Question
Download Solution PDFAt a point in a stressed body there are normal stresses of 1 N/mm² (tensile) on a vertical plane and 0.5 N/mm² (tensile) on a horizontal plane. The shearing stresses on these planes are zero. What will be the normal stress on a plane making an angle 50° with the vertical plane? [given, ]
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Normal stress on a plane inclined at angle \( \theta \) is given by:
\( \sigma_n = \sigma_v \cos^2 \theta + \sigma_h \sin^2 \theta \)
Given:
- \( \sigma_v = 1 \, \text{N/mm}^2 \)
- \( \sigma_h = 0.5 \, \text{N/mm}^2 \)
- \( \cos^2(50^\circ) = 0.413 \Rightarrow \sin^2(50^\circ) = 0.587 \)
Calculation:
\( \sigma_n = 1 \times 0.413 + 0.5 \times 0.587 = 0.413 + 0.2935 = 0.7065 \, \text{N/mm}^2 \)
Last updated on Jul 15, 2025
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