# Answers to Selected Even-Numbered Problems

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3 .10 6:57 mm..104 (a) and (b) 3.4 Nm :8 kn P 4:6 kn..108 (a) ER D. :9 O{ C :7 O 1:45 k/ O kn; (b) ET D.:9 O{/ kn..110 P? D. 0:970/P, P k D.0:43/ P..11 P k D 7:66 kn, P? D 6:43 kn, EP k D. 3:48 O{ 3:48 O 5:87 k/ O kn, EP? D.3:48 O{ C 3:48 O 4:13 k/ O kn..114 (a) x D 107 ı, y D 114 ı ; (b) 0:50 O{ C 0:693 O C 0:500 k. O.116 Answers given in problem statement. Chapter 3 3. R C D R D D 31:8 lb. 3.4 (a) T 1 D 51 lb, R D 141 lb; (b) T D 149 lb, R D 141 lb. 3.6 (a) F D 17:9 N; R A D 17:9 N; R B D 98:1 N; R D 5: N (b) Block A may move with constant velocity but it will not accelerate. (c) Block B will accelerate to the right. 3.8 T AB D 5:13 kn, T AC D 0:340 kn F AB D 1580 lb, T AC D 1500 lb. 3.1 (a) T AB D 175 N; (b) T CD D T CE D 109 N; (c) Essay-type answer (a) Answer provided in problem description; (b) F D T cos ; (c) Essay-type answer T ACB D 60 N; T CDE D 4530 N: 3.18 m D 547 kg. 3.0 T CD D 840 N, T BA D 6840 N, T CB D 7890 N, W D 110 N. 3. (a) System (a): T D W, system (b): T D 1 W, system (c): 1 W, system (d): 1 3 W, system (e): 1 3 W, system (f): 1 4 W ; (b) Essay-type answer. 3.4 W D 18 lb. 3.6 T BH D 540 lb; T HE D 540 lb; T BC D 717 lb; T EF D 586 lb; W A D 3 lb; W G D 369 lb; W D D 108 lb: 3 Last modified: March 19, 009

4 3.8 D 46:8 ı, T D 86: N W 1 D 3:6 N, W D 11:3 N. 3.3 D 30 ı ; intermediate answer: F BD D.5 lb/.cos 30ı Csin 30 ı / cos 30 ı cos sin 30 ı sin T AC D 3460 lb, T AE D 6000 lb, T AG D 440 lb, ı D 6:00 in: 3.36 (a) L AB D 94:3 cm; L AD D 10 cm; (b) x D 104 cm; y D 46:7 cm (a) 10:0 N, 8:00 N; (b) 7:00 N, 5:00 N k D 167 lb=in., L 0 D 1:0 in: 3.4 F 1 D 18 lb, F D 335 lb, F 3 D 18 lb ı D 3:4 mm (a) ı D 1:17 in:; (b) ı D 1:99 in: q.5 mm/ C. mm ı/ p 9mm 3.48 (a) F D. N=mm/. mm ı/ C.0:3 N=mm/ı; (b) Answer not provided; (c) q.5 mm/ C. mm ı/ F max D 0:600 N; (d) Essay-type answer m D D 91:7 kg. 3.5 W D 500 N (a) W D 3k 3.56 P D 800 lb P D 800 lb P D 1450 lb. 1 r p r Cd d; (b) d D 16:9 in: 3.6 (a) F AE D 40 lb, F AC D 10 lb, F AD D 10 lb; (b) F EB D 10 lb F AB D F AC D 780 lb, R D 800 lb W D 400 N, R 1 D 300 N, R D 13:7 N, ER 1 D. 7: O{ C18 O C04 O k/ N, ER D. 1:8 O{ C:7 O 4:13 O k/ N P D 107 N, ER D.3:0 O{ C 4:6 O 107 O k/n F ED D 80 lb, F EF D 360 lb, F FB D 10 lb, F EA D 380 lb, F F C D 10 lb, 60 lb must be added to traffic light E. 3.7 Essay-type answer T AE D 44:6 lb, T AD D 69:6 lb, T DF D 19 lb, T DG D 74 lb, T ABCD D 100 lb. 4 Last modified: March 19, 009

