3–1. a concrete cylinder having a diameter of 6.00 …...•8–1. a concrete cylinder having a...

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

50792

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Stress and Strain:

0 0

0.177 0.00005

0.336 0.00010

0.584 0.000167

0.725 0.000217

0.902 0.000283

1.061 0.000333

1.220 0.000375

1.362 0.000417

1.645 0.000517

1.768 0.000583

1.874 0.000625

Modulus of Elasticity: From the stress–strain diagram

Ans.Eapprox =1.31 - 0

0.0004 - 0= 3.275 A103 B ksi

e =dL

L(in./in.)s =

P

A(ksi)

•3–1. A concrete cylinder having a diameter of 6.00 in. andgauge length of 12 in. is tested in compression.The results ofthe test are reported in the table as load versus contraction.Draw the stress–strain diagram using scales of and From the diagram, determineapproximately the modulus of elasticity.

1 in. = 0.2110-32 in.>in.1 in. = 0.5 ksi

05.09.516.520.525.530.034.538.546.550.053.0

00.00060.00120.00200.00260.00340.00400.00450.00500.00620.00700.0075

Load (kip) Contraction (in.)

03 Solutions 46060 5/24/10 11:57 AM Page 92

•8–1. A concrete cylinder having a diameter of 150 mm and gauge length of 300 mm is tested in compression. The results of the test are reported in the table as load versus contraction. Draw the stress-strain diagram using scales of 10 mm = 2 MPa and 10 mm = 0.1(10–3) mm/mm. From the diagram, determine approximately the modulus of elasticity.

Stress and Strain:

s 5 P

A (MPa) e 5 dL

L (mm/mm)

0 0

1.41 0.00005

2.69 0.00010

4.67 0.000167

5.80 0.000217

7.22 0.000283

8.49 0.000333

9.76 0.000375

10.89 0.000417

13.16 0.000517

14.15 0.000583

15.00 0.000625

Eapprox 5 8.0 – 0

0.0003 – 0 = 26.67(103) MPa = 26.67 GPa

Load (kN) Contraction (mm)

025.047.582.5102.5127.5150.0172.5192.5232.5250.0265.0

00.01500.03000.05000.06500.08500.10000.11250.12500.17500.1850

s (MPa)

16

12

8

4

(mm/mm)

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Modulus of Toughness: The modulus of toughness is equal to the area under thestress–strain diagram (shown shaded).

Ans. = 85.0 in # lb

in3

+12

(12.3) A103 B ¢ lbin2 ≤(0.0004)¢ in.

in.≤

+12

(7.90) A103 B ¢ lbin2 ≤(0.0012)¢ in.

in.≤

+ 45.5 A103 B ¢ lbin2 ≤(0.0012)¢ in.

in.≤

(ut)approx =12

(33.2) A103 B ¢ lbin2 ≤(0.0004 + 0.0010)¢ in.

in.≤

3–3. Data taken from a stress–strain test for a ceramic aregiven in the table. The curve is linear between the originand the first point. Plot the diagram, and determineapproximately the modulus of toughness.The rupture stressis sr = 53.4 ksi.

033.245.549.451.553.4

00.00060.00100.00140.00180.0022

S (ksi) P (in./in.)


Ans.

Modulus of Resilience: The modulus of resilience is equal to the area under thelinear portion of the stress–strain diagram (shown shaded).

Ans.ut =12

(33.2) A103 B ¢ lbin2 ≤ ¢0.0006

in.in.≤ = 9.96

in # lbin3

E =33.2 - 0

0.0006 - 0= 55.3 A103 B ksi

3–2. Data taken from a stress–strain test for a ceramic aregiven in the table.The curve is linear between the origin andthe first point. Plot the diagram, and determine the modulusof elasticity and the modulus of resilience.

033.245.549.451.553.4

00.00060.00100.00140.00180.0022

S (ksi) P (in./in.)


(MPa) e (mm/mm)

0232.4318.5345.8360.5373.8

00.00060.00100.00140.00180.0022

E 5 232.4 – 00.0006 – 0

5 387.3(103) MPa = 387.3 GPa

ut = 12

(232.4)a Nmm2b a0.0006

mmmmb = 0.0697 N · mm/mm3 = 0.0697 MJ/m3

420

350

280

210

140

70

232.4

e (mm/mm)

(MPa) e (mm/mm)

0232.4318.5345.8360.5373.8

00.00060.00100.00140.00180.0022

373.8 MPa

(ut)approx = 12

(232.4)a Nmm2b(0.0004 + 0.0010)amm

mmb

+ 318.5 a Nmm2b(0.0012)amm

mmb

+ 12

(55.3)a Nmm2b(0.0012)amm

mmb

+ 12

(86.1)a Nmm2b(0.0004)amm

mmb

= 0.595 N · mm/mm3

= 0.595 MJ/m3

420

350

280

210

140

70

373.8

e (mm/mm)

318.5

232.4

s (MPa)

8–2.

8–3.

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Stress and Strain:

0 0

90.45 0.00035

259.9 0.00120

308.0 0.00204

333.3 0.00330

355.3 0.00498

435.1 0.02032

507.7 0.06096

525.6 0.12700

507.7 0.17780

479.1 0.23876


Ans.

Ultimate and Fracture Stress: From the stress–strain diagram

Ans.

Ans.(sf)approx = 479 MPa

(sm)approx = 528 MPa

(E)approx =228.75(106) - 0

0.001 - 0= 229 GPa

e =dL

L(mm/mm)s =

P

A(MPa)

*3–4. A tension test was performed on a specimen havingan original diameter of 12.5 mm and a gauge length of50 mm. The data are listed in the table. Plot the stress–straindiagram, and determine approximately the modulus ofelasticity, the ultimate stress, and the fracture stress. Use ascale of and Redraw the linear-elastic region, using the same stress scalebut a strain scale of 20 mm = 0.001 mm>mm.

20 mm = 0.05 mm>mm.20 mm = 50 MPa

0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8

0 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.890011.9380

Load (kN) Elongation (mm)


*8–4.

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51095


Stress and Strain:

0 0

90.45 0.00035

259.9 0.00120

308.0 0.00204

333.3 0.00330

355.3 0.00498

435.1 0.02032

507.7 0.06096

525.6 0.12700

507.7 0.17780

479.1 0.23876

Modulus of Toughness: The modulus of toughness is equal to thetotal area under the stress–strain diagram and can beapproximated by counting the number of squares. The totalnumber of squares is 187.

