Ans. 4 (32 ) = 2.25p in 2, Ans.

Transcription

1 04 Solutions /25/10 3:19 M age The shi is ushed through the water using an -36 steel roeller shaft that is 8 m long, measured from the roeller to the thrust bearing D at the engine. If it has an outer diameter of 400 mm and a wall thickness of 50 mm, determine the amount of axial contraction of the shaft when the roeller exerts a force on the shaft of 5 kn. The bearings at and are journal bearings. Internal Force: s shown on FD. 5 kn D Dislacement: d E (10 3 )(8) 4 ( ) 200(10 9 ) 8 m (10-6 ) m mm Negative sign indicates that end moves towards end D The coer shaft is subjected to the axial loads shown. Determine the dislacement of end with resect to end D. The diameters of each segment are d 3 in., d and d Take E cu in., D 1 in. 2 ksi. The normal forces develoed in segment, and D are shown in the FDS of each segment in Fig. a, b and c resectively. 6 ki 50 in. 75 in. 60 in. 2 ki 2 ki 3 ki D 1 ki The cross-sectional area of segment, and D are and D. 4 (12 ) 0.25 in 2 4 (22 ) in 2 4 (32 ) 2.25 in 2, Thus, d >D i i + + D D i E i E u E u D E u 6.00 (50) (2.25)18(10 3 )D (75) 18(10 3 )D (60) (0.25) 18(10 3 )D 0.766(10-3 ) in. The ositive sign indicates that end moves away from D. 122

2 04 Solutions /25/10 3:19 M age The -36 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 50 mm 2, determine the dislacement of its end D. Neglect the size of the coulings at,, and D. The normal forces develoed in segments, and D are shown in the FDS of each segment in Fig. a, b and c, resectively. 1 m 1.5 m 1.25 m 9 kn 4 kn D 2 kn The cross-sectional areas of all the segments are 2 50 mm 2 1 m a mm b 50.0(10-6 ) m 2 d D i i i E i 1 E S a + + D D b (10-6 ) 200(10 9 )D c -3.00(10 3 )(1) (10 3 )(1.5) (10 3 )(1.25) d 0.850(10-3 ) m mm The ositive sign indicates that end D moves away from the fixed suort. 123

3 04 Solutions /25/10 3:19 M age 124 *4 4. The -36 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 50 mm 2, determine the dislacement of. Neglect the size of the coulings at,, and D. 1 m 1.5 m 1.25 m 9 kn 4 kn D 2 kn The normal forces develoed in segments and are shown the FDS of each segment in Fig. a and b, resectively. The cross-sectional area of these two segments are 50 mm m a. Thus, mm b 50.0 (10-6 ) m 2 d i i i E i 1 E S (10-6 ) 200(10 9 )D c -3.00(10 3 )(1) (10 3 )(1.5) d (10-3 ) m mm The ositive sign indicates that couling moves away from the fixed suort The assembly consists of a steel rod and an aluminum rod, each having a diameter of 12 mm. If the rod is subjected to the axial loadings at and at the couling, determine the dislacement of the couling and the end. The unstretched length of each segment is shown in the figure. Neglect the size of the connections at and, and assume that they are rigid. E st 200 Ga, E al 70 Ga. 3 m 6 kn 2 m 18 kn d E 12(10 3 )(3) 4 (0.012)2 (200)( m 1.59 mm ) d E 12(10 3 )(3) 4 (0.012)2 (200)(10 9 ) + 18(10 3 )(2) 4 (0.012)2 (70)(10 9 ) m 6.14 mm 4 6. The bar has a cross-sectional area of 3 in 2, and x w 500x 1/3 lb/in. E ksi. Determine the dislacement of its end when it is subjected to the distributed loading. 4 ft x x (x) w dx 500 x 1 3 dx (x) dx d 0 E 1 (3)(35)(10 6 ) 0 0 4(12) x x dx a (3)(35)(10 8 )(4) ba3 7 b(48)1 3 d in. 124

4 04 Solutions /25/10 3:19 M age The load of 800 lb is suorted by the four 304 stainless steel wires that are connected to the rigid members and D. Determine the vertical dislacement of the load if the members were horizontal before the load was alied. Each wire has a cross-sectional area of 0.05 in 2. Referring to the FD of member,fig.a a+ M 0; F (5) - 800(1) 0 F 160 lb a+ M 0; 800(4) - F H (5) 0 F H 640 lb Using the results of F and F H, and referring to the FD of member D,Fig.b a+ M D 0; F F (7) - 160(7) - 640(2) 0 F F lb a+ M 0; 640(5) - F DE (7) 0 F DE lb Since E and F are fixed, d D F DE DE E st d F F F E st From the geometry shown in Fig. c, (4)(2) (10 6 )D (4)(12) (10 6 )D in T in T E H D 2 ft 5 ft 4.5 ft 800 lb 1 ft F 4 ft d H ( ) in T 7 Subsequently, Thus, d >H F H H E st d > F E st 160(4.5)(12) (10 6 )D +T d d H + d >H in T +T d d + d > in T From the geometry shown in Fig. d, 640(4.5)(12) (10 6 )D in T in T d ( ) in T 5 125

5 04 Solutions /25/10 3:19 M age 126 *4 8. The load of 800 lb is suorted by the four 304 stainless steel wires that are connected to the rigid members and D. Determine the angle of tilt of each member after the load is alied.the members were originally horizontal, and each wire has a cross-sectional area of 0.05 in 2. Referring to the FD of member,fig.a, a+ M 0; F (5) - 800(1) 0 F 160 lb a+ M 0; 800(4) - F H (5) 0 F H 640 lb Using the results of F and F H and referring to the FD of member D,Fig.b, a+ M D 0; F F (7) - 160(7) - 640(2) 0 F F lb a+ M 0; 640(5) - F DE (7) 0 F DE lb Since E and F are fixed, d D F DE DE E st d F F F E st From the geometry shown in Fig. c (4)(12) (10 6 )D (4)(12) (10 6 )D in T in T E H D 2 ft 5 ft 4.5 ft 800 lb 1 ft F 4 ft d H ( ) in T 7 u (12) 46.6(10-6 ) rad Subsequently, d >H F H H E st d > F E st 640 (4.5)(12) (10 6 )D 160 (4.5)(12) (10 6 )D in T in T Thus, +T d d H + d >H in T +T d d + d > in T From the geometry shown in Fig. d f (12) 0.355(10-3 ) rad 126

