Shaft Design for Combined Bending and Torsion Load By Equivalent Bending Moment and Torque Methodwritten by: Suvo edited by: Lamar Stonecypher updated: 8/22/2010This article will discuss drive shaft design concepts and shaft design formulas for shafts subjected to combined bending and torsion loading. In my previous article I have discussedhow to design a drive shaft only for bending. But, for many practical shaftdesign problemsyou have to consider bending as well as torsional load when designing the shaft. Such drive shaft design problem can be solved in two ways: By calculating the equivalent bending moment and toque of bending and torsion. By using the principle stress concept.In this article we will limit our discussion to the equivalent bending moment and torque method. Equations to Be Used Equivalent bending moment equation:Meqv. = 0.5 [M + SQRT (M^2 + T^2)]..eqn.1.1 Equivalent torque equation:Teqv. = SQRT (M^2 + T^2)..eqn.1.2 Equation of bending stress:Tb = (32*M/(pi*d^3)eqn.1.3 Equation for torsional shear stress:St = (16*T)/(pi*d^3).eqn.1.4Where,Meqv. Equivalent bending momentM Bending momentT TorqueTb Bending stressSt Shear stressd Diameter of the circular shaft Shaft Design Problem for Combined Bending and Torsion

Refer the picture above, apart from the self weight (1000N) of the pulley a torque (1000 N-mm) due to belt tension is also applied on the shaft. Assuming the maximum allowable stress in tension for the shaft material as 40 MPa, The following steps need to be performed for obtaining the minimum diameter of the shaft as per maximum principle stress theory: Maximum bending moment applied on the shaft = 1000 * 100 = 100000 N-mm Applied torque is given as 1000 N-mm. Equivalent bending moment (from eqn.1.1):Meqv.=1207 Mpa Equivalent torque ( from eqn1.2):Teqv.=1414 N-mm Maximum principle stress (from eqn 1.3):Sigma1 = Tb= (12301/d^3) Now for getting minimum shaft diameter as per maximum principle stress theory, we have to equateSigma1with allowable stress, hence:40 = (12301/d^3)Or, d = 6.7 mm