Ans.(ut)approx = 187(25) A106 B ¢ Nm2 ≤ a0.025

mmb = 117 MJ>m3

e =dL

L(mm/mm)s =

P

A(MPa)

3–5. A tension test was performed on a steel specimenhaving an original diameter of 12.5 mm and gauge lengthof 50 mm. Using the data listed in the table, plot thestress–strain diagram, and determine approximately themodulus of toughness. Use a scale of and20 mm = 0.05 mm>mm.

20 mm = 50 MPa

0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8

0 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.890011.9380

Load (kN) Elongation (mm)


•8–5.

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Normal Stress and Strain: Applying and .

Modulus of Elasticity:

Ans.E =¢s¢e

=9.167 - 2.546

0.000750= 8.83 A103 B ksi

¢e =0.009

12= 0.000750 in.>in.

s2 =1.80p4 (0.52)

= 9.167 ksi

s1 =0.500p4 (0.52)

= 2.546 ksi

e =dL

Ls =

P

A

3–6. A specimen is originally 1 ft long, has a diameter of0.5 in., and is subjected to a force of 500 lb. When the forceis increased from 500 lb to 1800 lb, the specimen elongates0.009 in. Determine the modulus of elasticity for thematerial if it remains linear elastic.

Allowable Normal Stress:

Ans.

Stress–Strain Relationship: Applying Hooke’s law with

Normal Force: Applying equation .

Ans.P = sA = 7.778 (0.2087) = 1.62 kip

s =P

A

s = Ee = 14 A103 B (0.000555) = 7.778 ksi

e =d

L=

0.023 (12)

= 0.000555 in.>in.

A = 0.2087 in2 = 0.209 in2

19.17 =4A

sallow =P

A

sallow = 19.17 ksi

3 =57.5sallow

F.S. =sy

sallow

3–7. A structural member in a nuclear reactor is made of azirconium alloy. If an axial load of 4 kip is to be supportedby the member, determine its required cross-sectional area.Use a factor of safety of 3 relative to yielding. What is theload on the member if it is 3 ft long and its elongation is0.02 in.? ksi, ksi. The material haselastic behavior.

sY = 57.5Ezr = 14(103)


8–6. A specimen is originally 300 mm long, has a diameter of 12 mm, and is subjected to a force of 2.5 kN. When the force is increased from 2.5 kN to 9 kN,, the specimen elongates 0.225 mm. Determine the modulus of elasticity for the material if it remains linear elastic.

s1 5 2.5(103)p4(12)2

5 22.10 MPa

s2 5 9(103)p4(12 2)

5 79.58 MPa

∆e 5 0.225300

5 0.000750 mm/mm

79.58 – 22.100.000750

5 76.64(103) MPa = 76.64 GPa

zirconium alloy. If an axial load of 20 kN is to be supported

0.5 mm? Ezr = 100 GPa, sY = 400 MPa. The material has

400

133.33 MPa

133.33 5 20(103)

A

A = 150 mm2

0.5

1(103) = 0.0005 mm/mm

s = Ee = 100(103)(0.0005) = 50 MPa

P = sA = 50(150) = 7500 N = 7.5 kN

load on the member if it is 1 m long and its elongation is

8–7.

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51297


Here, we are only interested in determining the force in wire AB.

a

The normal stress the wire is

Since , Hooke’s Law can be applied to determine the strainin wire.

The unstretched length of the wire is . Thus, the wirestretches

Ans. = 0.0821 in.

dAB = PAB LAB = 0.6586(10-3)(124.71)

LAB =9(12)

sin 60°= 124.71 in

PAB = 0.6586(10-3) in>in sAB = EPAB; 19.10 = 29.0(103)PAB

sAB 6 sy = 36 ksi

sAB =FABAAB

=600

p4 (0.22)

= 19.10(103) psi = 19.10 ksi

+©MC = 0; FAB cos 60°(9) -12

(200)(9)(3) = 0 FAB = 600 lb

*3–8. The strut is supported by a pin at C and an A-36steel guy wire AB. If the wire has a diameter of 0.2 in.,determine how much it stretches when the distributed loadacts on the strut.

9 ft

200 lb/ft

C

A

B

60�


steel guy wire AB. If the wire has a diameter of 5 mm,

2.7 m

3.4 kN/m

C

A

B

60

FAB cos 60°(2.7) – 12

(3.4)(2.7)(0.9) = 0 FAB = 3.06 kN

3.06(103)p4(52)

= 155.84 MPa

250 MPa, Hooke’s Law can be applied to determine the strain

155.84 = 200(103)eAB

eAB = 0.7792(10–3) mm/mm

2.7(103)sin 60°

= 3117.69. Thus, the wire

0.7792(10–3)(3117.69)

2.429 mm

1

2 (3.4)(2.7) kN

0.9 m 1.8 m

*8–8.

SM_CH08.indd 512 4/11/11 9:54:22 AM


513

From the graph

Ans.d = eL = 0.035(6.5) = 0.228 in.

e = 0.035 in.>in.

s =P

A=

343.750.229

= 1.50 ksi

•3–9. The diagram for a collagen fiber bundle fromwhich a human tendon is composed is shown. If a segmentof the Achilles tendon at A has a length of 6.5 in. and anapproximate cross-sectional area of determine itselongation if the foot supports a load of 125 lb, which causesa tension in the tendon of 343.75 lb.

0.229 in2,

s–P

98


125 lb

s (ksi)

P (in./in.)0.05 0.10

4.50

3.75

3.00

2.25

1.50

0.75

A

From the stress–strain diagram, Fig. a,

Ans.

Thus,

Ans.

Ans. Pu>t = su>t A = 100 Cp4 (0.52) D = 19.63 kip = 19.6 kip

PY = sYA = 60 Cp4 (0.52) D = 11.78 kip = 11.8 kip

sy = 60 ksi su>t = 100 ksi

E

1=

60 ksi - 00.002 - 0

; E = 30.0(103) ksi

3–10. The stress–strain diagram for a metal alloy having anoriginal diameter of 0.5 in. and a gauge length of 2 in. is givenin the figure. Determine approximately the modulus ofelasticity for the material, the load on the specimen that causesyielding, and the ultimate load the specimen will support.

0

105

90

75

60

45

30

15

000 0.350.05 0.10 0.15 0.20 0.25 0.30

0.0070.001 0.002 0.003 0.004 0.005 0.006

P (in./in.)

s (ksi)


of the Achilles tendon at A has a length of 165 mm and an approximate cross-sectional area of 145 mm2, determine its elongation if the foot supports a load of 625 N, which causes a tension in the tendon of 1718.75 N.

