A concrete cylinder having a a diameter of of in. mm and elasticity. Stress and Strain: Stress and Strain: 0.

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1 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently concrete cylinder having a a diameter of of in. mm and and gauge gauge length length of 12 of in. 300 is tested mm is in tested compression.the in compression. results The of results the test of are the reported test are in reported the table in as the load table versus as contraction. load versus contraction. Draw the stress strain Draw the diagram stress-strain using scales diagram of 1 using in. = 0.5 scales ksi of and 101 mm in. = Ma -3 and 2 in.>in. 10 mm From = 0.1(10 the diagram, 3 ) mm/mm. determine From the approximately diagram, determine the modulus approximately of elasticity. the modulus of elasticity. Stress and Strain: Stress and Strain: = e = dl s 5 (ksi) (in./in.) (Ma) e 5 dl L (mm/mm) Load Load (kn) (kip) Contraction (mm) (in.) Modulus of Elasticity: From the stress strain diagram = E 0 approx = B ksi = 26.67(103 ) Ma = Ga 16 s (Ma) (mm/mm) SM_CH08.indd 507 4/11/11 9:54:20 M

2 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 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 point. lot the diagram, and determine the modulus of elasticity and the modulus of resilience. Modulus of Elasticity: From the stress strain diagram E 5 = = B ksi (103 ) Ma = Ga S (Ma) (ksi) e (mm/mm) (in./in.) Modulus of Resilience: The modulus of resilience is equal to the area under the linear portion of the stress strain diagram (shown shaded). 420 u t = 1 2 (33.2)103 B lb in in in. = 9.96 in # lb u t = 1 2 (232.4)a N mm mm2b a mm b = N mm/mm3 = MJ/m 3 in e (mm/mm) 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 point. lot the diagram, and determine approximately the modulus of toughness. The rupture stress is s r = ksi. Ma Modulus of Toughness: The modulus of toughness is equal to the area under the stress strain diagram (shown shaded). (Ma) S (ksi) e (mm/mm) (in./in.) (33.2)103 B lb in. (u t ) approx = 1 2 ( ) 2 (232.4)a N mm2b( )amm in mm b in B lb in a N 2 (0.0012) mm 2b(0.0012)amm in in. b (7.90)103 B lb in (0.0012) 2 (55.3)a N mm 2b(0.0012)amm in b in. (12.3)103 B lb in (0.0004) 2 (86.1)a N mm 2b(0.0004)amm in b in. = in N # lb mm/mm 3 in 3 = MJ/m 3 s (Ma) e (mm/mm) SM_CH08.indd 508 4/11/11 9:54:20 M

3 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently *3 4. *8 4. tension test was performed on a specimen having an original diameter of 12.5 mm and a gauge length of 50 mm. The data are listed in the table. lot the stress strain diagram, and determine approximately the modulus of elasticity, the ultimate stress, and the fracture stress. Use a scale of 20 mm = 50 Ma and 20 mm = 0.05 mm>mm. Redraw the linear-elastic region, using the same stress scale but a strain scale of 20 mm = 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) approx = (106 ) = 229 Ga Ultimate and Fracture Stress: From the stress strain diagram (s m ) approx = 528 Ma (s f ) approx = 479 Ma SM_CH08.indd 509 4/11/11 9:54:21 M

4 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently tension test was performed on a steel specimen having an original diameter of 12.5 mm and gauge length of 50 mm. Using the data listed in the table, plot the stress strain diagram, and determine approximately the modulus of toughness. Use a scale of 20 mm = 50 Ma and 20 mm = 0.05 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 approximated by counting the number of squares. The total number of squares is 187. Load (kn) Elongation (mm) (u t ) approx = 187(25)10 6 B N m 2 a0.025 m b = 117 MJ>m3 m SM_CH08.indd 510 4/11/11 9:54:21 M

