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 $$2\,{\text{s}}l{\sigma _{\text{t}}}$$
A double fillet welded joint with parallel fillet weld of length '$$l$$' and leg 's' is subjected to a tensile force 'P'. Assuming uniform stress distribution, the shear stress in the weld is given by
A $$\frac{{\sqrt 2 {\text{P}}}}{{{\text{s}}l}}$$
B $$\frac{{\text{P}}}{{2{\text{s}}l}}$$
C $$\frac{{\text{P}}}{{\sqrt 2 {\text{s}}l}}$$
D $$\frac{{2{\text{P}}}}{{{\text{s}}l}}$$
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
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 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 $$\frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
The torque transmitted by a solid shaft of diameter (D) is (where $$\tau $$ = Maximum allowable shear stress)
A $$\frac{\pi }{4} \times \tau \times {{\text{D}}^3}$$
B $$\frac{\pi }{{16}} \times \tau \times {{\text{D}}^3}$$
C $$\frac{\pi }{{32}} \times \tau \times {{\text{D}}^3}$$
D $$\frac{\pi }{{64}} \times \tau \times {{\text{D}}^3}$$
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 $$4\tau $$
When a hole of diameter ‘d’ is punched in a metal of thickness 't', then the force required to punch a hole is equal to (where $${\tau _{\text{u}}}$$ = Ultimate shear strength of the material of the plate)
A $${\text{dt}}{\tau _{\text{u}}}$$
B $$\pi {\text{dt}}{\tau _{\text{u}}}$$
C $$\frac{\pi }{4} \times {{\text{d}}^2}{\tau _{\text{u}}}$$
D $$\frac{\pi }{4} \times {{\text{d}}^2} \times {\text{t}}{\tau _{\text{u}}}$$

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