2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 1. 3 1. concrete cylinder having a a diameter of of 6.00 150 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. = 0.2110 2 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) 0 0 0 1.41 0.177 0.00005 0.00005 2.69 0.336 0.00010 0.00010 4.67 0.584 0.000167 5.80 0.725 0.000217 0.000217 7.22 0.902 0.000283 0.000283 8.49 0.000333 1.061 0.000333 9.76 0.000375 1.220 0.000375 10.89 0.000417 1.362 0.000417 13.16 0.000517 14.15 1.645 0.000583 0.000517 15.00 1.768 0.000625 0.000583 1.874 0.000625 Load Load (kn) (kip) Contraction (mm) (in.) 0 0 25.05.0 47.59.5 82.516.5 20.5 102.5 25.5 127.5 30.0 150.0 34.5 172.5 38.5 192.5 46.5 232.5 50.0 53.0 250.0 265.0 00 0.0150 0.0006 0.0300 0.0012 0.0500 0.0020 0.0026 0.0650 0.0034 0.0850 0.0040 0.1000 0.0045 0.1125 0.0050 0.1250 0.0062 0.1750 0.0070 0.0075 0.1850 Modulus of Elasticity: From the stress strain diagram = 1.31 8.0-0 E 0 approx 5 0.0004-0 = 3.275103 B ksi 0.0003 0 = 26.67(103 ) Ma = 26.67 Ga 16 s (Ma) 12 8 4 (mm/mm) 50792 SM_CH08.indd 507 4/11/11 9:54:20 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 2. 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 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 = 232.4 33.2-00 - 0 = 55.3103 B ksi 0.0006 0 5 387.3(103 ) Ma = 387.3 Ga S (Ma) (ksi) 0 0 232.4 33.2 318.5 45.5 345.8 49.4 51.5 360.5 53.4 373.8 e (mm/mm) (in./in.) 0 0 0.0006 0.0010 0.0014 0.0018 0.0018 0.0022 0.0022 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. 2 0.0006 in in. = 9.96 in # lb u t = 1 2 (232.4)a N mm mm2b a0.0006 mm b = 0.0697 N mm/mm3 = 0.0697 MJ/m 3 in 3 350 280 232.4 210 140 70 e (mm/mm) 8 3. 3 3. 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 = 53.4 373.8 ksi. Ma Modulus of Toughness: The modulus of toughness is equal to the area under the stress strain diagram (shown shaded). (Ma) S (ksi) 0 232.4 33.2 318.5 45.5 345.8 49.4 360.5 51.5 53.4 373.8 e (mm/mm) (in./in.) 0 0 0.0006 0.0010 0.0014 0.0018 0.0022 0.0022 1 2 (33.2)103 B lb in. (u t ) approx = 1 2 (0.0004 + 0.0010) 2 (232.4)a N mm2b(0.0004 + 0.0010)amm in mm b in. 45.510 3 B lb in. + 318.5 a N 2 (0.0012) mm 2b(0.0012)amm in in. b (7.90)103 B lb in. + 1 2 (0.0012) 2 (55.3)a N mm 2b(0.0012)amm in b in. (12.3)103 B lb in. + 1 2 (0.0004) 2 (86.1)a N mm 2b(0.0004)amm in b in. = 0.595 85.0 in N # lb mm/mm 3 in 3 = 0.595 MJ/m 3 s (Ma) 420 373.8 350 318.5 280 232.4 210 140 70 e (mm/mm) 93 508 SM_CH08.indd 508 4/11/11 9:54:20 M
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 = 0.001 mm>mm. Stress and Strain: s = e = dl (Ma) L (mm/mm) Load (kn) 0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8 Elongation (mm) 0 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.8900 11.9380 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 Elasticity: From the stress strain diagram (E) approx = 228.75(106 ) - 0 0.001-0 = 229 Ga Ultimate and Fracture Stress: From the stress strain diagram (s m ) approx = 528 Ma (s f ) approx = 479 Ma 50994 SM_CH08.indd 509 4/11/11 9:54:21 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 5. 