5 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently specimen is originally ft mm long, long, has has a a diameter of of in., mm, and is and subjected is subjected to a force to a of force 500 lb. of When 2.5 kn. the When force the is increased force is increased from 500 lb from to kn lb, the to 9 specimen kn,, the elongates specimen elongates in Determine mm. Determine the modulus the modulus of elasticity of elasticity for the for the material material if it if remains it remains linear linear elastic. elastic. Normal Stress and Strain: pplying s = and e = dl. L s = (103 ) = ksi Ma p 4 (0.52 (12) 2 ) s = (103 ) 5 = Ma ksi p ( (12 2 ) e = e = mm/mm in.>in Modulus of Elasticity: E = s e = 5 = 76.64( ) B Ma ksi = Ga structural member in a nuclear reactor is made of a zirconium alloy. If an axial load of 20 4 kip kn is to be supported 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 13 m ft long and its elongation is mm? in.? E 14(10 3 zr = 100 Ga, ) ksi, s Y s Y = = Ma. ksi. The material has elastic behavior. llowable Normal Stress: F.S. = s y s allow 3 = s allow s allow = ksi Ma s allow = = 20(103 4 ) = mmin 2 2 = in 2 Stress Strain Relationship: pplying Hooke s law with e = d L = = mm/mm in.>in. 1(10 3 (12) 3 ) s = Ee = = 100( )(0.0005) 3 B ( ) = 50 Ma = ksi Normal Force: pplying equation s =. = s = = 50(150) (0.2087) = 7500 N = kn kip SM_CH08.indd 511 4/11/11 9:54:21 M

6 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently *3 8. *8 8. The strut is supported by a pin at C and an -36 steel guy wire B. If the wire has a diameter of mm, in., determine how much it stretches when the distributed load acts on the strut kn/m lb/ft Here, we are only interested in determining the force in wire B. C ftm B a+ M C = 0; F 60 (9) - B cos 60 (2.7) 1 (200)(9)(3) = lb 2 (3.4)(2.7)(0.9) = 0 F B = 3.06 kn The normal stress the wire is s B = F B B = 3.06( ) p 4 (0.22 p = 19.10(103 ) psi = ksi 4 (5 2 = Ma ) Since s B 6 s y = ksi Ma,, Hooke s Law Law can can be be applied to determine the strain in wire. s B = E B ; = 29.0(10 200(10 3 )e ) B e B B = ( ( ) mm/mm in>in The unstretched length of the wire is stretches 2.7(10 L B = 9(12) 3 ) = inthus,. the the wire wire sin 60 d B = B L B = ( ( )( ) )(124.71) = mm in. 1 (3.4)(2.7) kn m 1.8 m SM_CH08.indd 512 4/11/11 9:54:22 M

7 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently The s diagram for a collagen fiber bundle from which a human tendon is composed is shown. If a segment of the chilles tendon at has a a length of of mm in. and an approximate cross-sectional area of mm in 2, determine its elongation if the foot supports a load of lb, N, which causes a tension in the tendon of lb. N. s (Ma) (ksi) s 5 = = 1.50 ksi = Ma From the graph e = mm/mm in.>in Nlb (mm/mm) (in./in.) d = el = 0.035(6.5) 0.035(165) = mm in The The stress-strain stress strain diagram diagram for for a metal metal alloy alloy having having an an original original diameter diameter of of in. mm and and a gauge a gauge length length of 2 of in. 50 is mm given is in given the in figure. the figure. Determine Determine approximately approximately the the modulus modulus of elasticity of elasticity for the for material, the material, the load the on load the on specimen the specimen that causes that yielding, causes yielding, and the ultimate and the load ultimate the specimen load the will specimen support. will support. From the stress strain diagram, Fig. a, E ksi Ma - 0 = ; ; E = 30.0( Ga) ksi s y = ksi Ma s u>t s u/t = = ksi Ga Thus, Y = 60C p Y = s Y = 290[ p 4(12 4 ( )] )D = 32.80(10 = kip ) N = kip kn u>t u>t = 100C p u/t = s u/t = 500[ p 4(124 2 (0.52 )] = 62.20(10 )D = ) N kip = = 19.6 knkip s (ksi) (in./in.) (Ma) y 300 = E B 0 (mm/m) / 0.08/ / / / / / p Elastic Recovery (a) SM_CH08.indd 513 4/11/11 9:54:23 M

