Materials Select Torsionally Stressed Shaft

M.2 Strength Considerations—Torsionally Stressed Shaft • 3

M.2 STRENGTH CONSIDERATIONS—TORSIONALLY STRESSED SHAFTFor this portion of the design problem, we will establish a criterion for selection oflight and strong materials for this shaft. We will assume that the twisting moment andlength of the shaft are specified, whereas the radius (or cross-sectional area) may bevaried. We develop an expression for the mass of material required in terms of twist-ing moment, shaft length, and density and strength of the material. Using this expres-sion, it will be possible to evaluate the performance—that is, maximize the strengthof this torsionally stressed shaft with respect to mass and, in addition, relative tomaterial cost.

Consider the cylindrical shaft of length L and radius r, as shown in Figure M.1. Theapplication of twisting moment (or torque), Mt, produces an angle of twist �. Shear stress �at radius r is defined by the equation

(M.1)

Here, J is the polar moment of inertia, which for a solid cylinder is

(M.2)

Thus,

(M.3)

A safe design calls for the shaft to be able to sustain some twisting moment without frac-ture. In order to establish a materials selection criterion for a light and strong material,we replace the shear stress in Equation M.3 with the shear strength of the material �f

divided by a factor of safety N, as1

(M.4)tf

N�

2Mt

pr 3

t �2Mt

pr 3

J �pr 4

2

t �Mtr

J

properties. Selection of the proper material can reduce costs and improve performance.Elements of this materials selection process involve deciding on the constraints of theproblem and, from these, establishing criteria that can be used in materials selection tomaximize performance.

The component or structural element we have chosen to discuss is one that has rele-vance to a mechanical engineer: a solid cylindrical shaft that is subjected to a torsionalstress. Strength of the shaft will be considered in detail, and criteria will be developed formaximizing strength with respect to both minimum material mass and minimum cost.Other parameters and properties that may be important in this selection process are alsodiscussed briefly.

1The factor of safety concept as well as guidelines for selecting values are discussed in Section 7.20.

Figure M.1 A solid cylindrical shaft thatexperiences an angle of twist in response tothe application of a twisting moment Mt.

fr

�

L

Mt

4 • Online Support Module: Mechanical Engineering

It is now necessary to take into consideration material mass. The mass m of anygiven quantity of material is just the product of its density (�) and volume. Since thevolume of a cylinder is �r 2L, then

(M.5)

or, the radius of the shaft in terms of its mass is

(M.6)

Substituting this r expression into Equation M.4 leads to

(M.7)

Solving this expression for the mass m yields

(M.8)

The parameters on the right-hand side of this equation are grouped into three sets of paren-theses. Those contained within the first set (i.e., N and Mt) relate to the safe functioning ofthe shaft. Within the second parentheses is L, a geometric parameter. Finally, the materialproperties of density and strength are contained within the last set.

The upshot of Equation M.8 is that the best materials to be used for a light shaft thatcan safely sustain a specified twisting moment are those having low ratios. In termsof material suitability, it is sometimes preferable to work with what is termed a performanceindex, P, which is just the reciprocal of this ratio; that is

(M.9)

In this context we want to use a material having a large performance index.At this point it becomes necessary to examine the performance indices of a variety of

potential materials. This procedure is expedited by the use of materials selection charts.2

These are plots of the values of one material property versus those of another property.Both axes are scaled logarithmically and usually span about five orders of magnitude, soas to include the properties of virtually all materials. For example, for our problem, thechart of interest is logarithm of strength versus logarithm of density, which is shown inFigure M.2.3 It may be noted on this plot that materials of a particular type (e.g., woods,and engineering polymers) cluster together and are enclosed within an envelope delin-eated with a bold line. Subclasses within these clusters are enclosed using finer lines.

P �t2�3

f

r

r�t2�3f

m � 12NMt 22�3

1p1�3L 2 ar

t2�3f

b

� 2Mt BpL3r3

m3

tf

N�

2Mt

p aBmpLr

b3

r � BmpLr

m � pr2Lr

2A comprehensive collection of these charts may be found in M. F. Ashby, Materials Selection in Mechanical Design,4th edition, Butterworth-Heinemann, Woburn, UK, 2011.3Strength for metals and polymers is taken as yield strength; for ceramics and glasses, compressive strength; forelastomers, tear strength; and for composites, tensile failure strength.

For a cylindricalshaft of length L andradius r that isstressed in torsion,expression for massin terms of densityand shear strength ofthe shaft material

Strengthperformance indexexpression for atorsionally stressedcylindrical shaft

performance index

materials selectionchart

Now, taking the logarithm of both sides of Equation M.9 and rearranging yields

(M.10)

This expression tells us that a plot of log tf versus log r will yield a family of straight andparallel lines all having a slope of each line in the family corresponds to a different per-formance index, P.These lines are termed design guidelines, and four have been included inFigure M.2 for P values of 3, 10, 30, and 100 All materials that lie on one ofthese lines will perform equally well in terms of strength-per-mass basis; materials whose

(MPa)2�3m3/Mg.