6 4.56 The forces F are resolved into horizontal and vertical components as shown. The two vertical forces F cos 30 ı constitute one couple. The two horizontal forces F sin 30 ı and the horizontal force F constitute another couple EF A D.4 lb/. 0:976 O{ C 0:484 O 0:179 O k/, EF B D EF A D.4 lb/.0:976 O{ 0:484 O C 0:179 O k/; Additional result: the moment of the couple about line a is M a D 91 in:lb Force systems (a) and (d) are equivalent, and force systems (b) and (c) are equivalent. 4.6 For force system (b), F D kip and P D 4 kip; for force system (c), there are no values for F and P such that force systems (a) and (c) will be equivalent; for force system (d), F D 5 kip and P D 1 kip F Rx D 1490 lb, F Ry D 10 lb, M R 5110 in.lb F Rx D 0:500 kn, F Ry D 3:87 kn, :58 m to the right of the bearing at C (a) EF R D O k/ N, EM RO D.6:94 O{ 1:0 O / Nm; (b) EF R D O k/ N,.x; y/ D. 3:86 mm; 6:67 mm/: 4.70 (a) EF R D. 5:00 O{ C 50:0 O 10:0 O k/ nn, EM RB D.10 O{ 60:0 O k/ nnm; (b) EF R D. 5:00 O{ C 50:0 O 10:0 O k/ nn, EM RO D.0 O{ C 800 O C 3890 O k/ nnm. 4.7 (a) EF R D.10:0 O / kn, EM RA D.5:0 O{ C 4:00 O C 0:0 O k/ knm; (b) F Rx D 0, F Ry D 10:0 kn, F R D 0, M RAx D 5:0 knm, M RAy D 4:00 knm, M RA D 0:0 knm EF R D F O k, EM R D 3P r O k,.x; y/ D.0; r/ EF R D.:00 O{ C 3:00 O 4:00 O k/ N, EM RA D.9:00 O{ C 8:00 O 4:00 O k/ Nm EF R D F. O C k/, O EM R D F a 1. O C k/, O.x; y/ D. 1 a; b/ EM R D.150 O{ C 104 O 00 O k/ Nm: 4.8 (a) M a D 56 Nm; (b) Ou D 0:530 O{ C 0:66 O C 0:530 O k F D 300 lb P D 10 lb, Q D 5 lb, R D 0, S D 0 lb, M D D 10 in:lb (a) F Rx D 0:880 lb, F Ry D 0:660 lb, M O D 0:55 in. lb; (b) F Rx D 0:880 lb, F Ry D 0:660 lb, y D 0:67 in: 4.90 (a) F Rx D 0, F Ry D 0, F R D 1800 lb, M Ax D 8700 ftlb, M Ay D 0, M A D 6;700 ftlb; (b) F Rx D 0, F Ry D 0, F R D 1800 lb, x D 1: ft, D 7:83 ft. 4.9 EF R D. 30:0 O{ 10 O :5 O k/ lb, EM RO D. 990 O{ C 60:0 O C 1180 O k/ in.lb. Chapter 5 5. A D 3:5 N, B D 107 N. 5.4 F D 10 lb, B D 13 lb, A D 53:5 lb. 6 Last modified: March 19, 009

7 5.6 B D 4:9 N (force supported by one link), A x D 68:6 N, A y D 51:4 N. 5.8 d D 3:10 ft (a) A y D 3 F C 3 1 P, A x D 0, D y D 1 3 F C 3 P ; (b) A x D 0:19F C0:385P, A y D 3 F C 1 3 P, D D 0:385F C 0:770P ; (c) A y D F C P, D x D 0, M D D 3 1 L.F C P /; (d) A x D 0, A y D F C P, M A D 1 3 L.F C P /. 5.1 A y D 1:88 kn, E y D :1 kn, E x D :00 kn A x D 0, A y D 3:00 kip, P y D 1:00 kip C y D 11;300 lb, D y D 6170 lb A x D 0, B y D P, M A D P a. 5.0 Rollers B and C make contact, A y D 0, B y D 40:8 lb, C y D 31 lb, D y D Rollers A and C make contact, A y D 193 lb, B y D 0, C y D 198 lb, D y D (a) A D 498 lb; (b) Essay-type answer. 5.6 (a) > 75:5 ı ; (b) > 90:0 ı 5.8 Essay-type answer F x D 9:08 lb, F y D 4:8 lb, E y D 17:0 lb. 5.3 G D 4:66 N, N D 17:4 N, D 17:0 ı A x D 0, A y D 00 lb, T D 600 lb (a) A x D :59 kn, A y D 35: kn, B y D 54:8 kn; (b) A x D 7:59 kn, A y D 41: kn, B y D 59:5 kn T D 1:66 kip (tension in cable BCD), A x D 1:66 kip, A y D 0:55 kip T DE D :76 kip, F DF D :36 kip (a) Complete fixity, statically determinate; (b) T D 10:68F ; Additional results: C x D :000F, C y D 11:3F. 5.4 (a) Partial fixity, statically indeterminate; (b) The cable tension cannot be determined; Additional result: The equations of static equilibrium require F D 0. If F 0, then the equations of static equilibrium cannot be satisfied, and dynamic motion occurs F D 13:6 lb (a) D 4:58 ı ; (b) D 36:7 ı ; (c) Essay-type answer Essay-type answer 5.50 A x D 0, A y D F h 1 kl i F kl.k t C kl, C / y D.k t C kl /, M A D FL h 1 kl i k t C kl. 5.5 For structure (a), full fixity and statically determinate; for structure (b), full fixity and statically indeterminate; for structure (c), full fixity and statically determinate; for structure (d), partial fixity and statically determinate; for structure (e), full fixity and statically determinate; for structure (f), partial fixity and statically indeterminate D D 0:314 N, L D 1:34 N ı D :40 mm, D 1:49 ı (a) T EA D 1:33 kip, B x D 0, B y D 3:33 kip; (b) T EA D 1:3 kip, B x D 0, B y D 0:3 kip; (c) T EA D 15:9 kip, B x D 0, B y D 5:7 kip. 7 Last modified: March 19, 009