3 5. 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) 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 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) 0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8 Elongation (mm) 0 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.8900 11.9380 (u t ) approx = 187(25)10 6 B N m 2 a0.025 m b = 117 MJ>m3 m 95 510 SM_CH08.indd 510 4/11/11 9:54:21 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 6. 3 6. specimen is originally 300 1 ft mm long, long, has has a a diameter of of 0.512 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 1800 2.5 kn lb, the to 9 specimen kn,, the elongates specimen elongates 0.009 in. 0.225 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 = 0.500 1 5 2.5(103 ) = 5 2.546 22.10 ksi Ma p 4 (0.52 (12) 2 ) s = 1.80 2 5 9(103 ) 5 = 79.58 9.167 Ma ksi p (0.52 4 (12 2 ) e = 0.009 e 5 0.225 5 = 0.000750 mm/mm in.>in. 300 12 Modulus of Elasticity: E = s e 79.58 9.167-22.10 2.546 = 5 = 76.64(10 8.8310 33 ) B Ma ksi = 76.64 Ga 0.000750 8 7. 3 7. 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 0.5 0.02 mm? in.? E 14(10 3 zr = 100 Ga, ) ksi, s Y s Y = = 400 57.5 Ma. ksi. The material has elastic behavior. llowable Normal Stress: F.S. = s y s allow 3 = 57.5 400 s allow s allow = 19.17 133.33 ksi Ma s allow = 133.33 19.17 5 = 20(103 4 ) = 150 0.2087 mmin 2 2 = 0.209 in 2 Stress Strain Relationship: pplying Hooke s law with e = d L = 0.02 0.5 = 0.000555 mm/mm in.>in. 1(10 3 (12) 3 ) s = Ee = = 100(10 14 10 3 )(0.0005) 3 B (0.000555) = 50 Ma = 7.778 ksi Normal Force: pplying equation s =. = s = = 50(150) 7.778 (0.2087) = 7500 N = 1.62 7.5 kn kip 51196 SM_CH08.indd 511 4/11/11 9:54:21 M
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 0.2 5 mm, in., determine how much it stretches when the distributed load acts on the strut. 60 60 200 3.4 kn/m lb/ft Here, we are only interested in determining the force in wire B. C 2.7 9 ftm B a+ M C = 0; F 60 (9) - B cos 60 (2.7) 1 (200)(9)(3) = 0 600 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(10 600 3 ) p 4 (0.22 p = 19.10(103 ) psi = 19.10 ksi 4 (5 2 = 155.84 Ma ) Since s B 6 s y = 36 250 ksi Ma,, Hooke s Law Law can can be be applied to determine the strain in wire. s B = E B ; 19.10 155.84 = 29.0(10 200(10 3 )e ) B e B B = 0.6586(10 0.7792(10 3-3 ) mm/mm in>in The unstretched length of the wire is stretches 2.7(10 L B = 9(12) 3 ) = 124.71 3117.69. inthus,. the the wire wire sin 60 d B = B L B = 0.6586(10 0.7792(10 3-3 )(3117.69) )(124.71) = 0.0821 2.429 mm in. 1 (3.4)(2.7) kn 2 0.9 m 1.8 m 97 512 SM_CH08.indd 512 4/11/11 9:54:22 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 9. 3 9. 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 165 6.5 mm in. and an approximate cross-sectional area of 145 0.229 mm in 2, determine its elongation if the foot supports a load of 125 625 lb, N, which causes a tension in the tendon of 1718.75 343.75 lb. N. s (Ma) (ksi) 31.50 4.50 26.25 3.75 21.00 3.00 15.75 2.25 s 5 = 343.75 = 1.50 ksi = 1718.75 5 11.85 Ma 145 0.229 From the graph e = 0.035 mm/mm in.