32;

log tf �32

log r �32

log P


0.1 0.3 1 3 10 30

10,000

1000

100

10

1

0.1

MgAlloys

Ash

SiGlasses

Density (Mg /m3)

Engineeringceramics

Engineeringcomposites

Engineeringalloys

Porousceramics

Engineeringpolymers

Woods

Elastomers

Polymerfoams

Str

engt

h (M

Pa)

P = 100

P = 30

P = 10

P = 3

Diamond

Cermets

Sialons

B MgO

Al2O3 ZrO2

Si3N4SiC

Ge

NylonsPMMA

PS

PPMELPVC

EpoxiesPolyesters

HDPE

PU

PTFE

SiliconeLDPE

SoftButyl

WoodProducts

Ash

Balsa

Balsa

Oak

Oak

Pine

Pine

Fir

Fir

Parallelto Grain

Perpendicularto Grain

Cork

CementConcrete

Engineeringalloys

KFRPCFRPBe

GFRPLaminates

KFRP

Pottery TiAlloys

Steels

W Alloys

Mo Alloys

Ni Alloys

Cu Alloys

CastIrons

ZnAlloys

Stone,Rock

LeadAlloys

CFRPGFRP

UNIPLY

Al Alloys

Figure M.2 Strength versus density materials selection chart. Design guidelines for performance indices of 3,10, 30, and 100 (MPa)2�3m3/Mg have been constructed, all having a slope of (Adapted from M. F. Ashby, Materials Selection in Mechanical Design. Copyright © 1992. Reprinted by permission ofButterworth-Heinemann Ltd.)

32.

positions lie above a particular line will have higher performance indices, whereas thoselying below will exhibit poorer performances. For example, a material on the P � 30 linewill yield the same strength with one-third the mass as another material that lies alongthe P � 10 line.

The selection process now involves choosing one of these lines, a “selection line”that includes some subset of these materials; for the sake of argument let us pick P �10 m3/Mg, which is represented in Figure M.3. Materials lying along this line orabove it are in the “search region” of the diagram and are possible candidates for this1MPa 2 2�3


Engineeringalloys

KFRPCFRPBe

GFRPLaminates

KFRP

Pottery

MgAlloys

TiAlloys

Steels

W Alloys

Mo Alloys

Ni Alloys

Cu Alloys

CastIrons

ZnAlloys

Stone,Rock

Al Alloys

LeadAlloys

CFRPGFRP

UNIPLY

10,000

1000

100

10

1

Glasses

Engineeringcomposites

Engineeringalloys

Porousceramics

Engineeringpolymers

Woods

Elastomers

Polymerfoams

Str

engt

h (M

Pa)

P = 10

Diamond

Cermets

Sialons

B MgO

Al2O3 ZrO2

Si3N4SiC

Si Ge

NylonsPMMA

PS

PPMELPVC

EpoxiesPolyesters

HDPE

PU

PTFE

Silicone

CementConcrete

LDPE

SoftButyl

WoodProducts

Ash

AshBalsa

Balsa

Oak

Oak

Pine

Pine

Fir

Fir

Parallelto Grain

Perpendicularto Grain

Cork

300 MPa

(MPa)2/3 m3/Mg

0.1 0.3 1 3 10 300.1

Density (Mg /m3)

Engineeringceramics

Figure M.3 Strength versus density materials selection chart. Materials within the shaded region are acceptablecandidates for a solid cylindrical shaft that has a mass-strength performance index in excess of 10 and a strength of at least 300 MPa (43,500 psi).(Adapted from M. F. Ashby, Materials Selection in Mechanical Design. Copyright 1992. Reprinted by permission ofButterworth-Heinemann Ltd.)

©

(MPa)2�3m3/Mg


rotating shaft. These include wood products, some plastics, a number of engineeringalloys, the engineering composites, and glasses and engineering ceramics. On the basis offracture toughness considerations, the engineering ceramics and glasses are ruled out aspossibilities.

Let us now impose a further constraint on the problem—namely that the strength ofthe shaft must equal or exceed 300 MPa (43,500 psi).This may be represented on the ma-terials selection chart by a horizontal line constructed at 300 MPa, see Figure M.3. Nowthe search region is further restricted to the area above both of these lines.Thus, all woodproducts, all engineering polymers, other engineering alloys (viz., Mg and some Al alloys),and some engineering composites are eliminated as candidates; steels, titanium alloys,high-strength aluminum alloys, and the engineering composites remain as possibilities.

