Mechanics Of Materials - Formulas And Problems:... May 2026

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:

σ=Eϵwhere E is Young′s Modulussigma equals cap E epsilon space where cap E is Young prime s Modulus Mechanics of Materials - Formulas and Problems:...

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

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. Torsion (Circular Shafts) σmax=McIsigma sub m a x

δ=(80,000)(2)(400×10-6)(200×109)delta equals the fraction with numerator open paren 80 comma 000 close paren open paren 2 close paren and denominator open paren 400 cross 10 to the negative 6 power close paren open paren 200 cross 10 to the nineth power close paren end-fraction

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