3.5

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3.1 At the proportional limit, a 12-inch gage length of a 0.75-in.-diameter alloy bar has elongated 0.0325 in. and the diameter has been reduced 0.0006 in. The total tension force on the bar was 17.5 kips.Determine the following properties of the material: (a) the modulus of elasticity. (b) Poisson’s ratio. (c) the proportional limit.

Solution (a) The bar cross-sectional area is

2 2 2(0.75 in.) 0.441786 in.4 4

A Dπ π= = =

and thus, the normal stress corresponding to the 17.5-kip force is

2

17.5 kips 39.611897 ksi0.441786 in.

σ = =

The strain in the bar is

0.0325 in. 0.002708 in./in.12 in.

eL

ε = = =

The modulus of elasticity is therefore

39.611897 ksi 14,626 ksi 14,630 ksi0.002708 in./in.

E σε

= = = = Ans.

(b) The longitudinal strain in the bar was calculated previously as long 0.002708 in./in.ε = The lateral strain can be determined from the reduction of the diameter:

lat0.0006 in. 0.000800 in./in.0.75 in.

DD

ε Δ −= = = −

Poisson’s ratio for this specimen is therefore

lat

long

0.000800 in./in. 0.295421 0.2950.002708 in./in.

ενε

−= − = − = = Ans.

(c) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore, 39.6 ksiPLσ = Ans.

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3.2 At the proportional limit, a 20 mm thick × 75 mm wide bar elongates 6.8 mm under an axial load of 480 kN. The bar is 1.6-m long. If Poisson’s ratio is 0.32 for the material, determine: (a) the modulus of elasticity. (b) the proportional limit (c) the change in each lateral dimension.

Solution (a) The bar cross-sectional area is 2(20 mm)(75 mm) 1,500 mmA = = and thus, the normal stress corresponding to the 480-kN axial load is

2

(480 kN)(1,000 N/kN) 320 MPa1,500 mm

σ = =

The strain in the bar is

6.8 mm 0.004250 mm/mm(1.6 m)(1,000 mm/m)

eL

ε = = =

The modulus of elasticity is therefore

320 MPa 75,294 MPa 75.3 GPa0.004250 mm/mm

E σε

= = = = Ans.

(b) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore, 320 MPaPLσ = Ans. (c) Poisson’t ratio is given as ν = 0.32. The longitudinal strain in the bar was calculated previously as long 0.004250 mm/mmε = The corresponding lateral strain can be determined from Poisson’s ratio: lat long (0.32)(0.004250 mm/mm) 0.001360 mm/mmε νε= − = − = − Using this lateral strain, the change in bar width is latwidth (width) ( 0.001360 mm/mm)(75 mm) 0.1020 mmεΔ = = − = − Ans. and the change in bar thickness is latthickness (thickness) ( 0.001360 mm/mm)(20 mm) 0.0272 mmεΔ = = − = − Ans.

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3.3 At an axial load of 22 kN, a 15 mm thick × 45 mm wide polyimide polymer bar elongates 3.0 mm while the bar width contracts 0.25 mm. The bar is 200 mm long. At the 22-kN load, the stress in the polymer bar is less than its proportional limit. Determine: (a) the modulus of elasticity. (b) Poisson’s ratio (c) the change in the bar thickness.

Solution (a) The bar cross-sectional area is 2(15 mm)(45 mm) 675 mmA = = and thus, the normal stress corresponding to the 22-kN axial load is

2

(22 kN)(1,000 N/kN) 32.592593 MPa675 mm

σ = =

The strain in the bar is

3.0 mm 0.0150 mm/mm200 mm

eL

ε = = =

The modulus of elasticity is therefore

32.592593 MPa 2,173 MPa 2.17 GPa0.0150 mm/mm

E σε

= = = = Ans.

(b) The longitudinal strain in the bar was calculated previously as long 0.0150 mm/mmε = The lateral strain can be determined from the reduction of the bar width:

latwidth 0.25 mm 0.005556 mm/mm

width 45 mmε Δ −= = = −

Poisson’s ratio for this specimen is therefore

lat

long

0.005556 mm/mm 0.370370 0.3700.0150 mm/mm

ενε

−= − = − = = Ans.

(c) The change in bar thickness can be found from the lateral strain: latthickness (thickness) ( 0.005556 mm/mm)(15 mm) 0.0833 mmεΔ = = − = − Ans.

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3.4 A 0.75-in.-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width of the bar is w = 3.0 in. Strain gages bonded to the specimen measure the following strains in thelongitudinal (x) and transverse (y) directions: εx = 2,136 με and εy = −673 με. (a) Determine Poisson’s ratio for thisspecimen. (b) If the measured strains were producedby an axial load of P = 50 kips, what is themodulus of elasticity for this specimen?

Fig. P3.4

Solution (a) Poisson’s ratio for this specimen is

lat

long

673 με 0.3152,136 με

y

x

εενε ε

−= − = − = − = Ans.

(b) The bar area is 2(3.0 in.)(0.75 in.) 2.25 in.A = = and so the normal stress for an axial load of P = 50 kips is

2

50 kips 22.222222 ksi2.25 in.

