Solutions_Chapter_3(WǪ).pdf

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*3–8. The strut is supported by a pin at C and an A-36 AB. If the the wire wire has has aa diameter diameter of of 0.2 5 mm, steel guy wire AB. in., determine how much it stretches when the distributed load acts on the strut.

A

60� 60 200 lb/ft 3.4 kN/m

C

Here, we are only interested in determining the force in wire AB. 11 FAB cos 60°(2.7) 60°(9) -–  (3.4)(2.7)(0.9) (200)(9)(3) = 0= 0  F FAB 600kN lb AB cos AB ==3.06 22

a + ©MC = 0;

The normal stress the wire is sAB =

FAB 600 3) 3.06(10 3 = p p 22 ==19.10(10 ) psi = 19.10 ksi 155.84 MPa AAB (0.2 ) (5 ) 4 4

250ksi MPa, Hooke’s Lawcan canbe beapplied applied to determine Since sAB 6 sy = 36 , Hooke’s Law determine the thestrain strain in wire. sAB = EPAB;

155.84 = 200(1033)e 19.10 = 29.0(10 )PAB AB PeAB = 0.7792(10–3 - 3) mm/mm AB = 0.6586(10 ) in>in

9(12)3) 2.7(10 == 124.71 inThus, The unstretched length of the wire is LAB = . Thus, wire 3117.69. thethe wire sin sin 60° 60° stretches -3 0.7792(10–3 )(3117.69) )(124.71) dAB = PAB LAB = 0.6586(10

2.429 mm = 0.0821 in.

Ans.

1  (3.4)(2.7) kN 2

0.9 m

1.8 m

146

B 9 ftm 2.7

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3–9. The s-P diagram for elastic fibers that make up human skin and muscle is shown. Determine the modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience.

s (MPa) 0.385

0.077 1

2 2.25

P (mm/mm)

s (MPa) 0.385

E =

0.077 = 0.0385 MPa 2

Ans.

ut =

1 1 (2)(0.077) + (0.385 + 0.77)(2.25 - 2) = 0.13475 MPa 2 2

Ans.

1 (0.077)(11) = 0.077 MPa 2

Ans.

ur =

0.077 1

2 2.25

P (mm/mm)

Ans: E = 0.0385 MPa ut = 0.13475 MPa, ur = 0.077 M P a 147

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*3–12. The stress–strain diagram for a steel alloy having an original diameter diameter of of 12 0.5mm in. and and aagauge gaugelength lengthofof502mm in. is given in the figure. Determine approximately the modulus of resilience and the modulus of toughness for the material. The Modulus of resilience is equal to the area under the stress–strain diagram up to the proportional limit. MPa  PPL ePL= =0.002 0.001in>in. mm/mm sPL == 290 60 ksi Thus, (ui)r =

1 1 in # lb = 0.145 MPa 3 sPLPPL = [(290)](0.001) = 60.0 C 60(103) D (0.002) 2 2 in

Ans.

The modulus of toughness is equal to the area under the entire stress–strain diagram. This area can be approximated by counting the number of squares. The total number is 33. 38. Thus. Thus, lb in

in in

mm mm

in # lb in

3 3 = 33[100 b = 132 = i)38 c 15(10 ) 2 d MPa]a0.04 a 0.05 b = 28.5(10 ) MPa3 C (ui)t D approx[(u t]approx

s (ksi)  (MPa) 105

500

90

A 400

75 60 PL = 290

45

200

30

100

15 0

300

0 0

E 1

B P (in./in.) 0.050 00.100.05/ 0.20 0.250.20/ 0.30 0.350.30/ 0.35/  (mm/m) 0.150.08/ 0.001 0.002 0.003 0.0040.15/ 0.005 0.0060.25/ 0.007 0.001 0.002 0.003 0.004 0.005 0.006 0.007 (a)

150

Ans.

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*3–16. The wire has a diameter of 5 mm and is made from A-36 steel. If a 80-kg man is sitting on seat C, determine the elongation of wire DE.

E W 600 mm D A

B 800 mm

Equations of Equilibrium: The force developed in wire DE can be determined by writing the moment equation of equilibrium about A with reference to the freebody diagram shown in Fig. a, a + ©MA = 0;

3 FDE a b(0.8) - 80(9.81)(1.4) = 0 5 FDE = 2289 N

Normal Stress and Strain: sDE =

FDE 2289 = = 116.58 MPa p ADE (0.0052) 4

Since sDE < sY , Hooke’s Law can be applied sDE = EPDE 116.58(106) = 200(109)PDE PDE = 0.5829(10-3) mm>mm The unstretched length of wire DE is LDE = 26002 + 8002 = 1000 mm. Thus, the elongation of this wire is given by dDE = PDELDE = 0.5829(10-3)(1000) = 0.583 mm

Ans.