625 N

s (MPa)

P (mm/mm)0.05 0.10

31.50

26.25

21.00

15.75

10.50

5.25

A

s 5 PA

= 1718.75145

5 11.85 MPa

e = 0.035 mm/mm

0.035(165) = 5.775 mm

E

1 = 290 MPa – 0

0.001 – 0; E = 290 GPa

sy = 290 MPa su/t = 550 GPa

PY = sYA = 290[p4(122)] = 32.80(103) N = 32.80 kN

Pu/t = su/tA = 500[p4(122)] = 62.20(103) N = 62.20 kN

8–10 The stress-strain diagram for a metal alloy having an original diameter of 12 mm and a gauge length of 50 mm is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support.

100

200

300

400

500

A

E

B0 (mm/m)

0.05/0.001

0 0.08/0.002

0.15/0.003

0.20/0.004

0.25/0.005

0.30/0.006

0.35/0.007

(a)Elastic Recovery

p

1

= 290y

(MPa)

•8–9.

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51499


From the stress–strain diagram Fig. a, the modulus of elasticity for the steel alloy is

when the specimen is unloaded, its normal strain recovered along line AB, Fig. a,which has a gradient of E. Thus

Ans.

Thus, the permanent set is

Then, the increase in gauge length is

Ans.¢L = PPL = 0.047(2) = 0.094 in

PP = 0.05 - 0.003 = 0.047 in>in

Elastic Recovery =90E

=90 ksi

30.0(103) ksi= 0.003 in>in

E

1=

60 ksi - 00.002 - 0

; E = 30.0(103) ksi

3–11. The stress–strain diagram for a steel alloy having anoriginal diameter of 0.5 in. and a gauge length of 2 in. isgiven in the figure. If the specimen is loaded until it isstressed to 90 ksi, determine the approximate amount ofelastic recovery and the increase in the gauge length after itis unloaded.

0

105

90

75

60

45

30

15

000 0.350.05 0.10 0.15 0.20 0.25 0.30

0.0070.001 0.002 0.003 0.004 0.005 0.006

P (in./in.)

s (ksi)


original diameter of 12 mm and a gauge length of 50 mm is given in the figure. If the specimen is loaded until it is stressed to 500 MPa, determine the approximate amount of

E

1 = 290 MPa – 0

0.001 – 0; E = 290 GPa

Elastic Recovery 500E

= 500 MPa290(103) MPa

= 0.001724 mm/mm

Amount of Elastic Recovery = 0.001724(50 mm) = 0.0862 mm Ans.

Thus, the permanent set is

eP = 0.08 – 0.001724 = 0.078276 mm/mm

Then, the increase in gauge length is

∆L = ePL = 0.078276(50 mm) = 3.91379 mm Ans.

100

200

300

400

500

A

E

B0 (mm/mm)

0.05/0.001

0 0.08/0.002

0.15/0.003

0.20/0.004

0.25/0.005

0.30/0.006

0.35/0.007

(a)Elastic Recovery

p

1

(MPa)

8–11.

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515100


The Modulus of resilience is equal to the area under the stress–strain diagram up tothe proportional limit.

Thus,

Ans.

The modulus of toughness is equal to the area under the entire stress–straindiagram. This area can be approximated by counting the number of squares. Thetotal number is 38. Thus,

Ans.C(ui)t Dapprox = 38 c15(103) lbin2 d a0.05

ininb = 28.5(103)

in # lbin3

(ui)r =12

sPLPPL =12

C60(103) D(0.002) = 60.0 in # lb

in3

sPL = 60 ksi PPL = 0.002 in>in.

*3–12. The stress–strain diagram for a steel alloy having anoriginal diameter of 0.5 in. and a gauge length of 2 in.is given in the figure. Determine approximately the modulusof resilience and the modulus of toughness for the material.

0

105

90

75

60

45

30

15

000 0.350.05 0.10 0.15 0.20 0.25 0.30

0.0070.001 0.002 0.003 0.004 0.005 0.006

P (in./in.)

s (ksi)


original diameter of 12 mm and a gauge length of 50 mm

sPL = 290 MPa ePL = 0.001 mm/mm

[(290)](0.001) = 0.145 MPa

33. Thus.

[(ui)t]approx = 33[100 MPa]a0.04 mmmmb = 132 MPa

100

200

300

400

500

A

E

B0 (mm/m)

0.05/0.001

0 0.08/0.002

0.15/0.003

0.20/0.004

0.25/0.005

0.30/0.006

0.35/0.007

(a)

1

= 290PL

(MPa)

*8–12.

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Normal Stress and Strain:

Modulus of Elasticity:

Ans.E =s

e=

11.430.000400

= 28.6(103) ksi

e =dL

L=

0.0025

= 0.000400 in.>in.

s =P

A=

8.000.7

= 11.43 ksi

•3–13. A bar having a length of 5 in. and cross-sectionalarea of 0.7 is subjected to an axial force of 8000 lb. If thebar stretches 0.002 in., determine the modulus of elasticityof the material. The material has linear-elastic behavior.

in28000 lb8000 lb

5 in.

Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a

a

The normal stress developed in the wire is

Since , Hooke’s Law can be applied to determine the strain inthe wire.

The unstretched length of the wire is . Thus, thewire stretches

Ans. = 0.0632 in

dBD = PBD LBD = 1.054(10-3)(60)

LBD = 232 + 42 = 5ft = 60 in

PBD = 1.054(10-3) in.>in.

sBD = EPBD; 30.56 = 29.0(103)PBD

sBD 6 sy = 36 ksi

sBD =FBDABD

=1500

p4 (0.252)

= 30.56(103) psi = 30.56 ksi

+©MA = 0; FBD A45 B(3) - 600(6) = 0 FBD = 1500 lb

3–14. The rigid pipe is supported by a pin at A and anA-36 steel guy wire BD. If the wire has a diameter of0.25 in., determine how much it stretches when a load of

acts on the pipe.P = 600 lb

3 ft 3 ft

CDA

B

P4 ft

03 Solutions 46060 5/24/10 11:57 AM 1

•8–13. A bar having a length of 125 mm and cross-sectional area of 437.5 mm2 is subjected to an axial force of 40 kN. If the bar stretches 0.05 mm, determine the modulus of elasticity of the material. The material has linear-elastic behavior.

40 kN40 kN125 mm

e 5 dL

L = 0.05

125 5 0.000400 mm/mm

s 5 PA

= 40(103)437.5

5 95.81 MPa

E 5 se

= 95.810.000400

5 239.525 (103) MPa = 239.5 GPa

6 mm, determine how much it stretches when a load of P = 3 kN acts on the pipe.

