# approximately the modulus of elasticity. Stress and Strain:

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1 3 Solutions 466 5/7/1 8:45 M age 1 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently 3 1. concrete cylinder having a diameter of 6. in. and gauge length of 12 in. is tested in comression.the results of the test are reorted in the table as load versus contraction. Draw the stress strain diagram using scales of 1 in. =.5 ksi and 1 in. = in.>in. From the diagram, determine aroximately the modulus of elasticity. Stress and Strain: s = e = dl (ksi) L (in./in.) Modulus of Elasticity: From the stress strain diagram Load (ki) Contraction (in.) E arox = = B ksi 1

2 3 Solutions 466 5/7/1 8:45 M age 2 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently 3 2. Data taken from a stress strain test for a ceramic are given in the table.the curve is linear between the origin and the first oint. lot the diagram, and determine the modulus of elasticity and the modulus of resilience. Modulus of Elasticity: From the stress strain diagram E = = B ksi S (ksi) (in./in.) Modulus of Resilience: The modulus of resilience is equal to the area under the linear ortion of the stress strain diagram (shown shaded). u t = 1 2 (33.2)13 B lb in. 2.6 in in. = 9.96 in # lb in Data taken from a stress strain test for a ceramic are given in the table. The curve is linear between the origin and the first oint. lot the diagram, and determine aroximately the modulus of toughness. The ruture stress is s r = 53.4 ksi. Modulus of Toughness: The modulus of toughness is equal to the area under the stress strain diagram (shown shaded). (u t ) arox = 1 2 (33.2)13 B lb in. 2 (.4 +.1) in in B lb in. 2 (.12) in in (7.9)13 B lb in. 2 (.12) in in (12.3)13 B lb in. 2 (.4) in in. S (ksi) (in./in.) = 85. in # lb in 3 2

3 3 Solutions 466 5/7/1 8:45 M age 3 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently *3 4. tension test was erformed on a secimen having an original diameter of 12.5 mm and a gauge length of 5 mm. The data are listed in the table. lot the stress strain diagram, and determine aroximately the modulus of elasticity, the ultimate stress, and the fracture stress. Use a scale of 2 mm = 5 Ma and 2 mm =.5 mm>mm. Redraw the linear-elastic region, using the same stress scale but a strain scale of 2 mm =.1 mm>mm. Stress and Strain: s = e = dl (Ma) L (mm/mm) Load (kn) Elongation (mm) Modulus of Elasticity: From the stress strain diagram (E) arox = (16 ) = 229 Ga Ultimate and Fracture Stress: From the stress strain diagram (s m ) arox = 528 Ma (s f ) arox = 479 Ma 3

4 3 Solutions 466 5/7/1 8:45 M age 4 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently 3 5. tension test was erformed on a steel secimen having an original diameter of 12.5 mm and gauge length of 5 mm. Using the data listed in the table, lot the stress strain diagram, and determine aroximately the modulus of toughness. Use a scale of 2 mm = 5 Ma and 2 mm =.5 mm>mm. Stress and Strain: s = e = dl (Ma) L (mm/mm) Modulus of Toughness: The modulus of toughness is equal to the total area under the stress strain diagram and can be aroximated by counting the number of squares. The total number of squares is 187. Load (kn) Elongation (mm) (u t ) arox = 187(25)1 6 B N m 2 a.25 m b = 117 MJ>m3 m 4