σ = =

The modulus of elasticity is thus

22.222222 ksi 10,404 ksi 10,400 ksi1 in./in.(2,136 με)

1,000,000 με

E σε

= = = =⎛ ⎞⎜ ⎟⎝ ⎠

Ans.

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3.5 A 6-mm-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width of the bar is w = 30 mm. Strain gages bonded to the specimen measure the following strains in thelongitudinal (x) and transverse (y) directions: εx = 1,525 με and εy = −540 με. (a) Determine Poisson’s ratio for this specimen. (b) If the measured strains were producedby an axial load of P = 27.5 kN, what isthe modulus of elasticity for thisspecimen?

Fig. P3.5

Solution (a) Poisson’s ratio for this specimen is

lat

long

540 με 0.3541,525 με

y

x

εενε ε

−= − = − = − = Ans.

(b) The bar area is 2(30 mm)(6 mm) 180 mmA = = and so the normal stress for an axial load of P = 50 kips is

2

(27.5 kN)(1,000 N/kN) 152.777778 MPa180 mm

σ = =

The modulus of elasticity is thus

152.777778 MPa 100,182 MPa 100.2 GPa1 in./in.(1,525 με)

1,000,000 με

E σε

= = = =⎛ ⎞⎜ ⎟⎝ ⎠

Ans.

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3.6 A copper rod [E = 110 GPa] originally 600-mm long is pulled in tension with a normal stress of 275MPa. If the deformation is entirely elastic, what is the resulting elongation?

Solution Since the deformation is elastic, the strain in the rod can be determined from Hooke’s Law,

275 MPa 0.002500 mm/mm110,000 MPaE

σε = = =

The elongation in the rod is thus (0.002500 mm/mm)(600 mm) 1.500 mme Lε= = = Ans.

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3.7 A 6061-T6 aluminum tube [E = 10,000 ksi; ν = 0.33] has an outside diameter of 4.000 in. and a wallthickness of 0.065 in. (a) Determine the tension force that must be applied to the tube to cause its outside diameter to contractby 0.005 in. (b) If the tube is 84-in. long, what is the overall elongation?

Solution (a) The strain associated with the given diameter contraction is

lat0.005 in. 0.001250 in./in.

4.000 in.ε −= = −

From the given Poisson’s ratio, the longitudinal strain in the tube must be

latlong

0.001250 in./in. 0.003788 in./in.0.33

εεν

−= − = − =

and from Hooke’s Law, the normal stress can be calculated as (10,000 ksi)(0.003788 in./in.) 37.878788 ksiEσ ε= = = The area of the tube is needed to determine the tension force. Given that the outside diameter of the tube is 4.000 in. and the wall thickness is 0.065 in., the inside diameter of the tube is 3.870 in. The tube cross-sectional area is thus

2 2 2(4.000 in.) (3.870 in.) 0.803541 in.4

A π ⎡ ⎤= − =⎣ ⎦

and the force applied to the tube is 2(37.878788 ksi)(0.803541 in. ) 30.437154 kips 30.4 kipsF Aσ= = = = Ans. (b) The strain was calculated previously. Use the longitudinal strain to determine the overall elongation: (0.003788 in./in.)(84 in.) 0.318 in.e Lε= = = Ans.

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3.8 A metal specimen with an original diameter of 0.500 in. and a gage length of 2.000 in. is tested intension until fracture occurs. At the point of fracture, the diameter of the specimen is 0.260 in. and thefractured gage length is 3.08 in. Calculate the ductility in terms of percent elongation and percent reduction in area.

Solution Percent elongation is simply the longitudinal strain at fracture:

(3.08 in. 2.000 in.) 1.08 in. 0.54 in./in.2.000 in. 2.000 in.

percent elongation 54%

eL

ε −= = = =

∴ = Ans. The initial cross-sectional area of the specimen is

2 2 20 (0.500 in.) 0.196350 in.

4 4A Dπ π= = =

The final cross-sectional area at the fracture location is

2 2 2(0.260 in.) 0.053093 in.4 4fA Dπ π= = =

The percent reduction in area is

2 2

02

0

(0.196350 in. 0.053093 in. )percent reduction of area (100%) (100%) 73.0%0.196350 in.

fA AA− −= = = Ans.

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3.9 A portion of the stress-strain curve for astainless steel alloy is shown in Fig. P3.9. A 350-mm-long bar is loaded in tension until it elongates2.0 mm and then the load is removed. (a) What is the permanent set in the bar? (b) What is the length of the unloaded bar? (c) If the bar is reloaded, what will be theproportional limit?

Fig. P3.9

Solution

(a) The normal strain in the specimen is

2.0 mm 0.005714 mm/mm350 mm

eL

ε = = =

Construct a line parallel to the elastic modulus line that passes through the data curve at a strain of ε = 0.005714 mm/mm. The strain value at which this modulus line intersects the strain axis is the permanent set: permanent set 0.0035 mm/mm= Ans. (b) The length of the unloaded bar is therefore:

(0.0035 mm/mm)(350 mm) 1.225 mm

350 mm 2.225 mm 352.225 mmf

e L

L

ε= = =

= + = Ans. (c) From the stress-strain curve, the reload proportional limit is 444 MPa . Ans.

