Shaft Design

MOW227

Semester 2 2012

Shaft System

• A shaft is a rotating (or stationary) member usually having a circular cross-section with a diameter much smaller than its length.

• A shaft generally has power-transmitting elements such as gears, pulleys, belts, etc. mounted onto it.

Shaft System

• Loading on a shaft can be a combination of bending, torsion and axial loads which could be static or dynamic.

• The geometry of a shaft will generally have several diameters with many features along the length in order to accommodate all the necessary elements.

Shaft System

• Main considerations in shaft design:

– Strength

– Deflection

– Critical speed

– Fatigue

Shaft Design Procedure-Strength

• Develop a FBD by replacing elements with the equivalent loads.

• Draw the shear force diagram(s), bending moment diagram(s), and torque diagram.

• Identify one or more critical cross-sections (where bending moments and/or torque is maximum).

• Use appropriate failure theory to compute shaft diameter.

Shaft Design Procedure-Strength

• For the computed diameter, determine whether the design meets all other requirements.

• If all requirements are not met, iterate.

Static Loading-Bending and Torsion

• We will derive the governing equations for computing the diameter or the factor of safety for a critical element with normal stress due to bending and shear stress due to torsion.

• We will use two failure theories - MSST and DET – to check for yield. As a general practice, we should also check the corresponding factor of safety against brittle fracture – using MNST.

Static Loading-Bending and Torsion

4 4

3 3

0

2 64 32

32 160

x xy y

x xy y

M c T c

I J

d d dc I J

M T

d d

σ τ σ

π π

σ τ σπ π

= = =

= = =

= = =

2

2

1,2

2 2

3 3 3

2 2

3 3

2 2

3

2 2

16 16 16

16 16

16

x xxy

M M T

d d d

MM T

d d

M M Td

σ σσ τ

π π π

π π

π

= ± +

= ± +

= ± +

= ± +

For an element on the outer

edge of the shaft

Static Loading-Bending and Torsion

• Using plane stress condition to determine all three principal stresses.

2 2

1 3

2

2 2

3 3

16

0

16

M M Td

M M Td

σπ

σ

σπ

= + +

=

= − +

2 2

12 3

2 2

23 3

2 2

13 3

8

8

16

M M Td

M M Td

M Td

τπ

τπ

τπ

= + +

− = − +

= +

τ12 = (σ1-σ2)/2

Static Loading-Bending and Torsion

• Using DET:

( )22 2

1 3 1 3

2 2

1 3 1 3

2 2

3

1

2

164 3

v

M Td

σ σ σ σ σ

σ σ σ σ

π

= + + −

= + −

= +

2 2

3

1

32 2

3

2 2

164 3

164 3

16 4 3

y

v

y

y

y

SN

N M T Sd

Nd M T

S

S dN

M T

σ

π

π

π

=

+ =

= +

=+

Static Loading-Bending and Torsion

• Using MSST

1 3

2 2

3

3

2 2

1

32 2

32

32

32

y

y

y

y

SN

SN

M Td

S dN

M T

Nd M T

S

σ σ

π

π

π

=−

=

+

=+

= +

• MSST is more conservative than DET and will result in a higher diameter or a lower factor of safety.

Self study Example 1

An assembly of belts applies tensile forces tothe shaft as shown in the figure. The shaft issupported by journal bearings at A and B. Theshaft needs to be designed with a safety factorof 2 with a low steel alloy material (yieldstrength of 500 MPa). Determine the minimumshaft diameter using MSST and DET.

Self study Example 1

Static Loading-

Bending, Axial and Torsion

( ) ( )

( ) ( )

3 2 3

2 2

1 3

2

2 2

3 3

32 4 160

28 8 64

0

28 8 64

x xy y

M P T

d d d

M Pd M Pd Td

M Pd M Pd Td

σ τ σπ π π

σπ

σ

σπ

= + = =

= + + + +

=

= + − + +

Static Loading-

Bending, Axial and Torsion

• Using MSST

( )2 21 3

max 13 3

28 64

2M Pd T

d

σ στ τ

π

−= = = + +

( )

max

2 2

3

2

48 64

y

y

S

N

SM Pd T

d N

τ

π

=

+ + =

• Using DET

( )2 2

3

48 48

ySM Pd T

N dπ= + +

Static Loading-

Bending, Axial and Torsion

• Note that an explicit expression for diameter CANNOT be derived for bending, axial and torsion loading.

