Q#12 (Past Exam Paper – June 2014 Paper 12 Q21)
A composite rod is made by attaching a glass-reinforced plastic rod and a nylon rod end to end, as shown.
The rods have the same cross-sectional area and each rod is 1.00 m in length. The Young modulus Ep of the plastic is 40 GPa and the Young modulus En of the nylon is 2.0 GPa.
The composite rod will break when its total extension reaches 3.0 mm.
What is the greatest tensile stress that can be applied to the composite rod before it breaks?
A 7.1 × 10-14 Pa
B 7.1 × 10-2 Pa
C 5.7 × 106 Pa
D 5.7 × 109 Pa
Solution:
Answer: C.
The composite rod will break when its total extension reaches 3.0 mm.
For a material,
Stress = Force / Area = F/A
Strain = extension, e / original length, L = e/L
Young modulus, E = stress / strain
E = Stress / (e/L) = Stress×L / e
Extension, e = Stress×L / E
The same tensile stress (the same force and since the cross-sectional area is the same for both, the same tensile stress is therefore applied) is applied to the different materials (glass-reinforced plastic and nylon). The different materials will extend by different amounts when the same stress is applied to each.
The total extension cannot be equal to 3.0 mm (= 3.0×10-3 m) as the composite rod would then break.
For the plastic rod, extension ep = Stress × (1) / (40×109) = Stress / (40×109)
For the nylon rod, extension en = Stress × (1) / (2×109) = Stress / (2×109)
For the greatest tensile stress that can be applied to composite rod before it breaks, the sum of extensions of the plastic and nylon should be equal to 3.0 mm.
ep + en = 3.0×10-3
Stress/(40×109) + Stress/(2×109) = 3.0×10-3
Stress × [1/(40×109) + 1/(2×109)] = 3.0×10-3
Stress = (3.0×10-3) / [1/(40×109) + 1/(2×109)] = 5.71×106 Pa
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