Torsion: Shear Stress & Twist 1

Torsion: Shear Stress & Twist (3.1-3.5)

MAE 314 – Solid Mechanics

Yun Jing

Torsion: Shear Stress & Twist 2

Torsion of Circular Shafts In this chapter, we will examine uniaxial bars subject to torque.

Where does this occur?

Transmission Shaft

Force Couples

Torsion: Shear Stress & Twist 3

Torsion of Circular Shafts We assume

Bar is in pure torsion Small rotations (the length and radius will not change)

How does the bar deform? Cross-section of the bar remains the same shape, bar is simply rotating.

Cross-section remains perpendicular to axis of cylinder (cylinder does not warp). Not true for most non-circular bars

Torsion: Shear Stress & Twist 4

Angle of Twist Deformation of a circular shaft subjected to pure torsion

Fix left end of shaft A moves to A’ = angle of twist (in radians)

What are the boundary conditions on ? (x) = 0 at x = 0 (x) = at x = L

For pure torsion, is linear.L

xx

)(

x

Torsion: Shear Stress & Twist 5

Shearing Strain Calculate the surface shear strain in thecylinder. For pure torsion (x) = x / L, so

L

Torsion: Shear Stress & Twist 6

Shearing Strain The maximum shear strain on the surface ofthe cylinder occurs when ρ=c.

We can express the shearing strain at anydistance from the axis of the shaft as

Maximum shear strain on surface

L

c max

maxc

Torsion: Shear Stress & Twist 7

Shearing Strain We can also apply the equation for maximumsurface shear strain to a hollow circular tube.

This applies for all types of materials: elastic, linear, non-linear, plastic, etc.L

c 1min

L

c 2max

cc

Torsion: Shear Stress & Twist 8

Elastic Shearing Stress

Calculate shear stress in a bar made of linearly elastic material. Recall Hooke’s Law for shearing stress: τ=Gγ

L

GcG

maxmax maxc

Torsion: Shear Stress & Twist 9

Torque We still need to relate τ to the applied torque T, which is generally the known, applied load. First, find the resultant moment acting on a cross-section and set this equal to T.

c

maxc

dAc

dA max

2

dAc

dAc

TAA 2max

max

2

Torsion: Shear Stress & Twist 10

Torque Continuing from previous slide:

Where J is the polar moment of inertia of the cross section of the bar (see Appendix A.3 in your textbook). Plug this into the equation for τmax.

Jc

dAc

TA

max2max J

Tcmax

L

Gcmax J

Tc

L

GcGJ

TL

Torsion: Shear Stress & Twist 11

Torque For a non-uniform bar

For a continuously varying bar

n

i ii

iin

ii JG

LT

11

dxxGJ

xTL

0 )(

)(

Torsion: Shear Stress & Twist 12

Inclined Plane

Cut a rectangular element along the plane at an angle θ.

Torsion: Shear Stress & Twist 13

Inclined Plane Sum forces in x-direction.

Sum forces in y-direction.

y x

0costansinsec 000 AAA

cossincossin

2sincossin2

0sintancossec 000 AAA

2cos

22 sincos

Torsion: Shear Stress & Twist 14

Inclined Plane τmax occurs at θ = 0º, ±90º σmax occurs at θ = ±45º τmax = σmax

When σθ is max, τθ = 0, and when τθ is max, σθ =0.

Torsion: Shear Stress & Twist 15

Example ProblemPart 1. For the 60 mm diameter solid cylinder and loading shown,determine the maximum shearing stress.Part 2. Determine the inner diameter of the hollow cylinder , of 80 mmouter diameter, for which the maximum stress is the same as in part 1.

Torsion: Shear Stress & Twist 16

Example ProblemPart 1. For the aluminum shaft shown (G = 27 GPa), determine the torqueT that causes an angle of twist of 4o.Part 2. Determine the angle of twist caused by the same torque T in a solidcylindrical shaft of the same length and cross-sectional area.

Torsion: Statically Indeterminate Problems and Transmission Shafts

17

Torsion: Statically Indeterminate Problems and Transmission Shafts (3.6-3.8)

MAE 314 – Solid Mechanics

Yun Jing

Torsion: Statically Indeterminate Problems and Transmission Shafts

18

Statically Determinate Problems

Find the maximum shearing stress in each bar.

T1T2

T3

Torsion: Statically Indeterminate Problems and Transmission Shafts

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Statically Indeterminate Problems Method for torsion is the same as the method for statically indeterminate axial load deflection problems. Apply what you’ve already learned:

M = R – N M = number of compatibility equations needed R = number of unknown reactions (or internal stresses) N = number of equilibrium equations

Compatibility equations for a torsion problem are based on angle of twist.

Torsion: Statically Indeterminate Problems and Transmission Shafts

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Statically Indeterminate Problems

Find the largest torque T0 that can be applied to the end of shaftAB and the angle of rotation of the end A of shaft AB. Allowable shearing stress is

LCDdCD

dABLABrB

rC

allow

Torsion: Shear Stress & Twist 21

A circular shaft AB consists of a 10-in.-long, 7/8-in.-diameter steel cylinder, in which a 5-in.long,5/8-in.-diameter cavity has been drilled from end B. The shaft is attached to fixed supports at both ends, and a 90 lb.ft torque is applied at its midsection. Determine the torque exerted on the shaft by each of the supports.

Torsion: Statically Indeterminate Problems and Transmission Shafts

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Transmission Shafts In a transmission, a circular shaft transmits mechanical power from one device to another.

ω = angular speed of rotation of the shaft The shaft applies a torque T to another device To satisfy equilibrium the other device applies torque T to the shaft. The power transmitted by the shaft is

Generator

Turbine

TP

Torsion: Statically Indeterminate Problems and Transmission Shafts

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Transmission Shafts Units for P=Tω

ω = rad/s T = N·m (SI) T = ft·lb (English) P = Watts (1 W = 1 N·m/s) (SI) P = ft·lb/s (1 horsepower = hp = 550 ft·lb/s) (English)

We can also express power in terms of frequency.fTP 2

f 2 1 sHzf

Torsion: Statically Indeterminate Problems and Transmission Shafts

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Example ProblemA 1.5 meter long solid steel shaft of 22 mm diameter is totransmit 12 kW. Determine the minimum frequency at which theshaft can rotate, knowing that G = 77.2 GPa, that the allowableshearing stress is 30 MPa, and that the angle of twist must notexceed 3.5o.

Up to now, we assumed that transmission shafts are loaded at the ends through solidly attached, rigid end plates. In practice, torques are applied through flange couplings and fitted keyways, which produce high stress concentrations.

One way to reduce stress concentrations is through the use of a fillet.Torsion: Statically Indeterminate Problems and

Transmission Shafts25

Stress Concentrations in Circular Shafts

Flange coupling

Fitted keyway

Maximum shear stress at the fillet

Tc/J is calculated for the smaller-diameter shaft K = stress concentration factor

Torsion: Statically Indeterminate Problems and Transmission Shafts

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Stress Concentrations in Circular Shafts

FilletJ

TcKmax

Torsion: Statically Indeterminate Problems and Transmission Shafts

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Example ProblemThe stepped shaft shown rotates at 450 rpm. Knowing that r = 0.25 in,determine the maximum power that can be transmitted withoutexceeding an allowable shearing stress of 7500 psi.