>in. 10.50 1.50 5.25 0.75 0.05 0.10 625 125 Nlb (mm/mm) (in./in.) d = el = 0.035(6.5) 0.035(165) = 5.775 0.228 mm in. 3 10. 8 10 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 0.5 12 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 290 60 ksi Ma - 0 = 0.002 - ; ; E = 30.0(103 290 Ga) ksi 1 0.001 0 s y = 290 60 ksi Ma s u>t s u/t = = 100 550 ksi Ga Thus, Y = 60C p Y = s Y = 290[ p 4(12 4 (0.52 2 )] )D = 32.80(10 = 11.78 3 kip ) N = 32.80 11.8 kip kn u>t u>t = 100C p u/t = s u/t = 500[ p 4(124 2 (0.52 )] = 62.20(10 )D = 19.63 3 ) N kip = 62.20 = 19.6 knkip s (ksi) 105 90 75 60 45 30 15 0 (in./in.) 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 (Ma) 500 400 y 300 = 290 200 E 100 1 B 0 (mm/m) 0 0.05/ 0.08/ 0.001 0.002 0.003 0.15/ 0.004 0.20/ 0.005 0.25/ 0.006 0.30/ 0.007 0.35/ p Elastic Recovery (a) 513 98 SM_CH08.indd 513 4/11/11 9:54:23 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 11. 3 11. The stress strain diagram for a steel alloy having an original diameter of 12 0.5 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 500 90 ksi, Ma, determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded. s (ksi) 105 90 75 60 45 30 15 0 (in./in.) 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 From the stress strain diagram Fig. a, the modulus of elasticity for the steel alloy is E 290 60 ksi Ma - 0 = 0.002 - ; E = 290 30.0(103 Ga) ksi 1 0.001 0 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 500 30.0(10 3 = 0.003 in>in E = 500 Ma = 0.001724 mm/mm 290(10 3 ) Ma ) ksi Thus, the permanent mount of set Elastic is Recovery = 0.001724(50 mm) = 0.0862 mm Thus, the permanent = 0.05 -set 0.003 is = 0.047 in>in Then, the increase in gauge length e = 0.08 is 0.001724 = 0.078276 mm/mm Then, the L increase = L in gauge = 0.047(2) length = is 0.094 in L = e L = 0.078276(50 mm) = 3.91379 mm (Ma) 500 400 300 200 E 100 1 B 0 (mm/mm) 0 0.05/ 0.08/ 0.001 0.002 0.003 0.15/ 0.004 0.20/ 0.005 0.25/ 0.006 0.30/ 0.007 0.35/ p Elastic Recovery (a) 99 514 SM_CH08.indd 514 4/11/11 9:54:24 M
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 12 0.5 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 = 290 60 ksi Ma L e L = = 0.002 0.001 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) = 0.145 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 33. 38. Thus. Thus, C(u i ) t D approx = 38 c15(10 3 ) lb in d a0.05 2 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) 500 90 400 75 60 300 L = 290 45 200 30 E 15 100 1 B 0 0 0.05 0 (in./in.) 0.10 0.15 0.20 0.25 0.30 0.35 (mm/m) 0 0 0.001 0.002 0.05/ 0.003 0.08/ 0.004 0.005 0.006 0.007 0.001 0.002 0.003 0.15/ 0.004 0.20/ 0.005 0.25/ 0.006 0.30/ 0.007 0.35/ (a) 515 100 SM_CH08.indd 515 4/11/11 9:54:24 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 13. 3 13. 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 437.5 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 0.002 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. 40 8000 knlb 125 5 in. mm 40 8000 knlb Normal Stress and Strain: s 5 = 8.00 = 11.43 ksi = 40(103 ) 5 95.81 Ma 437.5 0.7 e 5 = dl 0.002 = in.