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3.10 The 16 × 22 × 25 mm rubber blocksshown in Fig. P3.10 are used in a double Ushear mount to isolate the vibration of amachine from its supports. An applied load ofP = 285 N causes the upper frame to bedeflected downward by 5 mm. Determine theshear modulus G of the rubber blocks.

Fig. P3.10

Solution Consider a FBD of the upper U frame. The downward force P is resisted by two upward shear forces V; therefore, V = 285 N / 2 = 142.5 N. Next, consider a FBD of one of the rubber blocks. The shear force acting on one rubber block is V = 142.5 N. The area of the rubber block that is parallel to the direction of V is 2(22 mm)(25 mm) 550 mmVA = = Consequently, the shear stress in one rubber block is

2

142.5 N 0.259091 MPa550 mmV

VA

τ = = =

The shear strain associated with the 5-mm downward displacement of the rubber blocks is given by:

5 mmtan 0.312500 0.302885 rad16 mm

γ γ= = ∴ =

From Hooke’s Law for shear stress and shear strain, the shear modulus G can be computed:

0.259091 MPa 0.855411 MPa 0.855 MPa0.302885 rad

G G ττ γγ

= ∴ = = = = Ans.

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3.11 Two hard rubber blocks are used in an anti-vibration mount to support a small machine, as shown in Fig. P3.11.An applied load of P = 150 lb causes a downward deflectionof 0.25 in. Determine the shear modulus of the rubber blocks.Assume a = 0.5 in., b = 1.0 in., and c = 2.5 in.

Fig. P3.11

Solution Determine the shear strain from the angle formed by the downward deflection and the block thickness a:

0.250 in.tan 0.500 0.463648 rad0.50 in.

γ γ= = ∴ =

Note: The small angle approximation tan γ ≈ γ is not applicable in this instance. Determine the shear stress from the applied load P and the block area. Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block is half of the total load: V = P/2 = 75 lb. To determine the area needed here, consider the surface that is bonded to the plate. This area has dimensions of b×c. The shear stress acting on a single block is therefore:

75 lb 30 psi(1.0 in.)(2.5 in.)

VA

τ = = =

The shear modulus G can be calculated from Hooke’s law for shear stress and strain:

30 psi 64.704 psi 64.7 psi0.463648 rad

G G ττ γγ

= ∴ = = = = Ans.

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3.12 Two hard rubber blocks [G = 350 kPa] are used in an anti-vibration mount to support a small machine, as shown in Fig.P3.12. Determine the downward deflection that will occur for anapplied load of P = 900 N. Assume a = 20 mm, b = 50 mm, and c = 80 mm.

Fig. P3.12

Solution Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block is half of the total load: V = P/2 = 450 N. Determine the shear stress from the shear force V and the block area. To determine the area needed here, consider the surface that is bonded to the plate. This area has dimensions of b×c. The shear stress acting on a single block is therefore:

450 N 0.112500 MPa(50 mm)(80 mm)

VA

τ = = =

Since the shear modulus G is given, the shear strain can be calculated from Hooke’s law for shear stress and shear strain:

0.112500 MPa 0.321429 rad0.350 MPa

GGττ γ γ= ∴ = = =

From the angle γ and the block thickness a, the downward deflection δ of the block can be determined from:

tan tan (20 mm) tan(0.321429 rad) 6.659512 mm 6.66 mmaaδγ δ γ= ∴ = = = = Ans.

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3.13 A load test on a 6 mm diameter by 150 mm long magnesium alloy rod found that a tension load of780 N caused an elastic elongation of 0.55 mm in the rod. Using this result, determine the elasticelongation that would be expected for a 19-mm-diameter rod of the same material if the rod were 1.2 mlong and subjected to a tension force of 2.6 kN.

Solution The area of the 6-mm-diameter rod is

2 2 2(6 mm) 28.274334 mm4 4

A Dπ π= = =

Thus, the normal stress in the rod due to a 780-N load is

2

780 N 27.586857 MPa28.274334 mm

FA

σ = = =

The strain in the 150-mm long rod associated with a 0.55-mm elongation is

0.55 mm 0.003667 mm/mm150 mm

eL

ε = = =

Therefore, the elastic modulus of the magnesium alloy is

27.586857 MPa 7,523.688217 MPa0.003667 mm/mm

E σε

= = =

The area of the 19-mm-diameter rod is

2 2 2(19 mm) 283.528737 mm4 4

A Dπ π= = =

Thus, the normal stress in the 19-mm-diameter rod due to a 2.6-kN load is

2

(2.6 kN)(1,000 N/kN) 9.170146 MPa283.528737 mm

FA

σ = = =

From Hooke’s Law, the strain that would be expected is

9.170146 MPa 0.001219 mm/mm7,523.688217 MPaE

σε = = =

Since the 19-mm-diameter rod is 1.2-m long, the expected elongation is (0.001219 mm/mm)(1.2 m)(1,000 mm/m) 1.462604 mm 1.463 mme Lε= = = = Ans.