8 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently The stress strain diagram for a steel alloy having an original diameter of mm in. and a gauge length of 50 2 in. mm is given is given in in the the figure. If If the the specimen is is loaded until it is stressed to ksi, Ma, determine the approximate 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 ksi Ma - 0 = ; E = (103 Ga) ksi when the specimen is unloaded, its normal strain recovered along line B, Fig. a, which has a gradient of E. Thus = 90 E = 90 ksi Elastic Recovery (10 3 = in>in E = 500 Ma = mm/mm 290(10 3 ) Ma ) ksi Thus, the permanent mount of set Elastic is Recovery = (50 mm) = mm Thus, the permanent = set is = in>in Then, the increase in gauge length e = 0.08 is = mm/mm Then, the L increase = L in gauge = 0.047(2) length = is in L = e L = (50 mm) = mm (Ma) E B 0 (mm/mm) / 0.08/ / / / / / p Elastic Recovery (a) SM_CH08.indd 514 4/11/11 9:54:24 M

9 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently *3 12. *8 12. The stress strain diagram for a steel alloy having an original diameter of mm in. and a a gauge length of of 502 mm 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. Thus, s L = ksi Ma L e L = = in>in. mm/mm (u i ) r = 1 2 s L L = 1 2 C60(103 )D(0.002) = 60.0 in # lb [(290)](0.001) = Ma in 3 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 Thus. Thus, C(u i ) t D approx = 38 c15(10 3 ) lb in d a in in b = 28.5(103 ) in # lb [(u i ) t ] approx = 33[100 Ma]a0.04 mm b = 132 Ma mm in 3 s (ksi) 105 (Ma) L = E B (in./in.) (mm/m) / / / / / / / (a) SM_CH08.indd 515 4/11/11 9:54:24 M

10 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently bar bar having having a a length length of of 5 in. 125 and mm cross-sectional and crosssectional area of 0.7 area in 2 is of subjected mm to 2 is an subjected axial force to an of 8000 axial lb. force If the of 40 bar kn. stretches If the bar stretches in., determine 0.05 mm, the determine modulus the of elasticity modulus of of the elasticity material. of the The material. material The has linear-elastic material has behavior. linear-elastic behavior knlb in. mm knlb Normal Stress and Strain: s 5 = 8.00 = ksi = 40(103 ) Ma e 5 = dl = in.>in. L = mm/mm 1255 Modulus of Elasticity: = E 5 s = 28.6(103 ) ksi e = (103 ) Ma = Ga The rigid pipe is supported by a pin at and an -36 steel guy wire BD. If the wire has a diameter of mm, in., determine how much it it stretches when a load of = 3600 kn acts lb acts on on the the pipe. pipe. B Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a ftm a+ M = 0; F BD 4 5 B(3) (0.9)- 600(6) 3(1.8) = 0 F BD BD = = kn lb The normal stress developed in the wire is s BD = F BD BD = 7.5( ) pp 4 (0.252 ) = 30.56(103 ) psi = ksi 4 ( Ma ) D ftm ftm C Since s BD 6 s y = ksi Ma,, Hooke s Law Law can can be be applied to to determine the the strain in in the wire. s BD = E BD ; = 200( (10 3 )e 3 ) B BD BD e = 1.054(10-3 BD = (10 3 ) ) in.>in. mm/mm The unstretched length of the wire is wire stretches 2 2 L BD = = = 5ft 1.5 = m. 60 Thus, in. Thus, the the d BD = BD L BD = ( ( )(60) )(1.5)(10 3 ) = inmm 3 kn 0.9 m 0.9 m SM_CH08.indd 516 4/11/11 9:54:25 M

