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# Lecture 11 shear stresses in beams

## on Aug 24, 2011

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## Lecture 11 shear stresses in beamsPresentation Transcript

• Unit 2- Stresses in BeamsTopics Covered  Lecture -1 – Review of shear force and bending moment diagram  Lecture -2 – Bending stresses in beams  Lecture -3 – Shear stresses in beams  Lecture -4- Deflection in beams  Lecture -5 – Torsion in solid and hollow shafts.
• Shear Stresses in Beams ofRectangular Cross Section  In the previous chapter we examined the case of a beam subjected to pure bending i.e. a constant moment along axis .  When a beam is in pure bending, the only stress resultants are the bending moments and the only stresses are the normal stresses acting on the cross sections.  Most beams are subjected to loads that produce both bending moments and shear forces (non-uniform bending)
• Shear Stresses in Beams ofRectangular Cross Section  In these cases, both normal and shear forces are developed in the beam.  Normal stresses are calculated with the Flexure Formula.  We will now look at the Shear Stresses
• Vertical & Horizontal  Shear Stresses subjected Consider a beam of rectangular cross section to a positive shear force.
• Shear Stresses A C Shear forces and bending moments are different across different sections. D Area A B σ σ +dσ dM − ×A×y − I y1 x y €M M+dM dx b − Ay € τ=F× I×b
• Shear stress distribution for different section Rectangular SectionA is the area of the x-section cut off by a lineparallel to the neutral axis. is the distance ofthe centroid of A from the neutral axis Parabolic distribution of shear stresses
• Shear stress distribution for different sectionRectangular Section The maximum value of shear stress would obviously beat the location y = 0.
• Shear stress distribution for different sectionRectangular Section
• Shear stress distributionSection Shear stress Max shear Shear Stress stress distribution F ⎛ d 2 2 ⎞ τ = ⎜ − y ⎟ τ max = 1.5τ avg 2I ⎝ 4 ⎠ € € F 2 4 τ = (R − y 2 ) τ max = τ avg 3I 3 € €