Chapter 3 Solid Materials

3.1 A 2-m long polycarbonate tensile rod has a cross-sectional diameter of 130 mm. It is usedto lift a tank weighing 65 tons from a 1.8 m deep ditch onto a road. The vertical motionof the cranes arc is limited to 4.2 m. Will it be possible to lift the tank onto the road?

Notes: There are two parts to this problem which must be considered: first, does the applied stressexceed that allowed for polycarbonate, and second, can the tank clear the ditch after including thedeflection of the polycarbonate. Since metric units are used in this problem, a ton is interpretedas 1000 kg.

Solution:From Table A.4 for polycarbonate, Su=65.5 MPa, and the total percent elongation is 110%. Theload lifted is P=65(1000kg)(9.81m/s

2)=637kN. The stress is given by Equation (2.7) as:

s =PA

=

P

pd2

4

=

637kN

p 0.13m( )24

= 48MPa

This stress is lower than the allowable stress for polycarbonate, so it is concluded that the tankcan be lifted.

3.2 Materials are normally classified according to their properties, processing routes, andapplications. Give examples of common metal alloys that do not show some of thetypical metal features in their applications.

Notes: There are many possible solutions to this problem, and students should be encouraged todescribe their own applications.

Solution:In terms of properties, gray cast iron does not have any ductility due to the plate-formed graphitedistributed in the matrix. Thus it has much greater compressive strength than tensile strength andis a good damping material, all of which are counter to usual metal trends. Often, powder metalsare sintered, using a processing route similar to ceramics. Mercury, being in molten form at roomtemperature, is not used in typical metal applications. Sodium is used as a heat transfer medium insome atomic power plants, where the metal is boiled at the hot end and condensed at the cool end.

3.3 Equation (B.56) gives the relationship between stresses and strains in isotropic materials.For a polyurethane rubber the elastic modulus at 100% elongation is 7 MPa. When therubber is exposed to a hydrostatic pressure of 10 MPa, the volume shrinks 0.5%.Calculate Poissons ratio for the rubber.

Notes: This problem is straightforward. Note that a hydrostatic pressure of p means thats x=s y=s z=-p. The volume change can be derived as shown here, or else a hint can be given to thestudent that the volume change is equal to:

DVV

= e1 + e2+ e3

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Solution:Consider, for a box-shaped element of sides l1,l2, and l3, so the volume is l1l2l3. If under a stressstate the strains are e 1, e 2 and e 3, then the new lengths are l1

=(1+e 1)l1, l2=(1+e 2)l2, and l3

=(1+e 3)l3.The new volume is given by:

V = l1 l2 l3 = 1 + e 1( ) 1+ e 2( ) 1+ e 3( ) l1l2l3

Therefore, the change in volume is given by:D VV= V - VV

=1+ e 1( ) 1+ e 2( ) 1+ e 3( ) l1l2l3 - l1l2l3

l1l2l3= 1+ e 1( ) 1+ e 2( ) 1+ e 3( ) - 1

If this is expanded, and if the strains are all small so that second order terms can be ignored, thevolume change is

D VV= e 1 + e 2 + e 3

From Equations (B.56), if the stress state is s x=s y=s z=-10MPa and all other stresses are zero,then

e x =1Es x - n s y + s z( )[ ] ; e y = 1E s y - n s z + s z( )[ ] ; e z =

1Es z - n s x + s y( )[ ]

Substituting into the expression for volume change,D VV= e 1 + e 2 + e 3 =

1Es x - n s y + s z( ) + s y - n s x + s z( ) + s z - n s x + s y( )[ ] = 3pE 1- 2n( )

For p=10MPa, D V/V=0.5%, soD VV=

3pE

1- 2n( ) ; 0.005= 3 1 0MPa( )7MPa( )

1- 2n( ) ; n = 0.4994

3.4 A fiber-reinforced plastic has fiber-matrix bond strength t f=10MPa and fiber ultimatestrength Su=1GPa. The fiber length is constant for all fibers at l=3mm. The fiber diameterd=30m. Find whether the fiber strength of the fiber-matrix bond will determine thestrength of the composite.

Notes: Equation (3.2) is needed to solve this problem.

Solution:From Equation (3.2),

lcr =Sud2t f=

1GPa( ) 30m m( )2 10MPa( )

= 0.0015m= 1.5mm

The actual fiber length is twice this critical length, so the fiber strength will determine thestrength of the composite.

3.5 Having the same material as in Problem 3.4, but with fiber length l=1mm, calculate if it ispossible to increase the fiber stress to Su=1GPa but making the fiber rectangular insteadof circular, maintaining the same cross sectional area of each fiber.

