Q25 of 29 Page 70

(a) A steel wire of mass μ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m-3(Young’s modules Y=2×1011 Nm-2).

(b) If the yield strength of steel is 2.5×108 Nm-2, what is the maximum weight that can be hung at the lower end of the wire?


(a)



Taking a cross-section of wire





At X=0, Tension=Mg and X=0, Tension= μgL + Mg. Integrating above equation will give



As force is a function of x, therefore relating force with young modulus we have to consider a small part in which force can be approximated to be constant .Let the length of that part is dx. Therefore, we can write



We have written the equation which is valid for only 1D,as it is given in question that lateral strains << longitudinal strains, therefore we can assume that only extension in length is happening only


Therefore change in radius is given by



Putting in the values from questions





(b) We have been given Yield Strength = 2.5×108 N/m2


We know from earlier equations that Maximum Tensile force at the bottom will be equal to





We know that,




Putting in the values



Solving the above equation we get,


.


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(a) For what angle is the tensile stress a maximum?


(b) For what angle is the shearing stress a maximum?


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