25741866-3-1-at-the-Proportional-Limit-A-12-Inch

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 A =

π

D2

=

π

(0.75 in.)2

in.2 = 0.441786 in

4 4 and thus, the normal stress corresponding to the 17.5-kip force is 17.5 kips 39.611897 ksi σ = = 39.611897 0.441786 in.2 The strain in the bar is e 0.0325 in. ε = = = 0.002708 in./in. L 12 in. The modulus of elasticity is therefore σ 39.611897 39.611897 ksi E = = ,626 ksi = 14,630 ,630 ksi = 14,626 ε 0.002708 in./in.

Ans.

(b) The longitudinal strain strain in the bar was calculated previously as ε long = 0.002708 in./in. The lateral strain can be determined from the reduction of the diameter: Δ D −0.0006 in. ε lat = = = −0.000800 in./in. D 0.75 in. Poisson’s ratio for this specimen is therefore ε −0.000800 in./in. ν = − lat = − = 0.295421 = 0.295 ε long 0.002708 in./in.

Ans.

(c) Based on the problem statement, the stress in the bar is equal to the proportional p roportional limit; therefore,

σ PL = 39.6 ksi

Ans.

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limit, a 20 mm thick × 75 mm wide bar elongates 6.8 mm under an axial load of 3.2 At the proportional limit, 480 kN. The bar ba r 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 A = (20 (20 mm mm)(75 )(75 mm) = 1,500 ,500 mm2 and thus, the normal stress corresponding to the 480-kN axial load is (480 kN)(1,000 N/kN) σ = = 320 MPa 1,500 mm2 The strain in the bar is e 6.8 mm ε = = = 0.004250 mm/mm L (1.6 m)(1,000 mm/m) The modulus of elasticity is therefore σ 320 MPa E = = ,294 MPa = 75.3 GPa = 75,294 ε 0.004250 mm/mm

Ans.

(b) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore,

σ PL = 320 MPa

Ans.

(c) Poisson’t ratio is given as ν = 0.32. The longitudinal strain in the the bar was calculated previously as ε long = 0.004250 mm/mm The corresponding lateral strain can be determined from Poisson’s ratio: (0.32) 2)(0 (0.0 .004 0425 250 0 mm mm/mm /mm) = − 0.00 0.0013 1360 60 mm/m mm/mm m ε lat = −νε long = −(0.3 Using this lateral strain, the change in bar width is

Δwidth = ε lat ( width) = (−0.001360 mm/mm)(75 mm) = − 0.1020 mm

Ans.

and the change in bar thickness is

Δthickness = ε lat ( thickness) = (−0.001360 mm/mm)(20 mm) = − 0.0272 mm

Ans.


limit, a 20 mm thick × 75 mm wide bar elongates 6.8 mm under an axial load of 3.2 At the proportional limit, 480 kN. The bar ba r 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 A = (20 (20 mm mm)(75 )(75 mm) = 1,500 ,500 mm2 and thus, the normal stress corresponding to the 480-kN axial load is (480 kN)(1,000 N/kN) σ = = 320 MPa 1,500 mm2 The strain in the bar is e 6.8 mm ε = = = 0.004250 mm/mm L (1.6 m)(1,000 mm/m) The modulus of elasticity is therefore σ 320 MPa E = = ,294 MPa = 75.3 GPa = 75,294 ε 0.004250 mm/mm

Ans.

(b) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore,

σ PL = 320 MPa

Ans.

(c) Poisson’t ratio is given as ν = 0.32. The longitudinal strain in the the bar was calculated previously as ε long = 0.004250 mm/mm The corresponding lateral strain can be determined from Poisson’s ratio: (0.32) 2)(0 (0.0 .004 0425 250 0 mm mm/mm /mm) = − 0.00 0.0013 1360 60 mm/m mm/mm m ε lat = −νε long = −(0.3 Using this lateral strain, the change in bar width is

Δwidth = ε lat ( width) = (−0.001360 mm/mm)(75 mm) = − 0.1020 mm

Ans.

and the change in bar thickness is


Ans.


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 A = (15 mm)(45 mm) = 675 675 mm mm2 and thus, the normal stress corresponding to the 22-kN axial load is (22 kN)(1,000 N/kN) σ = = 32.592593 MPa 675 mm2 The strain in the bar is e 3.0 mm ε = = = 0.0150 mm/mm L 200 mm The modulus of elasticity is therefore σ 32.592593 MPa E = = GPa = 2,173 MPa = 2.17 GP ε 0.0150 mm/mm

Ans.