>in. L = 0.05 5 0.000400 mm/mm 1255 Modulus of Elasticity: = 11.43 E 5 s = 28.6(103 ) ksi e = 95.81 0.000400 5 239.525 (103 ) Ma = 239.5 Ga 3 14. 8 14. The rigid pipe is supported by a pin at and an -36 steel guy wire BD. If the wire has a diameter of 60.25 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 1.2 4 ftm a+ M = 0; F BD 4 5 B(3) (0.9)- 600(6) 3(1.8) = 0 F BD BD = = 7.51500 kn lb The normal stress developed in the wire is s BD = F BD BD = 7.5(10 1500 3 ) pp 4 (0.252 ) = 30.56(103 ) psi = 30.56 ksi 4 (6 2 5 265.3 Ma ) D 0.9 3 ftm 0.9 3 ftm C Since s BD 6 s y = 36 250 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 ; 30.56 265.3 = 200(10 29.0(10 3 )e 3 ) B BD BD e = 1.054(10-3 BD = 1.3265(10 3 ) ) in.>in. mm/mm The unstretched length of the wire is wire stretches 2 2 L BD = 23 09. 2 + 12 4. 2 = = 5ft 1.5 = m. 60 Thus, in. Thus, the the d BD = BD L BD = 1.3265(10 1.054(10-3 3 )(60) )(1.5)(10 3 ) = 1.98975 0.0632 inmm 3 kn 0.9 m 0.9 m 101 516 SM_CH08.indd 516 4/11/11 9:54:25 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 3 15. 8 15. The rigid pipe is supported by a pin at and an -36 guy wire BD. If the wire has a a diameter of of 0.25 6 mm, in., determine the load if the end C is is displaced 1.875 0.075 mm in. downward. B 1.2 4 ftm D 0.9 3 ftm 0.9 3 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 = 23 09. 2 + 12 4. 2 = = 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 = 260 1500 2 + 0.075 1. 875 2-2(60)(0.075) ( 1500 1. 875) cos 143.13 = = 60.060017 1501.500 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(10-3 3 )D )] = 200 29.01 Ma ksi Since s BD 6 s y = 250 36 ksi Ma,, the the result result is valid. is valid. 2 2 60.060017 1501.500-1500 60 5 = 1.0000(10 1.0003(10 3 - ) 3 ) mm/mm in.>in. 1500 60 s BD = F BD ; 29.01(10 200 5 2.50 p 3 ) = BD 4 (6 2 ) 2.50 p 4 (0.252 ) = 2261.9 569.57 N lb = = 2.26 570 knlb L BD = 1.5 m 0.9 m 0.9 m 1.875 mm 517 102 SM_CH08.indd 517 4/11/11 9:54:26 M
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) 500 250 600 mm 50 mm 0.00125 0.05 5 mm (mm/mm) 50 mm 5 mm Normal Stress and Strain: The cross-sectional area of the hollow bar is = 0.05 2-0.04 2 = 0.9(10-3 )m 2. When = 100 kn, s 1 = = 100(103 ) 0.9(10-3 = 111.11 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 ) - 0 0.00125-0 = 200 Ga Since s 1 6 250 Ma, Hooke s Law can be applied. Thus s 1 = Ee 1 ; 111.11(10 6 ) = 200(10 9 )e 1 e 1 = 0.5556(10-3 ) mm>mm Thus, the elongation of the bar is d 1 = e 1 L = 0.5556(10-3 )(600) = 0.333 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 2-0.00125 400-250 = 0.05-0.00125 500-250 e 2 = 0.0305 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 = 0.002 mm>mm The permanent set is e = e 2 - e r = 0.0305-0.002 = 0.0285 mm>mm Thus, the permanent elongation of the bar is d = e L = 0.0285(600) = 17.1 mm 103 518 SM_CH08.indd 518 4/11/11 9:54:26 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 16. 3 16. Continued 519 104 SM_CH08.indd 519 4/11/11 9:54:26 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 17. 