Magnitude of torque - bending moment:
A prismatic bar of hollow circular cross-section along with outer and inner diameters 100 mm and 80 mm correspondingly, carries a bending moment of 5 kN-m. If the tensile, compressive and shear strengths of the material are given as 140 N/mm^{2}, 125 N/mm^{2} and 95 N/mm^{2} respectively, what is the magnitude of torque that may safely be applied in addition to the bending moment.
Solution
Here, the criteria to be considered are as follows:
σ1 › 140 N/mm^{2}
σ2 ›- 125 N/mm^{2}
τ_{max} › 95 N/mm^{2}
As the magnitudes of σ1 and σ2 for such bars will be equal, we need to satisfy only.
σ2 ›- 125 N/mm^{2}
and
τ_{max} › 95 N/mm^{2}
I for the section π/64(100^{4} -80^{4}) = 2.898 × 10^{6} mm^{4}
J for the section π/32(100^{4} -80^{4}) = 5.796 × 10^{6} mm^{4}
Maximum compression due to bending moment,
= - M/I y_{max }= -5×10^{6}× 50/2.898× 10^{6} = - 86.266 N/mm^{2}
If T be the torque applied (in kN-m units), then,
τ_{max }= T×10^{6}×60/5.796×10^{6} = 8.62664 T N/mm^{2 }(under torsion alone)
43.133^{2} + 8.62664^{2} T ^{2} = (- 81.867)^{2}
i.e. T = (81.867^{2} - 43.133^{2}/ 8.62664^{2})^{1/2} =8.066 kN-m
If the applied torque is within 8.066 kN-m, we are certain that the bar will be safe in compression as well as tension.
We should now analyse what torque will have to be applied if τ_{max } › 95 N/mm^{2}.
We know that
(under combined bending and torsion)
T= (95^{2} -(86.266/2)^{2}/8.62664^{2})^{1/2} = 9.8119 kN-m
Thus, we cannot apply this much torque, because it will cause compression failure. Therefore, the safe value of additional torque should be restricted to 8.066 kN-m.