Given: rod of circular cross section is subjected to uniaxial tension. Length, L=1500 mm radius, r = 10 mm E=2*10^5 N/mm^2 Force, F=20 kN = 20,000 N [note: newton (unit) in abbreviation is written in upper case, as in N ]
From given above, area of cross section = π r^2 = 100 π =314 mm^2 (i) Stress, σ =force/area = 20000 N / 314 mm^2 = 6366.2 N/mm^2 = 6370 N/mm^2 (to 3 significant figures)
(ii) Strain ε = ratio of extension / original length = σ / E = 6366.2 /(2*10^5) = 0.03183 = 0.0318 (to three significant figures)
(iii) elongation = ε * L = 0.03183*1500 mm = 47.746 mm = 47.7 mm (to three significant figures)