Notes: Equation (3.2) is needed to solve this problem.

Solution:

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Equation (3.2) gives the critical length of a fiber, but this is derived from force equilibrium.Equation (3.2) can be written as:

lcr =Sud2t f

; 2t f lcr = Sud; p dt flcr2 = Su

p d2

4; t f p d( )

lcr2 = SuAf

The first term in parentheses on the left side of the equation is the circumference; if the crosssection is square, then 2(wt+ht) should be substituted for p d, where wt is the width and ht is theheight of the cross section. Therefore, Equation (3.2) becomes:

t f 2 wt + ht( )( )lcr2 = Suwtht

Similarly, the cross-sectional area of the fiber is wtht, and since the cross sectional area isconstant, we know:

wtht =p4

30m m( ) 2 = 7.069 10- 10m2

To answer the question, we use these two simultaneous equations, along with lcr=1mm=0.001m,Su=1GPa, and t f=10MPa. This yields ht=12.06m and wt=58.63m or wt=12.06m and ht=58.63m (orientation is arbitrary).

3.6 A copper bar is stressed to its ultimate strength Su=250MPa. The cross sectional area ofthe bar before stressing is 100 mm

2, and the area at the deformed cross section where the

bar starts to break at the ultimate strength is 60 mm2. How large a force is needed to reach

the ultimate strength?

Notes: To solve this problem, one must realize that the ultimate strength is based on maximumload divided by initial area.

Solution:The force needed to reach Su is given by

P=(Su)A=(250MPa)(100mm2)=25kN

3.7 AISI 440C stainless steel has ultimate strength Su=807MPa and fracture strengthSfr=750MPa. At the ultimate strength the cross sectional area of a tension bar made ofAISI 440C is 80% of its undeformed value. At the fracture point the minimum cross-sectional area has shrunk to 70%. Calculate the real stresses at the point of ultimatestrength and at fracture.

Notes: One needs to realize that Su and Sfr are defined for a tension test specimen from its initialarea.

Solution:For an ultimate strength of 807MPa, the load that can be supported is (807MPa)(A). If the area is0.8A, then the real stress at this point is given by Equation (2.7) as:

s =PA=

807MPa( ) A0.8A

= 1009MPa

Similarly, at fracture, where the area is 0.70A, the real stress is:

s =PA=

750MPa( ) A0.7A

= 1071MPa

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3.8 According to sketch a a beam is supported at point A and at either B or C. At C thesilicon nitride tensile rod is lifting the beam end with force P=Stfr, where Ac is the crosssectional area of the rod. Find the distance A-B such that the silicon nitride rod would becrushed if it took up a compressive force at B instead of a tensile force at C. Note thatScfr=15S

tfr for silicon nitride. Also, find the reaction forces at A for the two load cases.

Notes: This is fairly straightforward. The reactions are found from equilibrium, and the reactionsyield the answer.

Solution:Assume that the distance is such that there is a compressive load on rod B as shown. Since thecompressive fracture strength is fifteen times larger than the tensile fracture strength, we knowP=15P. If the compressive rod exerts a force 15P and the tensile rod doesnt exist, then momentequilibrium about point A yields

MA = 0 = 15P l - mag l - x( ) ; l =mag l - x( )

15P=

mag l - x( )15SftAc( )

If there is a tensile rod at A and no compressive rod at B, and P=SftAc, then moment equilibriumgives the load mag that causes failure as:

MA = 0 = P l( ) - mag l - x( ) ; mag =Pl

l - xVertical force equilibrium gives the reaction at A as:

Fy = 0 = Ay - mag + P; Ay = mag - P =Pl

l - x- P =

Pxl - x

=SftAcx

l - x=

8P7

However, if there is a compressive rod at B, then moment equilibrium yields:

MA = 0 = 15P l ( ) - mag l - x( ) ; mag =15P l l - x

Vertical force equilibrium gives the reaction at A as:

Fy = 0 = Ay - mag - 15P; Ay = mag + 15P =15P l l - x

+ 15P = 15SftAc l + l - xl - x

=

120P7

The distance AB is l=l/15.

3.9 Polymers have different properties depending on the relationship between the localtemperature and the polymers glass transition temperature Tg. The rubber in a bicycletire has Tg=-12C. Could this rubber be used in tires for an Antarctic expedition attemperatures down to -70C?

Notes: This requires an understanding of the implications of glass transition temperature asshown in Figure 3.10 on page 103. This problem should consider the rubber to be a thermoplastic.

