Question 10.12: A solid circular bar is 45 mm in diameter and is subjected t...
A solid circular bar is 45 mm in diameter and is subjected to an axial tensile force of 120 kN along with a torque of 1150 N · m. Compute the maximum shear stress in the bar.
Objective Compute the maximum shear stress in the bar.
Given Diameter = D = 45 mm
Axial force = F = 120 kN = 120 000 N
Torque = T = 1150 N · m = 1 150 000 N · mm
Analysis Use Equation (10–2) to compute τ_{max} .
τ_{max} = \sqrt{(\frac{ \sigma }{2})^{2} + \tau^{2}} (10-2)
Results 1. First, the applied normal stress can be found using the direct stress formula:
σ = F/A
A = πD² /4 = π ( 45 mm )² /4 = 1590 mm²
σ = (120 000 N)/(1590 mm² ) = 75.5 N/mm² = 75.5 MPa
2. Next, the applied shear stress can be found from the torsional shear stress formula:
\tau = T/Z_{p}
Z_{p} = πD³ /16 = π(45 mm)³ /16 = 17 892 mm³
τ = (1 150 000 N·mm) / (17 892 mm³) = 64.3 N/mm² = 64.3 MPa
3. Then using Equation (10–2) yields
τ_{max} = \sqrt{\left\lgroup\frac{ 75.5 MPa }{2}\right\rgroup ^{2} +(64.3 MPa)^{2}} = 74.6 MPa
Comment This stress should be compared with the design shear stress.