3 17. 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) 490 70 420 60 350 280 40 210 30 140 20 10 70 00 0.02 0.04 0.06 0.08 0.10 0.002 0.004 0.006 0.008 0.010 (mm/mm) (in./in.) roportional Limit and Yield Strength: From the stress strain diagram, Fig. a, s pl = 308 44 ksi Ma s Y = 420 60 ksi Ma Modulus of Elasticity: From the stress strain diagram, the corresponding strain for s L = 308 44 ksi Ma is is e pl e pl = = 0.004 in.>in. mm/mm. Thus, Thus, E 5 = 308 44-0 - 0 = 11.0(103 ) ksi 0.004 0 5 77.0(103 ) Ma = 77.0 Ga Modulus of Resilience: The modulus of resilience is equal to the area under the s (Ma) 490 420 350 280 210 140 70 e (mm/mm) 105 520 SM_CH08.indd 520 4/11/11 9:54:27 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 3 18. 8 18. 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) 490 70 420 60 350 280 40 210 30 140 20 10 70 0 0.02 0.04 0.06 0.08 0.10 0.002 0.004 0.006 0.008 0.010 (mm/mm) (in./in.) stress strain diagram up to the proportional limit. From the stress strain diagram, s pl = 308 44 ksi Ma e pl = 0.004 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) = 0.616 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 3 19. 8 19. The stress strain diagram for a bone is shown, and can be described by the equation =0.45110-6 2 s 0.36110-12 2 s 3, where s is in ka. Determine the yield strength assuming a 0.3% offset. s 0.45(10 6 )s + 0.36(10 12 )s 3 e = 0.45(10-6 )s + 0.36(10-12 )s 3, d =0.45(10-6 ) + 1.08(10-12 ) s 2 Bds E = ds d 2 s = 0 1 = = 2.22 Ma 0.45(10-6 ) 521 106 SM_CH08.indd 521 4/11/11 9:54:27 M
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 = 0.45110-6 2 s 0.36110-12 2 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 + 0.36(10 12 )s 3 When e = 0.12 120(10 3 ) = 0.45 s + 0.36(10-6 )s 3 Solving for the real root: s = 6873.52 ka 6873.52 u t = d = (0.12 - e)ds L L 6873.52 u t = (0.12-0.45(10-6 )s - 0.36(10-12 )s 3 )ds L 0 0 = 0.12 s - 0.225(10-6 )s 2-0.09(10-12 )s 4 6873.52 0 = 613 kj>m 3 d = el = 0.12(200) = 24 mm 107 522 SM_CH08.indd 522 4/11/11 9:54:27 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 21. 3 25. 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 = 0.4. 300 N 200 mm 300 N s = = 300 p 4 (0.015)2 = 1.697 Ma e long = s E = 1.697(106 ) 2.70(10 9 ) = 0.0006288 d = e long L = 0.0006288 (200) = 0.126 mm e lat = -Ve long = -0.4(0.0006288) = -0.0002515 d = e lat d = -0.0002515 (15) = -0.00377 mm 523 SM_CH08.indd 523 4/11/11 9:54:28 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 22. 3 26. The short cylindrical block of 2014-T6 aluminum, having an original diameter of of 120.5 mm in. and a a length of of 37.5 1.5mm, in., is placed in the smooth jaws of a vise and squeezed until the axial load applied is 800 4 kn. lb. Determine (a) the decrease in its length and (b) its new diameter. 800 4 knlb 800 4 knlb a) = 800 s 5 = 4074.37 psi = 4(10 3 ) p 4 (12 (0.5)2 2 5 35.3678 Ma ) 4 e = -4074.37 10.6(10 6 ) = -0.0003844 long 5 s E = 35.3678 73.1(10 6 ) 5 0.00004838 d 5 = e = -0.0003844 (1.5) = -0.577 (10-3 long L 5 0.0004838(37.5) 5 0.0181 mm ) in. b) V = -e lat y = 0.35 e long e lat = -0.35 ( 0.0004838) (-0.0003844) = = 0.00016933 0.00013453 d = e lat d = 0.00013453 0.00016933(12) (0.5) = = 0.002032 0.00006727 mm d =d + d = 12.002032 0.5000673 mm in. 8 23. 