Question
Download Solution PDFA circular beam section is subjected to a shear force of 40π kN. The maximum shear stress allowed in the material is 6 MPa. Calculate the safe diameter of the section, assuming a factor of safety equal to 2.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Maximum shear stress in circular beam-
\(Maximum\, shear\, stress\,(\tau_{max})=({4\over 3})\tau_{ave}={4\over 3}×{Shear force\over Area}\)
Calculation:
Given data:
Shear force (F) = 40π kN
Maximum shear stress in material (\(\tau_{max}\)) = 6 MPa or 6 N/mm2
Safe diameter of the circular beam (D) =?
The factor of safety (FOS) = 2
Factored shear force (F') = Factor of safety × Shear force (F)
Factored shear force (F') = 2 × 40π = 80π kN
Factored shear force (F') = 80π × 103 N
\(Maximum\, shear\, stress\,(\tau_{max}) = {4\over 3}×{Shear force\over Area}\)
\(6={4\over 3}\times {80\pi \times 10^3\over {\pi\over 4}D^2}\)
\(6={4\over 3}\times {4\times 80 \times 10^3\over D^2}\)
\(D^2={4\over 3}\times {4\times 80 \times 10^3\over 6}\)
\(D^2=71.111\times 10^3\)
\(D=266.666\, mm\)
Last updated on May 28, 2025
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