6 04 Solutions /25/10 3:19 M age ontinued 127

7 04 Solutions /25/10 3:19 M age The assembly consists of three titanium (Ti-61-4V) rods and a rigid bar. The cross-sectional area of each rod is given in the figure. If a force of 6 ki is alied to the ring F, determine the horizontal dislacement of oint F. Internal Force in the Rods: a + M 0; F D (3) - 6(1) 0 F D 2.00 ki : + F x 0; F 0 F 4.00 ki D 4 ft D 1 in in 2 6 ft E 2 ft 1 ft F 6 ki 1 ft EF 2 in 2 Dislacement: d F D D D E 2.00(4)(12) (1)(17.4)( in. ) d F E 4.00(6)(12) (1.5)(17.4)( in. ) d F>E F EF EF EF E 6.00(1)(12) (2)(17.4)( in. ) de œ ; de œ in. 3 d E d + d œ E in. d F d E + d F>E in The assembly consists of three titanium (Ti-61-4V) rods and a rigid bar. The cross-sectional area of each rod is given in the figure. If a force of 6 ki is alied to the ring F, determine the angle of tilt of bar. Internal Force in the Rods: a + M 0; F D (3) - 6(1) 0 F D 2.00 ki : + F x 0; F 0 F 4.00 ki D 4 ft D 1 in in 2 6 ft E 2 ft 1 ft F 6 ki 1 ft EF 2 in 2 Dislacement: d F D D D E 2.00(4)(12) (1)(17.4)( in. ) d F E 4.00(6)(12) (1.5)(17.4)( in. ) u tan - 1 d - d 3(12) tan 3(12)

8 04 Solutions /25/10 3:19 M age The load is suorted by the four 304 stainless steel wires that are connected to the rigid members and D. Determine the vertical dislacement of the 500-lb load if the members were originally horizontal when the load was alied. Each wire has a cross-sectional area of in 2. E F G 3 ft Internal Forces in the wires: FD (b) a + M 0; F (4) - 500(3) 0 F lb D 1.8 ft H 1 ft 2 ft I 3 ft 1 ft 500 lb 5 ft + c F y 0; F H F H lb FD (a) a + M D 0; F F (3) (1) 0 F F lb + c F y 0; F DE F DE lb Dislacement: d D F DE DE DE E 83.33(3)(12) in (28.0)(10 6 ) d F F F F E 41.67(3)(12) in (28.0)(10 6 ) dh œ ; dh œ in. 3 d H in. d >H F H H H E 125.0(1.8)(12) in (28.0)(10 6 ) d d H + d >H in. d F G G G E 375.0(5)(12) in (28.0)(10 6 ) dl œ ; dl œ in. 4 d l in. 129

9 04 Solutions /25/10 3:19 M age 130 *4 12. The load is suorted by the four 304 stainless steel wires that are connected to the rigid members and D. Determine the angle of tilt of each member after the 500-lb load is alied. The members were originally horizontal, and each wire has a cross-sectional area of in 2. E F G 3 ft Internal Forces in the wires: FD (b) a + M 0; F G (4) - 500(3) 0 F G lb D 1.8 ft H 1 ft 2 ft I 3 ft 1 ft 500 lb 5 ft + c F y 0; F H F H lb FD (a) a + M D 0; F F (3) (1) 0 F F lb + c F y 0; F DE F DE lb Dislacement: d D F DE DE DE E 83.33(3)(12) in (28.0)(10 6 ) d F F F F E 41.67(3)(12) in (28.0)(10 6 ) dh œ ; dh œ in. 3 d H d œ H + d in. tan a ; a d >H F H H H E 125.0(1.8)(12) in (28.0)(10 6 ) d d H + d >H in. d F G G G E 375.0(5)(12) in (28.0)(10 6 ) tan b ; b

10 04 Solutions /25/10 3:19 M age The bar has a length and cross-sectional area. Determine its elongation due to the force and its own weight.the material has a secific weight g (weight>volume) and a modulus of elasticity E. (x) dx d (x) E 1 E 0 (gx + ) dx 1 E a g2 2 + b g2 2E + E The ost is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kn and the soil rovides a frictional resistance that is uniformly distributed along its sides of w 4 kn>m, determine the force F at its bottom needed for equilibrium.lso, what is the dislacement of the to of the ost with resect to its bottom? Neglect the weight of the ost. 2 m y 20 kn w Equation of Equilibrium: For entire ost [FD (a)] + c F y 0; F F 12.0 kn F Internal Force: FD (b) + c F y 0; -F(y) + 4y F(y) {4y - 20} kn Dislacement: F(y)dy d > 0 (y)e 1 E 0 2 m (4y - 20)dy 1 E 2y2-20y 2 m kn # m E 32.0(10 3 ) - 4 (0.062 ) 13.1 (10 9 ) m mm Negative sign indicates that end moves toward end. 131

11 04 Solutions /25/10 3:20 M age The ost is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kn and the soil rovides a frictional resistance that is distributed along its length and varies linearly from w 0 at y 0 to w 3 kn>m at y 2 m, determine the force F at its bottom needed for equilibrium. lso, what is the dislacement of the to of the ost with resect to its bottom? Neglect the weight of the ost. 2 m y 20 kn w F Equation of Equilibrium: For entire ost [FD (a)] + c F y 0; F F 17.0 kn Internal Force: FD (b) + c F y 0; -F(y) a 3y by F(y) e 3 4 y2-20 f kn Dislacement: F(y) dy d > 0 (y)e 1 E 0 2m 1 E a y3 2 m 4-20yb kn # m E 38.0(10 3 ) - 4 (0.062 ) 13.1 (10 9 ) m a 3 4 y2-20bdy mm Negative sign indicates that end moves toward end. 132

12 04 Solutions /25/10 3:20 M age 133 *4 16. The linkage is made of two in-connected -36 steel members, each having a cross-sectional area of 1.5 in 2. If a vertical force of 50 ki is alied to oint, determine its vertical dislacement at. 2 ft 1.5 ft 1.5 ft nalysing the equilibrium of Joint by referring to its FD,Fig.a, : + F x 0 ; F a 3 5 b - F a 3 5 b 0 F F F + c F y 0-2Fa 4 b F ki 5 The initial length of members and is in (2.50 ft)a b 30 in. The axial deformation of members 1 ft and is d F E (-31.25)(30) (1.5)29.0(10 3 )D in. The negative sign indicates that end moves toward and. From the geometry shown in Fig. b, u tan - 1 a 1.5. Thus, 2 b d g d cos u in. T cos

13 04 Solutions /25/10 3:20 M age The linkage is made of two in-connected -36 steel members, each having a cross-sectional area of 1.5 in 2. Determine the magnitude of the force needed to dislace oint in. downward. nalysing the equilibrium of joint by referring to its FD,Fig.a : + F x 0; F a 3 5 b - F a 3 5 b 0 F F F + c F y 0; -2Fa 4 5 b - 0 F ft 1.5 ft 1.5 ft The initial length of members and are in (2.50 ft)a b 30 in. The axial deformation of members 1 ft and is d F E (30) (1.5)29.0(10 3 )D (10-3 ) The negative sign indicates that end moves toward and. From the geometry shown in Fig. b, we obtain u tan - 1 a 1.5. Thus 2 b (d ) g d cos u (10-3 ) cos kis 134