11 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently The rigid pipe is supported by a pin at and an -36 guy wire BD. If the wire has a a diameter of of mm, in., determine the load if the end C is is displaced mm in. downward. B ftm D ftm ftm C Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a a+ M = 0; F BD 4 5 B(3) (0.9)- (6) (1.8) = = 00 F BD = 2.50 The unstretched length for wire BD is L BD = = = 51.5 ft m. = 60 From in. From the the geometry shown in Fig. b, the stretched length of wire BD is 2 2 L BD = (60)(0.075) ( ) cos = = mm Thus, the normal strain is BD = L BD - L BD L BD = Then, the normal stress can be obtain by applying Hooke s Law. s BD = E BD = 200(10 29(10 3 )C1.0003(10 3 )[1.0000( )D )] = Ma ksi Since s BD 6 s y = ksi Ma,, the the result result is valid. is valid = ( ( ) 3 ) mm/mm in.>in s BD = F BD ; 29.01( p 3 ) = BD 4 (6 2 ) 2.50 p 4 (0.252 ) = N lb = = knlb L BD = 1.5 m 0.9 m 0.9 m mm SM_CH08.indd 517 4/11/11 9:54:26 M

12 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently *3 16. *8 16. Determine the elongation of the square hollow bar when it is subjected to the axial force = 100 kn. If this axial force is increased to = 360 kn and released, find the permanent elongation of the bar. The bar is made of a metal alloy having a stress strain diagram which can be approximated as shown. s (Ma) mm 50 mm mm (mm/mm) 50 mm 5 mm Normal Stress and Strain: The cross-sectional area of the hollow bar is = = 0.9(10-3 )m 2. When = 100 kn, s 1 = = 100(103 ) 0.9(10-3 = Ma ) From the stress strain diagram shown in Fig. a, the slope of the straight line O which represents the modulus of elasticity of the metal alloy is E = 250(106 ) = 200 Ga Since s Ma, Hooke s Law can be applied. Thus s 1 = Ee 1 ; (10 6 ) = 200(10 9 )e 1 e 1 = (10-3 ) mm>mm Thus, the elongation of the bar is d 1 = e 1 L = (10-3 )(600) = mm When = 360 kn, s 2 = = 360(103 ) 0.9(10-3 = 400 Ma ) From the geometry of the stress strain diagram, Fig. a, e = e 2 = mm>mm When = 360 kn is removed, the strain recovers linearly along line BC, Fig. a, parallel to O. Thus, the elastic recovery of strain is given by s 2 = Ee r ; 400(10 6 ) = 200(10 9 )e r e r = mm>mm The permanent set is e = e 2 - e r = = mm>mm Thus, the permanent elongation of the bar is d = e L = (600) = 17.1 mm SM_CH08.indd 518 4/11/11 9:54:26 M

13 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently Continued SM_CH08.indd 519 4/11/11 9:54:26 M

14 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently tension test was performed on an aluminum 2014-T6 alloy specimen. The resulting stress strain diagram is shown in the figure. Estimate (a) the proportional limit, (b) the modulus of elasticity, and (c) the yield strength based on a 0.2% strain offset method. s (ksi) (Ma) (mm/mm) (in./in.) roportional Limit and Yield Strength: From the stress strain diagram, Fig. a, s pl = ksi Ma s Y = ksi Ma Modulus of Elasticity: From the stress strain diagram, the corresponding strain for s L = ksi Ma is is e pl e pl = = in.>in. mm/mm. Thus, Thus, E 5 = = 11.0(103 ) ksi (103 ) Ma = 77.0 Ga Modulus of Resilience: The modulus of resilience is equal to the area under the s (Ma) e (mm/mm) SM_CH08.indd 520 4/11/11 9:54:27 M