8 5.60 ABC is a 3-force member and BD is a -force member. 5.6 ABDE is a multiforce member and BC is a -force member force member force member ABCD is a multiforce member and DE is a -force member Multiforce member force member F AB D 318 kn to begin opening the dump when D 0 ı ; F AB D 0 when the center of gravity G of the dump and its contents is immediately above point O C x D 0:333 kn, C y D 0:667 kn, T AD D 1:5 kn, M Cx D 0, M Cy D :00 knm, M C D 3:33 knm (a) Partial fixity and statically determinate; (b) T D 7 lb, A x D 18:0 lb, A y D 9:00 lb, M Ax D 54:0 in.lb, M A D T AD D 50:75 kn. 5.8 (b), 0, 4, 1, 3; (a) 5.84 A x D 0, A y D 0, A D F, M Ax D F r, M Ay D 0, M A D P r. 8 Last modified: March 19, 009

9 5.86 F D 3:1 kn, C x D 3:00 kn, C y D :07 kn, C D 0:49 kn, M Cx D 0:343 knm, M Cy D 0:600 knm A D 5:77 kn, C D :89 kn, D y D 5:00 kn, D D 0, M Dy D 0, M D D 9:3 knm Essay-type answer; partial answer: the propellers rotate counterclockwise. 5.9 T D 163 N, C D 163 N, A D 83: N A x D 7:00 N, A y D 6:1 N, M A D 76 Ncm (a) F D 3:9 N; (b) F D 11:5 N; (c) F D 0:1 N For structure (a), full fixity and statically determinate; for structure (b), partial fixity and statically indeterminate; for structure (c), full fixity and statically indeterminate; for structure (d), partial fixity and statically determinate; for structure (e), partial fixity and statically determinate; for structure (f), full fixity and statically indeterminate (a) Essay-type answer; (b) T 1 D 15:7 lb, T D 1:5 lb, O x D 13:3 lb, O y D 11:5 lb, O D 19:4 lb (a) Full fixity and statically determinate; (b) A y D 0, A D 60:0 lb, B x D 40:0 lb, B D 40:0 lb, M Ay D 00 in.lb, M A D 80:0 in.lb. 9 Last modified: March 19, 009

10 Chapter 6 6. T AB D 500 lb, T AC D 0, T BC D 65 lb, T BD D 375 lb, T CD D 65 lb, T CE D 0, T DE D 500 lb. 6.4 P D 300 lb. 6.6 P D 8000 lb. 6.8 T CE D 3:75 kn, T DF D 3:66 kn T AB D 4:00 kn, T AC D 0, T BC D 7:1 kn, T BD D 6:00 kn, T CD D 5:00 kn, T CE D 6:00 kn, T DE D 7:1 kn. 6.1 P D 10:0 kn P D 6:66 kn :9: :77: 6.0 T AB D F, T AE D p F, T CD D 0, T CE D 0, T DB D p F, T DE D F, T EB D F. 6. T AB D 5:66 kn, T AD D 4:00 kn, T BC D 7:07 kn, T BD D :4 kn, T BE D 3:00 kn, T BF D 4:47 kn, T CF D 5:00 kn, T DE D 3:00 kn, T EF D 3:00 kn. 6.4 (a) F G, DG; (b) T F G D 0, T DG D 0, T AC D 1:0 kn, T AB D 10:4 kn, T CB D 4:00 kn, T BD D 10:4 kn, T GH D 8:00 kn, T FH D 6:93 kn, T EG D 8:00 kn, T CE D 8:00 kn, T DE D 8:00 kn, T CD D 16:0 kn, T DF D 6:93 kn. 6.6 (a) BG, DI, EJ, DE, CD, CH ; (b) T BG D 0, T CH D 0, T AB D 3:00 kip, T AF D 5:0 kip, T BC D 3:00 kip, T CF D 7:1 kip, T F G D 6:00 kip, T C G D 5:00 kip, T GH D 3:00 kip; (c) T DI D 0, T EJ D 0, T DE D 0, T CD D 0, T CJ D 14:4 kip, T IJ D 9:00 kip, T CI D 10:0 kip, T HI D 3:00 kip. 6.8 T BD D 5:77P, T CD D 5:77P, T CE D 8:66P 6.30 (a) DE, HI, JK, LM, NO; (b) eliminate DE, HI, LM, essay-type answer; (c) T GH D 5 8 Q. 6.3 (a) HI, JK, LM ; (b) eliminate JK, essay-type answer; (c) T F G D :00 kn (a) Statically determinate; (b) T CD D 8:00 kip, T DE D 0, T AD D 10:4 kip, T DB D 10:4 kip (a) T J T D 40: kn; (b) T HJ D 30:0 kn, T IJ D 14:1 kn, T JK D 0, T JM D 0:563 kn, T JL D 0:796 kn (a) Statically indeterminate; (b) mechanism; (c) statically indeterminate; (d) statically determinate (a) For truss system (a): T AB D 1:41W, T BD D W, T DF D 1:41W, T AC D W, T CE D W, T EF D W, T BC D W, T CD D 0, T DE D W ; for truss system (b): T AB D 1:1W, T BD D 0:500W, T DF D 1:1W, T AC D 0:500W, T CE D 0:500W, T EF D 0:500W, T BC D W, T CD D 0, T DE D W ; (b) Essay-type answer; (c) Essay-type answer; (d) Essay-type answer. 6.4 (a) Partial fixity; (b) statically determinate; (c) T AB D 0, T AC D 0, T AD D 4:00 kn, T BC D 0:417 kn, T BD D :13 kn, T BE D 0:838 kn, T CD D :13 kn, T CE D 0:838 kn, T DE D 1:33 kn. 10 Last modified: March 19, 009