(b) The longitudinal strain strain in the bar was calculated previously as ε long = 0.0150 mm/mm The lateral strain can be determined from the reduction of the bar width: Δwidth −0.25 mm ε lat = = = −0.005556 mm/mm width 45 mm Poisson’s ratio for this specimen is therefore ε −0.005556 mm/mm ν = − lat = − = 0.370370 = 0.370 0.0150 mm/mm ε long

Ans.

(c) The change in bar thickness can be found from from the lateral strain: strain:


Ans.


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 the longitudinal ( x) and transverse ( y) directions: ε x = 2,136 με and ε y = −673 με.

(a) Determine Poisson’s ratio for this specimen. (b) If the measured strains were produced by an axial load of P = 50 kips, what is the modulus of elasticity for this specimen? Fig. P3.4

Solution (a) Poisson’s ratio for this specimen is ε y ε −673 με ν = − lat = − =− = 0.315 ε long ε x 2,136 με (b) The bar area is A = (3.0 in.)(0.75 in.) = 2.25 in.2 and so the normal stress for an axial load of P = 50 kips is 50 kips σ = = 22.222222 ksi 2.25 in.2 The modulus of elasticity is thus 22.222222 ksi σ E = = = 10,404 ksi = 10,400 ksi ε ⎛ 1 in./in. ⎞ (2,136 με) ⎜ ⎟ ⎝ 1,000,000 με ⎠

Ans.

Ans.


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 the longitudinal ( x) and transverse ( y) directions: ε x = 1,525 με and ε y = −540 με.

(a) Determine Poisson’s ratio for this specimen. (b) If the measured strains were produced by an axial load of P = 27.5 kN, what is the modulus of elasticity for this specimen?

Fig. P3.5

Solution (a) Poisson’s ratio for this specimen is ε y ε −540 με =− = 0.354 ν = − lat = − ε long ε x 1,525 με (b) The bar area is A = (30 mm)(6 mm) = 180 mm2 and so the normal stress for an axial load of P = 50 kips is (27.5 kN)(1,000 N/kN) σ = = 152.777778 MPa 180 mm2 The modulus of elasticity is thus σ 152.777778 MPa E = = = 100,182 MPa = 100.2 GPa ε ⎛ 1 in./in. ⎞ (1,525 με) ⎜ ⎟ ⎝ 1,000,000 με ⎠

Ans.

Ans.


3.6 A copper rod [ E = 110 GPa] originally 600-mm long is pulled in tension with a normal stress of 275 MPa. 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, σ 275 MPa ε = = = 0.002500 mm/mm E 110,000 MPa The elongation in the rod is thus

e = ε L = (0.002500 mm/mm)(600 mm) = 1.500 mm

Ans.


3.7 A 6061-T6 aluminum tube [ E = 10,000 ksi; ν = 0.33] has an outside diameter of 4.000 in. and a wall thickness of 0.065 in. (a) Determine the tension force that must be applied to the tube to cause its outside diameter to contract by 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 −0.005 in. ε lat = = −0.001250 in./in. 4.000 in. From the given Poisson’s ratio, the longitudinal strain in the tube must be ε −0.001250 in./in. ε long = − lat = − = 0.003788 in./in. ν 0.33 and from Hooke’s Law, the normal stress can be calculated as σ = E ε = (10,000 ksi)(0.003788 in./in.) = 37.878788 ksi 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

A =

π

⎡⎣ (4.000 in.)2 − (3.870 in.)2 ⎤⎦ = 0.803541 in.2 4

and the force applied to the tube is

F

= σ A = (37.878788 ksi)(0.803541 in.2 ) = 30.437154 kips = 30.4 kips

Ans.

(b) The strain was calculated previously. Use the longitudinal strain to determine the overall elongation:

e = ε L = (0.003788 in./in.)(84 in.) = 0.318 in.

Ans.


3.8 A metal specimen with an original diameter of 0.500 in. and a gage length of 2.000 in. is tested in tension until fracture occurs. At the point of fracture, the diameter of the specimen is 0.260 in. and the fractured 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: e (3.08 in. − 2.000 in.) 1.08 in. ε = = = = 0.54 in./in. L 2.000 in. 2.000 in.