14 04 Solutions /25/10 3:20 M age The assembly consists of two -36 steel rods and a rigid bar D. Each rod has a diameter of 0.75 in. If a force of 10 ki is alied to the bar as shown, determine the vertical dislacement of the load. 2 ft 3 ft Here, F EF 10 ki. Referring to the FD shown in Fig. a, a+ M 0; F D (2) - 10(1.25) 0 F D 6.25 ki 1.25 ft E D 0.75 ft F 1 ft a+ M D 0; 10(0.75) - F (2) 0 F 3.75 ki 10 ki The cross-sectional area of the rods is. Since oints 4 (0.752 ) in 2 and are fixed, d F E st d D F D D E st 3.75 (2)(12) (10 3 )D 6.25(3)(12) (10 3 )D in. T in T From the geometry shown in Fig. b Here, d E ( ) in. T d F>E F EF EF E st 10 (1) (12) (10 3 )D in T Thus, +T d F d E + d F>E in T 135

15 04 Solutions /25/10 3:20 M age The assembly consists of two -36 steel rods and a rigid bar D. Each rod has a diameter of 0.75 in. If a force of 10 ki is alied to the bar, determine the angle of tilt of the bar. Here, F EF 10 ki. Referring to the FD shown in Fig. a, a+ M 0; F D (2) - 10(1.25) 0 F D 6.25 ki 2 ft 3 ft a+ M D 0; 10(0.75) - F (2) 0 F 3.75 ki E D The cross-sectional area of the rods is. Since oints 4 (0.752 ) in 2 and are fixed then, 1.25 ft 0.75 ft F 1 ft d F E st d D F D D E st From the geometry shown in Fig. b, 3.75 (2)(12) (10 3 )D 6.25 (3)(12) (10 3 )D in T in T 10 ki u (12) 0.439(10-3 ) rad 136

16 04 Solutions /25/10 3:20 M age 137 *4 20. The rigid bar is suorted by the in-connected rod that has a cross-sectional area of 500 mm 2 and is made of -36 steel. Determine the vertical dislacement of the bar at when the load is alied. Force In The Rod. Referring to the FD of member,fig.a 3 m 45 kn/m a+ M 0; F a 3 5 b (4) (45)(4) c 1 3 (4) d 0 F 50.0 kn 4 m Dislacement. The initial length of rod is m. The axial deformation of this rod is d F E st 50.0(10 3 )(5) 0.5(10-3 ) 200(10 9 )D 2.50 (10-3 ) m From the geometry shown in Fig. b, u tan - 1 a 3. Thus, 4 b (d ) g d sin u 2.50(10-3 ) sin (10-3 ) m 4.17 mm 137

17 04 Solutions /25/10 3:20 M age sring-suorted ie hanger consists of two srings which are originally unstretched and have a stiffness of k 60 kn>m, three 304 stainless steel rods, and D, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the ie and the fluid it carries have a total weight of 4 kn, determine the dislacement of the ie when it is attached to the suort. F k G 0.75 m E D k H 0.75 m Internal Force in the Rods: FD (a) 0.25 m 0.25 m a + M 0; F D (0.5) - 4(0.25) 0 F D 2.00 kn + c F y 0; F F 2.00 kn FD (b) + c F y 0; F EF F EF 4.00 kn Dislacement: d D d E F EF EF EF E 4.00(10 3 )(750) 4 (0.012)2 (193)( mm ) d > d >D D D D E 2(10 3 )(750) 4 (0.005)2 (193)( mm ) d d D + d >D mm Dislacement of the sring d s F s k m mm d lat d + d s mm 138

18 04 Solutions /25/10 3:20 M age sring-suorted ie hanger consists of two srings, which are originally unstretched and have a stiffness of k 60 kn>m, three 304 stainless steel rods, and D, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the ie is dislaced 82 mm when it is filled with fluid, determine the weight of the fluid. F k G 0.75 m E D k H 0.75 m Internal Force in the Rods: FD (a) a + M 0; F D (0.5) - W(0.25) 0 F D W m 0.25 m + c F y 0; F + W 2 - W 0 F W 2 FD (b) + c F y 0; F EF - W 2 - W 2 0 F EF W Dislacement: d D d E F EF EF EF E W(750) 4 (0.012)2 (193)(10 9 ) d > d >D F D D D E Dislacement of the sring d s F s k d lat d + d s (10-6 ) W W 2 (750) 4 (0.005)2 (193)(10 9 ) (10-6 ) W d d D + d >D (10-6 ) W (10-6 ) W (10-3 ) W W 2 60(10 3 (1000) W ) (10-3 ) W W W 9685 N 9.69 kn 139

19 04 Solutions /25/10 3:20 M age The rod has a slight taer and length. It is susended from the ceiling and suorts a load at its end. Show that the dislacement of its end due to this load is d >1Er 2 r 1 2. Neglect the weight of the material. The modulus of elasticity is E. r 2 r(x) r 1 + r 2 - r 1 x r 1 + (r 2 - r 1 )x (x) 2 (r 1 + (r 2 - r 1 )x) 2 dx d (x)e 2 E 0 dx [r 1 + (r 2 - r 1 )x] 2 r 1-2 E c 1 (r 2 - r 2 )(r 1 + (r 2 - r 1 )x) d ƒ 0-2 E(r 2 - r 1 ) c 1 r 1 + (r 2 - r 1 ) - 1 r 1 d 2 - E(r 2 - r 1 ) c 1 r 2-1 r 1 d - 2 E(r 2 - r 1 ) c r 1 - r 2 r 2 r 1 d 2 E(r 2 - r 1 ) c r 2 - r 1 r 2 r 1 d E r 2 r 1 QED *4 24. Determine the relative dislacement of one end of the taered late with resect to the other end when it is subjected to an axial load. d 2 t w d 1 + d 2 - d 1 h d (x) dx (x)e h E t 0 h E t d 1 h 0 h dx d 1 h + (d 2 - d 1 )x h x d 1 h + (d 2 - d 1 )x h E 0 dx d2 - d1 1 + d 1 h x h E t d 1 h a d 1 h bcln a1 + d 2 - d h 1 d 2 - d 1 d 1 h xbdƒ 0 h E t(d 2 - d 1 ) cln a1 + d 2 - d 1 d 1 bd h dx [d 1h + ( d 2 - d 1 )x ]t h h E t(d 2 - d 1 ) cln a d 1 + d 2 - d 1 bd d 1 d 1 h h E t(d 2 - d 1 ) cln d 2 d 1 d 140

20 04 Solutions /25/10 3:20 M age Determine the elongation of the -36 steel member when it is subjected to an axial force of 30 kn. The member is 10 mm thick. Use the result of rob mm 30 kn 30 kn 75 mm 0.5 m Using the result of rob by substituting d m, d m t 0.01 m and 0.5 m. d 2c E st t(d 2 - d 1 ) ln d 2 d d 1 30(10 3 ) (0.5) 2c 200(10 9 )(0.01)( ) ln a bd 0.360(10-3 ) m mm The casting is made of a material that has a secific weight g and modulus of elasticity E. If it is formed into a yramid having the dimensions shown, determine how far its end is dislaced due to gravity when it is susended in the vertical osition. b 0 b 0 Internal Forces: + c F z 0; (z) gz 0 (z) 1 3 gz Dislacement: (z) dz d (z) E 0 3 gz E dz 0 1 g z dz 3E 0 g2 6E 141

