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A steel bar with diameter d = 14 mm is subjected to a tensile load P = 10 kN (see figure). (a) What is the maximum normal stress σmax in the bar? (b) What is the maximum shear stress τmax?
Solution
(a) Maximum normal stress
σx = P / A = 10KN / ((π /4) * (14mm)2) = 64.9MPa
(b) Maximum shear stress
The maximum shear stress is on a 45o plane and equals σx / 2.
τmax = σmax / 2 = 64.9MPa / 2 = 32.45MPa
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