∴ percent elongation = 54%

Ans.

The initial cross-sectional area of the specimen is

A0

=

A f

=

π

D2

=

D2

=

π

(0.500 in.)2

= 0.196350 in.2

(0.260 in.)2

= 0.053093 in.2

4 4 The final cross-sectional area at the fracture location is

π

π

4 4 The percent reduction in area is A0 − A f percent reduction of area = (100%) A0

=

(0.196350 in.2 − 0.053093 in.2 ) (100%) 0.196350 in.2

= 73.0%

Ans.


3.9 A portion of the stress-strain curve for a stainless steel alloy is shown in Fig. P3.9. A 350mm-long bar is loaded in tension until it elongates 2.0 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 the proportional limit?

Fig. P3.9

Solution

(a) The normal strain in the specimen is e 2.0 mm ε = = = 0.005714 mm/mm L 350 mm 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: e = ε L = (0.0035 mm/mm)(350 mm) = 1.225 mm

L f

= 350 mm + 2.225 mm = 352.225 mm

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

Ans. Ans.


3.10 The 16 × 22 × 25 mm rubber blocks shown in Fig. P3.10 are used in a double U shear mount to isolate the vibration of a machine from its supports. An applied load of P = 285 N causes the upper frame to be deflected downward by 5 mm. Determine the shear 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 AV = (22 mm)(25 mm) = 550 mm2 Consequently, the shear stress in one rubber block is V 142.5 N τ = = = 0.259091 MPa AV 550 mm2 The shear strain associated with the 5-mm downwa rd displacement of the rubber blocks is given by: 5 mm = 0.312500 ∴ γ = 0.302885 rad tan γ = 16 mm From Hooke’s Law for shear stress and shear strain, the shear modulus G can be computed: τ 0.259091 MPa τ = Gγ ∴G = = = 0.855411 MPa = 0.855 MPa γ 0.302885 rad

Ans.


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 deflection of 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. = 0.500 ∴ γ = 0.463648 rad tan γ = 0.50 in. 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: V 75 lb τ = = = 30 psi A (1.0 in.)(2.5 in.) The shear modulus G can be calculated from Hooke’s law for shear stress and strain: τ 30 psi τ = Gγ ∴G = = = 64.704 psi = 64.7 psi γ 0.463648 rad

Ans.


3.12 Two hard rubber blocks [G = 350 kPa] are used in an antivibration mount to support a small machine, as shown in Fig. P3.12. Determine the downward deflection that will occur for an applied 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 V τ = = = 0.112500 MPa A (50 mm)(80 mm) Since the shear modulus G is given, the shear strain can be calculated from Hooke’s law for shear stress and shear strain: τ 0.112500 MPa τ = Gγ ∴ γ = = = 0.321429 rad G 0.350 MPa From the angle γ and the block thickness a, the downward deflection δ of the block can be determined from: tan γ

=

δ a

∴δ = a tan γ = (20 mm) tan(0.321429 rad) = 6.659512 mm = 6.66 mm

Ans.


3.13 A load test on a 6 mm diameter by 150 mm long magnesium alloy rod found that a tension load of 780 N caused an elastic elongation of 0.55 mm in the rod. Using this result, determine the elastic elongation that would be expected for a 19-mm-diameter rod of the same material if the rod were 1.2 m long and subjected to a tension force of 2.6 kN.

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

A =

π

D2

=

π

(6 mm)2

= 28.274334 mm2

4 4 Thus, the normal stress in the rod due to a 780-N load is F 780 N σ = = = 27.586857 MPa A 28.274334 mm2 The strain in the 150-mm long rod associated with a 0.55-mm elongation is e 0.55 mm ε = = = 0.003667 mm/mm L 150 mm Therefore, the elastic modulus of the magnesium alloy is 27.586857 MPa σ E = = = 7,523.688217 MPa ε 0.003667 mm/mm The area of the 19-mm-diameter rod is

A =

π

D2

=

π

(19 mm)2

= 283.528737 mm2

4 4 Thus, the normal stress in the 19-mm-diameter rod due to a 2.6-kN load is F (2.6 kN)(1,000 N/kN) σ = = = 9.170146 MPa A 283.528737 mm2 From Hooke’s Law, the strain that would be expected is 9.170146 MPa σ ε = = = 0.001219 mm/mm E 7,523.688217 MPa Since the 19-mm-diameter rod is 1.2-m long, the expected elongation is

e = ε L = (0.001219 mm/mm)(1.2 m)(1,000 mm/m) = 1.462604 mm = 1.463 mm

Ans.