21 04 Solutions /25/10 3:20 M age The circular bar has a variable radius of r r 0 e ax and is made of a material having a modulus of elasticity of E. Determine the dislacement of end when it is subjected to the axial force. Dislacements: The cross-sectional area of the bar as a function of x is (x) r 2 r 2 0 e 2ax. We have (x)dx d 0 (x)e r 2 0 E 0 r 2 0 E c - 1 2ae 2ax d 2 0 dx e 2ax x r 0 r r 0 e ax - 2ar 2 0 E a1 - e - 2a b *4 28. The edestal is made in a shae that has a radius defined by the function r 2>12 + y 1>2 2 ft, where y is in feet. If the modulus of elasticity for the material is E si, determine the dislacement of its to when it suorts the 500-lb load. y 500 lb 0.5 ft 4 ft r 2 (2 y 1/2 ) d (y) dy (y) E (10 3 )(144) 0 ( (10-3 ) (4 + 4y y) dy 0 dy y ) ft y r (10-3 )c4y + 4a y3 2 b + 2 y2 d (10-3 )(45.33) (10-3 ) ft in. 142

22 04 Solutions /25/10 3:20 M age The suort is made by cutting off the two oosite sides of a shere that has a radius r 0. If the original height of the suort is r 0 >2, determine how far it shortens when it suorts a load. The modulus of elasticity is E. r 0 r 0 2 Geometry: Dislacement: r 2 (r 0 cos u) 2 r 0 2 cos 2 u (y) dy d (y) E 0 2 E 0 y r 0 sin u; dy r 0 cos u du u r 0 cos u du r 2 0 cos 2 u R 2 r 0 E 0 u du cos u R 2 r 0 E [ln (sec u + tan u)] 2 u 0 2 [ln (sec u + tan u)] r 0 E When y r 0 4 ; u d 2 [ln (sec tan )] r 0 E r 0 E lso, Geometry: (y) x 2 (r y 2 ) Dislacement: (y) dy d (y) E 2 E dy r y 2 2 E 1 2r 0 ln r 0 + y [ln ln 1] r 0 E 0 r 0 - y R r 0 E 143

23 04 Solutions /25/10 3:20 M age The weight of the kentledge exerts an axial force of 1500 kn on the 300-mm diameter high strength concrete bore ile. If the distribution of the resisting skin friction develoed from the interaction between the soil and the surface of the ile is aroximated as shown, and the resisting bearing force F is required to be zero, determine the maximum intensity 0 kn>m for equilibrium. lso, find the corresonding elastic shortening of the ile. Neglect the weight of the ile m Internal oading: y considering the equilibrium of the ile with reference to its entire free-body diagram shown in Fig. a. We have 1 (12) c Fy 0; kn>m Thus, (y) 250 y 20.83y kn>m 12 The normal force develoed in the ile as a function of y can be determined by considering the equilibrium of a section of the ile shown in Fig. b. 1 (20.83y)y - (y) c Fy 0; (y) 10.42y2 kn Dislacement: The cross-sectional area of the ile is (0.32) m2. 4 We have d 12 m (y)dy 10.42(103)y2dy (29.0)(109) 0 0 (y)e 12 m (10-6)y2dy (10-6)y3 冷 0 12 m (10-3)m 2.93 mm 144 F

24 04 Solutions /25/10 3:20 M age The column is constructed from high-strength concrete and six -36 steel reinforcing rods. If it is subjected to an axial force of 30 ki, determine the average normal stress in the concrete and in each rod. Each rod has a diameter of 0.75 in. 4 in. 30 ki Equations of Equilibrium: + c F y 0; 6 st + con [1] 3 ft omatibility: d st d con st (3)(12) 4 (0.752 )(29.0)(10 3 ) con(3)(12) [ 4 (82 ) - 6( 4 )(0.75)2 ](4.20)(10 3 ) st con [2] Solving Eqs. [1] and [2] yields: st ki con ki verage Normal Stress: s con con con s st st ksi st 4 (0.752 ) (82 ) ( ) ksi *4 32. The column is constructed from high-strength concrete and six -36 steel reinforcing rods. If it is subjected to an axial force of 30 ki, determine the required diameter of each rod so that one-fourth of the load is carried by the concrete and three-fourths by the steel. 4 in. 30 ki Equilibrium: The force of 30 ki is required to distribute in such a manner that 3/4 of the force is carried by steel and 1/4 of the force is carried by concrete. Hence st 3 4 (30) 22.5 ki con 1 (30) 7.50 ki 4 omatibility: 3 ft d st d con st st E st con con E con st 22.5 con E con 7.50 E st 6 a 4 bd2 3 4 (82 ) d 2 D (4.20)(10 3 ) 29.0(10 3 ) d 1.80 in. 145

25 04 Solutions /25/10 3:20 M age The steel ie is filled with concrete and subjected to a comressive force of 80 kn. Determine the average normal stress in the concrete and the steel due to this loading. The ie has an outer diameter of 80 mm and an inner diameter of 70 mm. E st 200 Ga, E c 24 Ga. 80 kn + c F y 0; st + con st 4 ( ) (200) (10 9 ) con 4 (0.072 ) (24) (10 9 ) Solving Eqs. (1) and (2) yields st kn con kn s st st st d st d con st con (10 3 ) 4 ( ) 48.8 Ma s con con (103 ) 5.85 Ma con 4 (0.072 ) (1) (2) 500 mm The 304 stainless steel ost has a diameter of d 2 in. and is surrounded by a red brass tube. oth rest on the rigid surface. If a force of 5 ki is alied to the rigid ca, determine the average normal stress develoed in the ost and the tube. Equations of Equilibrium: 8 in. 5 ki 3 in. + c F y 0; st + br - 5 0[1] omatibility: d 0.5 in. d st d br st (8) 4 (22 )(28.0)(10 3 ) br(8) 4 ( )(14.6)(10 3 ) st br [2] Solving Eqs. [1] and [2] yields: verage Normal Stress: br ki st ki s br br br ( ) ksi s st st ksi st 4 (22 ) 146

26 04 Solutions /25/10 3:20 M age The 304 stainless steel ost is surrounded by a red brass tube. oth rest on the rigid surface. If a force of 5 ki is alied to the rigid ca, determine the required diameter d of the steel ost so that the load is shared equally between the ost and tube. Equilibrium: The force of 5 ki is shared equally by the brass and steel. Hence 8 in. 5 ki 3 in. st br 2.50 ki omatibility: d 0.5 in. d st d br st E st br E br st bre br E st a 4 bd2 4 ( )(14.6)(10 3 ) 28.0(10 3 ) d 2.39 in. *4 36. The comosite bar consists of a 20-mm-diameter -36 steel segment and 50-mm-diameter red brass end segments D and. Determine the average normal stress in each segment due to the alied load. 250 mm 500 mm 250 mm 50 mm 20 mm 75 kn 100 kn ; + F x 0; F - F D D 75 kn 100 kn F - F D (1) ; + 0 D - d D 150(0.5) 0 4 (0.02)2 (200)(10 9 ) - 50(0.25) 4 (0.052 )(101)(10 9 ) F D (0.5) - 4 (0.052 )(101)(10 9 ) - F D (0.5) 4 (0.022 )(200)(10 9 ) F D kn From Eq. (1), F kn s D D (103 ) 55.0 Ma D 4 (0.052 ) s 42.11(103 ) 134 Ma 4 (0.022 ) s (103 ) 80.4 Ma 4 (0.052 ) 147

