Question
Download Solution PDFIn a simply supported beam, maximum shear stress in a triangular cross-section (altitude h) occurs at a distance:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Maximum shear stress occurs at h/2 from base
Distance from the neutral axis \(= \frac{{\rm{h}}}{2} - \frac{{\rm{h}}}{3} = \frac{{\rm{h}}}{6}\)
Here
\({{\rm{\tau }}_{{\rm{avg}}}} = \frac{{{{\rm{V}}_{\rm{u}}}}}{{\frac{1}{2}{\rm{bh\;}}}}\)
where Vu = Maximum shear force
∴ The correct answer is h/6 from the neutral axis.
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