3.14 The stress-strain diagram for a particular stainless steel alloy is shown in Fig. P3.14. A rod made from this material is initially 800 mm long at a temperature of 20°C. After a tension force is applied to the rod and the temperature is increased by 200°C, the length of the rod is 804 mm. Determine the stress in the rod and state whether the elongation in the rod is elastic or inelastic. Assume the coefficient of thermal expansion for this material is 18 × 6 10− /°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: ε T = α ΔT = (18 ×10−6 / °C )(200°C ) = 0.003600 mm/mm which means that the rod elongates eT = ε T L = (0.003600 mm/mm)(800 mm) = 2.8800 mm The portion of the 4-mm total elongation due to load is therefore eσ = e − eT = 4 mm − 2.8800 mm = 1.1200 mm The strain corresponding to this elongation is e 1.1200 mm ε σ = σ = = 0.001400 mm/mm L 800 mm 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) Δσ E = = = 200,000 MPa Δε (0.002 mm/mm − 0) Using Hooke’s Law (or directly from the σ -ε diagram), the stress corresponding to the 0.001400 mm/mm strain is

σ = E ε σ = (200,000 MPa)(0.001400 mm/mm) = 280 MPa

Ans.


3.15 In Fig. P3.15, rigid bar ABC is supported by axial member (1), which has a cross-sectional area 2 of 400 mm , an elastic modulus of E = 70 GPa, and a coefficient of thermal expansion of α = 22.5 × 6 10− /°C. After load P is applied to the rigid bar and the temperature rises 40°C, a strain gage affixed to member (1) measures a strain increase of 1,650 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 ε T = α ΔT = (22.5 × 10−6 / °C )(40°C ) = 0.000900 mm/mm 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 ε σ = ε − ε T = 0.001650 mm/mm − 0.000900 mm/mm = 0.000750 mm/mm The elastic modulus of member (1) is E = 70 GPa; thus, from Hooke’s Law, the stress in the member is:

σ 1 = E ε σ = (70,000 MPa)(0.000750 mm/mm) = 52.5 MPa

Ans.

(b) If the normal stress in member (1) is 52.5 MPa, the axial force in the member is F1 = σ 1 A1 = (52.5 N/mm2 )(400 mm2 ) = 21,000 N Consider moment equilibrium of rigid bar ABC about joint A to determine the magnitude of P : ΣM A = (1.4 m)(21,000 N) − (2.4 m)P = 0

∴ P = 12, 250 N = 12.25 kN

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 e1 = ε 1 L1 = (0.001650 mm/mm)(2,500 mm) = 4.125 mm The deflection of the rigid bar at B is equal to this elongation; therefore, v B = e1 = 4.125 mm (downward). By similar triangles, the deflection of the rigid bar at C is given by: vC v B 1.4 m

=

2.4 m 2.4 m

∴ vC = vB

1.4 m

= (4.125 mm)

2.4 1.4

= 7.071429 mm = 7.07 mm ↓

Ans.


3.16 A tensile test specimen of 1045 hot-rolled steel having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Stress and 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 of the specimen at the location of the fracture was 0.392 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 σ 60 ksi E = = = 31, 600 ksi ε 0.0019 in./in.

Ans.

(b) From the diagram, the proportional limit is taken as

σ PL = 60 ksi

Ans.

(c) The ultimate strength is

σ ult = 105 ksi

Ans.

(d) The yield strength is

σ Y = 68 ksi

Ans.

(e) The fracture stress is

σ fracture = 98 ksi

Ans.

(f) The original cross-sectional area of the specimen is

A0

=

A f

=

π

D2

=

D2

=

π

(0.505 in.)2

= 0.200296 in.2

(0.392 in.)2

= 0.120687 in.2

4 4 The area of the specimen at the fracture location is

π

π

4 4 The true fracture stress is therefore 0.200296 in.2 true σ fracture = (98 ksi) 0.120687 in.2

= 162.6 ksi

Ans.