27 04 Solutions /25/10 3:20 M age The comosite bar consists of a 20-mm-diameter -36 steel segment and 50-mm-diameter red brass end segments D and. Determine the dislacement of with resect to due to the alied load. 250 mm 500 mm 250 mm 50 mm 20 mm 75 kn 100 kn D 75 kn 100 kn ; + 0 D - d D 0-150(10 3 )(500) 4 (0.022 )(200)(10 9 ) - 50(10 3 )(250) 4 (0.052 )(101)(10 9 ) F D (500) 4 (0.052 )(101)(10 9 ) - F D (500) 4 (0.02)2 (200)(10 9 ) F D kn Dislacement: d > 42.11(103 )(500) E st 4 (0.022 )200(10 9 ) mm The -36 steel column, having a cross-sectional area of 18 in 2, is encased in high-strength concrete as shown. If an axial force of 60 ki is alied to the column, determine the average comressive stress in the concrete and in the steel. How far does the column shorten? It has an original length of 8 ft. 16 in. 60 ki 9 in. + c F y 0; st + con d st d con ; st(8)(12) 18(29)(10 3 ) con (8)(12) [(9)(16) - 18](4.20)(10 3 ) st con Solving Eqs. (1) and (2) yields st ki; con ki s st st ksi st 18 s con con con ksi 9(16) - 18 Either the concrete or steel can be used for the deflection calculation. d st st E (8)(12) 18(29)( in. ) (1) (2) 8 ft 148

28 04 Solutions /25/10 3:20 M age The -36 steel column is encased in high-strength concrete as shown. If an axial force of 60 ki is alied to the column, determine the required area of the steel so that the force is shared equally between the steel and concrete. How far does the column shorten? It has an original length of 8 ft. 16 in. 60 ki 9 in. 8 ft The force of 60 ki is shared equally by the concrete and steel. Hence st con 30 ki d con d st ; con E con st E st st cone con E st [9(16) - st] 4.20(10 3 ) 29(10 3 ) d st st E st 18.2 in 2 30(8)(12) 18.2(29)(10 3 ) in. *4 40. The rigid member is held in the osition shown by three -36 steel tie rods. Each rod has an unstretched length of 0.75 m and a cross-sectional area of 125 mm 2. Determine the forces in the rods if a turnbuckle on rod EF undergoes one full turn. The lead of the screw is 1.5 mm. Neglect the size of the turnbuckle and assume that it is rigid. Note: The lead would cause the rod, when unloaded, to shorten 1.5 mm when the turnbuckle is rotated one revolution m 0.5 m E 0.5 m D 0.75 m a+ M E 0; -T (0.5) + T D (0.5) 0 F T T D T (1) +T F y 0; T EF - 2T 0 T EF 2T (2) Rod EF shortens 1.5mm causing (and D) to elongate. Thus: d > + d E>F T T(0.75) (125)(10-6 )(200)(10 9 ) + 2T(0.75) (125)(10-6 )(200)(10 9 ) T N T T D 16.7 kn T EF 33.3 kn 149

29 04 Solutions /25/10 3:20 M age The concrete ost is reinforced using six steel reinforcing rods, each having a diameter of 20 mm. Determine the stress in the concrete and the steel if the ost is subjected to an axial load of 900 kn. E st 200 Ga, E c 25 Ga. 900 kn 250 mm 375 mm Referring to the FD of the uer ortion of the cut concrete ost shown in Fig. a + c F y 0; con + 6 st (1) Since the steel rods and the concrete are firmly bonded, their deformation must be the same. Thus con con E con 0 con d st st st E st con 0.25(0.375) - 6( 4 )(0.022 )D25(10 9 )D con st Solving Eqs (1) and (2) yields st ( 4 )(0.022 )200(10 9 )D (2) st kn con kn Thus, s con con con (10 3 ) 0.15(0.375) - 6( 4 )(0.022 ) 8.42 Ma s st st 21.15(103 ) st 4 (0.022 ) 67.3 Ma 150

30 04 Solutions /25/10 3:20 M age The ost is constructed from concrete and six -36 steel reinforcing rods. If it is subjected to an axial force of 900 kn, determine the required diameter of each rod so that one-fifth of the load is carried by the steel and four-fifths by the concrete. E st 200 Ga, E c 25 Ga. The normal force in each steel rod is st 1 5 (900) 30 kn kn 250 mm 375 mm The normal force in concrete is con 4 (900) 720 kn 5 Since the steel rods and the concrete are firmly bonded, their deformation must be the same. Thus d con d st con con E con st st E st 720(10 3 ) 0.25(0.375) - 6( 4 d2 )D25(10 9 )D 49.5 d (10 3 ) 4 d2 200(10 9 )D d m 24.6 mm The assembly consists of two red brass coer alloy rods and D of diameter 30 mm, a stainless 304 steel alloy rod EF of diameter 40 mm, and a rigid ca G. If the suorts at, and F are rigid, determine the average normal stress develoed in rods, D and EF. 300 mm 450 mm 40 kn 30 mm E F 30 mm D G 40 kn 40 mm Equation of Equilibrium: Due to symmetry, F F D F. Referring to the freebody diagram of the assembly shown in Fig. a, : + F x 0; 2F + F EF - 240(10 3 )D 0 (1) omatibility Equation: Using the method of suerosition, Fig. b, : + 0 -d + d EF 40(10 3 )(300) 0-4 (0.032 )(101)(10 9 ) + c F EF (450) 4 (0.042 )(193)(10 9 ) + F EF>2(300) 4 (0.032 )(101)(10 9 ) d F EF N Substituting this result into Eq. (1), F N 151

31 04 Solutions /25/10 3:20 M age ontinued Normal Stress: We have, s s D F Ma D 4 (0.032 ) s EF F EF Ma EF 4 (0.042 ) 152

32 04 Solutions /25/10 3:20 M age 153 *4 44. The two ies are made of the same material and are connected as shown. If the cross-sectional area of is and that of D is 2, determine the reactions at and D when a force is alied at the junction. 2 2 D Equations of Equilibrium: ; + F x 0; F + F D - 0 [1] omatibility: : + 0 d - d 0 2 2E - F 2 E 0 4E - 3F 4E + F 2 2E S F 3 From Eq. [1] F D

33 04 Solutions /25/10 3:20 M age The bolt has a diameter of 20 mm and asses through a tube that has an inner diameter of 50 mm and an outer diameter of 60 mm. If the bolt and tube are made of -36 steel, determine the normal stress in the tube and bolt when a force of 40 kn is alied to the bolt.ssume the end cas are rigid. 40 kn 160 mm 150 mm 40 kn Referring to the FD of left ortion of the cut assembly, Fig. a : + F x 0; 40(10 3 ) - F b - F t 0 (1) Here, it is required that the bolt and the tube have the same deformation. Thus d t d b F t (150) 4 ( )200(10 9 )D F t F b F b (160) 4 (0.022 )200(10 9 )D (2) Solving Eqs (1) and (2) yields Thus, F b (10 3 ) N F t (10 3 ) N s b F b 10.17(103 ) 32.4 Ma b 4 (0.022 ) s t F t t (10 3 ) 4 ( ) 34.5 Ma 154