11 6.44 Answers to be provided Answers to be provided Answers to be provided FBDs to be provided. 6.5 Essay-type answer Horizontal force applied by the operator is Q D 0:8 lb, T AE D 6:67 lb, B x D 5:77 lb, B y D 13:3 lb, C x D 15:0 lb, C y D 13:3 lb Horizontal force applied by the operator is Q D 47:7 N, F BE D 63:0 N, A x D 55:7 N, A y D 10:7 N, C x D 103 N, C y D 10:7 N (b) F AD D 3390 lb; (a) 6.60 (b) T 1 D 83 N, T D 83 N; (c) from DE: A x D 60 N, A y D 5 N, from F G: A x D 60 N, A y D 5 N; (a) 6.6 (b) F CF D 7;00 lb; (a) 6.64 F FH D :5 N, F BF D 70:8 N. 11 Last modified: March 19, 009

12 6.66 (a) The structure is not a truss, essay-type answer; (b) T DG D :00 kn, T BE D 3:00 kn, T FH D 4:00 kn, T AD D 3:00 kn, T DE D 3:00 kn, T EF D 3:00 kn, T CF D 3:00 kn, B x D 3:00 kn, B y D 1:00 kn (a) Statically indeterminate; (b) QH and IQ, (c) F DG D p kn T BP D 813 lb, T BO D 813 lb, T CD D 100 lb. 6.7 T AD D 163 lb, T AB D 115 lb, T BE D 163 lb, T BD D 115 lb, T GI D 0, T HI D 85:0 lb, T EH D 184 lb, T DG D 184 lb, T EG D 35:0 lb (b) F AE D 55 lb; (a) 1 Last modified: March 19, 009

13 Chapter Nx; Ny/ D.0; 65:0/ mm Nx; Ny/ D.1:18; 1:18/ in Nx; Ny/ D.60:8; 30:0/ mm Nx; Ny; Ń/ D.:31; :1; 1:4/ in Nx; Ny/ D. 9 0 ; 9 0 / in Nx; Ny/ D. 6 5 ; 3 / mm Nx; Ny/ D ; Nx; Ny/ D.48:0; 16:0/ in Answer given in problem statement. 7.0 c 1 D b= p a, c D b=a 3,. Nx; Ny/ D. 1 5 a; 3 7 b/. R 1 in: 0 x p 1C9x 4 dx R 1 in: 0 R 1 in: 7. (a) Nx D p ; Ny D 0 x 3p R r1c 1 1C9x 4 in: dx 1C9x 4 R 1 in: p ; (b) Nx D 0 y1=3 13 y =3 dy r dx 0 1C9x 4 dx R 1 in: ; Ny D 0 1C 13 y =3 dy (c). Nx; Ny/ D.0:609; 0:366/ in. s s R r r x R x 1C p r p dx r 7.4 (a) Nx D x s r r x 1C p x dx r ; Ny D x s ; (b). Nx; Ny/ D.0; r/. x x 1C dx 1C dx R r r p r x 7.6 Nx D 11 8 R Nx; Ny; Ń/ D.133; 0; 0/ in Nx; Ny; Ń/ D.:98; 0; 0/ in. 7.3 m c D 680 g. R r r p r x r R 1 in: 0 y 1C 13 y =3 dy r R 1 in: ; 0 1C 13 y =3 dy Nx; Ny; Ń/ D.1:07; 0; 0:876/ in Nx; Ny/ D.1:46; 1:00/ m, mass of frame m D 44:6 kg, T D 639 N; B x D 553 N; B y D 118 N (a) Nx D R r= 0 x 0.r x /dxc R r r= x 0.r x /dx R r= 0 0.r x /dxc R r r= 0.r x /dx, Ny D 0; Ń D 0; (b) Nx D 5 56 r R a dx 0 (a) Nx D 0 h 1 x a R a dx ; (b) h x 16 5 a. a 7.4 R L 0 x 0 R. L x / =3 R x L dx (a) Nx D R L 0 0 R. L x / =3, Ny D 0; Ń D 0; (b) Nx D R x 14 L dx 7.44 (a) Ny D R 1 cm 0 1.1CxCx / 0 xœ1cx x dx R 1 cm 0 0 xœ1cx x dx ; (b) Nx D cm. 13 Last modified: March 19, 009