Solution:

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From Figure 3.10 on page 103, note that below the glass transition temperature, no permanentdeformation is allowed to the polymer. This should not be too great a concern, since a tire shouldhopefully remain elastic. However, the elastic strains which can be withstood at low temperaturesare much lower than those near or above the glass transition temperature. Therefore, it is notrecommended that the tire be used below its glass transition temperature.

3.10 Given an aluminum bronze with 20wt% aluminum and 80wt% copper, find the density ofthe aluminum bronze.

Notes: The rule of mixtures, described on Page 106, is needed to solve this problem.

Solution:The density of aluminum is, from the inside front cover, 2710kg/m3. The density of copper is8940kg/m3. The rule of mixtures (see page 106) suggests that the density can be calculated from alinear interpolation between the densities of the alloy components. Therefore, the density of thealuminum bronze is given by:

r bronze= 0.2r al + 0.8r cu = 0.2( ) 2710kg/m3( ) + 0.8( ) 8940kg/m3( ) = 7690kg/m3

3.11 The glass-fiber reinforced plastic in Example 3.8 (section 3.5.2) is used in an applicationwhere the bending deformations, caused by the applied static load, will crack the plasticby overstressing the fibers. Will a carbon-fiber reinforced plastic also crack if it has thesame elastic properties as the glass-reinforced plastic?

Notes: The following data is needed from Example 3.8: Ec=150GPa, Eg=70GPa, Sug=0.7GPa,Suc=1GPa. Since the strains are constant, one merely needs to apply Hookes law to obtain thestress in the carbon fibers.

Solution:The fiber reinforced composite has the same elastic properties whether the reinforcement is glassor carbon fiber (see Example 3.8 - volume percentages of reinforcement were calculated to ensurethis condition). Therefore, the total strain experienced by the composite is constant, as is thestrain seen by the fibers. Applying Hookes law,

s cEc=s gEg

;s c =EcEgs g =

150GPa70GPa

0.7GPa( ) = 1.5GPa

Since this stress still exceeds the ultimate strength for the carbon fibers, they will crack in thedescribed loading.

3.12 In Problem 3.11, carbon fibers were used to reinforce a polymer matrix. Theconcentration of fibers was decreased in Example 3.8 (section 3.5.2) to give the sameelastic properties for the carbon-fiber-reinforced polymer as for the glass-reinforcedpolymer. If instead the fiber concentration were kept constant at 10% when the glassfibers were changed to carbon fibers, how much smaller would the deformation be for thesame load, and would the fibers be overstressed or not? The material properties are thesame as in Example 3.8.

Page 3-6

Notes: Assume that the loading is a given load, not deformation-based. To solve this problem,one must determine the stiffness of the composite for the two cases, then express the strain ratioin terms of the elastic modulus. Substituting for the material properties of the glass fibers atfracture allows one to determine if the carbon fibers will survive. The following data is neededfrom Example 3.8: Ec=150GPa, Eg=70GPa, Sug=0.7GPa, Suc=1GPa, Em=2GPa.

Solution:For a fiber volume fraction of 10%, the stiffness of the carbon fiber reinforced polymer is givenby Equation (3.10) as:

Ecc=Emn m+Efn f=(2GPa)(0.9)+(150GPa)(0.1)=16.8GPaThe stiffness of the glass fiber reinforced polymer is similarly calulated as

Ecg=Emn m+Egn g=(2GPa)(0.9)+(70GPa)(0.1)=8.8GPaAssuming a constant applied stress for the two cases, the fiber strain ratio is given by

d cd g=

EcgEcc=

8.8GPa16.8GPa

= 0.524

Therefore, the stress in the carbon fiber, assuming the loading is sufficient to just cause failure ofglass fibers, is:

s c = 0.524EcEg

Sug = 0.524150GPa70GPa

0.7GPa( ) = 0.786GPa

Since the strength of the carbon fibers is 1GPa, they will survive the loading.

3.13 A bent beam, shown in sketch b, is loaded withP=128,000N. The beam has a square cross section a2. Thelength of a side a=30mm. The length l1=50mm andl2=100mm. The yield strength Sy=350MPa (medium-carbonsteel). Find whether the stresses in tension and shear arebelow the allowable stresses. Neglect bending.

Notes: The stresses calculated are average stresses, and do not reflectthe distribution of shear stress in beams as covered in Chapter 4.The allowable stresses given in Equations (3.13) and (3.14) are usedto solve this problem.

Solution:The average shear stress over length l1 is given by

t ave =P

a2=

128kN

0.03m( ) 2= 142.2MPa

The allowable shear stress is given by Equation (3.14) ast all=0.4Sy=0.4(350MPa)=140MPa

Therefore, the shear stress is larger than that allowed for the material.The normal stress in length l2 is:

s =P

a2= 142.2MPa

Since the lowest allowable stress is, from Equation (3.13), equal to 0.45Sy=157MPa, the normalstresses are acceptable.

