Three torques act on the shaft. Determine the internal torque at points A, B, C, and D.

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2 7. Three torques act on the shaft. Determine the internal torque at points,, C, and D. Given: M 1 M M Nm 400 Nm 00 Nm Solution: Section : x = 0; T M 1 M M 3 0 T M 1 M M 3 T Nm Section : M x = 0; T M 3 M 0 T M 3 M T Nm Section C: x = 0; T C M 3 0 T C M 3 T C Nm Section D: x = 0; T D 0

3 7 3. The strongback or lifting beam is used for materials handling. If the suspended load has a weight of kn and a center of gravity of G, determine the placement d of the padeyes on the top of the beam so that there is no moment developed within the length of the beam. The lifting bridle has two legs that are positioned at 45, as shown. SOLUTION 0. m 0. m 3m d E m d F Support Reactions: From FD (a), a+ M E = 0; F F = 0 F E = 1.00 kn G + c F y = 0; F F = 0 F F = 1.00 kn From FD (b), : + F x = 0; F C cos 45 - F C cos 45 = 0 F C = F C = F + c F y = 0; F sin = 0 F C = F C = F = kn Internal Forces: This problem requires M H = 0. Summing moments about point H of segment EH [FD (c)], we have a+ M H = 0; 1.001d + x sin 45 1x cos = 0 d = 0.00 m

4 7 4. Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: kn 10 3 N Given: w 0.75 kn m F 4 kn a b c Solution: 1.5 m d 1.5 m 1.5 m e 3.5 m f 4 M C = 0; f x ( c d) e f Fd 0 x fdf x 1. kn e f ( c d) M = 0; w( c d) c d y ( a b) x ( c d) 0 y ( c d) w x ( c d) a b y 0.40 kn F x = 0; N D x 0 N D x N D 1. kn F y = 0; V D y 0 V D y V D 0.4 kn M D = 0; M D y b 0 M D y b M D 0.6 knm

5 7 5. Determine the internal normal force, shear force, and moment at points and in the column. 8 kn 6 kn 0.4 m 0.4 m 30 SOLUTION 0.9 m pplying the equation of equilibrium to Fig. a gives 3 kn : + F x = 0; + c F y = 0; V - 6 sin 30 = 0 N - 6 cos 30-8 = 0 V = 3 kn N = 13. kn 1.5 m m a+ M = 0; 8(0.4) + 6 sin 30 (0.9) - 6 cos 30 (0.4) - M = 0 M = 3.8 kn # m and to Fig. b, : + F x = 0; + c F y = 0; a+ M = 0; V - 6 sin 30 = 0 N cos 30 = 0 V = 3 kn N = 16. kn 3(1.5) + 8(0.4) + 6 sin 30 (.9) - 6 cos 30 (0.4) - M = 0 M = 14.3 kn # m

6 7 6. Determine the distance a as a fraction of the beam s length L for locating the roller support so that the moment in the beam at is zero. P P SOLUTION a+ M -P a L = 0; 3 - ab + C y1l - a + Pa = 0 C y = P L 3 - a L - a a L/3 L C a+ M = 0; M = P L 3 - a L - a a L 3 b = 0 PL a L 3 - ab = 0 a = L 3

7 7 7. Determine the internal normal force, shear force, and moment at points E and D of the compound beam. Given: M F a 00 Nm c 4 m 800 N d m m e m b m Solution: Segment C : M M C y ( d e) 0 C y d e y C y 0 y C y Segment EC : N E 0 N E 0N N E 0.00 V E C y 0 V E C y V E N M E M C y e 0 M E C y e M M E Nm Segment D : N D 0 N D 0N N D 0.00 V D F y 0 V D F y V D N M D Fb y ( b c) 0 M D F b y ( b c) M D 1300 Nm

8 7 8. Determine the internal normal force, shear force, and moment at point C. 0. m 400 N 1.5 m C 1 m 3 m m...