0.9 m 0.9 m

CDA

B

P1.2 m

(0.9) – 3(1.8) = 0 FBD = 7.5 kN

7.5(103)p4(62)

5 265.3 MPa

250 MPa, Hooke’s Law can be applied to determine the strain in

265.3 = 200(103)eAB

eBD = 1.3265(10–3) mm/mm

0 9 1 22 2. .+ = 1.5 m. Thus, the

1.3265(10–3)(1.5)(103)

1.98975 mm

0.9 m 0.9 m

3 kN

8–14.

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517102


Here, we are only interested in determining the force in wire BD. Referring to theFBD in Fig. a

a

The unstretched length for wire BD is . From thegeometry shown in Fig. b, the stretched length of wire BD is

Thus, the normal strain is

Then, the normal stress can be obtain by applying Hooke’s Law.

Since , the result is valid.

Ans. P = 569.57 lb = 570 lb

sBD =FBDABD

; 29.01(103) =2.50 Pp4 (0.252)

sBD 6 sy = 36 ksi

sBD = EPBD = 29(103) C1.0003(10-3) D = 29.01 ksi

PBD =LBD¿ - LBDLBD

=60.060017 - 60

60= 1.0003(10-3) in.>in.

LBD¿ = 2602 + 0.0752 - 2(60)(0.075) cos 143.13° = 60.060017

LBD = 232 + 42 = 5 ft = 60 in

+©MA = 0; FBD A45 B(3) - P(6) = 0 FBD = 2.50 P

3–15. The rigid pipe is supported by a pin at A and an A-36 guy wire BD. If the wire has a diameter of 0.25 in.,determine the load P if the end C is displaced 0.075 in.downward.

3 ft 3 ft

CDA

B

P4 ft


A-36 guy wire BD. If the wire has a diameter of 6 mm, determine the load P if the end C is displaced 1.875 mm downward.

0.9 m 0.9 m

CDA

B

P1.2 m

(0.9) – P(1.8) = 0

0 9 1 22 2. .+ = 1.5 m. From the

1500 1 875 2 1500 1 8752 2+ −. ( )( . ) cos 143.13° = 1501.500 mm

1501.500 – 15001500

5 1.0000(10–3) mm/mm

200(103)[1.0000(10–3)] = 200 MPa

250 MPa, the result is valid.

200 5 2.50Pp4(62)

P = 2261.9 N = 2.26 kN

0.9 m 0.9 m

LBD = 1.5 m

1.875 mm

8–15.

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518103


Normal Stress and Strain: The cross-sectional area of the hollow bar is. When ,

From the stress–strain diagram shown in Fig. a, the slope of the straight line OAwhich represents the modulus of elasticity of the metal alloy is

Since , Hooke’s Law can be applied. Thus

Thus, the elongation of the bar is

Ans.

When ,

From the geometry of the stress–strain diagram, Fig. a,

When is removed, the strain recovers linearly along line BC, Fig. a,parallel to OA. Thus, the elastic recovery of strain is given by

The permanent set is

Thus, the permanent elongation of the bar is

Ans.dP = ePL = 0.0285(600) = 17.1 mm

eP = e2 - er = 0.0305 - 0.002 = 0.0285 mm>mm

er = 0.002 mm>mm

s2 = Eer; 400(106) = 200(109)er

P = 360 kN

e2 - 0.00125

400 - 250=

0.05 - 0.00125500 - 250

e2 = 0.0305 mm>mm

s2 =P

A=

360(103)

0.9(10-3)= 400 MPa

P = 360 kN

d1 = e1L = 0.5556(10-3)(600) = 0.333 mm

e1 = 0.5556(10-3) mm>mm

s1 = Ee1; 111.11(106) = 200(109)e1

s1 6 250 MPa

E =250(106) - 0

0.00125 - 0= 200 GPa

s1 =P

A=

100(103)

0.9(10-3)= 111.11 MPa

P = 100 kNA = 0.052 - 0.042 = 0.9(10-3)m2

*3–16. Determine the elongation of the square hollow barwhen it is subjected to the axial force If thisaxial force is increased to and released, findthe permanent elongation of the bar. The bar is made of ametal alloy having a stress–strain diagram which can beapproximated as shown.

P = 360 kNP = 100 kN.

P

P

600 mm

50 mm250

0.00125 0.05P (mm/mm)

500

50 mm 5 mm

5 mm

s (MPa)


*8–16.

SM_CH08.indd 518 4/11/11 9:54:26 AM


519104


3–16. Continued


8–16.

SM_CH08.indd 519 4/11/11 9:54:26 AM


520105


Proportional Limit and Yield Strength: From the stress–strain diagram, Fig. a,

Ans.

Ans.

Modulus of Elasticity: From the stress–strain diagram, the corresponding strain foris Thus,

Ans.

Modulus of Resilience: The modulus of resilience is equal to the area under the

E =44 - 0

0.004 - 0= 11.0(103) ksi

epl = 0.004 in.>in.sPL = 44 ksi

sY = 60 ksi

spl = 44 ksi

3–17. A tension test was performed on an aluminum2014-T6 alloy specimen. The resulting stress–strain diagramis shown in the figure. Estimate (a) the proportional limit,(b) the modulus of elasticity, and (c) the yield strengthbased on a 0.2% strain offset method.

P (in./in.)0.02 0.04 0.06 0.08 0.100.002 0.004 0.006 0.008 0.010

10

20

30

40

50

60

70

0

s (ksi)


P (mm/mm)0.02 0.04 0.06 0.08 0.100.002 0.004 0.006 0.008 0.010

70

140

210

280

350

420

490

0

s (MPa)

s (MPa)

e (mm/mm)

490

420

350

280

210

140

70

spl = 308 MPa

sY = 420 MPa

sPL = 308 MPa is epl = 0.004 mm/mm. Thus,

E 5 308 – 00.004 – 0

5 77.0(103) MPa = 77.0 GPa

•8–17.

SM_CH08.indd 520 4/11/11 9:54:27 AM


521106


stress–strain diagram up to the proportional limit. From the stress–strain diagram,

Thus,

Ans.

Modulus of Toughness: The modulus of toughness is equal to the area under theentire stress–strain diagram.This area can be approximated by counting the numberof squares. The total number of squares is 65. Thus,

Ans.

The stress–strain diagram for a bone is shown, and can be described by the equation

C AUi B t Dapprox = 65B10(103) lbin2R c0.01

in.in.d = 6.50(103)

in # lbin3

AUi B r =12splepl =

12

(44)(103)(0.004) = 88 in # lb

in3

spl = 44 ksi epl = 0.004 in.>in.

3–18. A tension test was performed on an aluminum2014-T6 alloy specimen. The resulting stress–straindiagram is shown in the figure. Estimate (a) the modulus ofresilience; and (b) modulus of toughness.

