Young\'s modulus of aluminum is 7x10^10 N/m^2 while that of uranium is 21x10^10

Question

Young's modulus of aluminum is 7x10^10 N/m^2 while that of uranium is 21x10^10 N/m^2. A pair of thin rods of aluminum and uranium, with length 2.5 m and cross-sectional area of 10^-3 m^2 have been prepared.

a) Suppose that a tensile force of 150 N is applied to the aluminum rod. Determine the amount by which the rod stretches.

b) Suppose that n tensile force of 350 N is applied to the uranium rod. Determine the amount by which the rod stretches.

c) Add the results obtained for (a) and (b).

d) Suppose that the two rods are attached end to end (i.e., in series). Compute the effectiveYoung's modulus for the series composite rod made with equal fractions of aluminium and uranium.

e) Suppose that a tensile force of 350 N is applied to the series composite rod, and determine the amount b which it stretches.

f) Suppose that the two rods are arranged side by side (i.e., in parallel) . Compute the effective Young's modulus for the parallel composite rod made with equal parts of aluminium and uranium.

with ne 25a and ose-sectional ares A10have b prepared o) Suppose thatile e of sos i pplinel to the alomi amount by which the rod strotchen dDo t (t) Suppose that a tenaile foee of 30 ie applied to the by which the rod stretches (e) Add the remults obtainet or (a) and th) d) Suppoe that the two ds sare atted nto d) Compate the offi Young's modulus for the seri (e) Suppo" that a tenale ce ofXON"applied to the wri-comp-, tod, and determine the ant boy which it stretche (1) Suppose that the two rods are ranged side boy de(e in paralel Compute the and i the amount by which ittretche Page I of

Explanation / Answer

a) Young's modulus(Y) = Stress/Strain = 7*10^10 N/m^2(given)

Length(l) = 2.5 m

Cross sectional area(A)= 10^-3 m^2

Applied Force(F) = 150 N

Let the amount by which the rod extends be del l

Y = [(F/A)/( del l/l)]

or, 7 * 10^10 = [(150/10^-3)/(del l/2.5)

On solving, we get del l = 5.357*10^-6 m Ans.

b) Young's modulus(Y) = Stress/Strain = 21*10^10 N/m^2(given)

Length(l) = 2.5 m

Cross sectional area(A)= 10^-3 m^2

Applied Force(F) = 350 N

Let the amount by which the rod extends be del l

Y = [(F/A)/( del l/l)]

or, 21* 10^10 = [(350/10^-3)/(del l/2.5)

On solving, we get del l = 4.167 * 10^-6 m Ans.

c) Sum of extensions calculated in a) and b) = (5.357+4.167) * 10^-6 m

= 9.524 * 10^-6 m Ans.

d) Equivalent Young's modulus(Yeq) for two rods of same length and equal area connected in series having Young's moduli Y1 and Y2 is given by: Yeq = Y1*Y2/(Y1+Y2)

= 7*10^10*21*10^10/28*10^10

= 5.25*10^10 N/m^2 Ans.

e) Young's modulus(Y) = Stress/Strain = 5.25*10^10 N/m^2(calculated above)

Length(l) = 2.5+2.5 = 5.0 m

Cross sectional area(A)= 10^-3 m^2

Applied Force(F) = 350 N

Let the amount by which the rod extends be del l

Y = [(F/A)/( del l/l)]

or, 5.25* 10^10 = [(350/10^-3)/(del l/5.0)

On solving, we get del l = 3.33*10^-5 m Ans.

f)  Equivalent Young's modulus(Yeq) for two rods of same length and equal area connected in parallel having Young's moduli Y1 and Y2 is given by: Yeq = (Y1+Y2)/2

= (7+21)/2*10^10

= 14*10^10 N/m^2 Ans.

g) Young's modulus(Y) = Stress/Strain = 14*10^10 N/m^2(calculated above)

Length(l) = 2.5 m

Cross sectional area(A)= 10^-3+10^-3 = 2*10^-3 m^2

Applied Force(F) = 350 N

Let the amount by which the rod extends be del l

Y = [(F/A)/( del l/l)]

or, 14* 10^10 = [(350/2*10^-3)/(del l/2.5)

On solving, we get del l = 3.125*10^-6 m Ans.

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