14 7.46 V D 83 mm 3, A D 6 mm V D 1: mm 3, A D 1: mm V D 8h 3, A D h.11 C 8 p /. 7.5 V D 19:8 in. 3, W c D 0:118 lb (a) V D 37 mm 3 ; (b) A outside D 34:610 3 mm ; (c) A inside D 1:10 3 mm V D 7 1 R3, A D 4.11 C p 5/R (a) w D :78 lb=in. for 0 x 36 in:; (b) w D 1:85 lb=in. for 0 x 18 in:, w D 3:70 lb=in. for 18 in: x 36 in:; (c) w D 1:85 lb=in. C.0:0514 lb=in. /x for 0 x 36 in:, (d) w D 1:85 lb=in. C.0:103 lb=in. /x for 0 x 18 in:, w D 5:56 lb=in..0:109 lb=in. /x for 18 in: x 36 in: 7.60 A x D 0, A y D 33:3 lb, B y D 66:7 lb. 7.6 A x D 0, A y D 49:1 kn, M A D 98:1 knm (a) Nx D 1:0 in:; (b) & (c) A x D 0, A y D : lb, B y D 77:8 lb (a) Nx D 1:67 m; (b) & (c) A x D 0, A y D 49:1 kn, M A D 81:8 knm (a) a D 800 N=m, b D 0, c D 8:00 N=m 3 ; (b) d D 800 N=m, f D 0 m. 7.7 A x D 0, A y D 4:00 kn, B y D 16:0 kn A x D 0, A y D 40:0 kip, B y D 40:0 kip A x D 0, A y D 6:08 kip, B y D 4:9 kip C x D 0, C y D 1:0 kn, M C D 4 knm A x D 0, A y D 80 kip, C y D 80 kip. 7.8 A x D 0, A y D w 1 a C w.l a/, M A D 1 w 1 a C 1 w.l a / a D 0:54 kn=m, b D 1:1 kn=m, c D 0:0556 kn=m 3, F D 3:0 kn, Nx D 4:05 m A x D 0, A y D 0, A D W, M Ax D WR T D d T D 39:0 lb D 0:35 rad D 13:5 ı (a) T D 40:7 lb; (b) T D 33 lb T B D 414 N, T C D 87 N, T D D 684 N F EG D 381 lb Nx; Ny/ D.0; 0:/ mm Nx; Ny/ D ; 1/ m Nx; Ny; Ń/ D.34:7; 0; 0/ mm (a) Let o D 0:00 g=mm 3, i D 0:003 g=mm 3, Nx D Ń D 0; (b). Nx; Ny; Ń/ D.3:0; 0; 0/ mm Nx D 4 3 m. R 1 cm 0 x.1cx Nx D / dx R, (b) Nx D 5 1 cm 0.1Cx / dx 8 cm (a) Nx D R km 1 km xp 1C4x R km 1 km V D 4 3 r3, A D 4r V D 4360 cm 3, A D 867 cm. /, M Ay D WR, M A D 0. R 6 mm 0 xf i. 6 1 x/ C o Œ.3C 6 1 x/. 1 6 x/ g dx R 6 mm 0 f i. 6 1 x/ C o Œ.3C 6 1 x/. 1 6 x/ g dx R km p, Ny D 1 km xp 1C4x p ; (b). Nx; Ny/ D.0:801; 1:48/ km. 1C4x 1C4x R km 1 km A x D 0, A y D 1700 lb, B y D 1300 lb B y D 640 lb, C x D 0, C y D 1460 lb where pin C is located 6 ft from the left-hand end of the beam. 7.1 A x D 0, A y D :17 kn, C y D 3:83 kn where roller C is located m from the right-hand end of the beam A x D 46:8 lb, A y D 81 lb, C x D 187 lb d D 1:5 mm., Ny D 0, 14 Last modified: March 19, 009