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3.14 The stress-strain diagram for aparticular stainless steel alloy is shown inFig. P3.14. A rod made from this material isinitially 800 mm long at a temperature of20°C. After a tension force is applied to therod and the temperature is increased by200°C, the length of the rod is 804 mm.Determine the stress in the rod and statewhether the elongation in the rod is elasticor inelastic. Assume the coefficient ofthermal expansion for this material is 18 ×10−6/°C.

Fig. P3.14

Solution The 4-mm total elongation of the rod is due to a combination of load and temperature increase. The 200°C temperature increase causes a normal strain of: 6(18 10 / )(200 ) 0.003600 mm/mmT T C Cε α −= Δ = × ° ° = which means that the rod elongates (0.003600 mm/mm)(800 mm) 2.8800 mmT Te Lε= = = The portion of the 4-mm total elongation due to load is therefore 4 mm 2.8800 mm 1.1200 mmTe e eσ = − = − = The strain corresponding to this elongation is

1.1200 mm 0.001400 mm/mm800 mm

eLσ

σε = = =

By inspection of the stress-strain curve, this strain is clearly in the linear region. Therefore, the rod is elastic in this instance. For the linear region, the elastic modulus can be determined from the lower scale plot:

(400 MPa 0) 200,000 MPa(0.002 mm/mm 0)

E σε

Δ −= = =Δ −

Using Hooke’s Law (or directly from the σ-ε diagram), the stress corresponding to the 0.001400 mm/mm strain is (200,000 MPa)(0.001400 mm/mm) 280 MPaE σσ ε= = = Ans.

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3.15 In Fig. P3.15, rigid bar ABC is supported byaxial member (1), which has a cross-sectional areaof 400 mm2, an elastic modulus of E = 70 GPa, and a coefficient of thermal expansion of α = 22.5 × 10−6 /°C. After load P is applied to the rigid barand the temperature rises 40°C, a strain gageaffixed to member (1) measures a strain increase of1,650 με. Determine: (a) the normal stress in member (1). (b) the magnitude of applied load P. (c) the deflection of the rigid bar at C.

Fig. P3.15

Solution (a) The strain measured in member (1) is due to both the internal force in the member and the temperature change. The strain caused by the temperature change is 6(22.5 10 / )(40 ) 0.000900 mm/mmT T C Cε α −= Δ = × ° ° = Since the total strain is ε = 1,650 με = 0.001650 mm/mm, the strain caused by the internal force in member (1) must be 0.001650 mm/mm 0.000900 mm/mm 0.000750 mm/mmTσε ε ε= − = − = The elastic modulus of member (1) is E = 70 GPa; thus, from Hooke’s Law, the stress in the member is: 1 (70,000 MPa)(0.000750 mm/mm) 52.5 MPaE σσ ε= = = Ans. (b) If the normal stress in member (1) is 52.5 MPa, the axial force in the member is 2 2

1 1 1 (52.5 N/mm )(400 mm ) 21,000 NF Aσ= = = Consider moment equilibrium of rigid bar ABC about joint A to determine the magnitude of P:

(1.4 m)(21,000 N) (2.4 m) 0

12, 250 N 12.25 kNAM P

P

Σ = − =

∴ = = Ans.

(c) The strain in member (1) was measured as ε = 1,650 με = 0.001650 mm/mm; therefore, the total elongation of member (1) is 1 1 1 (0.001650 mm/mm)(2,500 mm) 4.125 mme Lε= = = The deflection of the rigid bar at B is equal to this elongation; therefore, vB = e1 = 4.125 mm (downward). By similar triangles, the deflection of the rigid bar at C is given by:

1.4 m 2.4 m2.4 m 2.4(4.125 mm) 7.071429 mm 7.07 mm1.4 m 1.4

CB

C B

vv

v v

=

∴ = = = = ↓ Ans.

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3.16 A tensile test specimen of 1045 hot-rolled steel having a diameter of 0.505 in. and a gagelength of 2.00 in. was tested to fracture. Stressand strain data obtained during the test are shown in Fig. P3.16. Determine (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.20% offset). (e) the fracture stress. (f) the true fracture stress if the final diameter ofthe specimen at the location of the fracture was0.392 in.

Fig. P3.16

Solution From the stress-strain curve, the proportional limit will be taken as σ = 60 ksi at a strain of ε = 0.0019. (Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.) (a) The modulus of elasticity is

60 ksi 31,600 ksi0.0019 in./in.

E σε

= = = Ans.

(b) From the diagram, the proportional limit is taken as 60 ksiPLσ = Ans. (c) The ultimate strength is ult 105 ksiσ = Ans. (d) The yield strength is 68 ksiYσ = Ans. (e) The fracture stress is fracture 98 ksiσ = Ans. (f) The original cross-sectional area of the specimen is

2 2 20 (0.505 in.) 0.200296 in.

