Question

7) Given the state of stress at point A on the circular cross section a-a of the handle assembly is oz 47.11 MPa, ryz = 48.81 MPa. Show the results on the volume element shown. 800 N 400 mm 200 mm 1500 N 20 mm1000 N Section a-a

Answer 1

This is bi-axial state of stress where $\sigma _{y}=0.$

Maximum principal stress= $(\sigma _{y}+\sigma _{z})/2 + ((\sigma _{y}-\sigma _{z})/2 + \tau _{yz}^{2})^{0.5}$

Maximu principle stress = 0+47.11/2 + ((0-47.11)/2 + 48.81² ))0.5 = 23.55+54.19=77.74 Mpa

Minimum princile stress = $(\sigma _{y}+\sigma _{z})/2 - ((\sigma _{y}-\sigma _{z})/2 + \tau _{yz}^{2})^{0.5}$ = 23.55-54.19 = -30.64 Mpa.

Maximum shear stress = $((\sigma _{y}-\sigma _{z})/2 + \tau _{yz}^{2})^{0.5}$ = 54.19 Mpa.