MAE 322

Machine Design

Shafts -SummaryDr. Hodge Jenkins

Mercer University

Power Transmission

Nothing new, just calculate Torque, T, from power equation:

P = Tw

Careful with units!

Power (watts, ft-lb/s or hp)

Torque (N-m, lb-ft)

Angular velocity (RPM, rad/s or Hz)

Note: 1 HP = 550 ft-lb/s

fw 2 f = Hz or rev/s

Power, Speed and Torque

Shaft powered by 5 hp electric motor spins at 600 RPMS

[= 600revs/min = 10 Revs/sec = 10 Hz] ,

find Torque in shaft.

P = Tw

5 hp (550 ft-lb/s/hp) = 2,750 ft-lb/s

600 RPM=10 Hz (2 rad/rev) = 62.83 rad/s

T = 2750 ft-lb/s

62.83 rad/s

= 43.76 lb-ft

P

T

w

Shaft Stresses for Rotating Shaft

For rotating shaft with steady, alternating bending and torsion

◦ Bending stress is completely reversed (alternating), since a

stress element on the surface cycles from equal tension to

compression during each rotation (Ma) . Found from bending

moment diagrams

◦ Torsional stress is steady (constant or static Tm )

◦ Previous equations simplify with Mm and Ta equal to 0

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Shaft Stresses

Standard stress equations can be customized for shafts for

convenience

Axial loads are generally small and constant, so will be ignored

in this section

Standard alternating and midrange stresses

Customized for round shafts

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Shaft Stresses

Using modified Goodman line with DE,

Solving for d is convenient for design purposes

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Shaft Stresses

DE-ASME Elliptic Minimum Diameter Calculation

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Angular Deflection of Shafts

For stepped shaft with individual cylinder length li and torque Ti,

the angular deflection can be estimated from

For constant torque throughout homogeneous material

Experimental evidence shows that these equations slightly

underestimate the angular deflection.

Torsional stiffness of a stepped shaft is

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Critical Speeds for Shafts

For a simply supported shaft of uniform diameter, the

first critical speed is

For an ensemble of attachments, Rayleigh’s method

for lumped masses gives

◦ wi is the weight of the ith location and yi is the static deflection at the ith body

location

Or Finite Element Model for modal analysis (later)

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Standard Keys, Rectangular & Square

Shaft diameter

determines key

size

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Table 7–6

Keys

Failure of keys is by either direct shear or bearing stress

Key length is designed to provide desired factor of safety

Factor of safety should not be excessive, so the inexpensive key

is the weak link

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Shigley’s Mechanical Engineering Design

Example 7–6

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Fig. 7–19

Example 7–6 (continued)

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Example 7–6 (continued)

Shigley’s Mechanical Engineering Design