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:...
Torsion refers to the twisting of a structural member when loaded by couples (torques). Maximum at the outer surface ( 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. Maximum at the outer surface ( σmax=McIsigma sub
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
τ=TcJtau equals the fraction with numerator cap T c and denominator cap J end-fraction Measured in radians.