34 04 Solutions /25/10 3:20 M age If the ga between and the rigid wall at D is initially 0.15 mm, determine the suort reactions at and D when the force 200 kn is alied. The assembly is made of 36 steel. 600 mm 600 mm 0.15 mm D 25 mm 50 mm Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a, : + F x 0; 200(10 3 ) - F D - F 0 (1) omatibility Equation: Using the method of suerosition, Fig. b, : + d d - d FD 200(10 3 )(600) (0.052 )(200)(10 9 ) - F D (600) 4 (0.052 )(200)(10 9 ) + F D (600) 4 ( )(200)(10 9 ) S F D N 20.4 kn Substituting this result into Eq. (1), F N 180 kn 155

35 04 Solutions /25/10 3:20 M age Two -36 steel wires are used to suort the 650-lb engine. Originally, is 32 in. long and is in. long. Determine the force suorted by each wire when the engine is susended from them. Each wire has a crosssectional area of 0.01 in 2. + c F y 0; T + T (1) d d T (32) (0.01)(29)(10 6 ) T (32.008) (0.01)(29)(10 6 ) T T 2320 T 361 lb T 289 lb *4 48. Rod has a diameter d and fits snugly between the rigid suorts at and when it is unloaded. The 0 x modulus of elasticity is E. Determine the suort reactions at and if the rod is subjected to the linearly distributed axial load. x Equation of Equilibrium: Referring to the free-body diagram of rod shown in Fig. a, 0 : + F x 0; F - F 0 (1) omatibility Equation: Using the method of suerosition, Fig. b, : + 0 d - d F (x)dx 0 E 0 0 (x)dx - F 0 - F () E 156

36 04 Solutions /25/10 3:20 M age ontinued Here, (x) 1 2 a 0 xbx 0 2 x Thus, x3 3 ` x 2 dx - F 0 - F F 0 6 Substituting this result into Eq. (1), F

37 04 Solutions /25/10 3:20 M age The taered member is fixed connected at its ends and and is subjected to a load 7 ki at x 30 in. Determine the reactions at the suorts. The material is 2 in. thick and is made from 2014-T6 aluminum. 6 in. 3 in. x 60 in. y x y x : + F x 0; F + F (1) d > F 30 0 F dx 2( x)(2)(e) + 60 dx ( x) + F dx ( x) 0 40 F ln( x) F ln( x) F (0.2876) F F dx 2( x)(2)(e) 0 F F Thus, from Eq. (1). F 4.09 ki F 2.91 ki 158

38 04 Solutions /25/10 3:20 M age The taered member is fixed connected at its ends and and is subjected to a load. Determine the location x of the load and its greatest magnitude so that the average normal stress in the bar does not exceed s allow 4 ksi. The member is 2 in. thick. 6 in. 3 in. x 60 in. y x y x : + F x 0; F + F - 0 d > 0 - x 0 F dx 2( x)(2)(e) + x 60 F dx 2( x)(2)(e) 0 x 60 dx -F ( x) + F dx ( x) 0 0 x F (40) ln ( x) x 0 - F (40) ln ( x) 60 x 0 F ln ( x ) -F 3 ln ( x 1.5 ) For greatest magnitude of require, 4 F 2( x)(2) ; F x 4 F 2(3) ; F 24 ki Thus, ( x) ln a x 3 b -24 ln a x b 1.5 Solving by trial and error, x 28.9 in. Therefore, F 36.4 ki 60.4 ki 159

39 04 Solutions /25/10 3:20 M age The rigid bar suorts the uniform distributed load of 6 ki>ft. Determine the force in each cable if each cable has a cross-sectional area of 0.05 in 2, and E ksi. 6 ft 6 ki/ft 3 ft 3 ft 3 ft D a + M 0; T a 2 (1) 25 b(3) - 54(4.5) + T Da 2 25 b9 0 u tan (3) 2 + (8.4853) 2-2(3)(8.4853) cos u lso, 2 D (9) 2 + (8.4853) 2-2(9)(8.4853) cos u (2) Thus, eliminating cos u. - 2 ( ) D ( ) ( ) D D + 30 ut, d, D d D Neglect squares or d since small strain occurs. 2 D (245 + d ) d 2 D (245 + d D ) d D d 0.333( d D ) d 0.333(2245 d D ) Thus, d D 3d T D 245 E 3 T 245 E T D 3 T From Eq. (1). T D ki 27.2 ki T 9.06 ki 160

40 04 Solutions /25/10 3:20 M age 161 *4 52. The rigid bar is originally horizontal and is suorted by two cables each having a cross-sectional area of 0.05 in 2, and E ksi. Determine the slight rotation of the bar when the uniform load is alied. See solution of rob T D ki 6 ft d D Using Eq. (2) of rob. 4-51, T D (31)(10 3 ) (31)( ft ) ( ) 2 (9) 2 + (8.4852) 2-2(9)(8.4852) cos u 3 ft 6 ki/ft 3 ft 3 ft D u Thus, u The ress consists of two rigid heads that are held 1 together by the two -36 steel 2 -in.-diameter rods T6-solid-aluminum cylinder is laced in the ress and the screw is adjusted so that it just resses u against the cylinder. If it is then tightened one-half turn, determine the average normal stress in the rods and in the cylinder. The single-threaded screw on the bolt has a lead of 0.01 in. Note: The lead reresents the distance the screw advances along its axis for one comlete turn of the screw. 12 in. 2 in. 10 in. : + F x 0; 2F st - F al 0 d st d al F st (12) ( 4 )(0.5)2 (29)(10 3 ) F al (10) (1) 2 (10)(10 3 ) Solving, F st ki F al ki s rod F st st ( 9.28 ksi 4 )(0.5)2 s cyl F al ksi 2 al (1) 161

41 04 Solutions /25/10 3:20 M age The ress consists of two rigid heads that are held 12 in. 1 together by the two -36 steel 2 -in.-diameter rods T6-solid-aluminum cylinder is laced in the ress and the screw is adjusted so that it just resses u against the cylinder. Determine the angle through which the screw can be turned before the rods or the secimen begin to yield. The single-threaded screw on the bolt has a lead of 0.01 in. Note: The lead reresents the distance the screw advances along its axis for one comlete turn of the screw. 2 in. 10 in. : + F x 0; 2F st - F al 0 d st d - d al F st (12) ( 4 )(0.5)2 (29)(10 3 ) d - F al (10) (1) 2 (10)(10 3 ) (1) ssume steel yields first, s Y 36 Then F al ki; F st ( 4 )(0.5)2 ; F st ki s al (1) ksi 4.50 ksi 6 37 ksi steel yields first as assumed. From Eq. (1), Thus, d in. u u