3.17 A tensile test specimen of stainless steel alloy having a diameter of 0.495 in. and a gage length of 2.00 in. was tested to fracture. 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 final diameter of the specimen at the location of the 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 σ E = = = 30,000 ksi ε 0.002 in./in.

Ans.


σ PL = 60 ksi

Ans.


σ ult = 159 ksi

Ans.


σ Y = 80 ksi

Ans.


σ fracture = 135 ksi

Ans.


A0

=

A f

=

π

D2

=

D2

=

π

(0.495 in.)2

= 0.192442 in.2

(0.350 in.)2

= 0.096211 in.2


π

π

4 4 The true fracture stress is therefore 0.192442 in.2 true σ fracture = (135 ksi) 0.096211 in.2

= 270 ksi

Ans.


3.18 A bronze alloy specimen having a diameter of 12.8 mm and a gage length of 50 mm was tested to fracture. Stress and strain data obtained during the test are 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 final diameter of the specimen at the location of 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 σ E = = = 105,000 MPa ε 0.002 in./in.

Ans.


σ PL = 210 MPa

Ans.


σ ult = 380 MPa

Ans.


σ Y = 290 MPa

Ans.


σ fracture = 320 MPa

Ans.


A0

=

A f

=

π

D2

=

D2

=

π

(12.8 mm)2

= 128.679635 mm2

(10.5 mm)2

= 86.590148 mm2


π

π

4 4 The true fracture stress is therefore 128.679635 mm2 true σ fracture = (320 MPa) 86.590148 mm2

= 476 MPa

Ans.


3.19 An alloy specimen having a diameter of 12.8 mm and a gage length of 50 mm was tested to Load fracture. Load and deformation data obtained (kN) during the test are given. Determine: 0 (a) the modulus of elasticity. 7.6 (b) the proportional limit. 14.9 (c) the ultimate strength. 22.2 (d) the yield strength (0.05% offset). 28.5 (e) the yield strength (0.20% offset). 29.9 (f) the fracture stress. (g) the true fracture stress if the final diameter of 30.6 the specimen at the location of the fracture was 32.0 33.0 11.3 mm. 33.3 36.8 41.0

Change in Length

Load

Change in Length

(mm)

(kN)

(mm)

0 0.02 0.04 0.06 0.08 0.10 0.12 0.16 0.20 0.24 0.50 1.00

43.8 45.8 48.3 49.7 50.4 50.7 50.4 50.0 49.7 47.9 45.1

1.50 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 fracture

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


(a) The modulus of elasticity is 224.976 MPa σ E = = = 140,600 MPa ε 0.0016 mm/mm

Ans.


σ PL = 234 MPa

Ans.


σ ult = 400 MPa

Ans.

(d) The yield strength by the 0.05% offset method is

σ Y = 239 MPa

Ans.

(e) The yield strength by the 0.2% offset method is

σ Y = 259 MPa

Ans.

(f) The fracture stress is


Ans.


A0

=

A f

=

π

D2

=

D2

=

π

(12.8 mm)2

= 128.679635 mm2

(11.3 mm)2

= 100.287492 mm2


π

π


= 457 MPa

Ans.


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 2,690 5,670 8,360 11,050 12,540 13,150 13,140 12,530 12,540 12,840 12,840

0 0.0009 0.0018 0.0028 0.0037 0.0042 0.0046 0.0060 0.0079 0.0098 0.0121 0.0139

12,540 12,540 14,930 17,020 18,220 18,820 19,110 19,110 18,520 17,620 16,730 16,130 15,900

0.0209 0.0255 0.0487 0.0835 0.1252 0.1809 0.2551 0.2968 0.3107 0.3246 0.3339 0.3385 fracture



(a) The modulus of elasticity is 63.849 ksi σ E = = = 30,400 ksi ε 0.0021 in./in.

Ans.


σ PL = 63.8 ksi

Ans.


σ ult = 97.3 ksi

Ans.

(d) The yield strength using the 0.05% offset method is

σ Y = 65.4 ksi

Ans.

(e) The yield strength using the 0.2% offset method is

σ Y = 63.8 ksi

Ans.


σ fracture = 82.1 ksi

Ans.

(g) The original cross-sectional area of the specimen is

A0

=

A f

=

π

D2

=

D2

=

π

(0.500 in.)2

= 0.196350 in.2

(0.387 in.)2

= 0.117628 in.2


π

π

4 4 The true fracture stress is therefore 0.196350 in.2 true σ fracture = (82.1 ksi) 0.117628 in.2

= 137.0 ksi

Ans.