Page 3-7

3.14 A tough material, such as soft stainless steel (AISI 316), has yield strength Sy=207MPa,ultimate strength Su=552MPa, and 60% elongation. Find the ratio of the materialtoughness to the resilience assuming that the stress-strain curve consists of two straightlines according to sketch c.

Notes: The toughness is the ability to absorb energy up to fracture. Resilience is the amount ofenergy available in elastic recovery after yield.

Solution:Toughness is defined on page 113 as the ability to absorb energy up to fracture. Therefore, thetoughness is given by the area under the stress strain curve. For the material given, assuming thecurves are straight lines, this area is:

Toughness=12

207MPa( ) 0.002( ) +207MPa + 552MPa

2

0.60- 0.002( ) = 220MPa = 220MNm/m3

The resilience is the energy available for elastic recovery. Based on yield strength, the resilienceis:

R =12

207MPa( ) 0.002( ) = 0.207MNm/m3

The ratio of toughness to resilience is then 220/0.207=1062.

3.15 A steel cube has side length l=0.1m, modulus of elasticity E=206GPa, and Poissons ration =0.3. Find the compressive stresses needed on four of the cube faces to give the sameelongation perpendicular thereto as a stress s in that perpendicular direction.

Notes: This problem uses Hookes law to obtain a solution.

Solution:The elongation in the direction of stress is given by the strain and Hookes law as

d = e s l =sE

l

The average transverse strain for a compressive stress s c is

e t = - ns cE

The total transverse strain from the two perpendicular stresses is

e t,total= 2e t = - 2ns cE

The strain should be the same as if the stress s was applied in the direction:

d =sE

l = - 2ns cE

l;s c = -s2n= -

s2 0.3( )

= - 1.67s

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3.16 For the stressed steel cube in Problem 3.15, calculate the volume ratio (vtension/vcompression)when s =500MPa.

Notes: Realize that the volume of the stressed cube is v=[l(1+e 1)][l(1+e 2)][l(1+e 3)]. This problemuses the results of Problem 3.15.

Solution:The volume of the cube in tension is given by

vt = l 1+ e 1( ) l 1 + e 2( ) l 1 + e 3( ) = l3 1 +sE

1 - nsE

2

= 0.1m( ) 3 1 +500MPa206GPa

1 - 0.3( )500MPa206GPa

2= 1.00097 10- 3m3

The volume of the cube in compression is

vc = l 1 + e 1( ) l 1 + e 2( ) l 1+ e 3( ) = l3 1 - 2ns cE

1 +s cE- ns cE

2

= 0.1m( ) 3 1 - 2 0.3( )- 1.67( ) 500MPa

206GPa

1 +

- 1.67( ) 500MPa( )206GPa

- 0.3( )500MPa206GPa

2= 0.99675 10- 3m3

Therefore, the ratio of the volumes is vt/vc=1.0097/0.99675=1.004235.

3.17 Hookes law describes the relationship between uniaxial stress and uniaxial strain. Howlarge is the ratio of deformation for a given load within the group of materials consideredin Chapter 3?

Notes: The materials in Table 3.2 should be considered. Therefore, as the stiff material, usesilicon carbide, and as the compliant material, use natural rubber.

Solution:Reviewing Table 3.2 on page 108, we select silicon carbide as the stiffest material listed withEsic=450GPa, and natural rubber is the least stiff with Er=0.004GPa. From Hookes law, thedeformation is inversely proportional to applied stress. Therefore,

d rd sic=

1/Er( )1/Esic( )

=EsicEr=

450GPa0.004GPa

= 1.125 105

3.18 According to Archards wear equation, the wear depth is proportional to the slidingdistance and the contact pressure. How will the contact pressure be distributed radiallyfor a disk brake if the wear rate is the same for all radii?

Notes: Equation (3.26) is used to solve the problem.

Solution:From Equation (3.26) the wear rate per unit width is

Wr=KAplA=KAplf r=constant

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Therefore, solving for pl,

pl =C

KAf r=

C1r

Therefore, the pressure is inversely proportional to the radius.

3.19 Given a brake pad for a disk brake on a car, and using Archards wear constant,determine how the wear is distributed over the pad if the brake pressure is constant overthe pad.

Notes: Equation (3.26) is used to solve the problem.

Solution:The wear rate per width is given by Equation (3.26) as:

Wr=KAplA=KAplf r

If p1 is constant, then this can be rewritted as Wr=C1r, or that the wear rate across the pad is

directly proportional to the radius.