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Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 107 Pa? Assume that each rivet is to carry one-quarter of the load.

Diameter of the metal strip, d = 6.0 mm = 6.0 × 10–3 m

Radius, r = d/2 = 3 × 10-3 m 

Maximum shearing stress = 6.9 × 107 Pa 

Maximum stress = fraction numerator Minimum space load space or space force over denominator Area end fraction
Maximum force = Maximum stress × Area 

                     = 6.9 × 107 × π × (r) 2 

                     = 6.9 × 107 × π × (3 ×10–3)

                     = 1949.94 N 

Each rivet carries one quarter of the load. 

∴ Maximum tension on each rivet = 4 × 1949.94 = 7799.76 N.
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A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10–2 cmis stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint. 



Error converting from MathML to accessible text. 266 Views

The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 108 Pa. A steel ball of initial volume 0.32 m3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Water pressure at the bottom, = 1.1 × 108 Pa 

Initial volume of the steel ball, V = 0.32 m

Bulk modulus of steel, B = 1.6 × 1011 Nm–2 

The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface. 

Let the change in the volume of the ball on reaching the bottom of the trench be ΔV

Bulk modulus, B = p / (V/V

                ∆V  =  B / pV 

                     = 1.1 × 108 × 0.32 / (1.6 × 1011 ) 

                     =  
2.2 × 10-4 m

Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 × 10–4 m3

115 Views

A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass be suspended in order to produce (a) equal stresses and b) equal strains in both steel and aluminium wires.


Given, 

Cross-sectional area of wire A = a1 = 1.0 mm= 1.0 × 10–6m2

Cross-sectional area of wire Ba2 = 2.0 mm2 = 2.0 × 10–6m2

Young’s modulus for steel, Y1 = 2 × 1011 Nm–2

Young’s modulus for aluminium, Y2 = 7.0 ×1010 Nm–2

a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.

Stress in the wire = Force / Area  =  F / a

If the two wires have equal stresses, then 

F1 / a1  =  F2 / a

where,

F1Force exerted on the steel wire
F2 = Force exerted on the aluminum wire
rightwards double arrow space straight F subscript 1 over straight F subscript 2 space equals space straight a subscript 1 over straight a subscript 2 space equals space 1 half space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis

Now comma space taking space torque space about space the space point space of space suspension comma space

fraction numerator left parenthesis 1.05 space minus space straight y right parenthesis over denominator straight y end fraction space equals space 1 half space
2 left parenthesis left parenthesis 1.05 space minus space straight y right parenthesis space equals space straight y

straight y space equals space 0.7 space straight m

In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

b) Young apostrophe straight s space modulus space equals space Stress over strain
Strain space equals space fraction numerator Stress over denominator Young apostrophe straight s space Modulus end fraction equals fraction numerator begin display style bevelled fraction numerator left parenthesis straight F over denominator straight a end fraction end style right parenthesis over denominator straight Y end fraction

If space the space strain space in space the space two space wires space is space equal comma space then

fraction numerator bevelled fraction numerator left parenthesis straight F subscript 1 over denominator straight a subscript 1 end fraction right parenthesis over denominator straight Y subscript 1 end fraction space equals space fraction numerator open parentheses begin display style bevelled straight F subscript 2 over straight a subscript 2 end style close parentheses over denominator straight Y subscript 2 end fraction space

rightwards double arrow space straight F subscript 1 over straight F subscript 2 space equals space fraction numerator straight a subscript 1 straight Y subscript 1 over denominator straight a subscript 2 straight Y subscript 2 end fraction space

rightwards double arrow space straight a subscript 1 over straight a subscript 2 space equals space 1 half space
rightwards double arrow straight F subscript 1 over straight F subscript 2 space equals space open parentheses 1 half close parentheses cross times open parentheses fraction numerator 2 cross times 10 to the power of 11 over denominator 7 space cross times space 10 to the power of 10 end fraction close parentheses space equals space 10 over 7 space space space space space space space space space space space space space space space... space left parenthesis ii right parenthesis space

Taking space torque space about space the space point space where space mass space straight m subscript 1 space
is space suspended space at space straight a space distance space straight y subscript 1 space from space the space side space where space
wire space straight A space attached comma space we space get

straight F subscript 1 straight y subscript 1 space equals space straight F subscript 2 space end subscript left parenthesis 1.05 space – space straight y subscript 1 right parenthesis space

straight F subscript 1 over straight F subscript 2 space equals space fraction numerator left parenthesis 1.05 space minus space straight y subscript 1 right parenthesis over denominator space straight y subscript 1 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis iii right parenthesis

Using space equations space left parenthesis ii right parenthesis thin space and space left parenthesis iii right parenthesis comma space we space get
fraction numerator left parenthesis 1.05 space minus space straight y subscript 1 right parenthesis over denominator space straight y subscript 1 end fraction space equals space 10 over 7
rightwards double arrow space 7 left parenthesis 1.05 space minus space straight y subscript 1 right parenthesis space space equals space space 10 straight y subscript 1 space

rightwards double arrow space straight y subscript 1 space equals space 0.432 space straight m

In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached. 

225 Views

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Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?


Diameter of the cones at the narrow ends, d = 0.50 mm = 0.5 × 10–3 m

Radius, r = d/2  =  0.25 × 10-3 m 

Compressional force, F = 50000 N\

Pressure at the tip of the anvil is given by, 


straight P space space equals space straight F over straight A equals space fraction numerator 50000 over denominator straight pi space left parenthesis 0.25 cross times 10 to the power of negative 3 end exponent right parenthesis squared end fraction space

space space space space equals space 2.55 cross times 10 to the power of 11 space Pa space

Therefore, the pressure at the tip of the anvil is 2.55 × 1011Pa. 
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