154

C 600 mm

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3–18. A tension test was performed on a magnesium alloy specimen having a diameter 12 mm and gauge length of 50 mm. The resulting stress–strain diagram is shown in the figure. If the specimen is stressed to 210 MPa and unloaded, determine the permanent elongation of the specimen.

(MPa) 280 245 210 175 140 105 70 35 0

0.002

0.004

0.006

0.008

0.010

(mm/mm)

Permanent Elongation: From the stress–strain diagram, the strain recovered is along the straight line BC which is parallel to the straight line OA. Since 91 - 0 = 45.5(103) M Pa, then the permanent set for the specimen is Eapprox = 0.002 - 0 210(103) PP = 0.0078 = 0.00318 mm> mm. 45.5(106) Thus, dP = PPL = 0.00318(50) = 0.159 mm.

Ans.

Ans: dP = 0.159 mm. 156

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P

3–19. The stress–strain diagram for a bone is shown, and can be described by the equation P = 0.45110-62 s ⫹ 0.36110-122 s3, where s is in kPa. Determine the yield strength assuming a 0.3% offset.

s

P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3 P

P

P = 0.45(10-6)s + 0.36(10-12)s3, dP = A 0.45(10-6) + 1.08(10-12) s2 B ds

E =

ds 1 2 = = 2.22(106) kPa = 2.22 GPa dP 0.45(10 - 6) s=0

The equation for the recovery line is s = 2.22(106)(P - 0.003). This line intersects the stress–strain curve at sYS = 2027 kPa = 2.03 MPa

Ans.

Ans: sYS = 2.03 MPa 157

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3–22. The two bars are made of polystyrene, which has the stress–strain diagram shown. Determine the cross-sectional area of each bar so that the bars rupture simultaneously when the load P = 13.5 kN. Assume that buckling does not occur.

P 1.2 m

C

B

1m

A (MPa) 175

+ c ©Fy = 0; + : ©Fx = 0;

1 FBA a b - 13.5 = 0; 1.56 .06 a -FBC + 21.06

140

FBA = 21.06 kN

105

1.2 b = 0; FBC = 16.2 kN 1.56

compression

70 tension

35

For member BC:

0

(smax)t =

FBC ; ABC

ABC =

(smax)c =

FBA ; ABA

ABA =

16.2 kN = 462.85 mm2 35 MPa

0

0.20

0.40

0.60

0.80

(mm/mm)

Ans.

For member BA: 21.06 kN = 120.34 mm2 175 M Pa

Ans.

Ans: ABC = 462.85 mm2 , ABA = 120.34 mm2 160

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3–24. The wires AB and BC have original lengths of 0.6 m and 0.9 m, and diameters of 3 mm and 5 mm, respectively. If these wires are made of a material that has the approximate stress–strain diagram shown, determine the elongations of the wires after the 6750-N load is placed on the platform.

C

Equations of Equilibrium: The forces developed in wires AB and BC can be determined by analyzing the equilibrium of joint B, Fig. a, + : ©Fx = 0; + c ©Fy = 0;

FBC sin 30° - FAB sin 45° = 0

(1)

FBC cos 30° + FAB cos 45° = 6750

(2)

A

0.9 m 45⬚

30⬚

0.6 m B

Solving Eqs. (1) and (2), FAB = 3494.07 N

FBC = 4941.36 N

Normal Stress and Strain: sAB =

FAB 3494.07 = = 494.56 MPa p AAB (3)2 4 s (MPa)

sBC =

FBC 4941.36 = = 251.78 MPa p ABC 2 (5) 4

560 406

The corresponding normal strain can be determined from the stress–strain diagram, Fig. b. 251.78 406 = ; PBC 0.002

PBC = 0.00124 mm0> mm.

494 .56 - 406 560 - 406 = ; PAB - 0.002 0.01 - 0.002

PAB = 0.0066 mm> mm.

0.002

Thus, the elongations of wires AB and BC are dAB = PABLAB = 0.0066(600) = 3.96

Ans.

dBC = PBCLBC = 0.00124(900) = 0.1116

Ans.

162

0.01

P (mm/mm)

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3–26. The thin-walled tube is subjected to an axial force of 40 kN. If the tube elongates 3 mm and its circumference decreases 0.09 mm, determine the modulus of elasticity, Poisson’s ratio, and the shear modulus of the tube’s material. The material behaves elastically.