5 3 Solutions 466 5/7/1 8:45 M age 5 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently 3 6. secimen is originally 1 ft long, has a diameter of.5 in., and is subjected to a force of 5 lb. When the force is increased from 5 lb to 18 lb, the secimen elongates.9 in. Determine the modulus of elasticity for the material if it remains linear elastic. Normal Stress and Strain: lying s = and e = dl. L s 1 =.5 = ksi 4 (.52 ) s 2 = 1.8 = ksi 4 (.52 ) e =.9 12 =.75 in.>in. Modulus of Elasticity: E = s e = = B ksi 3 7. structural member in a nuclear reactor is made of a zirconium alloy. If an axial load of 4 ki is to be suorted by the member, determine its required cross-sectional area. Use a factor of safety of 3 relative to yielding. What is the load on the member if it is 3 ft long and its elongation is.2 in.? E zr = 14(1 3 ) ksi, s Y = 57.5 ksi. The material has elastic behavior. llowable Normal Stress: F.S. = s y s allow 3 = 57.5 s allow s allow = ksi s allow = = 4 =.287 in 2 =.29 in 2 Stress Strain Relationshi: lying Hooke s law with e = d L =.2 =.555 in.>in. 3 (12) s = Ee = B (.555) = ksi Normal Force: lying equation s =. = s = (.287) = 1.62 ki 5

6 3 Solutions 466 5/7/1 8:45 M age 6 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently *3 8. The strut is suorted by a in at C and an -36 steel guy wire B. If the wire has a diameter of.2 in., determine how much it stretches when the distributed load acts on the strut. 6 2 lb/ft Here, we are only interested in determining the force in wire B. C 9 ft B a+ M C = ; F B cos 6 (9) (2)(9)(3) = F B = 6 lb The normal stress the wire is s B = F B B = 6 4 (.22 ) = 19.1(13 ) si = 19.1 ksi Since s B 6 s y = 36 ksi, Hooke s Law can be alied to determine the strain in wire. s B = E B ; 19.1 = 29.(1 3 ) B B =.6586(1-3 ) in>in The unstretched length of the wire is stretches L B = 9(12) sin 6 = in. Thus, the wire d B = B L B =.6586(1-3 )(124.71) =.821 in. 6

7 3 Solutions 466 5/7/1 8:45 M age 7 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently 3 9. The s diagram for a collagen fiber bundle from which a human tendon is comosed is shown. If a segment of the chilles tendon at has a length of 6.5 in. and an aroximate cross-sectional area of.229 in 2, determine its elongation if the foot suorts a load of 125 lb, which causes a tension in the tendon of lb. s (ksi) s = = From the grah e =.35 in.>in. = 1.5 ksi lb (in./in.) d = el =.35(6.5) =.228 in The stress strain diagram for a metal alloy having an original diameter of.5 in. and a gauge length of 2 in. is given in the figure. Determine aroximately the modulus of elasticity for the material, the load on the secimen that causes yielding, and the ultimate load the secimen will suort. From the stress strain diagram, Fig. a, Thus, E 1 = 6 ksi ; E = 3.(13 ) ksi s y = 6 ksi s u>t = 1 ksi Y = s Y = 6C 4 (.52 )D = ki = 11.8 ki u>t = s u>t = 1C 4 (.52 )D = ki = 19.6 ki s (ksi) (in./in.)

8 3 Solutions 466 5/7/1 8:45 M age 8 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently The stress strain diagram for a steel alloy having an original diameter of.5 in. and a gauge length of 2 in. is given in the figure. If the secimen is loaded until it is stressed to 9 ksi, determine the aroximate amount of elastic recovery and the increase in the gauge length after it is unloaded. s (ksi) (in./in.) From the stress strain diagram Fig. a, the modulus of elasticity for the steel alloy is E 1 = 6 ksi ; E = 3.(13 ) ksi when the secimen is unloaded, its normal strain recovered along line B, Fig. a, which has a gradient of E. Thus Elastic Recovery = 9 E = 9 ksi 3.(1 3 ) ksi =.3 in>in Thus, the ermanent set is Then, the increase in gauge length is = =.47 in>in L = L =.47(2) =.94 in 8

9 3 Solutions 466 5/7/1 8:45 M age 9 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently *3 12. The stress strain diagram for a steel alloy having an original diameter of.5 in. and a gauge length of 2 in. is given in the figure. Determine aroximately 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 u to the roortional limit. Thus, s L = 6 ksi L =.2 in>in. (u i ) r = 1 2 s L L = 1 2 C6(13 )D(.2) = 6. in # lb in 3 The modulus of toughness is equal to the area under the entire stress strain diagram. This area can be aroximated by counting the number of squares. The total number is 38. Thus, C(u i ) t D arox = 38 c15(1 3 ) lb in d a.5 2 in in b = 28.5(13 ) in # lb in 3 s (ksi) (in./in.)