9 7 9. Determine the normal force, shear force, and moment at a section passing through point C. Take P = 8 kn. 0.1 m 0.5 m C 0.75 m 0.75 m 0.75 m SOLUTION P a+ M = 0; -T(0.6) + 8(.5) = 0 T = 30 kn : + F x = 0; x = 30 kn + c F y = 0; y = 8kN : + F x = 0; -N C - 30 = 0 N C = -30 kn + c F y = 0; V C + 8 = 0 V C = -8kN a+ M C = 0; -M C + 8(0.75) = 0 M C = 6kN# m

10 7 10. The cable will fail when subjected to a tension of kn. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at a section passing through point C for this loading. 0.1 m 0.5 m C 0.75 m 0.75 m 0.75 m SOLUTION P a+ M = 0; -(0.6) + P(.5) = 0 P = kn : + F x = 0; x = kn + c F y = 0; y = kn : + F x = 0; -N C - = 0 N C = -kn + c F y = 0; V C = 0 V C = kn a+ M C = 0; -M C (0.75) = 0 M C = kn # m

11 7 11. Determine the internal normal force, shear force, and moment at points D and E in the compound beam. Point E is located just to the left of the 10-kN concentrated load.ssume the support at is fixed and the connection at is a pin. kn/m 10 kn D E 1.5 m 1.5 m 1.5 m 1.5 m C

12 7 1. Determine the internal normal force, shear force, and the moment at points C and D. m C kn/m SOLUTION Support Reactions: FD (a). 6m 45 3m D 3m a+ M = 0; y cos cos 45 = 0 + c F y = 0; y = kn y = 0 y = kn : + F x = 0 x = 0 Internal Forces: pplying the equations of equilibrium to segment C [FD (b)], we have Q+ F x = 0; a+ F y = 0; a+ M C = 0; cos 45 - V C = 0 V C =.49 kn sin 45 - N C = 0 N C =.49 kn M C cos 45 1 = 0 M C = 4.97 kn # m pplying the equations of equilibrium to segment D [FD (c)], we have : + F x = 0; N D = 0 + c F y = 0; V D = 0 V D = -.49 kn a+ M D = 0; M D = 0 M D = 16.5 kn # m

13 7 13. Determine the internal normal force, shear force, and moment at point C in the simply supported beam. 4 kn/m 3 m C 3 m

14 7 14. Determine the normal force, shear force, and moment at a section passing through point D. Take w = 150 N>m. D w 4m 4m 3m C SOLUTION 4m a + M = 0; F C18 = 0 F C = 1000 N :+ F x = 0; x = 0 x = 800 N + c F y = 0; y = 0 y = 600 N : + F x = 0; N D = -800 N + c F y = 0; V D = 0 V D = 0 a+ M D = 0; M D = 0 M D = 100 N # m = 1.0 kn # m

15 7 15. The beam will fail if the maximum internal moment at D reaches 800 N # m or the normal force in member C becomes 1500 N. Determine the largest load w it can support. D w 4m 4m 3m C SOLUTION ssume maximum moment occurs at D; 4m a+ M D = 0; M D - 4w() = = 4w() w = 100 N/m a+ M = 0; - 800(4) + F C (0.6)(8) = 0 F C = N N w = 100 N/m (O.K.!)

16 7 16. Determine the internal normal force, shear force, and moment at point D in the beam. 600 N/m SOLUTION D 1 m 1 m 1 m 900 Nm Writing the equations of equilibrium with reference to Fig. a, we have C a + M = 0; F C 4 () - 600(3)(1.5) = 0 5 F C = 50 N a+ M = 0; 600(3)(0.5) y () = 0 y = 0 : + F x = 0; x = 0 x = 1350 N Using these results and referring to Fig. b, we have : + F x = 0; N D = 0 N D = N = kn + c F y = 0; -V D - 600(1) = 0 V D = -600 N a+ M D = 0; MD + 600(1)(0.5) = 0 M D = -300 N # m The negative sign indicates that N D, V D, and M D act in the opposite sense to that shown on the free-body diagram.