15 Chapter 8 8. N H D 34:7 lb, V H D 197 lb, M H D 1180 in.lb, N J D 480 lb, V J D 0, M J D N E D 00 lb, V E D 346 lb, M E D 0, N F D 00 lb, V F D 346 lb, M F D 8310 in.lb 8.6 N F D 3:0 kn, V F D 4:7 kn, M F D 0, N G D 3:0 kn, V G D 4:7 kn, M G D 10:7 kn m, N H D 0:800 kn, V H D 1:07 kn, M H D 10:7 knm, N I D 0:800 kn, V I D 1:07 kn, M I D 13:3 knm. 8.8 N F D 11: kn, V F D 1:60 kn, M F D 0, N G D 11: kn, V G D 1:60 kn, M G D 4:00 kn m, N H D 8:80 kn, V H D 1:60 kn, M H D 4:00 knm, N I D 8:80 kn, V I D 1:60 kn, M I D N D D 0, V D D 100 lb, M D D 0, N E D 0, V E D 100 lb, M E D 150 in.lb, N F D 70:7 lb, V F D 70:7 lb, M F D 56 in.lb. 8.1 N H D :00 kn, V H D :00 kn, M H D 0, N I D :00 kn, V I D :00 kn, M I D 5:60 knm: 8.14 N L D :00 kn, V L D :00 kn, M L D 3:0 knm, N O D :00 kn, V O D :00 kn, M O D N R D 5:83 kn, V R D 0 kn, M R D 0, N A D 4:00 kn, V A D 0, M A D 4:0 knm N H D 400 lb, V H D 693 lb, M H D 1390 ftlb, N I D 400 lb, V I D 693 lb, M I D N L D 46 lb, V L D 653 lb, M L D 1600 ftlb, N M D 0, V M D 800 lb, M M D Essay-type answer. 8.4 V Ax D 0, V Ay D 150 lb, N A D 0, M Ax D 400 in.lb, M Ay D 0, M A D 3600 in.lb. 8.6 N Bx D 0, V By D 0, V B D 10 N, M Bx D 8:40 Nm, M By D 1:6 Nm, M B D V Dx D 0, N Dy D 0, V D D 10 N, M Dx D 13:8 Nm, M Dy D 14:4 Nm, M D D V D :50 kn, M D.:50 kn/x for 0 x 3 m; V D 7:50 kn, M D.7:50 kn/.4 m x/ for 3 m x 4 m. 8.3 (a) V D P, M D P x for 0 x 3 1 L; V D 0, M D 3 1 PL for 1 3 L x 3 L; V D P, M D P.L x/ for 3 L x L; (b) Essay-type answer V D 15 lb, M D.15 lb/ x for 0 x 8 ft; V D 15 lb, M D. 15 lb/.16 ft x/ for 8 ft x 16 ft V D 1000 lb, M D lb/.0 ft x/ V D. 8:00 kn=m/ x, M D. 4:00 kn=m/ x V D 100 N.600 N=m/ x C.50 N=m / x, M D.100 N/ x.300 N=m/ x C.16:7 N=m / x Essay-type answer V D 3:00 N, M D 8:00 N mm C.3:00 N/ x for 0 x mm; V D 1:00 N, M D.1:00 N/ x for mm x 4 mm; V D :00 N, M D 1 Nmm C.:00 N/ x for 4 mm x 6 mm V D 3:00 N, M D 8:00 Nmm C.3:00 N/ x for 0 x mm; V D 0, M D 0 for mm x 6 mm: 8.48 Essay-type answer V D 100 lb=ft.6:00 ft x/ C 1000 lb, M D 50 lb=ftœ.1:0 ft/x x C.1000 lb/ x for 0 x 6 ft; V D 100 lb=ft.6:00 ft x/ 1000 lb, M D 50 lb=ftœ.1:0 ft/x x C.1000 lb/.1 ft x/ for 6 ft x 1 ft. 8.5 V D 1000 N, M D.1000 N/ x for 0 x 1 m; V D 100 N, M D.100 N/.9:00 m C x/ for 1 m x m; V D 1100 N, M D.1100 N/.3:00 m x/ for m x 3 m V D.0:0 kip=in./ x, M D.0:101 kip=in./x for 0 x 0 in:; V D 10:0 kip C.0:0 kip=in./ x, M D 00 in. lb.10:0 kip/ x C.0:101 kip=in./x for 0 in: x 79 in:; V D. 0:0 kip=in./.99:0 in: x/, M D.0:101 kip=in./.99:0 in: x/ for 79 in: x 99 in: 8.56 V D. 8:00 kn=m/ x, M D. 4:00 kn=m/ x V D 60 kn.8:00 kn=m/ x. 1 3 kn=m / x, M D 19 knm C.60:0 kn/ x.4 kn=m/ x. 1 9 kn=m / x V D 4000 lb.15:0 lb=ft/œ.40:0 ft/x x, M D.4000 lb/ x.5:00 lb=ft /Œ.60 ft/x x 3, M max D 15;400 ftlb at x D 8:45 ft. 8.6 V D 1 6 w 0 L w 0.x 1 L x /, M D w 0. L 6 x 1 x C 1 3L x3 /, M max D.0:0160/w 0 L at x D.0:11/L and M max D. 0:0160/w 0 L at x D.0:789/L V D P, M D 3 PL C P x for 0 x L ; V D P, M D 1 PL C P.x 1 L/ for L x L V D. 5:0 lb=ft / x 5, M D. 3 lb=ft / x 3 for 0 x 4 ft; V D 1800 lb.5:0 lb=ft / x, M D 700 ftlb C.1800 lb/ x. 5 3 lb=ft / x 3 for 4 ft x 1 ft V D. 1:00 kn=m/ x, M D. 0:500 kn=m/ x for 0 x 1:3 m; V D 6:86 kn.1:00 kn=m/.x 1:30 m/, M D 7: knm.5:56 kn/ x.0:500 kn=m/ x for 1:3 m x 4 m; V D 14:0 kn.1:00 kn=m/.x 4:00 m/, M D 5:0 knm.10:0 kn/ x.0:500 kn=m/ x for 4 m x 5:5 m. 15 Last modified: March 19, 009