3 27. 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 12.99265 mm. Determine oisson s ratio for the material. s(ma) 400 Normal Stress: s = = 50(103 ) p = 376.70 Mpa 4 (0.0132 ) 0.002 (mm/mm) Normal Strain: From the stress strain diagram, the modulus of elasticity E = 400(106 ) = 200 Ga. pplying Hooke s law 0.002 e long = s E = 376.70(106 ) 200(10 4 ) = 1.883510-3 B mm>mm e lat = d - d 0 12.99265-13 = = -0.5653810-3 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 = - -0.56538(10-3 ) 1.8835(10-3 ) = 0.300 111 524 SM_CH08.indd 524 4/11/11 9:54:29 M
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 (0.0132 ) = 150.68Mpa 0.002 (mm/mm) Normal Strain: From the Stress Strain diagram, the modulus of elasticity E = 400(106 ) = 200 Ga. pplying Hooke s Law 0.002 e long = s E = 150.68(106 ) 200(10 9 ) = 0.753410-3 B mm>mm Thus, dl = e long L 0 = 0.753410-3 B(50) = 0.03767 mm L = L 0 + dl = 50 + 0.03767 = 50.0377 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 = -0.301410-3 B mm>mm dd = e lat d = -0.301410-3 B(13) = -0.003918 mm d = d 0 + dd = 13 + (-0.003918) = 12.99608 mm 8 25. 3 29. 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 1.500132 37.5033 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. 37.51.5 mmin. 8 kip 2 in. 40 kn 50 mm 40 8 kip kn 753 mm in. = 8 s 5 = 2.667 ksi = 40(10 3 ) 5 21.33 Ma (50)(37.5) (2)(1.5) e = -2.667 long 5 s E = 21.33 70(10 10(10 3 ) 3 ) 5 = 0.00030476-0.0002667 37.5033 1.500132 37.5-1.5 e lat 5 = 5 = 0.0000880 37.5 1.5 v 5 = -0.0000880 0.0000880 0.00030476-0.0002667 5 = 0.289 0.330 h h9 5 = 50 2 + 0.0000880(50) 0.0000880(2) = 52.000176 50.0044 mm in. 525 112 SM_CH08.indd 525 4/11/11 9:54:34 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 26. 3 30. The block is made of titanium Ti-61-4V and is subjected to a compression of 1.5 0.06 mm in. along the y axis, and its shape is given a tilt of u = 89.7. Determine x, y, and g xy. y Normal Strain: e y = dl y 1.5 = -0.06 5 = 0.0150-0.0150 mm/mm in.>in. L y 100 4 oisson s Ratio: The lateral and longitudinal strain can be related using oisson s ratio. 1004 mm in. u 125 5 in. mm x e x = -ve y = -0.36(-0.0150) = 0.00540 in. mm/mm >in. Shear Strain: b = 180-89.7 = 90.3 = 1.576032 rad g xy = p 2 - b = p 2-1.576032 = -0.00524 rad 8 27. 3 31. The shear stress strain diagram for a steel alloy is shown in the figure. If a bolt having a diameter of 0.75 20 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 = 60 420 ksi Ma.. Thus, Thus, t y = V y ; 60 = /2 >2 420 5 p p 4 (20 2 4 0.752 ) B t(ma) t(ksi) 420 60 0.00545 /2 /2 g(rad) = 263893.8 53.01 kip N = 53.0 263.89 kipkn From the shear stress strain diagram, the shear modulus is Thus, the modulus of elasticity is 60 420 ksi G 5 = = 11.01(103 ksi 0.00545 5 77.064(103 ) Ma 5 77.064 Ga G = E 2(1 + y) ; 11.01(103 77.064) = E 2(1 + 0.3) E = E 28.6(10 5 200.4 3 ) Ga ksi 113 526 SM_CH08.indd 526 4/11/11 9:54:38 M