17 7 17. Determine the normal force, shear force, and moment at a section passing through point E of the two-member frame. 400 N/m.5 m 3m D SOLUTION C E a+ M = 0; :+ F x = 0; F C16 = 0 F C = 080 N -N E = 0 6m N E = -190 N = -1.9 kn + c F y = 0; V E = 0 V E = 800 N a+ M E = 0; M E = 0 M E = 400 N # m =.40 kn # m

18 7 18. Determine the internal normal force, shear force, and moment at point C in the cantilever beam. w 0 C L L

19 7 19. Determine the internal normal force, shear force, and moment at points E and F in the beam. C SOLUTION E D 45 F With reference to Fig. a, a+ M = 0; T(6) + T sin 45 (3) - 300(6)(3) = 0 T = N : + F x = 0; cos 45 - x = 0 x = N 1.5 m 1.5 m 1.5 m 1.5 m 300 N/m + c F y = 0; y sin (6) = 0 y = N Use these result and referring to Fig. b, : + F x = 0; N E = 0 N E = 470 N + c F y = 0; (1.5) - V E = 0 V E = 15 N a+ M E = 0; M E + 300(1.5)(0.75) (1.5) = 0 M E = 660 N # m lso, by referring to Fig. c, : + F x = 0; N F = 0 + c F y = 0; V F = 0 V F = -15 N a+ M F = 0; 664.9(1.5) - 300(1.5)(0.75) - M F = 0 M F = 660 N # m The negative sign indicates that V F acts in the opposite sense to that shown on the free-body diagram.

20 7 0. Determine the internal normal force, shear force, and moment at points C and D in the simply supported beam. Point D is located just to the left of the 5-kN force. 3 kn/m 5 kn C 1.5 m 1.5 m D 3 m

21 7 1. Determine the internal normal force, shear force, and moment at points D and E in the overhang beam. Point D is located just to the left of the roller support at, where the couple moment acts. kn/m 6 kn m D E 3 m 1.5 m 1.5 m C kn

22 7. Determine the internal normal force, shear force, and moment at points E and F in the compound beam. Point F is located just to the left of the 15-kN force and 5-kN # m couple moment..5 m 3 kn/m E C F m.5 m 1.5 m 15 kn 5 knm D m SOLUTION With reference to Fig. b, we have : + F x = 0; a+ M C = 0; a+ M D = 0; C x = 0 D y (4) - 15() - 5 = 0 15() C y (4) = 0 D y = kn C y = 1.5 kn Using these results and referring to Fig. a, we have : + F x = 0; a+ M = 0; x = 0 3(6)(1.5) - 1.5(1.5) - y (4.5) = 0 y = kn With these results and referring to Fig. c, : + F x = 0; + c F y = 0; a+ M E = 0; N E = (.5) - V E = 0 V E = kn M E + 3(.5) - 3(.5)(8.15) = 0 M E = 4.97 kn # m lso, using the result of D y referring to Fig. d, we have : + F x = 0; N F = 0 + c F y = 0; V F = 0 V F = 1.5 kn a+ M F = 0; 13.75() MF = 0 M F =.5 kn # m The negative sign indicates that V E acts in the opposite sense to that shown in the free-body diagram.

23 7 3. Determine the internal normal force, shear force, and moment at points D and E in the frame. Point D is located just above the 400-N force. m 00 N/m SOLUTION 400 N 1 m D E 1 m 30.5 m With reference to Fig. a, we have 1.5 m a+ M = 0; F cos 30 () + F sin 30 (.5) - 00()(1) - 400(1.5) = 0 F = N C Using this result and referring to Fig. b, we have : + F x = 0; + c F y = 0; a+ M D = 0; V D sin 30 = cos 30-00() - N D = 0 V D = 168 N N D = -110 N cos 30 () sin 30 (1) - 00()(1) - M D = 0 M D = 348 N # m lso, by referring to Fig. c, we can write : + F x = 0; + c F y = 0; a+ M E = 0; -N E sin 30 = 0 N E = -168 N V E cos 30-00(1) = 0 V E = N cos 30 (1) - 00(1)(0.5) - M E = 0 M E = 190 N # m The negative sign indicates that N D, N E, and V E acts in the opposite sense to that shown in the free-body diagram.