16 8.70 V max D 4P at x D 0, M max D 10P d at x D V max D 45:0 N just to the left of x D 30 mm, M max D 1130 Nmm at x D 30 mm V max D 10:0 lb just to the right of x D 17 in:, M max D 16:0 in.lb at x D 4 in: 8.76 Beam 1: Shear diagram (b), Moment diagram (g); Beam : Shear diagram (a), Moment diagram (g); Beam 3: Shear diagram (d), Moment diagram (f) N F D 6:19 kip, V F D 0:940 kip, M F D 0, N G D 6:19 kip, V G D 0:940 kip, M G D 4:00 kipft N E D 833 N, V E D 833 N, M E D 0, N F D 1180 N, V F D 0, M F D 69:0 N m, N G D 833 N, V G D 833 N, M G D V D 1:0 kn.0:750 kn=m / x, M D.1:0 kn/ x.0:50 kn=m / x 3 for 0 x 4 m; V D 36:0 kn.1:0 kn=m/ x C.0:750 kn=m / x, M D 3:0 knm C.36:0 kn/ x.6:00 kn=m/ x C.0:50 kn=m / x 3 for 4 m x 8 m V D 1:0 kn.0:750 kn=m / x, M D.1:0 kn/ x.0:50 kn=m / x 3 for 0 x 4 m; V D 36:0 kn.1:0 kn=m/ x C.0:750 kn=m / x, M D 3:0 knm C.36:0 kn/ x.6:00 kn=m/ x C.0:50 kn=m / x 3 for 4 m x 8 m V D 60:0 N, M D 1;000 Nmm C.60:0 N/ x V D 60:0 N, M D 1;000 Nmm C.60:0 N/ x for 0 x 150 mm; V D 40 N, M D 48;000 Nmm C.40 N/ x for 150 mm x 00 mm V max D 1750 lb for 15 ft x 0 ft, M max D 8750 ftlb at x D 15 ft. 8.9 (a) V D 4:00 kn. 1 3 kn=m / x ; (b) M D. 4:00 kn/ x. 1 9 kn=m / x 3 ; (c) V max D 9:00 kn at x D 3 m, M max D 15:0 knm at x D 3 m (a) M D. :00 kip=ft/ x ; (b) V D 56:0 kip.8:00 kip=ft/ x C. 3 1 kip=ft / x ; (c) V max D 0:0 kip at x D 6 ft, M max D 7:0 ftkip at x D 6 ft (a) M D 13 knm.17:0 kn/ x C.0:500 kn=m/ x ; (b) V D 16:0 kn.4:00 kn=m/ x C. 1 3 kn=m / x ; (c) V max D 16:0 kn at x D 0, M max D 48:0 knm at x D 6 m. Chapter 9 9. N D 5:1 lb, F D 10:3 lb, the box remains at rest on the ramp : D tan h D 60:0 cm The dam will not slide or tip when the reservoir is completely full. 9.1 The dam will not tip and will fail by sliding N D 1: lb, F D 8: lb P D 47:4 N (a) min D 0:89; (b) The answer to Part (a) will change, essay-type answer; (c) The answer to Part (a) does not change, essay-type answer. 9.0 s D 0: P D 0:900 lb will cause all books to slide. 9.4 s D 0: Q D 7:70 lb. 9.8 P D 10:6 N P D 7 kn. 9.3 s D The truck is not capable of pulling the dumpster (a) P D 4:56 N; (b) T CD D 6:58 N, essay-type answer (a), (b), & (c) P D 73:0 lb M A D 3:1 Nm. 9.4 M A D 575 in.lb. 16 Last modified: March 19, 009

