Engineering Board Exam Reviewer · Strength of Materials · easy
A solid circular rod 20 mm in diameter carries an axial tensile load of 50 kN. What is the normal stress in the rod?
- 39.79
- 159.15
- 125
- 0.16
Answer
B. 159.15
Solution
A = π d2/4 = π (20)2/4 = 314.16 mm2. = P/A = (50 × 1000 N) / 314.16 mm2 = 159.15 MPa.
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