24 7 4. Determine the internal normal force, shear force, and bending moment at point C. 8 kn/m 40 kn 60 3m C 3m 3m 0.3 m SOLUTION Free body Diagram: The support reactions at need not be computed. Internal Forces: pplying equations of equilibrium to segment C, we have : + F x = 0; + c F y = 0; a+ M C = 0; -40 cos 60 - N C = 0 N C =-0.0 kn V C sin 60 = 0 V C = 70.6 kn sin M C = 0 M C = -30 kn # m

25 7 5. Determine the internal normal force, shear force, and moment at point C in the double-overhang beam. 3 kn/m C 1.5 m 1.5 m 1.5 m 1.5 m

26 7 6. Determine the ratio of a>b for which the shear force will be zero at the midpoint C of the beam. w C C SOLUTION a+ M = 0; - w (a + b)c 3 (a + b) - (a + b) d + y (b) = 0 a b/ b/ a y = w (a + b)(a - b) 6b : + F x = 0; x = 0 + c F y = 0; - w 6b (a + b)(a - b) - w 4 aa + b b - V C = 0 Since V C = 0, - 1 6b (a + b)(a - b) = 1 4 (a + b)a 1 b - 1 6b (a - b) = a + b = 3 4 b a b = 1 4

27 7 7. Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. 00 N/m 400 N/m.5 m 3m D SOLUTION a+ M = 0; -100(3) - 600(4) F C (6) = 0 F C = 600 N : + F x = 0; x = 1 (600) = 400 N 13 C 6m + c F y = 0; y (600) = 0 y = 800 N : + F x = 0; N D = 400 N =.40 kn + c F y = 0; V D = 0 V D = 50 N a+ M D = 0; -800(3) + 600(1.5) + 150(1) + M D = 0 M D = 1350 N # m = 1.35 kn # m

28 7 8. Determine the internal normal force, shear force, and moment at point C in the simply supported beam. Point C is located just to the right of the 1500-lb.5 kn m ft couple moment kn/m lb/ft C kn lb mft 30 6 mft 6 mft Writing the moment equation of equilibrium about point with reference to Fig. a, + ΣM = 0; F cos 30 (4) 10 (4) ().5 = 0 F = kn Using the result of F and referring to Fig. b, + ΣF x = 0; N C sin 30 = 0 N C = kn + ΣF y = 0; V C cos () = 0 V C = 0.65 kn + ΣM C = 0; cos 30 () 10 () (1) M C = 0 M C = 1.5 kn m The negative sign indicates that N C and V C act in the opposite sense to that shown on the free body diagram. 10 (4) kn 10 () kn.5 kn m m m 1 m 1 m F = kn

29 7 9. Determine the internal normal force, shear force, and moment at point D of the two-member frame. 1.5 m m D 50 N/m C E 4 m 300 N/m...

30 7 30. Determine the internal normal force, shear force, and moment at point E of the two-member frame. 1.5 m m D 50 N/m C E 4 m 300 N/m...

31 7 31. The hook supports the 4-kN load. Determine the internal normal force, shear force, and moment at point. 4 kn SOLUTION With reference to Fig. a, +Q F x = 0; +a F y = 0; V - 4 cos 45 = 0 N - 4 sin 45 = 0 V =.83 kn N =.83 kn mm a+ M = 0; 4 sin 45 (0.075) - M = 0 M = 0.1 kn # m = 1 N # m 4 kn

32 7 3. Determine the internal normal force, shear force, and moment at points C and D in the simply supported beam. Point D is located just to the left of the 10-kN concentrated load. 6 kn/m 10 kn C D 1.5 m 1.5 m 1.5 m 1.5 m

33 7 33. Determine the distance a in terms of the beam s length L between the symmetrically placed supports and so that the internal moment at the center of the beam is zero. w 0 w 0 a L a

34 7 34. The beam has a weight w per unit length. Determine the internal normal force, shear force, and moment at point C due to its weight. L L C u