P (in./in.)0.02 0.04 0.06 0.08 0.100.002 0.004 0.006 0.008 0.010

10

20

30

40

50

60

70

0

s (ksi)

,

Ans.E =ds

dP 2s= 0

=1

0.45(10- 6)= 2.22 MPa

dP = A0.45(10-6) + 1.08(10-12) s2 Bdse = 0.45(10-6)s + 0.36(10-12)s3

3–19. The stress–strain diagram for a bone is shown, andcan be described by the equation �

where is in kPa. Determine the yieldstrength assuming a 0.3% offset.

s0.36110-122 s3,P = 0.45110-62 s P

P

P � 0.45(10�6)s + 0.36(10�12)s3

P

s


P (mm/mm)0.02 0.04 0.06 0.08 0.100.002 0.004 0.006 0.008 0.010

70

140

210

280

350

420

490

0

s (MPa)

spl = 308 MPa epl = 0.004 mm/mm

(308)(0.004) = 0.616 MJ/m3

65[70 MPa]c 0.01mmmm

d = 45.5 MJ/m3

8–18.

8–19.

SM_CH08.indd 521 4/11/11 9:54:27 AM


522107


When

Solving for the real root:

Ans.

Ans. d = eL = 0.12(200) = 24 mm

= 613 kJ>m3

= 0.12 s - 0.225(10-6)s2 - 0.09(10-12)s4|6873.52

0

ut = L6873.52

0(0.12 - 0.45(10-6)s - 0.36(10-12)s3)ds

ut = LA dA = L6873.52

0(0.12 - e)ds

s = 6873.52 kPa

120(103) = 0.45 s + 0.36(10-6)s3

e = 0.12

*3–20. The stress–strain diagram for a bone is shown andcan be described by the equation �

where is in kPa. Determine the modulusof toughness and the amount of elongation of a 200-mm-long region just before it fractures if failure occurs atP = 0.12 mm>mm.

ss3,0.36110-122 0.45110-62 sP =P

P

P � 0.45(10�6)s + 0.36(10�12)s3

P

s


*8–20.

SM_CH08.indd 522 4/11/11 9:54:27 AM


523

110


Choose,

Ans.

Ans.k = 4.23(10-6)

n = 2.73

ln (0.3310962) = n ln (0.6667)

0.3310962 = (0.6667)n

0.29800 = k(60)n

0.098667 = k(40)n

0.3 =60

30(103)+ k(60)n

0.1 =40

30(103)+ k(40)n

s = 60 ksi, e = 0.3

s = 40 ksi, e = 0.1

*3–24. The stress–strain diagram for many metal alloyscan be described analytically using the Ramberg-Osgoodthree parameter equation where E, k, andn are determined from measurements taken from thediagram. Using the stress–strain diagram shown in thefigure, take and determine the other twoparameters k and n and thereby obtain an analyticalexpression for the curve.

E = 3011032 ksi

P = s>E + ksn,

s (ksi)

P (10–6)0.1 0.2 0.3 0.4 0.5

80

60

40

20

Ans.

Ans.¢d = elatd = -0.0002515 (15) = -0.00377 mm

elat = -Velong = -0.4(0.0006288) = -0.0002515

d = elong L = 0.0006288 (200) = 0.126 mm

elong =s

E=

1.697(106)

2.70(109)= 0.0006288

s =P

A=

300p4(0.015)2 = 1.697 MPa

•3–25. The acrylic plastic rod is 200 mm long and 15 mm indiameter. If an axial load of 300 N is applied to it, determinethe change in its length and the change in its diameter.

np = 0.4.Ep = 2.70 GPa,

300 N

200 mm

300 N


•8–21.

SM_CH08.indd 523 4/11/11 9:54:28 AM


524111


Normal Stress:

Normal Strain: From the stress–strain diagram, the modulus of elasticity

. Applying Hooke’s law

Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’sratio.

Ans.V = -elat

elong= -

-0.56538(10-3)

1.8835(10-3)= 0.300

elat =d - d0

d0=

12.99265 - 1313

= -0.56538 A10-3 B mm>mm

elong =s

E=

376.70(106)

200(104)= 1.8835 A10-3 B mm>mm

E =400(106)

0.002= 200 GPa

s =P

A=

50(103)p4 (0.0132)

= 376.70 Mpa

3–27. The elastic portion of the stress–strain diagram for asteel alloy is shown in the figure. The specimen from whichit was obtained had an original diameter of 13 mm and agauge length of 50 mm. When the applied load on thespecimen is 50 kN, the diameter is 12.99265 mm. DeterminePoisson’s ratio for the material.

a)

Ans.

b)

Ans.d¿ = d + ¢d = 0.5000673 in.

¢d = elat d = 0.00013453 (0.5) = 0.00006727

elat = -0.35 (-0.0003844) = 0.00013453

V =-elat

elong= 0.35

d = elong L = -0.0003844 (1.5) = -0.577 (10-3) in.

elong =s

E=

-4074.37

10.6(106)= -0.0003844

s =P

A=

800p4 (0.5)2 = 4074.37 psi

3–26. The short cylindrical block of 2014-T6 aluminum,having an original diameter of 0.5 in. and a length of 1.5 in.,is placed in the smooth jaws of a vise and squeezed until theaxial load applied is 800 lb. Determine (a) the decrease in itslength and (b) its new diameter.

800 lb 800 lb

400

P(mm/mm)0.002

s(MPa)


having an original diameter of 12 mm and a length of 37.5 mm,

axial load applied is 4 kN. Determine (a) the decrease in its

4 kN 4 kN

s 5 PA

= 4(103)p4(122)

5 35.3678 MPa

elong 5 sE

= –35.367873.1(106)

5 –0.00004838

d 5 elong L 5 –0.0004838(37.5) 5 –0.0181 mm

y

(–0.0004838) = 0.00016933

0.00016933(12) = 0.002032 mm

12.002032 mm

8–22.

8–23.

SM_CH08.indd 524 4/11/11 9:54:29 AM


525112


Normal Stress:

Normal Strain: From the Stress–Strain diagram, the modulus of elasticity

. Applying Hooke’s Law

Thus,

Ans.

Poisson’s Ratio: The lateral and longitudinal can be related using poisson’s ratio.