17 9.44 (a) P D 14:1 N; (b) P D 4:54 N W D 35:3 lb W B D 53:8 lb, surface C will stick while motion is impending at A and B D 0: D 0:0943, the normal and friction forces between the dam and foundation are located 0:730 m to the left of point A P D 1:60 lb, the roll of paper will tip P D 56: lb, slip is impending at E and there is no slip at B and C T 0 D 57:4 lb. Chapter Essay-type answer I x D 568 a 4, I y D 38 a I x D 0:0404 in. 4, I y D 0:0570 in I x D 1 1, I y D 1 1 hb Answer provided in problem statement I y D 16 r4 0, J O D 8 r4 o, essay-type answer J A D 4 r4 o Essay-type answer (a) I x D :74 mm 4 ; (b) I y D 11:0 mm (a) I x D 0:305 mm 4 ; (b) I y D 1:60 mm (a) I x D 0:500 in. 4 ; (b) I y D 0:375 in c 1 D b= p a, c D b=a 3, I x D 1 10 ab I x D 0:305 mm I x D 4: mm 4, I y D 4: mm 4, k x D 105 mm, k y D 33:1 mm I x D 117 in. 4, essay-type answer k x D 3:1 in: For the rectangular cross section: I x D 108 in. 4, I y D 108 in. 4 ; For the I cross section: I x D 48 in. 4, I y D 76:0 in I x D 4: mm 4, I y D 1: mm I x D 37:6 in. 4, I y D 10:9 in (a) I x1 D 180 mm 4 ; (b) I x D 550 mm (a) d D 4:00 in:; (b) I x D 18 in. 4 ; (c) I y D 704 in (a) d D :00 mm; (b) I x D 88:0 mm 4 ; (c) I y D 48:0 mm (a) d D 65:0 mm; (b) I x D 1: mm 4 ; (c) I y D 1: mm Answer provided in problem statement I y D 1 4 mr where m D r t I y D m. 1 4 R C 1 3 L / where m D R L I D 3 5 m. 1 4 R C L / where m D 1 3 R L I y D 3 5 m. 1 3 a C h / where m D 4 3 abh Essay-type answer (a) I x D 1 R r= 0 0.r x / dx C I x D 4:810 5 slugin. R r r= 0.r x / dx; (b) I x D r Last modified: March 19, 009

18 10.64 I y D 1 3 ma where m D 1 ha (a) I y D R L 0 where m D LR (a) I x D R L 0 t 0. x L I y D 1: kgcm. 4 R4 Œ. x L /4=3. x L /4 dx C R L 0 R x Œ. x L /=3. x L / dx; (b) I y D m L C7689 R I x D 5 56 mr where m D 7 1 R I D 7 16 mr where m D 7 1 R I O D 0:017 slugft, essay-type answer I y D 1 3 m.a C b / I B D 0;700 kgmm I A D 9:410 4 slugin I x D 0:497 slugin I B D 0:0778 kgm I x D 1:61 slugin I D 0:619 slugin I y D 3:53 slugin I y D 5: slugft J O D 4.r4 o r 4 i / (a) I y D R 4 m 0 x. p x q Rx 3 / L q1 C R L dx; (b) I x D 0 9 mr where m D 3 4 t 0 RL 1 C R L. 1 4 x/ dx; (b) I y D 0:6 m c 1 D 6, c D 1 3, c 3 D 1; (a) I y D R 9 in: 0 x (a) I y D R 1 in: 0 x.x 1/ dx; (b) I y D 0:0667 in I x D : mm 4, I y D 3:510 5 mm (a) d D 1:183 in:; (b) I x D 5:56 in I D 3 10 mr where m D 1 3 r h I x D 1700 kgmm : I D 3040 kgmm : I x D 0:0113 kgmm : I x D 1 R h 0 1 x dx C I y D 1: kgm : R h h x dx (a) I x D 1: mm 4 ; (b) I x D 5: kgmm I x D 13:3 slugin I D 16:7 slugin x p x/ dx; (b) Iy D 86 in Last modified: March 19, 009

COURSE: CE 201 (STATICS) LECTURE NO.: 28 to 30 FACULTY: DR. SHAMSHAD AHMAD DEPARTMENT: CIVIL ENGINEERING UNIVERSITY: KING FAHD UNIVERSITY OF PETROLEUM & MINERALS, DHAHRAN, SAUDI ARABIA TEXT BOOK: ENGINEERING

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