10 3 Solutions 466 5/7/1 8:45 M age 1 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently bar having a length of 5 in. and cross-sectional area of.7 in 2 is subjected to an axial force of 8 lb. If the bar stretches.2 in., determine the modulus of elasticity of the material. The material has linear-elastic behavior. 8 lb 5 in. 8 lb Normal Stress and Strain: s = = 8..7 e = dl L =.2 5 = ksi =.4 in.>in. Modulus of Elasticity: E = s e = = 28.6(13 ) ksi The rigid ie is suorted by a in at and an -36 steel guy wire BD. If the wire has a diameter of.25 in., determine how much it stretches when a load of = 6 lb acts on the ie. B Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a 4 ft a+ M = ; F BD 4 5 B(3) - 6(6) = F BD = 15 lb The normal stress develoed in the wire is s BD = F BD BD = 15 4 (.252 ) = 3.56(13 ) si = 3.56 ksi D 3 ft 3 ft C Since s BD 6 s y = 36 ksi, Hooke s Law can be alied to determine the strain in the wire. s BD = E BD ; 3.56 = 29.(1 3 ) BD BD = 1.54(1-3 ) in.>in. The unstretched length of the wire is wire stretches L BD = = 5ft = 6 in. Thus, the d BD = BD L BD = 1.54(1-3 )(6) =.632 in 1

11 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently The rigid ie is suorted by a in at and an -36 guy wire BD. If the wire has a diameter of.25 in., determine the load if the end C is dislaced.75 in. downward. B 4 ft D 3 ft 3 ft C Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a a+ M = ; F BD 4 5 B(3) - (6) = F BD = 2.5 The unstretched length for wire BD is L BD = = 5 ft = 6 in. From the geometry shown in Fig. b, the stretched length of wire BD is L BD = (6)(.75) cos = Thus, the normal strain is BD = L BD - L BD = = 1.3(1-3 ) in.>in. L BD 6 Then, the normal stress can be obtain by alying Hooke s Law. s BD = E BD = 29(1 3 )C1.3(1-3 )D = 29.1 ksi Since s BD 6 s y = 36 ksi, the result is valid. s BD = F BD BD ; 29.1(1 3 ) = (.252 ) = lb = 57 lb 11

12 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently *3 16. Determine the elongation of the square hollow bar when it is subjected to the axial force = 1 kn. If this axial force is increased to = 36 kn and released, find the ermanent elongation of the bar. The bar is made of a metal alloy having a stress strain diagram which can be aroximated as shown. s (Ma) mm 5 mm mm (mm/mm) 5 mm 5 mm Normal Stress and Strain: The cross-sectional area of the hollow bar is = =.9(1-3 )m 2. When = 1 kn, s 1 = = 1(13 ).9(1-3 = Ma ) From the stress strain diagram shown in Fig. a, the sloe of the straight line O which reresents the modulus of elasticity of the metal alloy is E = 25(16 ) = 2 Ga Since s Ma, Hooke s Law can be alied. Thus s 1 = Ee 1 ; (1 6 ) = 2(1 9 )e 1 e 1 =.5556(1-3 ) mm>mm Thus, the elongation of the bar is d 1 = e 1 L =.5556(1-3 )(6) =.333 mm When = 36 kn, s 2 = = 36(13 ).9(1-3 = 4 Ma ) From the geometry of the stress strain diagram, Fig. a, e = e 2 =.35 mm>mm When = 36 kn is removed, the strain recovers linearly along line BC, Fig. a, arallel to O. Thus, the elastic recovery of strain is given by s 2 = Ee r ; 4(1 6 ) = 2(1 9 )e r e r =.2 mm>mm The ermanent set is e = e 2 - e r = =.285 mm>mm Thus, the ermanent elongation of the bar is d = e L =.285(6) = 17.1 mm 12

