δ=PLAEdelta equals the fraction with numerator cap P cap L and denominator cap A cap E end-fraction 2. Torsion (Circular Shafts)

δ=160,00080,000,000=0.002 m or 2 mmdelta equals the fraction with numerator 160 comma 000 and denominator 80 comma 000 comma 000 end-fraction equals 0.002 m or 2 mm Practice Problem: Bending Stress A rectangular beam ( ) experiences a maximum bending moment of . Determine the maximum bending stress. Solution: Find : Find : Apply Formula: Result:

σmax=McIsigma sub m a x end-sub equals the fraction with numerator cap M c and denominator cap I end-fraction 4. Transverse Shear Internal shear forces ( ) result in shear stresses across the cross-section.

σ=Eϵwhere E is Young′s Modulussigma equals cap E epsilon space where cap E is Young prime s Modulus

Bending moments cause internal stresses that vary linearly from the neutral axis.

ϵ=δLepsilon equals the fraction with numerator delta and denominator cap L end-fraction For materials in the elastic region.

ϕ=TLGJphi equals the fraction with numerator cap T cap L and denominator cap G cap J end-fraction (Note: is the polar moment of inertia; for solid shafts). 3. Pure Bending