Solid Mechanics Spring 2007

Chapter 4 - Torsion

pITc

=τ GI

TL

p

=θ

Objectives - Develop relationship between shear strain, γ, and angle of twist, φ - Determine variation in shear stress, τ, as a function of the radius of a shaft - Calculate angle of twist and shear stress in a shaft - Design shafts considering maximum desired angle of twist and maximum allowable shear

stress - Solve statically indeterminate problems under torsional loading Torsion Now, we consider shear stresses and twisting of a rod subjected to torsion about its axis Examples of machines in torsion Socket wrench Transmission shafts

T

Solid Mechanics Spring 2007

Deformations in a Circular Shaft Goals - Find shear strain, γ, then find shear stress τ = G γ. - Relate angle of twist θ and torque T. Distribution of Shear Strains Angle of twist, θ, is proportional to the magnitude of the torque, T, and shaft length, L. Consider a square element. Applying torsion to the shaft causes the square element to deform Shear strain, γ, is How are shear strain, γ, and angle of twist, θ, related? If is the strain is small, then the arc length AA' is related to Stresses in Elastic Range Shear stresses, τ, in a circular shaft vary linearly with the distance, ρ, measured from the centerline to the point of interest. Solid circular shaft: Hollow circular shaft:

Solid Mechanics Spring 2007

Relation between shear stress, τ, and applied torque, T Consider shaft JK subjected to equal and opposite torque T and T’. Pass a cut through point C. C J K FBD of left side C J Equilibrium requires that the sum of the torques caused by the shearing forces integrate to equal the applied torque, T. Substitute Integrate Maximum shear stress, τmax Shear stress, τ, on cross-section at any radial distance, ρ

Solid Mechanics Spring 2007

Example Given - Solid shaft of uniform diameter 1.5 in. is supported by 2 bearings Find - On the cross-section at pt. E a) shear stress at outer radius, b) shear stress at radius of 0.15 inches Assumptions - linearly elastic 42.5 kip-in

30 kip-in

12.5 kip-in E

Solid Mechanics Spring 2007

Angle of Twist Now relate angle of twist, φ, and applied torque, T. Shaft of length L, radius r, torque T applied at free end Hooke's law Geometry Combine Example Known - Hollow aluminum pipe, torque T causes 0.2 radians angle of twist Given - G = 27 GPa, ro = 50 mm, ri = 40 mm, L = 2.5 m Find - a) magnitude of T b) angle of twist for solid cylinder with same mass

Solid Mechanics Spring 2007

If both ends of shaft rotate, then angle of twist, φ, equals the angle that one end rotates with respect to the other end. Example Known - steel shafts coupled with meshing gears Given - radius of gear C rc = 75 mm, gear B rb = 150 mm, radius of shafts = 10 mm, G = 80 GPa, torque T = 45 N-m Find - φC angle of twist at C and φA angle of twist at A

FBD of each shaft and equilibrium in order to determine reaction forces between gears and torque in the shafts AB and CD Angle of twist φC Angle of twist φA

Solid Mechanics Spring 2007

Known - Solid steel shaft is locked at A, free to rotate in bearing D and loaded at B and C by torques T1 and T2 Given - T1 = 20,000 lb-in., T2 = 12,000 lb-in., L1 = 20 in., L2 = 30 in., L3 = 20 in., G = 11,500 ksi Find - the maximum shear stress in each section, the angle of twist of D relative to A

FBD and equilibrium Shear stresses Angles of twist

Solid Mechanics Spring 2007

Statically Indeterminate Shafts Many problems exist where the torques on a shaft can not be determined from statics alone. Again, 1) draw the FBD and equilibrium equation and 2) use deformation (angle of twist) to write a compatibility equation. Use both equations to solve for the 2 reaction torques. Example Given - Shaft made of brass sections BC and BD is loaded at center C with torque T = 100 lb.-ft., G = 6.4x106 psi (brass) Find - The torque exerted on the shaft by walls at the fixed ends FBD and equilibrium Angle of twist in each section Compatibility (Geometry)

B C D

5 in 5 in

d = 2 ind = 3 in