13 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently Continued 13

14 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently tension test was erformed on an aluminum 214-T6 alloy secimen. The resulting stress strain diagram is shown in the figure. Estimate (a) the roortional limit, (b) the modulus of elasticity, and (c) the yield strength based on a.2% strain offset method. s (ksi) (in./in.) roortional Limit and Yield Strength: From the stress strain diagram, Fig. a, s l = 44 ksi s Y = 6 ksi Modulus of Elasticity: From the stress strain diagram, the corresonding strain for s L = 44 ksi is e l =.4 in.>in. Thus, E = = 11.(13 ) ksi Modulus of Resilience: The modulus of resilience is equal to the area under the 14

15 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently tension test was erformed on an aluminum 214-T6 alloy secimen. The resulting stress strain diagram is shown in the figure. Estimate (a) the modulus of resilience; and (b) modulus of toughness. s (ksi) (in./in.) stress strain diagram u to the roortional limit. From the stress strain diagram, Thus, s l = 44 ksi e l =.4 in.>in. U i B r = 1 2 s le l = 1 2 (44)(13 )(.4) = 88 in # lb in 3 Modulus of Toughness: The modulus of toughness is equal to the area under the entire stress strain diagram. This area can be aroximated by counting the number of squares. The total number of squares is 65. Thus, CU i B t D arox = 65B1(1 3 ) lb in R 2 c.1in. in. d = 6.5(13 ) in # lb in 3 The stress strain diagram for a bone is shown, and can be described by the equation The stress strain diagram for a bone is shown, and can be described by the equation = s s 3, where s is in ka. Determine the yield strength assuming a.3% offset. s.45(1 6 )s +.36(1 12 )s 3 e =.45(1-6 )s +.36(1-12 )s 3, d =.45(1-6 ) + 1.8(1-12 ) s 2 Bds E = ds d 2 s = 1 = = 2.22 Ma.45(1-6 ) 15

16 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently *3 2. The stress strain diagram for a bone is shown and can be described by the equation = s s 3, where s is in ka. Determine the modulus of toughness and the amount of elongation of a 2-mmlong region just before it fractures if failure occurs at =.12 mm>mm. s.45(1 6 )s +.36(1 12 )s 3 When e =.12 12(1 3 ) =.45 s +.36(1-6 )s 3 Solving for the real root: s = ka u t = d = (.12 - e)ds L L u t = ( (1-6 )s -.36(1-12 )s 3 )ds L =.12 s -.225(1-6 )s 2 -.9(1-12 )s = 613 kj>m 3 d = el =.12(2) = 24 mm 16

17 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently The stress strain diagram for a olyester resin is given in the figure. If the rigid beam is suorted by a strut B and ost CD, both made from this material, and subjected to a load of = 8 kn, determine the angle of tilt of the beam when the load is alied.the diameter of the strut is 4 mm and the diameter of the ost is 8 mm. 2 m B C From the stress strain diagram,.75 m.75 m.5 m D E = 32.2(1)6.1 Thus, s B = F B = 4(13 ) = Ma B (.4)2 e B = s B E = 31.83(16 ) 3.22(1 9 =.9885 mm>mm ) s CD = F CD = 4(13 ) = Ma CD (.8)2 e CD = s CD E = 7.958(16 ) 3.22(1 9 =.2471 mm>mm ) d B = e B L B =.9885(2) = mm d CD = e CD L CD =.2471(5) = mm ngle of tilt a: = 3.22(1 9 ) a 4 4 s (Ma) comression tension (mm/mm) tan a = ; a =.78 17