40 kN 900 mm

10 mm

40 kN 12.5 mm

Normal Stress and Strain: s =

40(103) P = 226.35 MPa = A p(0.01252 - 0.012)

Pa =

d 3 = = 3.3333 (10-3) mm>mm L 900

Applying Hooke’s law, s = EPa;

226.35(106) = E [3.3333(10-3)] E = 67.91(106) Pa = 67.9 GPa

Ans.

Poisson’s Ratio: The circumference of the loaded tube is 2p(12.5) - 0.09 = 78.4498 mm. Thus, the outer radius of the tube is r =

78.4498 = 12.4857 mm 2p

The lateral strain is Plat =

r - r0 12.4857 - 12.5 = = -1.1459(10-3) mm>mm r0 12.5

n = -

-1.1459(10-3) Plat d = 0.3438 = 0.344 = -c Pa 3.3333(10-3)

G =

Ans.

67.91(109) E = = 25.27(109) Pa = 25.3 GPa 2(1 + n) 2(1 + 0.3438)

Ans.

Ans: E = 67.9 GPa, v = 0.344, G = 25.3 GPa 164

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3–27. When the two forces are placed on the beam, the diameter of the A-36 steel rod BC decreases from 40 mm to 39.99 mm. Determine the magnitude of each force P.

C P 1m

A

P 1m

1m

1m

B 0.75 m

Equations of Equilibrium: The force developed in rod BC can be determined by writing the moment equation of equilibrium about A with reference to the free-body diagram of the beam shown in Fig. a. 4 FBC a b (3) - P(2) - P(1) = 0 5

a + ©MA = 0;

FBC = 1.25P

Normal Stress and Strain: The lateral strain of rod BC is Plat =

d - d0 39.99 - 40 = = -0.25(10 - 3) mm>mm d0 40

Plat = -nPa;

-0.25(10-3) = -(0.32)Pa Pa = 0.78125(10-3) mm>mm

Assuming that Hooke’s Law applies, sBC = EPa;

sBC = 200(109)(0.78125)(10-3) = 156.25 MPa

Since s 6 sY, the assumption is correct. sBC =

FBC ; ABC

156.25(106) =

1.25P p A 0.042 B 4

P = 157.08(103)N = 157 kN

Ans.

Ans: P = 157 kN 165

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3–29. The friction pad A is used to support the member, which is subjected to an axial force of P = 2 kN. The pad is made from a material having a modulus of elasticity of E = 4 MPa and Poisson’s ratio n = 0.4. If slipping does not occur, determine the normal and shear strains in the pad. The width is 50 mm. Assume that the material is linearly elastic. Also, neglect the effect of the moment acting on the pad.

P

60⬚

25 mm

A

100 mm

Internal Loading: The normal force and shear force acting on the friction pad can be determined by considering the equilibrium of the pin shown in Fig. a. + : ©Fx = 0;

V - 2 cos 60° = 0

V = 1 kN

+ c ©Fy = 0;

N - 2 sin 60° = 0

N = 1.732 kN

Normal and Shear Stress: t =

1(103) V = = 200 kPa A 0.1(0.05)

s =

1.732(103) N = = 346.41 kPa A 0.1(0.05)

Normal and Shear Strain: The shear modulus of the friction pad is G =

4 E = = 1.429 MPa 2(1 + n) 2(1 + 0.4)

Applying Hooke’s Law, s = EP;

346.41(103) = 4(106)P

P = 0.08660 mm>mm

Ans.

t = Gg;

200(103) = 1.429(106)g

g = 0.140 rad

Ans.

Ans: P = 0.08660 mm>mm, g = 0.140 rad 167

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3–30. The lap joint is connected together using a 30 mm diameter bolt. If the bolt is made from a material having a shear stress–strain diagram that is approximated as shown, determine the shear strain developed in the shear plane of the bolt when P = 340 kN.

P 2

P

P 2 t (MPa) 525 350

Internal Loadings: The shear force developed in the shear planes of the bolt can be determined by considering the equilibrium of the free-body diagram shown in Fig. a. + : ©Fx = 0;

340 - 2V = 0

0.005

0.05

g (rad)

V = 170 kN

Shear Stress and Strain: t =

V 170 = p A A 302 B 4

= 240.62 MPa

Using this result, the corresponding shear strain can be obtained from the shear stress–strain diagram, Fig. b. 240.62 350 = ; g 0.005

g = 3.43(10-3) rad

Ans.

