2} $$
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 $$4\tau $$
The change in length due to a tensile or compressive force acting on a body is given by (where P = Tensile or compressive force acting on the body, $$l$$ = Original length of the body, A = Cross-sectional area of the body and E = Young's modulus for the material of the body)
A $$\frac{{{\text{P}}l{\text{A}}}}{{\text{E}}}$$
B $$\frac{{{\text{P}}l}}{{{\text{AE}}}}$$
C $$\frac{{\text{E}}}{{{\text{P}}l{\text{A}}}}$$
D $$\frac{{{\text{AE}}}}{{{\text{P}}l}}$$

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