42 04 Solutions /25/10 3:20 M age The three susender bars are made of -36 steel and have equal cross-sectional areas of 450 mm 2. Determine the average normal stress in each bar if the rigid beam is subjected to the loading shown. 2 m 50 kn 80 kn D E F 1 m 1 m 1 m 1 m Referring to the FD of the rigid beam, Fig. a, + c F y 0; F D + F E + F F - 50(10 3 ) - 80(10 3 ) 0 (1) a + M D 0; F E (2) + F F (4) - 50(10 3 )(1) - 80(10 3 )(3) 0 (2) Referring to the geometry shown in Fig. b, d E d D + a d F - d D b(2) 4 d E 1 2 d D + d F F E E 1 2 a F D E + F F E b F D + F F 2 F E (3) Solving Eqs. (1), (2) and (3) yields F E 43.33(10 3 ) N F D 35.83(10 3 ) N F F 50.83(10 3 ) N Thus, s E F E 43.33(103 ) 0.45( Ma ) s D F D 35.83(103 ) 0.45( Ma ) s F 113 Ma 163

43 04 Solutions /25/10 3:20 M age 164 *4 56. The rigid bar suorts the 800-lb load. Determine the normal stress in each -36 steel cable if each cable has a cross-sectional area of 0.04 in ft 800 lb D 5 ft 5 ft 6 ft Referring to the FD of the rigid bar, Fig. a, a + M 0; F a 12 (1) 13 b(5) + F D a 3 b(16) - 800(10) 0 5 The unstretched length of wires and D are ft and D ft. The stretches of wires and D are d F E F (13) d E D F D D F D(20) E E Referring to the geometry shown in Fig. b, the vertical dislacement of the oints on d the rigid bar is dg. For oints and D, cos u and cos u D Thus, cos u 13 5 the vertical dislacement of oints and D are d g d D g The similar triangles shown in Fig. c give d F (13)>E 169 F cos u 12>13 12E d D F D (20)>E 100 F D cos u D 3>5 3 E d g 5 d D g a 169 F 12 E b 1 16 a 100 F D 3E b F F D (2) Solving Eqs. (1) and (2), yields Thus, F D lb F lb s D F D D (103 ) si 15.4 ksi s F (103 ) si 11.4 ksi 164

44 04 Solutions /25/10 3:20 M age ontinued 165

45 04 Solutions /25/10 3:20 M age The rigid bar is originally horizontal and is suorted by two -36 steel cables each having a crosssectional area of 0.04 in 2. Determine the rotation of the bar when the 800-lb load is alied. 12 ft 800 lb D Referring to the FD of the rigid bar Fig. a, 5 ft 5 ft 6 ft a + M 0; F a 12 (1) 13 b(5) + F D a 3 b(16) - 800(10) 0 5 The unstretched length of wires and D are ft and D ft. The stretch of wires and D are d F E F (13) d D F D D F D(20) E E E Referring to the geometry shown in Fig. b, the vertical dislacement of the oints on d the rigid bar is d. For oints and D, cos u and cos u D 3 12 g. Thus, cos u 13 5 the vertical dislacement of oints and D are d g d D g The similar triangles shown in Fig. c gives d F (13)>E 169 F cos u 12>13 12E d D F D (20)>E 100 F D cos u D 3>5 3 E d g 5 d D g a 169 F 12 E b 1 16 a 100 F D 3 E b F F D (2) Solving Eqs (1) and (2), yields Thus, F D lb F lb d D g 100(614.73) 3(0.04)29.0 (10 6 )D ft Then u a ft ba ft b

46 04 Solutions /25/10 3:20 M age The horizontal beam is assumed to be rigid and suorts the distributed load shown. Determine the vertical reactions at the suorts. Each suort consists of a wooden ost having a diameter of 120 mm and an unloaded (original) length of 1.40 m. Take E w 12 Ga. 18 kn/m 1.40 m 2 m 1 m a + M 0; F (1) - F (2) 0 (1) + c F y 0; F + F + F (2) d - d 2 d - d 3 ; 3d - d 2d 3F E - F E 2F E ; 3F - F 2F (3) Solving Eqs. (1) (3) yields : F 5.79 kn F 9.64 kn F 11.6 kn The horizontal beam is assumed to be rigid and suorts the distributed load shown. Determine the angle of tilt of the beam after the load is alied. Each suort consists of a wooden ost having a diameter of 120 mm and an unloaded (original) length of 1.40 m.take E w 12 Ga. a + M 0; F (1) - F (2) 0 (1) c + F y 0; F + F + F (2) 18 kn/m 1.40 m d - d 2 d - d 3 ; 3d - d 2d 2 m 1 m 3F E - F E 2F E ; 3F - F 2F (3) Solving Eqs. (1) (3) yields : F kn; F kn; F kn d F E (103 )(1.40) 4 (0.122 )12( (10-3 ) m ) d F E (103 )(1.40) 4 (0.122 )12( (10-3 ) m ) tan u (10-3 ) u (10-3 ) rad 1.14(10-3 ) 167

47 04 Solutions /25/10 3:20 M age 168 *4 60. The assembly consists of two osts D and F made of -36 steel and having a cross-sectional area of 1000 mm 2, and a 2014-T6 aluminum ost E having a crosssectional area of 1500 mm 2. If a central load of 400 kn is alied to the rigid ca, determine the normal stress in each ost. There is a small ga of 0.1 mm between the ost E and the rigid member. 400 kn 0.5 m 0.5 m 0.4 m D E F Equation of Equilibrium. Due to symmetry, F D F F F. Referring to the FD of the rigid ca, Fig. a, + c F y 0; F E + 2F - 400(10 3 ) 0 (1) omatibility Equation. Referring to the initial and final osition of rods D (F) and E,Fig.b, d d E F(400) 1(10-3 )200(10 9 )D F E (399.9) 1.5(10-3 )73.1(10 9 )D F F E + 50(10 3 ) (2) Solving Eqs (1) and (2) yield Normal Stress. F E 64.56(10 3 ) N F (10 3 ) N s D s F F st (103 ) 1(10-3 ) 168 Ma s E F E 64.56(103 ) al 1.5( Ma ) 168

48 04 Solutions /25/10 3:20 M age The distributed loading is suorted by the three susender bars. and EF are made of aluminum and D is made of steel. If each bar has a cross-sectional area of 450 mm 2, determine the maximum intensity w of the distributed loading so that an allowable stress of 1s allow 2 st 180 Ma in the steel and 1s allow 2 al 94 Ma in the aluminum is not exceeded. E st 200 Ga, E al 70 Ga. ssume E is rigid. 1.5 m 1.5 m D F al st al 2 m E w a+ M 0; F EF (1.5) - F (1.5) 0 F EF F F + c F y 0; 2F + F D - 3w 0 (1) omatibility condition : d d F (70)(10 9 ) F D (200)(10 9 ) ; F 0.35 F D (2) ssume failure of and EF: F (s allow ) al 94(10 6 )(450)(10-6 ) N From Eq. (2) F D N From Eq. (1) w 68.5 kn>m ssume failure of D: F D (s allow ) st 180(10 6 )(450)(10-6 ) N From Eq. (2) F N From Eq. (1) w 45.9 kn>m (controls) 169

