Solve Problem 1.44 by assuming the wire diameters of springs 1 and 2 to be \(0.125 \mathrm{~m}\) and \(0.0125 \mathrm{~m}\) instead of \(0.05 \mathrm{~m}\) and \(0.025 \mathrm{~m}\), respectively.

Data From Problem 1.44:-

Consider two helical springs with the following characteristics:

Spring 1: material—steel; number of turns- 10 ; mean coil diameter \(-0.3 \mathrm{~m}\); wire diameter\(0.05 \mathrm{~m}\); free length- \(0.4 \mathrm{~m}\); shear modulus- \(80 \times 10^{9} \mathrm{~Pa}\).

Spring 2: material-aluminum; number of turns- 10 ; mean coil diameter- \(0.25 \mathrm{~m}\); wire diameter \(-0.025 \mathrm{~m}\); free length- \(0.4 \mathrm{~m}\); shear modulus \(-24 \times 10^{9} \mathrm{~Pa}\).

Determine the equivalent spring constant when

(a) spring 2 is placed inside spring 1 , and

(b) spring 2 is placed on top of spring 1 .

Transcribed Image Text:

25 cm, 15 cm 10 cm -5 m FIGURE 1.92 A composite propeller shaft. T Steel Aluminum Section AA