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 527 114 SM_CH08.indd 527 4/11/11 9:54:39 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 29. 3 33. 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 = 4166.7 a (0.03)(0.02) 5 N g = t G = 4166.7 = 0.02083 rad 0.2(10 6 ) d = 40(0.02083) = 0.833 mm 8 30. 3 34. 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 115 528 SM_CH08.indd 528 4/11/11 9:54:40 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 31. 3 35. 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 0.5 12.5 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 0.49935 is 12.48375 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 = 11400.65 ksi e 5 500 0.00614 5 81.433(103 ) Ma When specimen is loaded with a 945-kN - kip load, = = 9 s 5 = 45.84 ksi 5 45(103 ) p 2 5 366.69 Ma 4 (12.5) (0.5)2 = = 45.84 e long 5 s = 0.0040208 in.>in. E 5 366.69 81.433(10 11400.65 3 5 0.0045030 mm/mm ) = d -d 0.49935-0.5 e lat 5 d9 d 12.48375 12.5 5 = 5 = 0.0013 - mm/mm in.>in. d 12.5 0.5 e lat 4-0.0013 V 5 = - e lat 0.0013 e 5 = - e long 0.0040208 = 0.32332 long 0.0045030 5 0.28870 s(ksi) s (Ma) 500 70 0.00614 (mm/mm) (in./in.) E at = v) = 11.4(10 at G 2(1 + 0.32332) = 4.31(103 al 5 ) ksi 2(1 + v) 5 81.433(10 3 ) 2(1 + 0.28870) 5 31.60(103 ) Ma 5 31.60 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 12.5 0.5 in. mm. If If the the applied load is is 10 50 kip, kn, determine the new diameter of the specimen. The shear modulus is G 3.8110 3 al = 28 Ga. 2 ksi. = = s 5 = 50.9296 ksi 5 50(103 ) p 2 5 407.44 Ma 4 (12.5) (0.5)2 4 From the stress-strain stress strain diagram 500 70 E 5 = = 11400.65 ksi 0.00614 5 81.433(103 ) Ma e = 50.9296 long 5 s = 0.0044673 in.>in. E = 407.44 81.433(10 11400.65 3 5 0.0050033 mm/mm ) E G 5 = v) ; 3.8(103 ) = 11400.65 2(1 + v) ; 28(103 )5 81.433(103 ) ; v 2(1 + v) ; 5 0.45416 2(1 + v) v = 0.500 e lat 5 = ve - ve long long 5 = 0.45416(0.0050033) - 0.500(0.0044673) 5 = 0.002272-0.002234 in.>in. d d 5 = ee lat lat d d 5 = 0.002272(12.5) - 0.002234(0.5) 5 = 0.0284-0.001117 mm in. s(ksi) s (Ma) 500 70 0.00614 (mm/mm) (in./in.) d9 d 5 =d d + + d d 5 = 12.5 0.5-0.0284 0.001117 5 12.4716 = 0.4989 mmin. 529 116 SM_CH08.indd 529 4/11/11 9:54:41 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 33. 3 37. 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) 385 55 11 77 1 2 2.25 (in./in.) (mm/mm) E = 11 al 5 77 = 55.5 38.5 psi Ma 22 u = (2)(11) 1 t 5 1 (55 + 11)(2.25-2) = 19.25 psi 2 (2)(77) + 1 (385 + 77)(2.25 2) 5 134.75 MJ/m3 2 2 u = t 5 1 (2)(77) (2)(11) 5 = 77 11 MJ/m3 psi 2 3 38. 8 34. 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 = - 15.915 Ma s = E e long ; - 15.915(10 6 ) = 68.9(10 9 ) e long e long = - 0.0002310 mm>mm d = e long L = - 0.0002310(75) = - 0.0173 mm b) e lat e lat v = - ; 0.35 = - e long -0.0002310 e lat = 0.00008085 mm>mm d = e lat d = 0.00008085(20) = 0.0016 mm d =d + d = 20 + 0.0016 = 20.0016 mm 117 530 SM_CH08.indd 530 4/11/11 9:54:42 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 35. 3 39. 