Ans: g = 3.43(10-3) rad 168

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*3–32. A shear spring is made by bonding the rubber annulus to a rigid fixed ring and a plug. When an axial load P is placed on the plug, show that the slope at point y in the rubber is dy>dr = -tan g = -tan1P>12phGr22. For small angles we can write dy>dr = -P>12phGr2. Integrate this expression and evaluate the constant of integration using the condition that y = 0 at r = ro. From the result compute the deflection y = d of the plug.

P

h

ro

y

d

ri r y

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

g =

P = . 2p r h

tA P = G 2p h G r

dy P = -tan g = -tan a b dr 2p h G r

(Q.E.D)

If g is small, then tan g = g. Therefore, dy P = dr 2p h G r

At r = ro,

y = -

dr P 2p h G L r

y = -

P ln r + C 2p h G

0 = -

P ln ro + C 2p h G

y = 0

C =

Then, y =

ro P ln r 2p h G

At r = ri,

y = d d =

P ln ro 2p h G

ro P ln ri 2p h G

Ans.

170

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3–33. The aluminum block has a rectangular cross section and is subjected to an axial compressive force of 36 kN. If the 37-mm side changed its length to 37.5033 mm, determine Poisson’s ratio and the new length of the 50-mm. Eal = 70 GPa.

s =

50 mm

36 kN

36 kN 75 mm

P 36 = = 19.45 MPa A (50)(37)

Plong =

Plat =

n =

37 mm

s - 19.45 = -0.0002778 = E 70000

37.5033 - 37 37

= 0.00008918

-0.00008918 = 0.321 -0. 0002778

Ans.

h¿ = 50 + 0.00008918(2) = 50.000178 mm.

Ans.

Ans: n = 0.321, h¿ = 50.000178 mm. 171

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3–37. The rigid beam rests in the horizontal position on two 2014-T6 aluminum cylinders having the unloaded lengths shown. If each cylinder has a diameter of 30 mm. determine the placement x of the applied 80-kN load so that the beam remains horizontal. What is the new diameter of cylinder A after the load is applied? nal = 0.35.

80 kN x

A

B

220 mm

210 mm

3m

a +©MA = 0;

FB(3) - 80(x) = 0;

a +©MB = 0;

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

FB =

80x 3 FA =

(1) 80(3 - x) 3

(2)

Since the beam is held horizontally, dA = dB s =

P

P ; A

s A = E E

P =

d = PL = a

P A

E

bL =

80(3 - x) 3

dA = dB;

PL AE (220)

AE

=

80x 3

(210)

AE

80(3 - x)(220) = 80x(210) x = 1.53 m

Ans.

From Eq. (2), FA = 39.07 kN sA =

39.07(103) FA = = 55.27 MPa p A (0.032) 4

Plong =

sA E

55.27(106) = -

73.1(109)

= -0.000756

Plat = -nPlong = -0.35(-0.000756) = 0.0002646 dA¿ = dA + d Plat = 30 + 30(0.0002646) = 30.008 mm

Ans.

Ans: x = 1.53 m, dA ¿ = 30.008 mm 175

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3–42. The pipe with two rigid caps attached to its ends is subjected to an axial force P. If the pipe is made from a material having a modulus of elasticity E and Poisson’s ratio n, determine the change in volume of the material.

ri ro L P

a

Section a – a

a P

Normal Stress: The rod is subjected to uniaxial loading. Thus, slong =

P and slat = 0. A

dV = AdL + 2prLdr = APlong L + 2prLPlatr Using Poisson’s ratio and noting that AL = pr2L = V, dV = PlongV - 2nPlongV = Plong (1 - 2n)V slong =

E

(1 - 2n)V

Since slong = P>A, dV =

=

P (1 - 2n)AL AE PL (1 - 2n) E

Ans.

Ans: dV =

180

PL (1 - 2n) E

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3–43. The 8-mm-diameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has an inner diameter of 12 mm and an outer diameter of 20 mm. If the original lengths of the bolt and sleeve are 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN. Assume the material at A is rigid. Eal = 70 GPa, Emg = 45 GPa.

50 mm A

30 mm

Normal Stress: 8(103)

sb =

P = Ab

p 2 4 (0.008 )

ss =

P = As

p 2 4 (0.02

= 159.15 MPa

8(103) - 0.0122)

= 39.79 MPa

Normal Strain: Applying Hooke’s Law Pb =

159.15(106) sb = 0.00227 mm>mm = Eal 70(109)

Ans.

Ps =

39.79(106) ss = 0.000884 mm>mm = Emg 45(109)

Ans.

Ans: Pb = 0.00227 mm>mm, Ps = 0.000884 mm>mm 181