4 4A Dπ π= = =

The area of the specimen at the fracture location is

2 2 2(0.392 in.) 0.120687 in.4 4fA Dπ π= = =

The true fracture stress is therefore

2

fracture 2

0.200296 in.true (98 ksi) 162.6 ksi0.120687 in.

σ = = Ans.

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3.17 A tensile test specimen of stainlesssteel alloy having a diameter of 0.495 in.and a gage length of 2.00 in. was tested tofracture. Stress and strain data obtained during the test are shown in Fig. P3.17.Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.20% offset). (e) the fracture stress. (f) the true fracture stress if the finaldiameter of the specimen at the location ofthe fracture was 0.350 in.

Fig. P3.17

Solution From the stress-strain curve, the proportional limit will be taken as σ = 60 ksi at a strain of ε = 0.002. (Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.) (a) The modulus of elasticity is

60 ksi 30,000 ksi0.002 in./in.

E σε

= = = Ans.

(b) From the diagram, the proportional limit is taken as 60 ksiPLσ = Ans. (c) The ultimate strength is ult 159 ksiσ = Ans. (d) The yield strength is 80 ksiYσ = Ans. (e) The fracture stress is fracture 135 ksiσ = Ans. (f) The original cross-sectional area of the specimen is

2 2 20 (0.495 in.) 0.192442 in.

4 4A Dπ π= = =

The area of the specimen at the fracture location is

2 2 2(0.350 in.) 0.096211 in.4 4fA Dπ π= = =

The true fracture stress is therefore

2

fracture 2

0.192442 in.true (135 ksi) 270 ksi0.096211 in.

σ = = Ans.

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3.18 A bronze alloy specimen having adiameter of 12.8 mm and a gage lengthof 50 mm was tested to fracture. Stressand strain data obtained during the testare shown in Fig. P3.18. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.20% offset). (e) the fracture stress. (f) the true fracture stress if the finaldiameter of the specimen at the locationof the fracture was 10.5 mm.

Fig. P3.18

Solution From the stress-strain curve, the proportional limit will be taken as σ = 210 MPa at a strain of ε = 0.002. (Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.) (a) The modulus of elasticity is

210 MPa 105,000 MPa0.002 in./in.

E σε

= = = Ans.

(b) From the diagram, the proportional limit is taken as 210 MPaPLσ = Ans. (c) The ultimate strength is ult 380 MPaσ = Ans. (d) The yield strength is 290 MPaYσ = Ans. (e) The fracture stress is fracture 320 MPaσ = Ans. (f) The original cross-sectional area of the specimen is

2 2 20 (12.8 mm) 128.679635 mm

4 4A Dπ π= = =

The area of the specimen at the fracture location is

2 2 2(10.5 mm) 86.590148 mm4 4fA Dπ π= = =

The true fracture stress is therefore

2

fracture 2

128.679635 mmtrue (320 MPa) 476 MPa86.590148 mm

σ = = Ans.

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3.19 An alloy specimen having a diameter of 12.8mm and a gage length of 50 mm was tested tofracture. Load and deformation data obtainedduring the test are given. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.05% offset). (e) the yield strength (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter ofthe specimen at the location of the fracture was11.3 mm.

Load Change in Length Load Change in

Length (kN) (mm) (kN) (mm) 0 0 7.6 0.02 43.8 1.50 14.9 0.04 45.8 2.00 22.2 0.06 48.3 3.00 28.5 0.08 49.7 4.00 29.9 0.10 50.4 5.00 30.6 0.12 50.7 6.00 32.0 0.16 50.4 7.00 33.0 0.20 50.0 8.00 33.3 0.24 49.7 9.00 36.8 0.50 47.9 10.00 41.0 1.00 45.1 fracture

Solution The plot of the stress-strain data is shown below.

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(a) The modulus of elasticity is

224.976 MPa 140,600 MPa0.0016 mm/mm

E σε

= = = Ans.

(b) From the diagram, the proportional limit is taken as 234 MPaPLσ = Ans. (c) The ultimate strength is ult 400 MPaσ = Ans. (d) The yield strength by the 0.05% offset method is 239 MPaYσ = Ans. (e) The yield strength by the 0.2% offset method is 259 MPaYσ = Ans. (f) The fracture stress is fracture 356 MPaσ = Ans. (f) The original cross-sectional area of the specimen is

2 2 20 (12.8 mm) 128.679635 mm

4 4A Dπ π= = =

The area of the specimen at the fracture location is

2 2 2(11.3 mm) 100.287492 mm4 4fA Dπ π= = =

The true fracture stress is therefore

2

fracture 2

128.679635 mmtrue (356 MPa) 457 MPa100.287492 mm

σ = = Ans.

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3.20 A 1035 hot-rolled steel specimen with a diameter of 0.500 in. and a 2.0-in. gage length was tested to fracture. Load and deformation data obtained during the test are given. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.05% offset). (e) the yield strength (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.387 in.

