Question
Download Solution PDF20 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
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
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
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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