15 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently tension test was performed on an aluminum 2014-T6 alloy specimen. The resulting stress strain diagram is shown in the figure. Estimate (a) the modulus of resilience; and (b) modulus of toughness. s (Ma) (ksi) (mm/mm) (in./in.) stress strain diagram up to the proportional limit. From the stress strain diagram, s pl = ksi Ma e pl = mm/mm in.>in. Thus, U i B r = 1 2 s ple pl = 1 2 (44)(103 )(0.004) = 88 in # lb (308)(0.004) = MJ/m 3 in 3 Modulus of Toughness: 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 of squares is 65. Thus, CU i B t D approx = 65B10(10 3 ) lb in R 2 c0.01in. in. d = 6.50(103 ) in # lb 65[70 Ma]c 0.01 mm d = 45.5 MJ/m3 mm 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 0.3% offset. s 0.45(10 6 )s (10 12 )s 3 e = 0.45(10-6 )s (10-12 )s 3, d =0.45(10-6 ) (10-12 ) s 2 Bds E = ds d 2 s = 0 1 = = 2.22 Ma 0.45(10-6 ) SM_CH08.indd 521 4/11/11 9:54:27 M

16 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently *3 20. *8 20. 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 200-mmlong region just before it fractures if failure occurs at =0.12 mm>mm. s 0.45(10 6 )s (10 12 )s 3 When e = (10 3 ) = 0.45 s (10-6 )s 3 Solving for the real root: s = ka u t = d = ( e)ds L L u t = ( (10-6 )s (10-12 )s 3 )ds L 0 0 = 0.12 s (10-6 )s (10-12 )s = 613 kj>m 3 d = el = 0.12(200) = 24 mm SM_CH08.indd 522 4/11/11 9:54:27 M

17 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. E p = 2.70 Ga, n p = N 200 mm 300 N s = = 300 p 4 (0.015)2 = Ma e long = s E = 1.697(106 ) 2.70(10 9 ) = d = e long L = (200) = mm e lat = -Ve long = -0.4( ) = d = e lat d = (15) = mm 523 SM_CH08.indd 523 4/11/11 9:54:28 M

18 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently The short cylindrical block of 2014-T6 aluminum, having an original diameter of of mm in. and a a length of of mm, in., is placed in the smooth jaws of a vise and squeezed until the axial load applied is kn. lb. Determine (a) the decrease in its length and (b) its new diameter knlb knlb a) = 800 s 5 = psi = 4(10 3 ) p 4 (12 (0.5) Ma ) 4 e = (10 6 ) = long 5 s E = (10 6 ) d 5 = e = (1.5) = (10-3 long L (37.5) mm ) in. b) V = -e lat y = 0.35 e long e lat = ( ) ( ) = = d = e lat d = (12) (0.5) = = mm d =d + d = mm in The elastic portion of the stress strain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. When the applied load on the specimen is 50 kn, the diameter is mm. Determine oisson s ratio for the material. s(ma) 400 Normal Stress: s = = 50(103 ) p = Mpa 4 ( ) (mm/mm) Normal Strain: From the stress strain diagram, the modulus of elasticity E = 400(106 ) = 200 Ga. pplying Hooke s law e long = s E = (106 ) 200(10 4 ) = B mm>mm e lat = d - d = = B mm>mm d 0 13 oisson s Ratio: The lateral and longitudinal strain can be related using oisson s ratio. V = - e lat e long = (10-3 ) (10-3 ) = SM_CH08.indd 524 4/11/11 9:54:29 M

19 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently *3 28. *8 24. The elastic portion of the stress strain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. If a load of = 20 kn is applied to the specimen, determine its diameter and gauge length. Take n = 0.4. s(ma) 400 Normal Stress: s = = 20(103 ) p 4 ( ) = Mpa (mm/mm) Normal Strain: From the Stress Strain diagram, the modulus of elasticity E = 400(106 ) = 200 Ga. pplying Hooke s Law e long = s E = (106 ) 200(10 9 ) = B mm>mm Thus, dl = e long L 0 = B(50) = mm L = L 0 + dl = = mm oisson s Ratio: The lateral and longitudinal can be related using poisson s ratio. e lat = -ve long = -0.4(0.7534)10-3 B = B mm>mm dd = e lat d = B(13) = mm d = d 0 + dd = 13 + ( ) = mm The aluminum block has a rectangular cross section and is subjected to an axial compressive force of 840 kip. kn. If the 37.5-mm 1.5-in. side side changed its its length to mm, in., determine oisson s ratio and the new length of of the the 502-in. mm side. E 10(10 3 al = 70 Ga. ) ksi mmin. 8 kip 2 in. 40 kn 50 mm 40 8 kip kn 753 mm in. = 8 s 5 = ksi = 40(10 3 ) Ma (50)(37.5) (2)(1.5) e = long 5 s E = (10 10(10 3 ) 3 ) 5 = e lat 5 = 5 = v 5 = = h h9 5 = (50) (2) = mm in SM_CH08.indd 525 4/11/11 9:54:34 M