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 = 0.35. 220 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 = 39.07 kn s = F = 39.07(103 ) p = 55.27 Ma 4 (0.032 ) e long = s E = -55.27(106 ) 73.1(10 9 ) = -0.000756 e lat = -ve long = -0.35(-0.000756) = 0.0002646 d œ = d + d e lat = 30 + 30(0.0002646) = 30.008 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 29110 3 16 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 = = 800 3 16 B = 28.97 ksi 6 2 = 40 ksi 5 4(103 ) p 4 (5) 2 5 203.72 Ma < s g 5 280 Ma Normal Strain: Since s 6 s g, Hooke s law is still valid. e 5 = s = 28.97 29(10 3 = 0.000999 in.>in. E 5 203.72 3 5 0.0010186 mm/mm 200(10) ) If the nut is unscrewed, the load is zero. Therefore, the strain e = 0 531 118 SM_CH08.indd 531 4/11/11 9:54:42 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 37. 3 41. 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 = 0.35. 0.4 m B G 1.25 m 1.5 m 0.3 m Equations of Equilibrium: a + M B = 0; F (2.75) - 7848(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 = 3126.69 N. Modulus of Rigidity: F = 3908.37 N F = = 3126.69 N G = t = V = 3126.69 = 148.89 ka (0.14)(0.15) E 2(1 + v) = 4 = 1.481 Ma 2(1 + 0.35) Shear Strain: pplying Hooke s law for shear g = t G = 148.89(103 ) 1.481(10 6 ) = 0.1005 rad Thus, d h = hg = 30(0.1005) = 3.02 mm 119 532 SM_CH08.indd 532 4/11/11 9:54:43 M
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 38. 3 42. 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 0.625 0.025 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 0.002 1.25 mm in 22,, determine the weight W. 0.6 2 ft m 0.9 3 ft m 3 0.025 5 D B 0.9 0.625 = 1.5 d 1.2 4 ft m d = 1.04167 0.0417 in mm C e DE = d = 0.00116 in.>in. L = 1.04167 = 0.0011574 mm/mm 0.9(1000) 3(12) W s = 29(10 3 DE = Ee DE = 200(10 3 )(0.0011574) )(0.00116) = = 33.56 231.48 ksi Ma F DE = s DE DE DE DE = = 231.48(1.25) 33.56 (0.002) = 289.35 = 0.0672 N kip a+ M = 0; (289.35)(1.5) -(0.0672) (5) + 0.9(W) 3(W) = 0 1.04167 0.6 m 0.9 m 0.625 W = 482.25 0.112 kip N = 112 lb s BC = 0.112 BC = W = 482.25 = = 385.8 55.94 Ma ksi BC 1.25 BC 0.002 e BC BC = 55.94 BC = s BC E = 385.8 200(10 29 (10 3 = 3 = 0.00193 ) 0.00193 mm/mm in.>in. ) F DE = 289.35 N 0.6 m 0.9 m 8 39. 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. 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 = 159.15 Ma 4 (0.0082 ) s s = 8(10 3 ) = p s 4 (0.022-0.012 2 = 39.79 Ma ) Normal Strain: pplying Hooke s Law e b = s b E al = 159.15(106 ) 70(10 9 ) e s = = 0.00227 mm>mm s s = 39.79(106 ) E mg 45(10 9 = 0.000884 mm>mm ) 533 120 SM_CH08.indd 533 4/11/11 9:54:44 M
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 = 45.0. Determine the applied load needed to cause u = 44.9. 400 mm u 400 mm B L B sin 90.2 = 400 sin 44.9 L B = 566.67 mm L B = e = L B - L B 566.67-565.69 = = 0.001744 L B 565.69 s = Ee = 200(10 9 ) (0.001744) = 348.76 Ma a+ M = 0 400 sin 45 = 565.69 (400 cos 0.2 ) - F B sin 44.9 (400) = 0 (1) However, F B = s = 348.76(10 6 )(10)(10-6 ) = 3.488 kn From Eq. (1), = 2.46 kn 121 534 SM_CH08.indd 534 4/11/11 9:54:44 M