Ans.d = d0 + dd = 13 + (-0.003918) = 12.99608 mm

dd = elat d = -0.3014 A10-3 B(13) = -0.003918 mm

= -0.3014 A10-3 B mm>mm

elat = -velong = -0.4(0.7534) A10-3 B

L = L0 + dL = 50 + 0.03767 = 50.0377 mm

dL = elong L0 = 0.7534 A10-3 B(50) = 0.03767 mm

elong =s

E=

150.68(106)

200(109)= 0.7534 A10-3 B mm>mm

= 200 GPaE =400(106)

0.002

s =P

A=

20(103)p4 (0.0132)

= 150.68Mpa

*3–28. The elastic portion of the stress–strain diagram fora steel alloy is shown in the figure. The specimen fromwhich it was obtained had an original diameter of 13 mmand a gauge length of 50 mm. If a load of kN isapplied to the specimen, determine its diameter and gaugelength. Take n = 0.4.

P = 20400

P(mm/mm)0.002

s(MPa)

Ans.

Ans.h¿ = 2 + 0.0000880(2) = 2.000176 in.

v =-0.0000880-0.0002667

= 0.330

elat =1.500132 - 1.5

1.5= 0.0000880

elong =s

E=

-2.66710(103)

= -0.0002667

s =P

A=

8(2)(1.5)

= 2.667 ksi

•3–29. The aluminum block has a rectangular crosssection and is subjected to an axial compressive force of8 kip. If the 1.5-in. side changed its length to 1.500132 in.,determine Poisson’s ratio and the new length of the 2-in.side. Eal � 10(103) ksi.

3 in.

1.5 in.

8 kip8 kip 2 in.


40 kN. If the 37.5-mm side changed its length to 37.5033 mm, determine Poisson’s ratio and the new length of the 50 mm side. Eal = 70 GPa.

s 5 PA

= 40(103)

(50)(37.5) 5 21.33 MPa

elong 5 sE

= –21.3370(103)

5 –0.00030476

elat 5 37.5033 – 37.5

37.5 5 0.0000880

v 5 –0.0000880–0.00030476

5 0.289

h9 5 50 + 0.0000880(50) 5 50.0044 mm

75 mm

37.5 mm

40 kN40 kN 50 mm

*8–24.

•8–25.

SM_CH08.indd 525 4/11/11 9:54:34 AM


526113


The shear force developed on the shear planes of the bolt can be determined byconsidering the equilibrium of the FBD shown in Fig. a

From the shear stress–strain diagram, the yield stress is . Thus,

Ans.

From the shear stress–strain diagram, the shear modulus is

Thus, the modulus of elasticity is

Ans. E = 28.6(103) ksi

G =E

2(1 + y) ; 11.01(103) =

E

2(1 + 0.3)

G =60 ksi

0.00545= 11.01(103) ksi

P = 53.01 kip = 53.0 kip

ty =Vy

A ; 60 =

P>2p4 A0.752 B

ty = 60 ksi

:+ ©Fx = 0; V + V - P = 0 V = = P

2

Normal Strain:

Ans.

Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio.

Ans.

Shear Strain:

Ans.gxy =p

2- b =

p

2- 1.576032 = -0.00524 rad

b = 180° - 89.7° = 90.3° = 1.576032 rad

= 0.00540 in. >in.

ex = -vey = -0.36(-0.0150)

ey =dLy

Ly=

-0.064

= -0.0150 in.>in.

3–30. The block is made of titanium Ti-6A1-4V and issubjected to a compression of 0.06 in. along the y axis, and itsshape is given a tilt of Determine and gxy.Py,Px,u = 89.7°.

4 in. u

y

x5 in.

P

0.00545

60

g(rad)

t(ksi)

P/2P/2

3–31. The shear stress–strain diagram for a steel alloy isshown in the figure. If a bolt having a diameter of 0.75 in.is made of this material and used in the double lap joint,determine the modulus of elasticity E and the force Prequired to cause the material to yield. Take n = 0.3.


subjected to a compression of 1.5 mm along the y axis, and its

100 mm u

y

x125 mm

–1.5100

5 –0.0150 mm/mm

mm/mm

shown in the figure. If a bolt having a diameter of 20 mm P

0.00545

420

g(rad)

t(MPa)

P/2P/2

P

2

420 MPa. Thus,

420 5 P/2

p4(202)

P = 263893.8 N = 263.89 kN

G 5 420

0.00545 5 77.064(103) MPa 5 77.064 GPa

77.064

E 5 200.4 GPa

8–26.

8–27.

SM_CH08.indd 526 4/11/11 9:54:38 AM


527114


Shear Stress–Strain Relationship: Applying Hooke’s law with .

(Q.E.D)

If is small, then tan . Therefore,

At

Then,

At

Ans.d =P

2p h G ln rori

r = ri, y = d

y =P

2p h G ln ror

C =P

2p h G ln ro

0 = - P

2p h G ln ro + C

r = ro, y = 0

y = - P

2p h G ln r + C

y = - P

2p h G Ldrr

dy

dr= -

P

2p h G r

g = gg

dy

dr= - tan g = - tan a P

2p h G rb

g =tA

G=

P

2p h G r

tA =P

2p r h

*3–32. A shear spring is made by bonding the rubberannulus to a rigid fixed ring and a plug. When an axial loadP is placed on the plug, show that the slope at point y inthe rubber is For smallangles we can write Integrate thisexpression and evaluate the constant of integration usingthe condition that at From the result computethe deflection of the plug.y = d

r = ro.y = 0

dy>dr = -P>12phGr2.- tan1P>12phGr22.dy>dr = - tan g =

P

y

rori

y

r

h

d


*8–28.

SM_CH08.indd 527 4/11/11 9:54:39 AM


528115


Ans.d = 40(0.02083) = 0.833 mm

g =t

G=

4166.7

0.2(106)= 0.02083 rad

tavg =V

A=

2.5(0.03)(0.02)

= 4166.7 Pa

•3–33. The support consists of three rigid plates, whichare connected together using two symmetrically placedrubber pads. If a vertical force of 5 N is applied to plateA, determine the approximate vertical displacement ofthis plate due to shear strains in the rubber. Each padhas cross-sectional dimensions of 30 mm and 20 mm.Gr = 0.20 MPa.

C B

40 mm40 mm

A

5 N

Average Shear Stress: The rubber block is subjected to a shear force of .

Shear Strain: Applying Hooke’s law for shear

Thus,

Ans.d = a g = =P a

2 b h G

g =t

G=

P2 b h

G=

P

2 b h G

t =V

A=

P2

b h=P

2 b h

V =P

2

3–34. A shear spring is made from two blocks of rubber,each having a height h, width b, and thickness a. Theblocks are bonded to three plates as shown. If the platesare rigid and the shear modulus of the rubber is G,determine the displacement of plate A if a vertical load P isapplied to this plate. Assume that the displacement is smallso that d = a tan g L ag.

P

h

aa

Ad


•8–29.

8–30.

