Question
Download Solution PDFWhat will be the tensile force required to cause an elongation of 0.045 mm in a steel rod of 1000 mm length and 12 mm diameter, (where E = 2 × 106 kg/cm2)?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFElongation of rod is given by:
\(\Delta = \frac{{PL}}{{AE}}\)
P = 3000 kg
E = 2 × 106 kg/cm2
D = diameter = 12 mm = 1.2 cm
A = Area = \(\frac{\pi }{4} \times 1.2 \times 1.2 = 1.13\;c{m^2}\)
L = length = 1 m = 1000 mm
\(0.045 = \frac{{P \times 1000}}{{1.13 \times 2 \times 1000000}}\)
\( P= 102\;kg\)
Last updated on Jun 4, 2025
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