20 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently The block is made of titanium Ti-61-4V and is subjected to a compression of mm in. along the y axis, and its shape is given a tilt of u = Determine x, y, and g xy. y Normal Strain: e y = dl y 1.5 = = mm/mm in.>in. L y oisson s Ratio: The lateral and longitudinal strain can be related using oisson s ratio mm in. u in. mm x e x = -ve y = -0.36( ) = in. mm/mm >in. Shear Strain: b = = 90.3 = rad g xy = p 2 - b = p = rad The shear stress strain diagram for a steel alloy is shown in the figure. If a bolt having a diameter of mm in. is made of this material and used in the double lap joint, determine the modulus of elasticity E and the force required to cause the material to yield. Take n = 0.3. The shear force developed on the shear planes of the bolt can be determined by considering the equilibrium of the FBD shown in Fig. a : + F x = 0; V + V - = 0 V = = 2 2 From the shear stress strain diagram, the yield stress is t y = ksi Ma.. Thus, Thus, t y = V y ; 60 = /2 > p p 4 ( ) B t(ma) t(ksi) /2 /2 g(rad) = kip N = kipkn From the shear stress strain diagram, the shear modulus is Thus, the modulus of elasticity is ksi G 5 = = 11.01(103 ksi (103 ) Ma Ga G = E 2(1 + y) ; 11.01( ) = E 2( ) E = E 28.6( ) Ga ksi SM_CH08.indd 526 4/11/11 9:54:38 M

21 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently *8 28. *3 32. shear spring is made by bonding the rubber annulus to a rigid fixed ring and a plug. When an axial load is placed on the plug, show that the slope at point y in the rubber is dy>dr =-tan g = - tan1>12phgr22. For small angles we can write dy>dr = ->12phGr2. Integrate this expression and evaluate the constant of integration using the condition that y = 0 at r = r o. From the result compute the deflection y = d of the plug. r o d r i r y h y Shear Stress Strain Relationship: pplying Hooke s law with t =. 2p r h g = t G = 2p h G r dy dr = -tan g = -tan a 2p h G r b (Q.E.D) If g is small, then tan g = g. Therefore, t r = r o, y = 0 Then, y = 2p h G ln r o r t r = r i, y = d dy dr = - 2p h G r dr y = - 2p h G L r y = - 0 = - C = 2p h G ln r + C 2p h G ln r o + C 2p h G ln r o d = 2p h G ln r o r i SM_CH08.indd 527 4/11/11 9:54:39 M

22 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently The support consists of three rigid plates, which are connected together using two symmetrically placed rubber pads. If a vertical force of 5 N is applied to plate, determine the approximate vertical displacement of this plate due to shear strains in the rubber. Each pad has cross-sectional dimensions of 30 mm and 20 mm. G r = 0.20 Ma. C 40 mm B 40 mm t avg = V = 2.5 = a (0.03)(0.02) 5 N g = t G = = rad 0.2(10 6 ) d = 40( ) = mm shear spring is made from two blocks of rubber, each having a height h, width b, and thickness a. The blocks are bonded to three plates as shown. If the plates are rigid and the shear modulus of the rubber is G, determine the displacement of plate if a vertical load is applied to this plate. ssume that the displacement 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: pplying 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 SM_CH08.indd 528 4/11/11 9:54:40 M