18 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently The stress strain diagram for a olyester resin is given in the figure. If the rigid beam is suorted by a strut B and ost CD made from this material, determine the largest load that can be alied to the beam before it rutures. The diameter of the strut is 12 mm and the diameter of the ost is 4 mm. 2 m B Ruture of strut B: s R = F B B ; 5(1 6 ) = >2 4 (.12)2; = 11.3 kn (controls).75 m.75 m C.5 m D Ruture of ost CD: s R = F CD CD ; 95(1 6 ) = >2 4 (.4)2 = 239 kn s (Ma) comression tension (mm/mm) By adding lasticizers to olyvinyl chloride, it is ossible to reduce its stiffness. The stress strain diagrams for three tyes of this material showing this effect are given below. Secify the tye that should be used in the manufacture of a rod having a length of 5 in. and a diameter of 2 in., that is required to suort at least an axial load of 2 ki and also be able to stretch at most 1 4 in. s (ksi) 15 unlasticized 1 coolymer Normal Stress: s = = 2 = ksi 4 (22 ) 5 flexible (lasticized) (in./in.) Normal Strain: e =.25 5 =.5 in.>in. From the stress strain diagram, the coolymer will satisfy both stress and strain requirements. 18

19 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently *3 24. The stress strain diagram for many metal alloys can be described analytically using the Ramberg-Osgood three arameter equation =s>e + ks n, where E, k, and n are determined from measurements taken from the diagram. Using the stress strain diagram shown in the figure, take E = ksi and determine the other two arameters k and n and thereby obtain an analytical exression for the curve. s (ksi) Choose, (1 6 ) s = 4 ksi, e =.1 s = 6 ksi, e =.3.1 =.3 = 4 3(1 3 ) + k(4)n 6 3(1 3 ) + k(6)n = k(4) n.298 = k(6) n = (.6667) n ln ( ) = n ln (.6667) n = 2.73 k = 4.23(1-6 ) The acrylic lastic rod is 2 mm long and 15 mm in diameter. If an axial load of 3 N is alied to it, determine the change in its length and the change in its diameter. E = 2.7 Ga, n =.4. 3 N 2 mm 3 N s = = 3 4 (.15)2 = Ma e long = s E = 1.697(16 ) 2.7(1 9 ) =.6288 d = e long L =.6288 (2) =.126 mm e lat = -Ve long = -.4(.6288) = d = e lat d = (15) = mm 19

20 3 Solutions 466 5/7/1 8:45 M age 2 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently The short cylindrical block of 214-T6 aluminum, having an original diameter of.5 in. and a length of 1.5 in., is laced in the smooth jaws of a vise and squeezed until the axial load alied is 8 lb. Determine (a) the decrease in its length and (b) its new diameter. 8 lb 8 lb a) s = = 8 4 (.5)2 = si e long = s E = (1 6 ) = d = e long L = (1.5) = (1-3 ) in. b) V = -e lat e long =.35 e lat = -.35 (-.3844) = d = e lat d = (.5) =.6727 d =d + d =.5673 in The elastic ortion of the stress strain diagram for a steel alloy is shown in the figure. The secimen from which it was obtained had an original diameter of 13 mm and a gauge length of 5 mm. When the alied load on the secimen is 5 kn, the diameter is mm. Determine oisson s ratio for the material. s(ma) 4 Normal Stress: s = = 5(13 ) = Ma 4 (.132 ).2 (mm/mm) Normal Strain: From the stress strain diagram, the modulus of elasticity E = 4(16 ) = 2 Ga. lying Hooke s law.2 e long = s E = 376.7(16 ) 2(1 4 ) = B mm>mm e lat = d - d = = B mm>mm d 13 oisson s Ratio: The lateral and longitudinal strain can be related using oisson s ratio. V = - e lat e long = (1-3 ) (1-3 ) =.3 2