• Use an iterative method or a numerical method to solve for diameter.

Shaft Design

Incorporating Stress Concentration

• We will use the same formulation as we have so far.

• But we will account for stress concentrations for all normal and shear stresses.

• We will use the following symbols:– kt – Stress concentration for normal stress due

to bending load

– kts – Stress concentration for shear stress due to torsion load

– kta – Stress concentration for normal stress due to purely axial load

Bending and Torsion Loads

1/3

2 2 2 2164 3t ts

y

Nd k M k T

Sπ

= +

3

2 2 2 216 4 3

y

t ts

S dN

k M k T

π=

+

1/3

2 2 2 232t ts

y

Nd k M k T

Sπ

= +

3

2 2 2 232

y

t ts

S dN

k M k T

π=

+

Using DET Using MSST

Bending, Axial & Torsion Loads

Using DET

Using MSST

( )2 2 2

3

48 48

y

t ta ts

Sk M k Pd k T

N dπ= + +

( )2 2 2

3

48 64

y

t ta ts

Sk M k Pd k T

N dπ= + +

Keys

• Many power-transmitting elements such as gears, pulleys, cams, etc. are mounted on rotating shafts.

• Portion of mounted member in contact with shaft is called the ‘hub’.

• Keys can be used to attach the ‘hub’ to the shaft.

Keys

Keys

• There are many kinds of keys, the simplest type is the flat key.

• Purpose of a key is to prevent any relative rotation between the shaft and the connected member through which the torque is transmitted.

• Purpose of a key is to transmit full torque.

Keys

• Keys are generally made of low-carbon steel, AISI 1020, with cold-drawn finish. A tight fit is designed between the shaft and the hub connected through the key.

• Keys are also used as a safety device (mechanical fuse) to avoid damage to the power transmitting elements in case of overloads.

• Failure due to key shear:

2

2

,

2

s

s

T TP

d d

Shear area A wl

P T

A d wlτ

= =

=

= =

• Failure criteria: ysS

Nτ =

Shear yield strength

(Force)

Flat Key - Design

Flat Key - Design

• For low-carbon steel: 0.4ys y

S S=

0.42 yST

d wl N=• Design criteria (shear):

• Failure due to bearing/compression stress:

4

c

c

A lh

P T

A d lhσ

=

= =

/2 h

L compression area

Flat Key - Design

• Failure criteria:yc

S

Nσ =

• For low-carbon steel: 0.9yc y

S S=

• Design criteria (compression):0.94 y

ST

d lh N=

Self study Example 2

• Design a flat key for a 100 mm diameter shaft carrying a maximum torque of 1560 Nm. The key needs to be designed so as to have a minimum factor of safety of 2 and needs to be capable of carrying the maximum torque. The key is to be made of low-carbon steel (yield strength of 300 MPa) and should have a cross-section of 25 mm x 25 mm.

Critical Speed – Rotating Shafts

• All rotating shafts deflect during rotation.

• Magnitude of deflection depends on stiffness of the shaft and its supports, mass of the whole shaft and the amount of damping.

• Critical speed of a rotating shaft is the speed at which the rotating shaft becomes dynamically unstable.

7-6

Critical Speed – Rotating Shafts• Mathematically, a shaft has an infinite

number of critical speeds. Often, only the first few are important for design.

• For a designer, it is important to determine whether a critical speed of the shaft lies within the operating range of the shaft speed, usually it is designed such that the first natural frequency is at least twice the operating speed

Critical Speed – Rotating Shafts

• There are two approximate methods that can be used to determine the first critical speed (also called the lowest natural frequency) of the shaft system.

• The two methods are called the Rayleigh equation and the Dunkerley equation respectively.

Critical Speed – Rotating Shafts

• For uniform shaft simply supported

Where m is the mass per unit length, A the cross

sectional area and γ the specific weight

γ

ππω

A

gEI

lm

EI

l

22

1

=

=

Fatigue Failure

• Another aspect of design for rotating shafts that needs to be taken into account is Fatigue Failure.

• Any component that undergoes dynamic loading must be designed to withstand fatigue failure (next topic of discussion)

• Fatigue = number of cycles of operation before failure occurs.