20 mm × 20 mm वर्ग अनुप्रस्थ काट के स्टील बार को 200 kN के अक्षीय संपीडक भार के अधीन किया जाता है। यदि बार की लंबाई 1 m है और E = 100 GPa, the contraction of the bar will be

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ISRO LPSC Technical Assistant Mechanical 23 Feb 2020 Official Paper
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  1. 0.2 mm
  2. 2.5 mm
  3. 5 mm
  4. 0.5 m

Answer (Detailed Solution Below)

Option 3 : 5 mm
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संकल्पना:

Elongation  or Contraction in the rod when an axial force is applied is given by:

\(δ L = \frac{{PL}}{{AE}}\)

जहाँ P अक्षीय भार, L बार की लंबाई, δL बार की लंबाई में दीर्घीकरण, A बार का अनुप्रस्थ काट क्षेत्रफल, E प्रत्यास्थता मापांक है।

गणना:

दिया गया  है:

P = 200 kN, L = 1 m, A = 20 mm × 20 mm, E = 100 GPa

F2 Ashik 14.12.20 Pallavi D2

बार का दीर्घीकरण,

\(δ L = \frac{{PL}}{{AE}} = \frac{{200 \times 1000 }}{{0.02 \times 0.02 \times 100 \times {{10}^9}}} = 5 \times {10^{ - 3}}\;m = 5~mm\)

Hence the required contraction of the bar will be 5 mm.

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