21 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently *3 28. The elastic ortion of the stress strain diagram for a steel alloy is shown in the figure. The secimen from which it was obtained had an original diameter of 13 mm and a gauge length of 5 mm. If a load of = 2 kn is alied to the secimen, determine its diameter and gauge length. Take n =.4. s(ma) 4 Normal Stress: s = = 2(13 ) 4 (.132 ) = 15.68Ma.2 (mm/mm) Normal Strain: From the Stress Strain diagram, the modulus of elasticity E = 4(16 ) = 2 Ga. lying Hooke s Law.2 e long = s E = 15.68(16 ) 2(1 9 ) = B mm>mm Thus, dl = e long L = B(5) =.3767 mm L = L + dl = = mm oisson s Ratio: The lateral and longitudinal can be related using oisson s ratio. e lat = -ve long = -.4(.7534)1-3 B = B mm>mm dd = e lat d = B(13) = mm d = d + dd = 13 + (-.3918) = mm The aluminum block has a rectangular cross section and is subjected to an axial comressive force of 8 ki. If the 1.5-in. side changed its length to in., determine oisson s ratio and the new length of the 2-in. side. E al 1(1 3 ) ksi. 8 ki 3 in. 1.5 in. 2 in. 8 ki s = = 8 = ksi (2)(1.5) e long = s E = (1 3 ) = e lat = =.88 v = =.33 h = (2) = in. 21

22 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently 3 3. The block is made of titanium Ti-61-4V and is subjected to a comression of.6 in. along the y axis, and its shae is given a tilt of u = Determine x, y, and g xy. y Normal Strain: e y = dl y = -.6 = -.15 in.>in. L y 4 oisson s Ratio: The lateral and longitudinal strain can be related using oisson s ratio. e x = -ve y = -.36(-.15) 4 in. u 5 in. x =.54 in. >in. Shear Strain: b = = 9.3 = rad g xy = 2 - b = = rad The shear stress strain diagram for a steel alloy is shown in the figure. If a bolt having a diameter of.75 in. is made of this material and used in the double la joint, determine the modulus of elasticity E and the force required to cause the material to yield. Take n =.3. The shear force develoed on the shear lanes of the bolt can be determined by considering the equilibrium of the FBD shown in Fig. a : + F x = ; V + V - = V = = 2 t(ksi) /2 /2 g(rad) From the shear stress strain diagram, the yield stress is t y = 6 ksi. Thus, t y = V y ; 6 = > B = 53.1 ki = 53. ki From the shear stress strain diagram, the shear modulus is Thus, the modulus of elasticity is G = G = 6 ksi.545 = 11.1(13 ) ksi E 2(1 + y) ; 11.1(13 ) = E 2(1 +.3) E = 28.6(1 3 ) ksi 22

23 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently *3 32. shear sring is made by bonding the rubber annulus to a rigid fixed ring and a lug. When an axial load is laced on the lug, show that the sloe at oint y in the rubber is dy>dr =-tan g = - tan1>12hgr22. For small angles we can write dy>dr = ->12hGr2. Integrate this exression and evaluate the constant of integration using the condition that y = at r = r o. From the result comute the deflection y = d of the lug. r o d r i r y h y Shear Stress Strain Relationshi: lying Hooke s law with t =. 2 r h g = t G = 2 h G r dy dr = -tan g = -tan a 2 h G r b (Q.E.D) If g is small, then tan g = g. Therefore, t r = r o, y = Then, y = 2 h G ln r o r t r = r i, y = d dy dr = - 2 h G r dr y = - 2 h G L r y = - = - C = 2 h G ln r + C 2 h G ln r o + C 2 h G ln r o d = 2 h G ln r o r i 23