3.21 A 2024-T4 aluminum test specimen with a diameter of 0.505 in. and a 2.0-in. gage length was Load tested to fracture. Load and deformation data (lb) obtained during the test are given. Determine: 0 (a) the modulus of elasticity. 1,300 (b) the proportional limit. 2,390 (c) the ultimate strength. 3,470 (d) the yield strength (0.05% offset). 4,560 (e) the yield strength (0.20% offset). 5,640 (f) the fracture stress. 6,720 (g) the true fracture stress if the final diameter of 7,380 the specimen at the location of the fracture was 8,240 0.452 in. 8,890 9,330 9,980 10,200 10,630

Change in Length (in.) 0.0000 0.0014 0.0023 0.0032 0.0042 0.0051 0.0060 0.0070 0.0079 0.0088 0.0097 0.0107 0.0116 0.0125

Load (lb) 11,060 11,500 12,360 12,580 12,800 13,020 13,230 13,450 13,670 13,880 14,100 14,100 14,100 14,100 14,100

Change in Length (in.) 0.0139 0.0162 0.0278 0.0394 0.0603 0.0788 0.0974 0.1159 0.1391 0.1623 0.1994 0.2551 0.3200 0.3246 fracture



(a) The modulus of elasticity is 33.55 ksi σ E = = = 11,180 ksi ε 0.003 in./in.

Ans.


σ PL = 33.6 ksi

Ans.


σ ult = 70.4 ksi

Ans.

(d) The yield strength using the 0.05% offset method is

σ Y = 44.4 ksi

Ans.

(e) The yield strength using the 0.2% offset method is

σ Y = 54.5 ksi

Ans.


σ fracture = 70.4 ksi

Ans.

(g) The original cross-sectional area of the specimen is

A0

=

A f

=

π

D2

=

D2

=

π

(0.505 in.)2

= 0.200296 in.2

(0.452 in.)2

= 0.160460 in.2


π

π

4 4 The true fracture stress is therefore 0.200296 in.2 true σ fracture = (70.4 ksi) 0.160460 in.2

= 87.9 ksi

Ans.


3.22 A 1045 hot-rolled steel tension test specimen has a diameter of 6.00 mm and a gage length of 25 Load mm. In a test to fracture, the stress and strain data (kN) below were obtained. Determine: 0.00 (a) the modulus of elasticity. 2.94 (b) the proportional limit. 5.58 (c) the ultimate strength. 8.52 (d) the yield strength (0.05% offset). 11.16 (e) the yield strength (0.20% offset). 12.63 (f) the fracture stress. (g) the true fracture stress if the final diameter of 13.02 the specimen at the location of the fracture was 13.16 13.22 4.65 mm. 13.22 13.25 13.22

Change in Length

Load

Change in Length

(mm)

(kN)

(mm)

0.00 0.01 0.02 0.03 0.05 0.05 0.06 0.08 0.08 0.10 0.14 0.17

13.22 16.15 18.50 20.27 20.56 20.67 20.72 20.61 20.27 19.97 19.68 19.09 18.72

0.29 0.61 1.04 1.80 2.26 2.78 3.36 3.83 3.94 4.00 4.06 4.12 fracture



(a) The modulus of elasticity is 394.765 MPa σ E = = = 247,000 MPa ε 0.0016 mm/mm

Ans.


σ PL = 400 MPa

Ans.


σ ult = 732 MPa

Ans.

(d) The yield strength by the 0.05% offset method is

σ Y = 465 MPa

Ans.

(e) The yield strength by the 0.2% offset method is

σ Y = 465 MPa

Ans.



Ans.


A0

=

A f

=

π

D2

=

D2

=

π

(6 mm)2

= 28.274334 mm2


π

π

(4.65 mm)2

= 16.982272 mm2


= 1,124 MPa

Ans.


3.23 Rigid bar BCD in Fig. P3.23 is supported by a pin at C and by aluminum rod (1). A concentrated load P is applied to the lower end of aluminum rod (2), which is attached to the rigid bar at D. The cross2 sectional area of each rod is A = 0.20 in. and the elastic modulus of the aluminum material is E = 10,000 ksi. After the load P is applied at E , the strain in rod (1) is measured as 900 με (tension). (a) Determine the magnitude of load P . (b) Determine the total deflection of point E relative to its initial position.

