Round bars, otherwise known as shafts and axles, transmit torque and rotational power as well as carry radial and thrust loads when used as axles. Most power-transmitting shafts are composed of steel or stainless steel because of the balance of strength, rigidity, and hardness of this metal that go along with its relative economy compared to other metals. Designers of shafts transmitting torque or torsional forces must consider both the maximum and allowable yield stresses, tensile stresses, and shear stresses that a metal can withstand when selecting a shaft for a specific application.

### Other People Are Reading

### Things you need

- Calculator

Show More

## Instructions

- 1
Define the round bar torsion application. In this example, a 1-inch-diameter axle shaft (round bar) will drive two wheels on a lawn tractor from a common drive sprocket. If the axle is composed of alloy steel with an ultimate tensile strength of 80,000 psi and a yield strength of 27216 Kilogram, you can calculate both the maximum torsion the shaft will handle before deforming or breaking, as well as the maximum allowable (operational) torsion.

- 2
Determine the formula for maximum torsion the shaft will sustain without breaking. The maximum torsional stress is expressed as T max = (pi/16) x Su (sigma) x D^3, where T max is the maximum torsional stress in inch-pounds (in.-lb.), Su (sigma) is the maximum shear stress in psi, and D is the shaft or round bar diameter in inches.

- 3
Determine and substitute actual values for this alloy and solve for Tmax ultimate (breaking strength). The ultimate tensile strength of 80,000 psi should be multiplied by 0.75 according to the approximation formula Ssu = 0.75 x Su, where Ssu is the ultimate shear strength and Su is the ultimate tensile strength. This yields a value of 60,000 psi for ultimate shear strength Su. Substituting values, Tmax (breaking) = (pi/16) x 60,000 x D^3 = 0.1963 x 60,000 x 1 in.^3 = 11,778 in.-lb. torsion/12 in./ft. = 981.5ft.-lb. torsion.

- 4
Calculate the maximum torsion the shaft will sustain without permanent deformation. Multiply the tensile yield strength of 60,000 psi by 0.58 according to the approximation formula Ssyp = 0.585 Syp to yield 34,800 psi. Substituting values, Tmax (deformation) = 0.1963 x 34,800 psi x 1 inch^3 = 6821 in.-lb. torsion/12 in./ft. = 568.41ft.-lb. torsion.

- 5
Calculate the maximum allowable torsion under which the shaft should be operated (for longevity and safety). Generally, the industry considers 60 per cent of the yield torsion with a 1.5 safety factor or 40 per cent of yield torsion. Therefore, 568.41ft.-lb. x 0.40 factor = 227.36ft.-lb. for the 1-inch alloy steel shaft as an operational limit.