49 04 Solutions /25/10 3:20 M age The rigid link is suorted by a in at, a steel wire having an unstretched length of 200 mm and crosssectional area of 22.5 mm 2, and a short aluminum block having an unloaded length of 50 mm and cross-sectional area of 40 mm 2. If the link is subjected to the vertical load shown, determine the average normal stress in the wire and the block. E st 200 Ga, E al 70 Ga. Equations of Equilibrium: a+ M 0; 450(250) - F (150) - F D (150) F - F D 0 omatibility: [1] 200 mm 100 mm 150 mm 150 mm 450 N D 50 mm Solving Eqs. [1] and [2] yields: verage Normal Stress: d d D F (200) 22.5(10-6 )200(10 9 ) F D (50) 40(10-6 )70(10 9 ) F F D F D N F N s D F D Ma D 40(10-6 ) [2] s F (10-6 ) 9.55 Ma 170

50 04 Solutions /25/10 3:20 M age The rigid link is suorted by a in at, a steel wire having an unstretched length of 200 mm and crosssectional area of 22.5 mm 2, and a short aluminum block having an unloaded length of 50 mm and cross-sectional area of 40 mm 2. If the link is subjected to the vertical load shown, determine the rotation of the link about the in. Reort the answer in radians. E st 200 Ga, E al 70 Ga. Equations of Equilibrium: a+ M 0; 450(250) - F (150) - F D (150) F - F D 0 omatibility: [1] 200 mm 100 mm 150 mm 150 mm 450 N D 50 mm Solving Eqs. [1] and [2] yields : Dislacement: d d D F (200) 22.5(10-6 )200(10 9 ) F D (50) 40(10-6 )70(10 9 ) F F D F D N F N [2] d D F D D D E al (50) 40(10-6 )(70)(10 9 ) mm tan u d D u 63.7(10-6 ) rad

51 04 Solutions /25/10 3:20 M age 172 *4 64. The center ost of the assembly has an original length of mm, whereas osts and have a length of 125 mm. If the cas on the to and bottom can be considered rigid, determine the average normal stress in each ost. The osts are made of aluminum and have a cross-sectional area of 400 mm 2. E al 70 Ga. 100 mm 100 mm 800 kn/m 125 mm a+ M 0; -F (100) + F (100) 0 F F F + c F y 0; 2F + F d d F (0.125) 400 (10-6 )(70)(10 6 ) F (0.1247) 400 (10-6 )(70)(10 6 ) F F 8.4 Solving Eqs. (2) and (3) F kn F kn (1) (2) (3) 800 kn/m s s (103 ) 400(10-6 ) 189 Ma s (103 ) 400 (10-6 ) 21.4 Ma The assembly consists of an -36 steel bolt and a red brass tube. If the nut is drawn u snug against the tube so that 75 mm, then turned an additional amount so that it advances 0.02 mm on the bolt, determine the force in the bolt and the tube.the bolt has a diameter of 7 mm and the tube has a cross-sectional area of 100 mm 2. Equilibrium: Since no external load is alied, the force acting on the tube and the bolt is the same. omatibility: 0.02 d t + d b 0.02 (75) 100(10-6 )(101)(10 9 ) + (75) 4 ( )(200)(10 9 ) N 1.16 kn 172

52 04 Solutions /25/10 3:20 M age The assembly consists of an -36 steel bolt and a red brass tube. The nut is drawn u snug against the tube so that 75 mm. Determine the maximum additional amount of advance of the nut on the bolt so that none of the material will yield. The bolt has a diameter of 7 mm and the tube has a cross-sectional area of 100 mm 2. llowable Normal Stress: (s g ) st st 4 (0.007)2 st kn (s g ) br br 100(10-6 ) br 7.00 kn Since st 7 br, by comarison he brass will yield first. omatibility: a d t + d b 7.00(10 3 )(75) 100(10-6 )(101)(10 9 ) (10 3 )(75) 4 (0.007)2 (200)(10 9 ) mm 173

53 04 Solutions /25/10 3:20 M age The three susender bars are made of the same material and have equal cross-sectional areas. Determine the average normal stress in each bar if the rigid beam E is subjected to the force. D F a + M 0; F D (d) + F EF (2d) - a d 2 b 0 F D + 2F EF 2 (1) E + c F y 0; F + F D + F EF - 0 (2) d 2 d 2 d d - d E d d - d E 2d 2d d + d E 2F D E F E + F EF E 2F D - F - F EF 0 (3) Solving Eqs. (1), (2) and (3) yields F 7 12 F D 3 F EF 12 s s D s EF *4 68. steel surveyor s tae is to be used to measure the length of a line. The tae has a rectangular cross section of 0.05 in. by 0.2 in. and a length of 100 ft when T 1 60 F and the tension or ull on the tae is 20 lb. Determine the true length of the line if the tae shows the reading to be ft when used with a ull of 35 lb at T 2 90 F. The ground on which it is laced is flat. a st > F, E st ksi. 0.2 in in. d T a T 9.6(10-6 )(90-60)(463.25) ft d (35-20)(463.25) ft E (0.2)(0.05)(29)(10 6 ) ft 174

54 04 Solutions /25/10 3:20 M age Three bars each made of different materials are connected together and laced between two walls when the temerature is T Determine the force exerted on the (rigid) suorts when the temerature becomes T The material roerties and cross-sectional area of each bar are given in the figure. Steel E st 200 Ga a st 12(10 6 )/ st 200 mm 2 rass E br 100 Ga a br 21(10 6 )/ br 450 mm 2 oer E cu 120 Ga a cu 17(10 6 )/ cu 515 mm mm 200 mm 100 mm ( ; + ) 0 T - d 0 12(10-6 )(6)(0.3) + 21 (10-6 )(6)(0.2) + 17 (10-6 )(6)(0.1) F(0.3) - 200(10-6 )(200)(10 9 ) - F(0.2) 450(10-6 )(100)(10 9 ) - F(0.1) 515(10-6 )(120)(10 9 ) F 4203 N 4.20 kn The rod is made of -36 steel and has a diameter of 0.25 in. If the rod is 4 ft long when the srings are comressed 0.5 in. and the temerature of the rod is T 40 F, determine the force in the rod when its temerature is T 160 F. k 1000 lb/in. 4 ft k 1000 lb/in. omatibility: : + x d T - d F x 6.60(10-6 )(160-40)(2)(12) (0.5)(2)(12) 4 (0.252 )(29.0)(10 3 ) x in. F 1.00( ) ki ft-long steam ie is made of -36 steel with s Y 40 ksi. It is connected directly to two turbines and as shown. The ie has an outer diameter of 4 in. and a wall thickness of 0.25 in. The connection was made at T 1 70 F. If the turbines oints of attachment are assumed rigid, determine the force the ie exerts on the turbines when the steam and thus the ie reach a temerature of T F. 6 ft omatibility: : + 0 d T - d F (10-6 F(6)(12) )(275-70)(6)(12) - 4 ( )(29.0)(10 3 ) F 116 ki 175