Fig. P3.23

Solution (a) From the measured strain, the stress in rod (1) is σ 1 = E 1ε 1 = (10,000 ksi)(900 × 10−6 in./in.) = 9 ksi and thus, the force in rod (1) is F1 = σ 1 A1 = (9 ksi)(0.20 in.2 ) = 1.8 kips (T) Consider the equilibrium of the rigid bar, and write a moment equilibrium equation about C to determine the magnitude of load P : ΣM C = (20 in.)(1.8 kips) − (30 in.)P = 0

∴ P = 1.2 kips

Ans.

(b) From the measured strain, the elongation of rod (1) is e1 = ε 1 L1 = (900 × 10−6 in./in.)(50 in.) = 0.0450 in. From similar triangles, the deflection of the rigid bar at D can be expressed in terms of the deflection at B: v B vD 20 in.

=

30 in. 30 in.

∴ v D = vB

20 in.

= (0.045 in.)

30 in. 20 in.

= 0.0675 in.


The elongation of rod (2) due to the 1.2-kip load must be determined. The stress in rod (2) is F 1.2 kips σ 2 = 2 = = 6 ksi A2 0.2 in.2 and consequently, the strain in rod (2) is σ 6 ksi ε 2 = 2 = = 0.000600 in./in. E 2 10,000 ksi

From the strain, the elongation in rod (2) can be computed: e2 = ε 2 L2 = (0.000600 in./in.)(100 in.) = 0.06 in. The deflection of joint E is the sum of the rigid bar deflection at D and the elongation in rod (2): v E

= vD + e2 = 0.0675 in. + 0.06 in. = 0.1275 in. ↓

Ans.


3.24 The rigid bar AC in Fig. P3.24 is supported by two axial bars (1) and (2). Both axial bars are made of bronze [ E = 6 100 GPa; α = 18 × 10− mm/mm/°C]. The cross-sectional area of bar (1) is A1 = 240 2 mm and the cross-sectional area of bar (2) 2 is A2 = 360 mm . After load P has been applied and the temperature of the entire assembly has increased by 20°C, the total strain 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: ε T = α ΔT = (18 ×10−6 mm/mm/°C)(20°C) = 0.000360 mm/mm The strain caused by the axial force in the bar is thus: ε 2,σ = ε 2 − ε 2,T = 0.000800 mm/mm − 0.000360 mm/mm = 0.000440 mm/mm The stress in bar (2) is σ 2 = E 2ε 2,σ = (100, 000 MPa)(0.000440 mm/mm) = 44 MPa and the force in bar (2) is F2 = σ 2 A2 = (44 N/mm2 )(360 mm2 ) = 15,840 N Next, consider a FBD of the rigid bar AC . Equilibrium equations for this FBD are: ΣF y = F1 + F2 − P = 0

ΣM A = (1, 400 mm)F2 − (500 mm)P = 0 which can be solve simultaneously to give: 500 mm F2 = P = 0.357143P 1,400 mm and F1 = 0.642857 P The applied load P can be expressed in terms of F 2 as 1 P= F2 = 2.8F2 0.357143 Excerpts from this work may be reproduced by instructors for distribution on a not-for-pro fit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this w ork beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

and so the magnitude of load P is P = 2.8F 2

= 2.8(15,840 N) = 44,352 N = 44.4 kN

Ans.

(b) The force in bar (1) is F1 = 0.642857 P = (0.642857)(44,352 N) = 28,512 N Thus, the stress in bar (1) is F 28,512 N = 118.800 MPa σ 1 = 1 = A1 240 mm2 The normal strain due to the axial force in bar (1) is σ 118.800 MPa ε 1,σ = 1 = = 0.001188 mm/mm E 1 100,000 MPa The normal strain caused by the 20°C temperature increase is: ε1,T = α ΔT = (18 × 10−6 mm/mm/°C)(20°C) = 0.000360 mm/mm Therefore, the total strain in bar (1) is ε1 = ε1,σ + ε 1,T = 0.001188 mm/mm + 0.000360 mm/mm = 0.001548 mm/mm and the elongation in bar (1) is e1 = ε 1,σ L1 = (0.001548 mm/mm)(1,300 mm) = 2.012400 mm 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,

v A

= e1 = 2.01 mm ↓

Ans.


