Question
Download Solution PDFA rectangular beam of uniform strength and subjected to a bending moment ‘M’ has a constant width. The variation in depth will be proportional to
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
We have the bending equation for a beam, as
\(\frac{\text{ }\!\!\sigma\!\!\text{ }}{\text{y}}=\frac{\text{M}}{\text{I}}=\frac{\text{E}}{\text{R}}\)
Where,
σ = Bending stress, y = distance from the neutral axis, M = Bending moment of any section, I = Moment of inertia, E = Modulus of Elasticity of material, and R = Radius of curvature.
When a beam is designed such that the extreme fibers are loaded to the maximum permissible stress ρmax by varying c/s, it will be known as beam of uniform strength.
\(\therefore \frac{\sigma }{y}=\frac{M}{I}\)
\(\text{ }\!\!\sigma\!\!\text{ }=\frac{\text{M}\times \text{y}}{\text{b}{{\text{d}}^{3}}/12}\)
\(\text{ }\!\!\sigma\!\!\text{ }=\frac{\text{M}\times {{\text{d}}_{\text{x}}}\times 12}{\text{bd}_{\text{x}}^{2}\times 2}\therefore \text{y}={{\text{d}}_{\text{x}}}/2\)
\(\text{ }\!\!\sigma\!\!\text{ }=\frac{6\text{M}}{\text{bd}_{\text{x}}^{2}}\)
\(\text{d}_{\text{x}}^{2}=\frac{6\text{M}}{\text{ }\!\!\sigma\!\!\text{ b}}\Rightarrow {{\text{d}}_{\text{x}}}=\sqrt{\frac{6\text{M}}{\text{ }\!\!\sigma\!\!\text{ b}}}\)
\(\therefore {{\text{d}}_{\text{x}}}\propto \sqrt{\text{M}}\)
Last updated on May 28, 2025
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