23 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently The elastic portion of the tension stress strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 50 2 in. mm and and a a diameter of of in. mm. When When the applied the applied load load is 9 kip, is 45 the kn, new the diameter new diameter of the of specimen the specimen is is in. Compute mm. Compute the shear the modulus shear modulus for Gthe al for aluminum. the aluminum. G al From the stress strain diagram, = = 70 E al 5 s = ksi e (103 ) Ma When specimen is loaded with a 945-kN - kip load, = = 9 s 5 = ksi 5 45(103 ) p Ma 4 (12.5) (0.5)2 = = e long 5 s = in.>in. E ( mm/mm ) = d -d e lat 5 d9 d = 5 = mm/mm in.>in. d e lat V 5 = - e lat e 5 = - e long = long s(ksi) s (Ma) (mm/mm) (in./in.) E at = v) = 11.4(10 at G 2( ) = 4.31(103 al 5 ) ksi 2(1 + v) (10 3 ) 2( ) (103 ) Ma Ga *3 36. *8 32. The elastic portion of the tension stress strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a a gauge length of of 502 mm in. and a diameter of in. mm. If If the the applied load is is kip, kn, determine the new diameter of the specimen. The shear modulus is G al = 28 Ga. 2 ksi. = = s 5 = ksi 5 50(103 ) p Ma 4 (12.5) (0.5)2 4 From the stress-strain stress strain diagram E 5 = = ksi (103 ) Ma e = long 5 s = in.>in. E = ( mm/mm ) E G 5 = v) ; 3.8(103 ) = (1 + v) ; 28(103 ) (103 ) ; v 2(1 + v) ; (1 + v) v = e lat 5 = ve - ve long long 5 = ( ) ( ) 5 = in.>in. d d 5 = ee lat lat d d 5 = (12.5) (0.5) 5 = mm in. s(ksi) s (Ma) (mm/mm) (in./in.) d9 d 5 =d d + + d d 5 = = mmin SM_CH08.indd 529 4/11/11 9:54:41 M

24 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently The s 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(ma) s(psi) (in./in.) (mm/mm) E = 11 al 5 77 = psi Ma 22 u = (2)(11) 1 t 5 1 ( )(2.25-2) = psi 2 (2)(77) + 1 ( )(2.25 2) MJ/m3 2 2 u = t 5 1 (2)(77) (2)(11) 5 = MJ/m3 psi short cylindrical block of 6061-T6 aluminum, having an original diameter of 20 mm and a length of 75 mm, is placed in a compression machine and squeezed until the axial load applied is 5 kn. Determine (a) the decrease in its length and (b) its new diameter. a) s = = -5(103 ) p 4 (0.02)2 = Ma s = E e long ; (10 6 ) = 68.9(10 9 ) e long e long = mm>mm d = e long L = (75) = mm b) e lat e lat v = - ; 0.35 = - e long e lat = mm>mm d = e lat d = (20) = mm d =d + d = = mm SM_CH08.indd 530 4/11/11 9:54:42 M

25 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 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 after the load is applied? n al = mm x 80 kn B 210 mm a + M = 0; F B (3) - 80(x) = 0; F B = 80x (1) 3 3 m 80(3 - x) a + M B = 0; -F (3) + 80(3 - x) = 0; 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 80(3 - x) 3 (220) d = d B ; = E 80(3 - x)(220) = 80x(210) 80x 3 (210) E x = 1.53 m From Eq. (2), F = kn s = F = 39.07(103 ) p = Ma 4 (0.032 ) e long = s E = (106 ) 73.1(10 9 ) = e lat = -ve long = -0.35( ) = d œ = d + d e lat = ( ) = mm *3 40. *8 36. The head H is connected to the cylinder of a compressor using six steel bolts. If the clamping force in each bolt is is 4800 kn, lb, determine the the normal normal strain strain the in bolts. the 3 bolts. Each Each bolt has bolt a has diameter a diameter of 5 of mm. If s Y = 40 ksi If in. s Y = 280 Ma and H E st = 200 Ga, 2 ksi, what what is the is strain the strain each in each bolt when bolt when the nut the is nut unscrewed is unscrewed so that so the that clamping the clamping force is force released? is released? LC Normal Stress: s 5 = = B = ksi 6 2 = 40 ksi 5 4(103 ) p 4 (5) Ma < s g Ma Normal Strain: Since s 6 s g, Hooke s law is still valid. e 5 = s = (10 3 = in.>in. E mm/mm 200(10) ) If the nut is unscrewed, the load is zero. Therefore, the strain e = SM_CH08.indd 531 4/11/11 9:54:42 M

