The elongation delta of a solid circular rod under tensile loading is given by d

Question

The elongation delta of a solid circular rod under tensile loading is given by delta = PL/EA where P is the applied force. L is the length of the bar. E is Young's modulus, and A is the cross-sectional area of the bar. Suppose that we want to determine Young's modulus from the results of a tensile test. The following values have been either calculated or estimated: P = 5000 Plusminus 50 N delta = 0.05 plusminus 0.001 mm L = 50 plusminus 0.1 mm d = 7 plusminus 0.05 mm where d is the bar diameter. Determine E plusminus delta E. The cantilever beam shown at right is fabricated from a thin-walled circular tube of radius r and wall thickness t. The beam is subjected to a point load P at its end. resulting in a F deflection delta at this point. delta = PL3/3 pi Er3t Given that P = 100 plusminus 5 N. r= 10 plusminus 0.05 mm t = 0.2 plusminus 0.01 mm L = 100 plusminus 0.5 mm and E = 200 plusminus 10 GPa. determine the uncertainty in delta.

Explanation / Answer

1.) E = PL/(delta*A) = 5000*50*10^-3/(0.05*10^-3*pi*(3.5x10^-3)^2) = 1.3 x 10^11 N/m^2

E = 4*P*L/(delta*pi*d^2)

So dE/E = (dP/P) + (dL/L) + (d(delta)/delta) + 2*(d(d)/d)

dE/E = (50/5000) + (0.1/50) + (0.001/0.05) + 2*(0.05/7)

dE/E = 0.046286

So dE = (1.3x 10^11)(0.046286) = 0.0602 x 10^11 N/m^2

So E +/- dE = (1.3 +/- 0.0602) x 10^11 N/m^2

2.) delta = P*L^3/(3pi*E*r^3*t) = 100*(10^-1)^3/(3pi*2*10^11*(10^-2)^3*0.2*10^-3) = 2.6526 x 10^-4 m = 0.2653 mm

delta = P*L^3/(3pi*E*r^3*t)

So d(delta)/delta = (dP/P) + 3*(dL/L) + (dE/E) + 3*(dr/r) + (dt/t)

d(delta)/delta = (5/100) + 3*(0.5/100) + (10/200) + 3*(0.05/10) + (0.01/0.2)

d(delta)/delta = 0.18

So d(delta) = (0.26526)(0.18) = 0.04775 mm

So the uncertainity in delta = 0.04775 mm

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