Load Change in Length Load Change

in Length (lb) (in.) (lb) (in.) 0 0 12,540 0.0209 2,690 0.0009 12,540 0.0255 5,670 0.0018 14,930 0.0487 8,360 0.0028 17,020 0.0835 11,050 0.0037 18,220 0.1252 12,540 0.0042 18,820 0.1809 13,150 0.0046 19,110 0.2551 13,140 0.0060 19,110 0.2968 12,530 0.0079 18,520 0.3107 12,540 0.0098 17,620 0.3246 12,840 0.0121 16,730 0.3339 12,840 0.0139 16,130 0.3385 15,900 fracture

Solution The plot of the stress-strain data is shown below.

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(a) The modulus of elasticity is

63.849 ksi 30,400 ksi0.0021 in./in.

E σε

= = = Ans.

(b) From the diagram, the proportional limit is taken as 63.8 ksiPLσ = Ans. (c) The ultimate strength is ult 97.3 ksiσ = Ans. (d) The yield strength using the 0.05% offset method is 65.4 ksiYσ = Ans. (e) The yield strength using the 0.2% offset method is 63.8 ksiYσ = Ans. (f) The fracture stress is fracture 82.1 ksiσ = Ans. (g) The original cross-sectional area of the specimen is

2 2 20 (0.500 in.) 0.196350 in.

4 4A Dπ π= = =

The area of the specimen at the fracture location is

2 2 2(0.387 in.) 0.117628 in.4 4fA Dπ π= = =

The true fracture stress is therefore

2

fracture 2

0.196350 in.true (82.1 ksi) 137.0 ksi0.117628 in.

σ = = Ans.

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3.21 A 2024-T4 aluminum test specimen with adiameter of 0.505 in. and a 2.0-in. gage length wastested to fracture. Load and deformation dataobtained during the test are given. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.05% offset). (e) the yield strength (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter ofthe specimen at the location of the fracture was 0.452 in.

Load Change in Length Load Change in

Length (lb) (in.) (lb) (in.) 0 0.0000 11,060 0.0139 1,300 0.0014 11,500 0.0162 2,390 0.0023 12,360 0.0278 3,470 0.0032 12,580 0.0394 4,560 0.0042 12,800 0.0603 5,640 0.0051 13,020 0.0788 6,720 0.0060 13,230 0.0974 7,380 0.0070 13,450 0.1159 8,240 0.0079 13,670 0.1391 8,890 0.0088 13,880 0.1623 9,330 0.0097 14,100 0.1994 9,980 0.0107 14,100 0.2551 10,200 0.0116 14,100 0.3200 10,630 0.0125 14,100 0.3246 14,100 fracture

Solution The plot of the stress-strain data is shown below.

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(a) The modulus of elasticity is

33.55 ksi 11,180 ksi0.003 in./in.

E σε

= = = Ans.

(b) From the diagram, the proportional limit is taken as 33.6 ksiPLσ = Ans. (c) The ultimate strength is ult 70.4 ksiσ = Ans. (d) The yield strength using the 0.05% offset method is 44.4 ksiYσ = Ans. (e) The yield strength using the 0.2% offset method is 54.5 ksiYσ = Ans. (f) The fracture stress is fracture 70.4 ksiσ = Ans. (g) The original cross-sectional area of the specimen is

2 2 20 (0.505 in.) 0.200296 in.

4 4A Dπ π= = =

The area of the specimen at the fracture location is

2 2 2(0.452 in.) 0.160460 in.4 4fA Dπ π= = =

The true fracture stress is therefore

2

fracture 2

0.200296 in.true (70.4 ksi) 87.9 ksi0.160460 in.

σ = = Ans.

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3.22 A 1045 hot-rolled steel tension test specimenhas a diameter of 6.00 mm and a gage length of 25mm. In a test to fracture, the stress and strain databelow were obtained. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.05% offset). (e) the yield strength (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter ofthe specimen at the location of the fracture was4.65 mm.

Load Change in Length Load Change in

Length (kN) (mm) (kN) (mm) 0.00 0.00 13.22 0.29 2.94 0.01 16.15 0.61 5.58 0.02 18.50 1.04 8.52 0.03 20.27 1.80 11.16 0.05 20.56 2.26 12.63 0.05 20.67 2.78 13.02 0.06 20.72 3.36 13.16 0.08 20.61 3.83 13.22 0.08 20.27 3.94 13.22 0.10 19.97 4.00 13.25 0.14 19.68 4.06 13.22 0.17 19.09 4.12 18.72 fracture

Solution The plot of the stress-strain data is shown below.

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(a) The modulus of elasticity is

394.765 MPa 247,000 MPa0.0016 mm/mm

E σε

= = = Ans.

(b) From the diagram, the proportional limit is taken as 400 MPaPLσ = Ans. (c) The ultimate strength is ult 732 MPaσ = Ans. (d) The yield strength by the 0.05% offset method is 465 MPaYσ = Ans. (e) The yield strength by the 0.2% offset method is 465 MPaYσ = Ans. (f) The fracture stress is fracture 675 MPaσ = Ans. (f) The original cross-sectional area of the specimen is

2 2 20 (6 mm) 28.274334 mm

4 4A Dπ π= = =

The area of the specimen at the fracture location is

2 2 2(4.65 mm) 16.982272 mm4 4fA Dπ π= = =

The true fracture stress is therefore

2

fracture 2

28.274334 mmtrue (675 MPa) 1,124 MPa16.982272 mm

σ = = Ans.