SM_CH08.indd 528 4/11/11 9:54:40 AM


529116


From the stress–strain diagram,

When specimen is loaded with a 9 - kip load,

Ans.Gal =Eat

2(1 + v)=

11.4(103)

2(1 + 0.32332)= 4.31(103) ksi

V = -

elat

elong= -

-0.00130.0040208

= 0.32332

elat =d¿ - dd

=0.49935 - 0.5

0.5= - 0.0013 in.>in.

elong =s

E=

45.8411400.65

= 0.0040208 in.>in.

s =P

A=

9p4 (0.5)2 = 45.84 ksi

Eal =s

e=

700.00614

= 11400.65 ksi

3–35. The elastic portion of the tension stress–straindiagram for an aluminum alloy is shown in the figure. Thespecimen used for the test has a gauge length of 2 in. and adiameter of 0.5 in. When the applied load is 9 kip, the newdiameter of the specimen is 0.49935 in. Compute the shearmodulus for the aluminum.Gal

0.00614

70

s(ksi)

P (in./in.)

From the stress–strain diagram

Ans.d¿ = d + ¢d = 0.5 - 0.001117 = 0.4989 in.

¢d = elat d = - 0.002234(0.5) = - 0.001117 in.

elat = - velong = - 0.500(0.0044673) = - 0.002234 in.>in.

G =E

2(1 + v) ; 3.8(103) =

11400.652(1 + v)

; v = 0.500

elong =s

E=

50.929611400.65

= 0.0044673 in.>in.

E =70

0.00614= 11400.65 ksi

s =P

A=

10p4 (0.5)2 = 50.9296 ksi

*3–36. The elastic portion of the tension stress–straindiagram for an aluminum alloy is shown in the figure. Thespecimen used for the test has a gauge length of 2 in. and adiameter of 0.5 in. If the applied load is 10 kip, determinethe new diameter of the specimen. The shear modulus isGal = 3.811032 ksi.

0.00614

70

s(ksi)

P (in./in.)


specimen used for the test has a gauge length of 50 mm and a diameter of 12.5 mm. When the applied load is 45 kN, the new diameter of the specimen is 12.48375 mm. Compute the shear modulus Gal for the aluminum.

0.00614

500

s (MPa)

P (mm/mm)

Eal 5 se

5 500

0.00614 5 81.433(103) MPa

When specimen is loaded with a 45-kN load,

s 5 PA

5 45(103)p4(12.5)2 5 366.69 MPa

elong 5 sE

5 366.69

81.433(103) 5 0.0045030 mm/mm

elat 5 d9 – d

d 5

12.48375 – 12.512.5

5 –0.0013 mm/mm

V 5 elatelong

5 –0.0013

0.0045030 5 0.28870

Gal 5 Eat

2(1 + v) 5

81.433(103)2(1 + 0.28870)

5 31.60(103) MPa 5 31.60 GPa

specimen used for the test has a gauge length of 50 mm and a diameter of 12.5 mm. If the applied load is 50 kN, determine the new diameter of the specimen. The shear modulus is Gal = 28 GPa.

s 5 PA

5 50(103)p4(12.5)2 5 407.44 MPa

From the stress-strain diagram

E 5 500

0.00614 5 81.433(103) MPa

elong 5 sE

= 407.44

81.433(103) 5 0.0050033 mm/mm

G 5 E

2(1 + v); 28(103)5

81.433(103)2(1 + v)

; v 5 0.45416

elat 5 –velong 5 –0.45416(0.0050033) 5 –0.002272

∆d 5 elatd 5 –0.002272(12.5) 5 –0.0284 mm

d9 5 d + ∆d 5 12.5 – 0.0284 5 12.4716 mm

0.00614

500

s (MPa)

P (mm/mm)

8–31.

*8–32.

SM_CH08.indd 529 4/11/11 9:54:41 AM


530117


Ans.

Ans.

Ans.ut =12

(2)(11) = 11 psi

ut =12

(2)(11) +12

(55 + 11)(2.25 - 2) = 19.25 psi

E =112

= 5.5 psi

3–37. The diagram for elastic fibers that make uphuman skin and muscle is shown. Determine the modulus of elasticity of the fibers and estimate their modulus oftoughness and modulus of resilience.

s–P

21 2.25

11

55

P(in./in.)

s(psi)

a)

Ans.

b)

Ans.d¿ = d + ¢d = 20 + 0.0016 = 20.0016 mm

¢d = elat d = 0.00008085(20) = 0.0016 mm

elat = 0.00008085 mm>mm

v = -

elat

elong; 0.35 = -

elat

-0.0002310

d = elong L = - 0.0002310(75) = - 0.0173 mm

elong = - 0.0002310 mm>mm

s = E elong ; - 15.915(106) = 68.9(109) elong

s =P

A=

-5(103)p4 (0.02)2 = - 15.915 MPa

3–38. A short cylindrical block of 6061-T6 aluminum,having an original diameter of 20 mm and a length of75 mm, is placed in a compression machine and squeezeduntil the axial load applied is 5 kN. Determine (a) thedecrease in its length and (b) its new diameter.


21 2.25

77

385

P(mm/mm)

s(MPa)

Eal 5 772

5 38.5 MPa

ut 5 12

(2)(77) + 12

(385 + 77)(2.25 – 2) 5 134.75 MJ/m3

ut 5 12

(2)(77) 5 77 MJ/m3

•8–33.

8–34.

SM_CH08.indd 530 4/11/11 9:54:42 AM


531118


a (1)

a (2)

Since the beam is held horizontally,

Ans.

From Eq. (2),

Ans.dAœ = dA + d elat = 30 + 30(0.0002646) = 30.008 mm

elat = -velong = -0.35(-0.000756) = 0.0002646

elong =sA

E= -

55.27(106)

73.1(109)= -0.000756

sA =FAA

=39.07(103)p4(0.032)

= 55.27 MPa

FA = 39.07 kN

x = 1.53 m

80(3 - x)(220) = 80x(210)

dA = dB ; 80(3 - x)3 (220)

AE=

80x3 (210)

AE

d = eL = aPA

Eb L =

PL

AE

s =P

A ; e =

s

E=PA

E

dA = dB

+ ©MB = 0; -FA(3) + 80(3 - x) = 0; FA =80(3 - x)

3

+ ©MA = 0; FB(3) - 80(x) = 0; FB =80x

3

3–39. The rigid beam rests in the horizontal position ontwo 2014-T6 aluminum cylinders having the unloaded lengthsshown. If each cylinder has a diameter of 30 mm, determinethe placement x of the applied 80-kN load so that the beamremains horizontal. What is the new diameter of cylinder Aafter the load is applied? nal = 0.35.