24 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently The suort consists of three rigid lates, which are connected together using two symmetrically laced rubber ads. If a vertical force of 5 N is alied to late, determine the aroximate vertical dislacement of this late due to shear strains in the rubber. Each ad has cross-sectional dimensions of 3 mm and 2 mm. G r =.2 Ma. C 4 mm B 4 mm t avg = V = 2.5 = a (.3)(.2) 5 N g = t G = =.283 rad.2(1 6 ) d = 4(.283) =.833 mm shear sring is made from two blocks of rubber, each having a height h, width b, and thickness a. The blocks are bonded to three lates as shown. If the lates are rigid and the shear modulus of the rubber is G, determine the dislacement of late if a vertical load is alied to this late. ssume that the dislacement is small so that d = a tan g L ag. d h verage Shear Stress: The rubber block is subjected to a shear force of V =. 2 Shear Strain: lying Hooke s law for shear Thus, t = V = g = t G = 2 b h = 2 b h 2 b h G = 2 b h G a a d = a g = = a 2 b h G 24

25 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently The elastic ortion of the tension stress strain diagram for an aluminum alloy is shown in the figure. The secimen used for the test has a gauge length of 2 in. and a diameter of.5 in. When the alied load is 9 ki, the new diameter of the secimen is in. Comute the shear modulus for the aluminum. G al From the stress strain diagram, E al = s e = 7 = ksi.614 When secimen is loaded with a 9 - ki load, s = = 9 4 (.5)2 = ksi e long = s E = =.428 in.>in s(ksi) (in./in.) e lat = d -d d = = -.13 in.>in. e lat -.13 V = - = - e long.428 = G al = E at 2(1 + v) = 11.4(1 3 ) 2( ) = 4.31(13 ) ksi *3 36. The elastic ortion of the tension stress strain diagram for an aluminum alloy is shown in the figure. The secimen used for the test has a gauge length of 2 in. and a diameter of.5 in. If the alied load is 1 ki, determine the new diameter of the secimen. The shear modulus is G al = ksi. s = = 1 From the stress strain diagram E = G = 4 (.5)2 = ksi 7 = ksi.614 e long = s E = = in.>in E 2(1 + v) ; 3.8(13 ) = (1 + v) ; v =.5 e lat = - ve long = -.5(.44673) = in.>in. d = e lat d = (.5) = in. s(ksi) (in./in.) d =d + d = =.4989 in. 25

26 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently The s diagram for elastic fibers that make u human skin and muscle is shown. Determine the modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience. 55 s(si) (in./in.) E = 11 2 = 5.5 si u t = 1 2 (2)(11) + 1 ( )(2.25-2) = si 2 u t = 1 (2)(11) = 11 si short cylindrical block of 661-T6 aluminum, having an original diameter of 2 mm and a length of 75 mm, is laced in a comression machine and squeezed until the axial load alied is 5 kn. Determine (a) the decrease in its length and (b) its new diameter. a) s = = -5(13 ) 4 (.2)2 = Ma s = E e long ; (1 6 ) = 68.9(1 9 ) e long e long = mm>mm d = e long L = -.231(75) = mm b) e lat e lat v = - ;.35 = - e long e lat =.885 mm>mm d = e lat d =.885(2) =.16 mm d =d + d = = 2.16 mm 26

27 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently The rigid beam rests in the horizontal osition on two 214-T6 aluminum cylinders having the unloaded lengths shown. If each cylinder has a diameter of 3 mm, determine the lacement x of the alied 8-kN load so that the beam remains horizontal. What is the new diameter of cylinder after the load is alied? n al = mm x 8 kn B 21 mm a + M = ; F B (3) - 8(x) = ; F B = 8x (1) 3 3 m 8(3 - x) a + M B = ; -F (3) + 8(3 - x) = ; F = (2) 3 Since the beam is held horizontally, d = d B s = ; e = s E = E d = el = a E b L = L E 8(3 - x) 8x 3 (22) 3 d = d B ; = (21) E E 8(3 - x)(22) = 8x(21) x = 1.53 m From Eq. (2), F = 39.7 kn s = F = 39.7(13 ) = Ma 4 (.32 ) e long = s E = (16 ) 73.1(1 9 ) = e lat = -ve long = -.35(-.756) =.2646 d œ = d + d e lat = 3 + 3(.2646) = 3.8 mm *3 4. The head H is connected to the cylinder of a comressor using six steel bolts. If the claming force in each bolt is 8 lb, determine the normal strain in the 3 bolts. Each bolt has a diameter of If and E st = in. s Y = 4 ksi H 2 ksi, what is the strain in each bolt when the nut is unscrewed so that the claming force is released? LC Normal Stress: s = = B 2 = ksi 6 s g = 4 ksi Normal Strain: Since s 6 s g, Hooke s law is still valid. e = s E = (1 3 =.999 in.>in. ) If the nut is unscrewed, the load is zero. Therefore, the strain e = 27