35 7 35. Determine the internal normal force, shear force, and bending moment at points E and F of the frame. C 1m 1m D 0.5 m 0.5 m G SOLUTION Support Reactions: Members HD and HG are two force members. Using method of joint [FD (a)], we have : + F x = 0 G cos D cos 6.57 = 0 D = G = F 1m E H 800 N 0.5 m F 0.5 m 1m + c F y = 0; F sin = 0 D = G = F = N From FD (b), a + M = 0; C x 1 cos C y 1 sin = 0 (1) From FD (c), a + M = 0; C x 1 cos C y 1 sin 6.57 = 0 () Solving Eqs. (1) and () yields, C y = 0 C x = 500 N Internal Forces: pplying the equations of equilibrium to segment DE [FD (d)], we have Q+ F x = 0; Q + F y = 0; a+ M E = 0; V E = N E = 0 N E = 894 N M E = 0 pplying the equations of equilibrium to segment CF [FD (e)], we have +Q F x = 0; V F cos = 0 V F = 447 N Q + F y = 0; -N F sin 6.57 = 0 N F = 4 N a+ M F = 0; M F cos = 0 M F = 4 N # m

36 7 36. Determine the distance a between the supports in terms of the shaft s length L so that the bending moment in the symmetric shaft is zero at the shaft s center. The intensity of the distributed load at the center of the shaft is w 0. The supports are journal bearings. w 0 SOLUTION Support reactions: FD(a) a L Moments Function: a+ M = 0; (w 0)a L ba1 3 bal b w 0La a b = 0 a = L 3

37 7 37. Determine the internal normal force, shear force, and moment acting at point C. The cooling unit has a total mass of 5 kg with a center of mass at G. F 0. m D E 3 m C 3 m G...

38 7 38. Determine the internal normal force, shear force, and moment at points D and E of the frame which supports the 100-kg crate. Neglect the size of the smooth peg at C m C 1. m E 0.6 m 0.45 m 0.45 m D N 981 (1.35) = N m = 1.3 kn m N = 589 N N = 1.77 kn 981 (1.35) 981 (0.6) = N m = 1.68 kn m 1.35 m 1.35 m 1. m 981 N 0.6 m 100 (9.81) N =981 N 100 (9.81) N

39 7 39. The semicircular arch is subjected to a uniform distributed load along its axis of w 0 per unit length. Determine the internal normal force, shear force, and moment in the arch at u = 45. O r u w 0 SOLUTION Resultants of distributed load: u u F Rx = w 0 (rdu) sin u = r w 0 (-cos u) ` = r w 0 (1 - cos u) L 0 0 F Ry = L u 0 u M Ro = w 0 (rdu) r = r w 0 u L 0 t u = 45 u w 0 (rdu) cos u = r w 0 (sinu) ` = r w 0 (sinu) 0 +b F x = 0; -V + F Rx cos u - F Ry sin u = 0 V = 0.99 r w 0 cos r w 0 sin 45 V = r w 0 +a F y = 0; N + F Ry cos u + F Rx sin u = 0 N = rw 0 cos rw 0 sin 45 N = rw 0 a+ M o = 0; -M + r w 0 a p 4 b + ( r w 0)(r) = 0 M = r w 0

40 7 40. The semicircular arch is subjected to a uniform distributed load along its axis of w 0 per unit length. Determine the internal normal force, shear force, and moment in the arch at u = 10. O r u w 0 SOLUTION Resultants of distributed load: u u F Rx = w 0 (rdu) sinu = r w 0 (-cos u) ` = r w 0 (1 - cosu) L 0 u u F Ry = w 0 (rdu) cos u = r w 0 (sin u) ` = r w 0 (sin u) L 0 u M Ro = w 0 (rdu) r = r w 0 u L 0 t u = 10, 0 0 F Rx = r w 0 (1 - cos 10 ) = 1.5 r w 0 F Ry = r w 0 sin 10 = r w 0 +b F x = 0; N r w 0 cos r w 0 sin 30 = 0 N = r w 0 +a F y = 0; V r w 0 sin r w 0 cos 30 = 0 V = -1.5 rw 0 a + M o = 0; -M + r w 0 (p)a b + ( r w 0)r = 0 M = 1.3 r w 0

41 7 41. Determine the x, y, z components of internal loading at a section passing through point C in the pipe assembly. Neglect the weight of the pipe.take F 1 = 5350j - 400k6 kn and F = 5150i - 300k6 kn. z kn. kn m x 1.5 m F 1 C m F y kn. 3m 3 m kn. kn m MN m. MN m. kn m.