3.25 The rigid bar in Fig. P3.25 is supported by axial bar (1) and by a pin connection at C . Axial bar (1) has a cross-sectional area of 2 A1 = 275 mm , an elastic modulus of E = 200 GPa, and a coefficient of thermal −6 expansion of α = 11.9 × 10 mm/mm/°C. The pin at C has a diameter of 25 mm. After load P has been applied and the temperature of the entire assembly has been increased by 20°C, the total strain in bar (1) is measured as 675 με (elongation). Determine: (a) the magnitude of load P . (b) the shear stress in pin C .

Fig. P3.25

Solution The total strain in bar (1) consists of thermal strain as well a s normal strain caused by normal stress:

ε = ε σ + ε T The normal strain due to the increase in temperature is: ε T = α ΔT = (11.9 × 10−6 mm/mm/°C)(20°C) = 0.000238 mm/mm Therefore, the normal stress in bar (1) causes a normal strain of: ε σ = ε − ε T = 0.000675 mm/mm − 0.000238 mm/mm = 0.000437 mm/mm From Hooke’s law, the normal stress in bar (1) ca n be calculated as: σ 1 = E ε σ = (200,000 MPa)(0.000437 mm/mm) = 87.4 MPa and thus the axial force in bar (1) must be: F1 = σ 1 A1 = (87.4 N/mm2 )(275 mm2 ) = 24,035 N Next, consider a free-body diagram of the rigid bar. Write a moment equilibrium equation about pin C : ΣM C = (200 mm)F1 − (380 mm)P

= (200 mm)(24, 035 N) − (380 mm) P = 0 ∴ P = 12, 650 N = 12.65 kN

Ans.

Now that P is known, the horizontal and vertical reactions at C can be calculated: 035 N ΣFx = Cx − F1 = 0 ∴ Cx = F1 = 24,

ΣFy = C y − P = 0

∴ C y = P = 12, 650 N

The resultant force acting on pin C is:

C

= C x2 + C y2 = (24, 035 N)2 + (12, 650 N)2 = 27,160.702 N


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 atC (in other words, the cross-sectional area of the pin) is:

A pin

=

π

(25 mm)2

= 490.874 mm2

4 and thus the shear stress in pin C is: V 13,580.351 N τ C = = = 27.666 N/mm2 2 AV 490.874 mm

= 27.7 MPa

Ans.


3.26 The rigid bar in Fig. P3.26 is supported by axial bar (1) and by a pin connection at C . Axial bar (1) has a cross2 sectional area of A1 = 275 mm , an elastic modulus of E = 200 GPa, and a coefficient −6 of thermal expansion of α = 11.9 × 10 mm/mm/°C. The pin at C has a diameter of 25 mm. After load P has been applied and the temperature of the entire assembly has been decreased by 30°C, the total strain 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 a s normal strain caused by normal stress:

ε = ε σ + ε T The normal strain due to the decrease in temperature is: ε T = α ΔT = (11.9 × 10−6 mm/mm/°C)( − 30°C) = −0.000357 mm/mm Therefore, the normal stress in bar (1) causes a normal strain of: ε σ = ε − ε T = 0.000675 mm/mm − (− 0.000357 mm/mm) = 0.001032 mm/mm From Hooke’s law, the normal stress in bar (1) ca n be calculated as: σ 1 = E ε σ = (200,000 MPa)(0.001032 mm/mm) = 206.4 MPa and thus the axial force in bar (1) must be: F1 = σ 1 A1 = (206.4 N/mm2 )(275 mm2 ) = 56,760 N Next, consider a free-body diagram of the rigid bar. Write a moment equilibrium equation about pin C : ΣM C = (200 mm)F1 − (380 mm)P

= (200 mm)(56, 760 N) − (380 mm) P = 0 ∴ P = 29, 874 N = 29.9 kN

Ans.

Now that P is known, the horizontal and vertical reactions at C can be calculated: 760 N ΣFx = Cx − F1 = 0 ∴ Cx = F1 = 56,

ΣFy = C y − P = 0

N ∴ C y = P = 29,874

The resultant force acting on pin C is:

C

= C x2 + C y2 = (56, 760 N)2 + (29, 874 N)2 = 64,142 N


25741866-3-1-at-the-Proportional-Limit-A-12-Inch

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