28 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently The stone has a mass of 8 kg and center of gravity at G. It rests on a ad at and a roller at B.The ad is fixed to the ground and has a comressed height of 3 mm, a width of 14 mm, and a length of 15 mm. If the coefficient of static friction between the ad and the stone is m s =.8, determine the aroximate horizontal dislacement of the stone, caused by the shear strains in the ad, before the stone begins to sli. ssume the normal force at acts 1.5 m from G as shown. The ad is made from a material having E = 4 Ma and n = m B G 1.25 m 1.5 m.3 m Equations of Equilibrium: a + M B = ; F (2.75) (1.25) - (.3) = [1] : + F x = ; - F = [2] Note: The normal force at does not act exactly at. It has to shift due to friction. Friction Equation: F = m s F =.8 F [3] Solving Eqs. [1], [2] and [3] yields: verage Shear Stress: The ad is subjected to a shear force of V = F = N. Modulus of Rigidity: F = N F = = N G = t = V = = ka (.14)(.15) E 2(1 + v) = 4 = Ma 2(1 +.35) Shear Strain: lying Hooke s law for shear g = t G = (13 ) 1.481(1 6 ) =.15 rad Thus, d h = hg = 3(.15) = 3.2 mm 28

29 3 Solutions 466 5/7/1 8:45 M age earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently The bar D is rigid and is originally held in the horizontal osition when the weight W is suorted from C. E If the weight causes B to be dislaced downward.25 in., determine the strain in wires DE and BC. lso, if the wires 3 ft are made of -36 steel and have a cross-sectional area of.2 in 2, determine the weight W. 2 ft 3 ft 3.25 = 5 d d =.417 in e DE = d L =.417 3(12) =.116 in.>in. s DE = Ee DE = 29(1 3 )(.116) = ksi D B C 4 ft W F DE = s DE DE = (.2) =.672 ki a+ M = ; -(.672) (5) + 3(W) = W =.112 ki = 112 lb s BC = W =.112 = ksi BC.2 e BC = s BC E = (1 3 =.193 in.>in. ) 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 2 mm. If the original lengths of the bolt and sleeve are 8 mm and 5 mm, resectively, 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. ssume the material at is rigid. E al = 7 Ga, E mg = 45 Ga. 5 mm 3 mm Normal Stress: s b = b = 8(1 3 ) = Ma 4 (.82 ) s s = 8(1 3 ) = s 4 ( = Ma ) Normal Strain: lying Hooke s Law e b = s b = (16 ) E al 7(1 9 =.227 mm>mm ) e s = s s = 39.79(16 ) E mg 45(1 9 =.884 mm>mm ) 29

30 3 Solutions 466 5/7/1 8:45 M age 3 21 earson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently *3 44. The -36 steel wire B has a cross-sectional area of 1 mm 2 and is unstretched when u = 45.. Determine the alied load needed to cause u = mm u 4 mm B L B sin 9.2 = 4 sin 44.9 L B = mm L B = e = L B - L B = =.1744 L B s = Ee = 2(1 9 ) (.1744) = Ma a+ M = 4 sin 45 = (4 cos.2 ) - F B sin 44.9 (4) = (1) However, F B = s = (1 6 )(1)(1-6 ) = kn From Eq. (1), = 2.46 kn 3

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