42 7 4. Determine the x, y, z components of internal loading at a section passing through point C in the pipe assembly. Neglect the weight of the pipe. Take F 1 = 5-80i + 00j - 300k6 kn and F = 550i - 150j - 00k6 kn. z x 1.5 m C F F 1 m y 3 m kn m kn kn.. 3m kn. kn m MN m. kn m kn m

43 7 43. Determine the x, y, z components of internal loading in the rod at point D. Units Used: kn Given: M 10 3 N 3 knm F kn a b c d e 0.75 m 0. m 0. m 0.6 m 1 m Solution: Guesses C x z 1 N C y 1 N x 1 N 1 N y 1 N z 1 N Given 0 y z x 0 z C x C y 0 F 0 e b c d 0 0 y z 0 b c 0 x 0 z 0 0 a C x C y 0 0 b c d 0 0 F 0 0 M

44 y z x z C x C y Find y z x z C x C y y z x z C x C y kn Guesses V Dx 1 N N Dy 1 N V Dz M Dy 1N M Dx 1 Nm 1 Nm M Dz 1 Nm Given C x C y 0 0 b a V Dx N Dy 0 V Dz C x C y 0 M Dx M Dy M Dz M V Dx N Dy V Dz M Dx M Dy M Dz Find V Dx N Dy V Dz M Dx M Dy M Dz V Dx N Dy V Dz M Dx M Dy M Dz kn knm

45 Determine the x, y, z components of internal loading in the rod at point E. Units Used: kn 10 3 N Given: M 3 kn m F kn a 0.75 m b 0.4 m c 0.6 m d 0.5 m e 0.5 m Solution: Guesses C x 1 N C y 1 N x 1 N z 1 N y 1 N z 1 N Given 0 y z x 0 z C x C y 0 F 0 d e b c 0 0 y z 0 b 0 x 0 z 0 0 a C x C y 0 0 b c 0 F 0 0 M

46 y z x z C x C y Find y z x z C x C y y z x z C x C y kn Guesses N Ex 1 N V Ey 1 N V Ez M Ey 1 N M Ex 1 Nm 1 Nm M Ez 1 Nm Given 0 y z e 0 0 N Ex V Ey 0 V Ez 0 y z M Ex M Ey 0 M Ez N Ex V Ey V Ez M Ex M Ey M Ez Find N Ex V Ey V Ez M Ex M Ey M Ez N Ex V Ey V Ez M Ex M Ey M Ez kn knm

47 7 45. Draw the shear and moment diagrams for the overhang beam. P SOLUTION Since the loading discontinues at, the shear stress and moment equation must be written for regions 0 x 6 b and b 6 x a + b of the beam. The free-body diagram of the beam s segment sectioned through an arbitrary point in these two regions are shown in Figs. b and c. b a C Region 0 x 6 b, Fig. b + c F y = 0; - Pa b - V = 0 V = - Pa b (1) a + M = 0; M + Pa M = - Pa () b x = 0 b x Region b 6 x a + b, Fig. c F y = 0; V - P = 0 V = P (3) a + M = 0; -M - P(a + b - x) = 0 M = -P(a + b - x) (4) The shear diagram in Fig. d is plotted using Eqs. (1) and (3), while the moment diagram shown in Fig. e is plotted using Eqs. () and (4). The value of moment at is evaluated using either Eqs. () or (4) by substituting x = b; i.e., M ƒ x = b = - Pa b (b) = -Pa or M ƒ x = b = -P(a + b - b) = -Pa

48 ....