At this point we are in a position to evaluate and compare the strength performancebehavior of specific materials. Table M.1 presents the density, strength, and strength per-formance index for three engineering alloys and two engineering composites, whichwere deemed acceptable candidates from the analysis using the materials selectionchart. In this table, strength was taken as 0.6 times the tensile yield strength (for thealloys) and 0.6 times the tensile strength (for the composites); these approximationswere necessary because we are concerned with strength in torsion, and torsionalstrengths are not readily available. Furthermore, for the two engineering composites, itis assumed that the continuous and aligned glass and carbon fibers are wound in a helicalfashion (Figure 15.15) and at a 45� angle referenced to the shaft axis. The five materialsin Table M.1 are ranked according to strength performance index, from highest to lowest:carbon fiber-reinforced and glass fiber-reinforced composites, followed by aluminum,titanium, and 4340 steel alloys.

Materials cost is another important consideration in the selection process. In real-life engineering situations, economics of the application often is the overriding issue andnormally will dictate the material of choice. One way to determine materials cost is bytaking the product of the price (on a per-unit mass basis) and the required mass ofmaterial.

Cost considerations for these five remaining candidate materials—steel, aluminum,and titanium alloys, and two engineering composites—are presented in Table M.2. Inthe first column is tabulated The next column lists the approximate relativecost, denoted as this parameter is simply the per-unit mass cost of material divided bythe per-unit mass cost for low-carbon steel, one of the common engineering materials.The underlying rationale for using is that although the price of a specific material willvary over time, the price ratio between that material and another will, most likely,change more slowly.

c

c;r/t2�3

f .

� � f � f2�3/� � P

Material (Mg/m3) (MPa) [(MPa)2/3m3/Mg]

Carbon fiber-reinforced composite 1.5 1140 72.8(0.65 fiber fraction)a

Glass fiber-reinforced composite 2.0 1060 52.0(0.65 fiber fraction)a

Aluminum alloy (2024-T6) 2.8 300 16.0

Titanium alloy (Ti-6Al-4V) 4.4 525 14.8

4340 Steel (oil-quenched 7.8 780 10.9and tempered)

a The fibers in these composites are continuous, aligned, and wound in a helical fashion ata 45� angle relative to the shaft axis.

Table M.1

Density (�), Strength

(�f), and Strength

Performance Index (P)

for Five Engineering

Materials

Finally, the right-hand column of Table M.2 shows the product of and . Thisproduct provides a comparison of these materials on the basis of the cost of materialsfor a cylindrical shaft that would not fracture in response to the twisting moment Mt. Weuse this product inasmuch as is proportional to the mass of material required(Equation M.8) and is the relative cost on a per-unit mass basis. Now the most econom-ical is the 4340 steel, followed by the glass fiber-reinforced composite, the carbon fiber-reinforced composite, 2024-T6 aluminum, and the titanium alloy. Thus, when the issue ofeconomics is considered, there is a significant alteration within the ranking scheme. Forexample, inasmuch as the carbon fiber-reinforced composite is relatively expensive, it issignificantly less desirable; in other words, the higher cost of this material may not out-weigh the enhanced strength it provides.

cr�t2�3

f

r�t2�3fc


Table M.2 Tabulation of the ��f2�3 Ratio, Relative Cost (c–), and Product of ��f

2�3 and c– for Five Engineering

Materialsa

�� f2�3 c– c–(��f

2�3 )Material ��10�2 [Mg�(MPa)2�3m3]�� ($�$) ��10�2($�$)[Mg�(MPa)2�3m3]��

4340 Steel (oil-quenched 9.2 3.0 27and tempered)

Glass fiber-reinforced composite 1.9 28.3 54(0.65 fiber fraction)b

Carbon fiber-reinforced composite 1.4 43.1 60(0.65 fiber fraction)b

Aluminum alloy (2024-T6) 6.2 12.4 77

Titanium alloy (Ti-6Al-4V) 6.8 94.2 641aThe relative cost is the ratio of the price per unit mass of the material and a low-carbon steel.bThe fibers in these composites are continuous, aligned, and wound in a helical fashion at a angle relative to theshaft axis.

45�

M.3 OTHER PROPERTY CONSIDERATIONS AND THE FINAL DECISIONTo this point in our materials selection process we have considered only the strength ofmaterials. Other properties relative to the performance of the cylindrical shaft may beimportant—for example, stiffness, and, if the shaft rotates, fatigue behavior (Sections M.9through M.13). Furthermore, fabrication costs should also be considered; in our analysisthey have been neglected.

Relative to stiffness, a stiffness-to-mass performance analysis similar to the one just dis-cussed could be conducted. For this case, the stiffness performance index is

(M.11)

where G is the shear modulus. The appropriate materials selection chart (log G versuslog r) would be used in the preliminary selection process. Subsequently, performanceindex and per-unit-mass cost data would be collected on specific candidate materials;from these analyses the materials would be ranked on the basis of stiffness perform-ance and cost.

In deciding on the best material, it may be worthwhile to make a table employing theresults of the various criteria that were used.The tabulation would include, for all candidatematerials, performance index, cost, and so forth for each criterion, as well as comments rel-ative to any other important considerations. This table puts in perspective the importantissues and facilitates the final decision process.

Ps �1Gr

Ps

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Materials Select Torsionally Stressed Shaft