26 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently The stone has a mass of 800 kg and center of gravity at G. It rests on a pad at and a roller at B.The pad is fixed to the ground and has a compressed height of 30 mm, a width of 140 mm, and a length of 150 mm. If the coefficient of static friction between the pad and the stone is m s = 0.8, determine the approximate horizontal displacement of the stone, caused by the shear strains in the pad, before the stone begins to slip. ssume the normal force at acts 1.5 m from G as shown. The pad is made from a material having E = 4 Ma and n = m B G 1.25 m 1.5 m 0.3 m Equations of Equilibrium: a + M B = 0; F (2.75) (1.25) - (0.3) = 0 [1] : + F x = 0; - F = 0 [2] Note: The normal force at does not act exactly at. It has to shift due to friction. Friction Equation: F = m s F = 0.8 F [3] Solving Eqs. [1], [2] and [3] yields: verage Shear Stress: The pad is subjected to a shear force of V = F = N. Modulus of Rigidity: F = N F = = N G = t = V = = ka (0.14)(0.15) E 2(1 + v) = 4 = Ma 2( ) Shear Strain: pplying Hooke s law for shear g = t G = (103 ) 1.481(10 6 ) = rad Thus, d h = hg = 30(0.1005) = 3.02 mm SM_CH08.indd 532 4/11/11 9:54:43 M

27 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently The bar D is rigid and is originally held in the horizontal position when the weight W is supported from C. E If the weight causes B to be displaced downward mm, in., determine the strain in wires DE and BC. lso, if the wires 0.93 mft are made of -36 steel and have a cross-sectional area of mm in 22,, determine the weight W ft m ft m D B = 1.5 d ft m d = in mm C e DE = d = in.>in. L = = mm/mm 0.9(1000) 3(12) W s = 29(10 3 DE = Ee DE = 200(10 3 )( ) )( ) = = ksi Ma F DE = s DE DE DE DE = = (1.25) (0.002) = = N kip a+ M = 0; (289.35)(1.5) -(0.0672) (5) + 0.9(W) 3(W) = m 0.9 m W = kip N = 112 lb s BC = BC = W = = = Ma ksi BC 1.25 BC e BC BC = BC = s BC E = (10 29 (10 3 = 3 = ) mm/mm in.>in. ) F DE = N 0.6 m 0.9 m 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. ssume the material at is rigid. E al = 70 Ga, E mg = 45 Ga. 50 mm 30 mm Normal Stress: s b = b = 8(10 3 ) p = Ma 4 ( ) s s = 8(10 3 ) = p s 4 ( = Ma ) Normal Strain: pplying Hooke s Law e b = s b E al = (106 ) 70(10 9 ) e s = = mm>mm s s = 39.79(106 ) E mg 45(10 9 = mm>mm ) SM_CH08.indd 533 4/11/11 9:54:44 M

28 2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently *3 44. *8 40. The -36 steel wire B has a cross-sectional area of 10 mm 2 and is unstretched when u = Determine the applied load needed to cause u = mm u 400 mm B L B sin 90.2 = 400 sin 44.9 L B = mm L B = e = L B - L B = = L B s = Ee = 200(10 9 ) ( ) = Ma a+ M = sin 45 = (400 cos 0.2 ) - F B sin 44.9 (400) = 0 (1) However, F B = s = (10 6 )(10)(10-6 ) = kn From Eq. (1), = 2.46 kn SM_CH08.indd 534 4/11/11 9:54:44 M