3 m

210 mm220 mm

x

A B

80 kN

Normal Stress:

Normal Strain: Since , Hooke’s law is still valid.

Ans.

If the nut is unscrewed, the load is zero. Therefore, the strain Ans.e = 0

e =s

E=

28.9729(103)

= 0.000999 in.>in.

s 6 sg

s =P

A=

800p4 A 3

16 B2 = 28.97 ksi 6 sg = 40 ksi

*3–40. The head H is connected to the cylinder of acompressor using six steel bolts. If the clamping force ineach bolt is 800 lb, determine the normal strain in thebolts. Each bolt has a diameter of If and

what is the strain in each bolt when thenut is unscrewed so that the clamping force is released?Est = 2911032 ksi,

sY = 40 ksi316 in. H

LC


each bolt is 4 kN, determine the normal strain in the bolts. Each bolt has a diameter of 5 mm. If sY = 280 MPa and Est = 200 GPa, what is the strain in each bolt when the nut is unscrewed so that the clamping force is released?

s 5 PA

5 4(103)p4(5)2 5 203.72 MPa < sg 5 280 MPa

e 5 sE

5 203.72

200(10)3 5 0.0010186 mm/mm

8–35.

*8–36.

SM_CH08.indd 531 4/11/11 9:54:42 AM


532119


Equations of Equilibrium:

a [1]

[2]

Note: The normal force at A does not act exactly at A. It has to shift due to friction.

Friction Equation:

[3]

Solving Eqs. [1], [2] and [3] yields:

Average Shear Stress: The pad is subjected to a shear force of .

Modulus of Rigidity:

Shear Strain: Applying Hooke’s law for shear

Thus,

Ans.dh = hg = 30(0.1005) = 3.02 mm

g =t

G=

148.89(103)

1.481(106)= 0.1005 rad

G =E

2(1 + v)=

42(1 + 0.35)

= 1.481 MPa

t =V

A=

3126.69(0.14)(0.15)

= 148.89 kPa

V = F = 3126.69 N

FA = 3908.37 N F = P = 3126.69 N

F = ms FA = 0.8 FA

:+ ©Fx = 0; P - F = 0

+©MB = 0; FA(2.75) - 7848(1.25) - P(0.3) = 0

•3–41. The stone has a mass of 800 kg and center of gravityat G. It rests on a pad at A and a roller at B.The pad is fixedto the ground and has a compressed height of 30 mm, awidth of 140 mm, and a length of 150 mm. If the coefficientof static friction between the pad and the stone is determine the approximate horizontal displacement of thestone, caused by the shear strains in the pad, before thestone begins to slip. Assume the normal force at A acts1.5 m from G as shown. The pad is made from a materialhaving MPa and n = 0.35.E = 4

ms = 0.8,

0.4 m

1.25 m 1.5 m

0.3 mP

B A

G


•8–37.

SM_CH08.indd 532 4/11/11 9:54:43 AM


533120


Normal Stress:

Normal Strain: Applying Hooke’s Law

Ans.

Ans.es =ss

Emg=

39.79(106)

45(109)= 0.000884 mm>mm

eb =sb

Eal=

159.15(106)

70(109)= 0.00227 mm>mm

ss =P

As=

8(103)p4 (0.022 - 0.0122)

= 39.79 MPa

sb =P

Ab=

8(103)p4 (0.0082)

= 159.15 MPa

3–43. The 8-mm-diameter bolt is made of an aluminumalloy. It fits through a magnesium sleeve that has an innerdiameter of 12 mm and an outer diameter of 20 mm. If theoriginal lengths of the bolt and sleeve are 80 mm and50 mm, respectively, determine the strains in the sleeve andthe bolt if the nut on the bolt is tightened so that the tensionin the bolt is 8 kN. Assume the material at A is rigid.

Emg = 45 GPa.Eal = 70 GPa,

Ans.

a

Ans.

Ans.eBC =sBC

E=

55.9429 (103)

= 0.00193 in.>in.

sBC =W

ABC=

0.1120.002

= 55.94 ksi

W = 0.112 kip = 112 lb

+ ©MA = 0; -(0.0672) (5) + 3(W) = 0

FDE = sDEADE = 33.56 (0.002) = 0.0672 kip

sDE = EeDE = 29(103)(0.00116) = 33.56 ksi

eDE =d

L=

0.04173(12)

= 0.00116 in.>in.

d = 0.0417 in

30.025

=5d

3–42. The bar DA is rigid and is originally held in thehorizontal position when the weight W is supported from C.If the weight causes B to be displaced downward 0.025 in.,determine the strain in wires DE and BC. Also, if the wiresare made of A-36 steel and have a cross-sectional area of0.002 in2, determine the weight W. 2 ft 3 ft

4 ft

3 ft

D AB

E

C

W

50 mm

30 mm

A


If the weight causes B to be displaced downward 0.625 mm,

1.25 mm2, determine the weight W. 0.6 m 0.9 m

1.2 m

0.9 m

D AB

E

C

W

0.6 m 0.9 m

1.041670.625

FDE = 289.35 N

0.6 m 0.9 m

0.90.625

= 1.5d

d = 1.04167 mm

eDE = d

L = 1.04167

0.9(1000) = 0.0011574 mm/mm

sDE = EeDE = 200(103)(0.0011574) = 231.48 MPa

FDE = sDEADE = 231.48(1.25) = 289.35 N

–(289.35)(1.5) + 0.9(W) = 0

W = 482.25 N

sBC = W

ABC =

482.251.25

= 385.8 MPa

eBC = sBC

E =

385.8200(103)

= 0.00193 mm/mm

8–38.

8–39.

SM_CH08.indd 533 4/11/11 9:54:44 AM


534121


a

(1)

However,

From Eq. (1),

Ans.P = 2.46 kN

FAB = sA = 348.76(106)(10)(10-6) = 3.488 kN

P(400 cos 0.2°) - FAB sin 44.9° (400) = 0

+©MA = 0

s = Ee = 200(109) (0.001744) = 348.76 MPa

e =L¿AB - LABLAB

=566.67 - 565.69

565.69= 0.001744

LAB =400

sin 45°= 565.69

L¿AB = 566.67 mm

L¿AB

sin 90.2°=

400sin 44.9°

*3–44. The A-36 steel wire AB has a cross-sectional areaof and is unstretched when Determinethe applied load P needed to cause u = 44.9°.

u = 45.0°.10 mm2

400 mm

A

B

P

400 mm

u


*8–40.

SM_CH08.indd 534 4/11/11 9:54:44 AM

3–1. a concrete cylinder having a diameter of 6.00 …...•8–1. a concrete cylinder having a...

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