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3.23 Rigid bar BCD in Fig. P3.23 issupported by a pin at C and by aluminumrod (1). A concentrated load P is applied tothe lower end of aluminum rod (2), which isattached to the rigid bar at D. The cross-sectional area of each rod is A = 0.20 in.2and the elastic modulus of the aluminummaterial is E = 10,000 ksi. After the load Pis applied at E, the strain in rod (1) ismeasured as 900 με (tension). (a) Determine the magnitude of load P. (b) Determine the total deflection of point Erelative to its initial position.

Fig. P3.23

Solution (a) From the measured strain, the stress in rod (1) is 6

1 1 1 (10,000 ksi)(900 10 in./in.) 9 ksiEσ ε −= = × = and thus, the force in rod (1) is 2

1 1 1 (9 ksi)(0.20 in. ) 1.8 kips (T)F Aσ= = = Consider the equilibrium of the rigid bar, and write a moment equilibrium equation about C to determine the magnitude of load P:

(20 in.)(1.8 kips) (30 in.) 0

1.2 kipsCM P

P

Σ = − =

∴ = Ans.

(b) From the measured strain, the elongation of rod (1) is 6

1 1 1 (900 10 in./in.)(50 in.) 0.0450 in.e Lε −= = × = From similar triangles, the deflection of the rigid bar at D can be expressed in terms of the deflection at B:

20 in. 30 in.30 in. 30 in.(0.045 in.) 0.0675 in.20 in. 20 in.

B D

D B

v v

v v

=

∴ = = =

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The elongation of rod (2) due to the 1.2-kip load must be determined. The stress in rod (2) is

22 2

2

1.2 kips 6 ksi0.2 in.

FA

σ = = =

and consequently, the strain in rod (2) is

22

2

6 ksi 0.000600 in./in.10,000 ksiE

σε = = =

From the strain, the elongation in rod (2) can be computed: 2 2 2 (0.000600 in./in.)(100 in.) 0.06 in.e Lε= = = The deflection of joint E is the sum of the rigid bar deflection at D and the elongation in rod (2): 2 0.0675 in. 0.06 in. 0.1275 in.E Dv v e= + = + = ↓ Ans.

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3.24 The rigid bar AC in Fig. P3.24 issupported by two axial bars (1) and (2).Both axial bars are made of bronze [E = 100 GPa; α = 18 × 10−6 mm/mm/°C]. Thecross-sectional area of bar (1) is A1 = 240mm2 and the cross-sectional area of bar (2)is A2 = 360 mm2. After load P has beenapplied and the temperature of the entireassembly has increased by 20°C, the totalstrain in bar (2) is measured as 800 με(elongation). Determine: (a) the magnitude of load P. (b) the vertical displacement of pin A.

Fig. P3.24

Solution (a) The total strain in bar (2) is caused partly by the axial force in the bar and partly by the increase in temperature. The strain caused by the 20°C temperature increase is: 6(18 10 mm/mm/ C)(20 C) 0.000360 mm/mmT Tε α −= Δ = × ° ° = The strain caused by the axial force in the bar is thus: 2, 2 2, 0.000800 mm/mm 0.000360 mm/mm 0.000440 mm/mmTσε ε ε= − = − = The stress in bar (2) is 2 2 2, (100,000 MPa)(0.000440 mm/mm) 44 MPaE σσ ε= = = and the force in bar (2) is 2 2

2 2 2 (44 N/mm )(360 mm ) 15,840 NF Aσ= = = Next, consider a FBD of the rigid bar AC. Equilibrium equations for this FBD are:

1 2

2

0

(1,400 mm) (500 mm) 0y

A

F F F P

M F P

Σ = + − =

Σ = − =

which can be solve simultaneously to give:

2500 mm 0.357143

1,400 mmF P P= =

and 1 0.642857F P= The applied load P can be expressed in terms of F2 as

2 21 2.8

0.357143P F F= =

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and so the magnitude of load P is 22.8 2.8(15,840 N) 44,352 N 44.4 kNP F= = = = Ans. (b) The force in bar (1) is 1 0.642857 (0.642857)(44,352 N) 28,512 NF P= = = Thus, the stress in bar (1) is

11 2

1

28,512 N 118.800 MPa240 mm

FA

σ = = =

The normal strain due to the axial force in bar (1) is

11,

1

118.800 MPa 0.001188 mm/mm100,000 MPaEσ

σε = = =

The normal strain caused by the 20°C temperature increase is: 6

1, (18 10 mm/mm/ C)(20 C) 0.000360 mm/mmT Tε α −= Δ = × ° ° = Therefore, the total strain in bar (1) is 1 1, 1, 0.001188 mm/mm 0.000360 mm/mm 0.001548 mm/mmTσε ε ε= + = + = and the elongation in bar (1) is 1 1, 1 (0.001548 mm/mm)(1,300 mm) 2.012400 mme Lσε= = = Since rigid bar ABC is connected to bar (1) (with a perfect connection), joint A displaces downward by an amount equal to the elongation of bar (1); therefore, 1 2.01 mmAv e= = ↓ Ans.

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3.25 The rigid bar in Fig. P3.25 is supportedby axial bar (1) and by a pin connection atC. Axial bar (1) has a cross-sectional area ofA1 = 275 mm2, an elastic modulus of E = 200 GPa, and a coefficient of thermalexpansion of α = 11.9 × 10−6 mm/mm/°C.The pin at C has a diameter of 25 mm. Afterload P has been applied and the temperatureof the entire assembly has been increased by 20°C, the total strain in bar (1) is measuredas 675 με (elongation). Determine: (a) the magnitude of load P. (b) the shear stress in pin C.

Fig. P3.25

Solution The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress: Tσε ε ε= + The normal strain due to the increase in temperature is: 6(11.9 10 mm/mm/ C)(20 C) 0.000238 mm/mmT Tε α −= Δ = × ° ° = Therefore, the normal stress in bar (1) causes a normal strain of: 0.000675 mm/mm 0.000238 mm/mm 0.000437 mm/mmTσε ε ε= − = − = From Hooke’s law, the normal stress in bar (1) can be calculated as: 1 (200,000 MPa)(0.000437 mm/mm) 87.4 MPaE σσ ε= = = and thus the axial force in bar (1) must be: 2 2

1 1 1 (87.4 N/mm )(275 mm ) 24,035 NF Aσ= = = Next, consider a free-body diagram of the rigid bar. Write a moment equilibrium equation about pin C:

1(200 mm) (380 mm)(200 mm)(24,035 N) (380 mm) 0

12,650 N 12.65 kN

CM F PP

P

Σ = −= − =

∴ = = Ans. Now that P is known, the horizontal and vertical reactions at C can be calculated:

1 10 24,035 N0 12,650 N

x x x

y y y

F C F C FF C P C P

Σ = − = ∴ = =Σ = − = ∴ = =

The resultant force acting on pin C is:

2 2 2 2(24,035 N) (12,650 N) 27,160.702 Nx yC C C= + = + =

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Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the resultant force: V = 13,580.351 N. The area of one shear plane of the 25-mm-diameter pin at C (in other words, the cross-sectional area of the pin) is:

2 2pin (25 mm) 490.874 mm

4A π= =

and thus the shear stress in pin C is:

22

13,580.351 N 27.666 N/mm 27.7 MPa490.874 mmC

V

VA

τ = = = = Ans.

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3.26 The rigid bar in Fig. P3.26 is supported by axial bar (1) and by a pinconnection at C. Axial bar (1) has a cross-sectional area of A1 = 275 mm2, an elasticmodulus of E = 200 GPa, and a coefficientof thermal expansion of α = 11.9 × 10−6

mm/mm/°C. The pin at C has a diameter of 25 mm. After load P has been appliedand the temperature of the entire assemblyhas been decreased by 30°C, the totalstrain in bar (1) is measured as 675 με(elongation). Determine: (a) the magnitude of load P. (b) the shear stress in pin C.

Fig. P3.26

Solution The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress: Tσε ε ε= + The normal strain due to the decrease in temperature is: 6(11.9 10 mm/mm/ C)( 30 C) 0.000357 mm/mmT Tε α −= Δ = × ° − ° = − Therefore, the normal stress in bar (1) causes a normal strain of: 0.000675 mm/mm ( 0.000357 mm/mm) 0.001032 mm/mmTσε ε ε= − = − − = From Hooke’s law, the normal stress in bar (1) can be calculated as: 1 (200,000 MPa)(0.001032 mm/mm) 206.4 MPaE σσ ε= = = and thus the axial force in bar (1) must be: 2 2

1 1 1 (206.4 N/mm )(275 mm ) 56,760 NF Aσ= = = Next, consider a free-body diagram of the rigid bar. Write a moment equilibrium equation about pin C:

1(200 mm) (380 mm)(200 mm)(56,760 N) (380 mm) 0

29,874 N 29.9 kN

CM F PP

P

Σ = −= − =

∴ = = Ans. Now that P is known, the horizontal and vertical reactions at C can be calculated:

1 10 56,760 N0 29,874 N

x x x

y y y

F C F C FF C P C P

Σ = − = ∴ = =Σ = − = ∴ = =

The resultant force acting on pin C is:

2 2 2 2(56,760 N) (29,874 N) 64,142 Nx yC C C= + = + =

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Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the resultant force: V = 32,071 N. The area of one shear plane of the 25-mm-diameter pin at C (in other words, the cross-sectional area of the pin) is:

2 2pin (25 mm) 490.874 mm

4A π= =

and thus the shear stress in pin C is:

22

32,071 N 65.3345 N/mm 65.3 MPa490.874 mmC

V

VA

τ = = = = Ans.