49 7 47. Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set P = 4800 kn, lb, a = a 1.5 = 5 m, ft, L L = = 3.61 m. ft. P P a L a (a) For 0 x < a + ΣF y = 0; V = P + ΣM = 0; M = Px For a < x < L a + ΣF y = 0; V = 0 + ΣM = 0; Px + P(x a) + M = 0 M = Pa For L a < x L + ΣF y = 0; V = P + ΣM = 0; M + P(L x) = 0 M = P(L x) (b) Set P = 4 kn, a = 1.5 m, L = 3.6 m For 0 x < 1.5 m + ΣF y = 0; V = 4 kn 4 kn 3.6-x 4 kn 1.5 m 4 kn + ΣM = 0; M = 4x kn m 4 kn For For 1.5 m < x <.1 m + ΣF y = 0; V = 0 + ΣM = 0; 4(x) + 4(x 1.5) + M = 0 M = 6 kn m.1 m < x 3.6 m + ΣF y = 0; V = 4 kn + ΣM = 0; M + 4(3.6 x) = 0 4 kn 4 kn 4 kn 4 kn 1.5 m 0.6 m 1.5 m V (kn) 4 M (kn m) 4 6 M = (14.4 4x) kn m

50 7 48. Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set M 0 = 500 N # m, L = 8m. M 0 M 0 L/3 L/3 L/3 SOLUTION (a) For 0 x L 3 + c F y = 0; V = 0 a+ M = 0; M = 0 For L 3 6 x 6 L 3 + c F y = 0; V = 0 a+ M = 0; M = M 0 For L 3 6 x L + c F y = 0; V = 0 a+ M = 0; M = 0 (b) Set M 0 = 500 N # m, L = 8m For 0 x m + c F y = 0; V = 0 c+ M = 0; M = 0 For 8 3 m 6 x m + c F y = 0; V = 0 c+ M = 0; M = 500 N # m For 16 3 m 6 x 8m + c F y = 0; V = 0 c+ M = 0; M = 0

51 7 49. If L = 9m, the beam will fail when the maximum shear force is V max = 5kNor the maximum bending moment is M max = kn# m. Determine the magnitude M 0 of the largest couple moments it will support. M 0 M 0 L/3 L/3 L/3 SOLUTION See solution to Prob a. M max = M 0 = kn# m

52 7 50. Draw the shear and moment diagrams for the cantilever beam. kn/m m 6 kn m

53 ....

54 7 5. Draw the shear and moment diagrams for the beam..4 kn/m C m 3m SOLUTION Support Reactions: a+ M = 0; C y = 0 C y =.00 kn c F y = 0; y = 0 y = 0.40 kn Shear and Moment Functions: For 0 x< m[fd (a)],.4 + c F y = 0; a+ M = 0; V = 0 V = 0.4 kn M - 0.4x = 0 M = 10.4x kn m # 0.40 For m<x 3 m [FD (b)], + c F y = 0; x - - V = V = x6 kn a+ M = 0; 0.4x -.41x - a x - b - M = 0 M = -1.x + 5. x kn m #

55 7 53. Draw the shear and moment diagrams for the beam. 40 kn/m 0 kn 8m 3m C 100 kn m SOLUTION 0 x c F y = 0; x - V = 0 V = x 100 kn m a+ M = 0; M + 40xa x b - 140x = kn 00 kn M = 140x - 0x 8 6 x 11 + c F y = 0; V - 0 = 0 V = m 180 c+ M = 0; M + 0(11 - x) = 0 45 M = 0x m

56 7 54. Draw the shear and moment diagrams for the simply supported beam. 300 N m 300 N/m 4 m

57

58 7 55. Draw the shear and moment diagrams for the simply supported beam. 100 N m 300 N/m 4 m x x m m M x = m = 150(1.917 ) + 575(1.917) = 651 N m

59 100 N m 575 N 65 N 100 N m 575 N

60 7 56. Draw the shear and bending-moment diagrams for beam C. Note that there is a pin at. w C L L SOLUTION Support Reactions: From FD (a), a+ M C = 0; From FD (b), wl a L 4 b - y a L b = 0 y = wl 4 wl + c F y = 0; y - - a+ M = 0; wl M a L - 4 b - wl 4 = 0 y = 3wL 4 wl 4 a L b = 0 M = wl 4 Shear and Moment Functions: For 0 x L [FD (c)], + c F y = 0; 3wL 4 - wx - V = 0 V = w (3L - 4x) 4 3wL c+ M = 0; 4 (x) - wxa x b - wl 4 - M = 0 M = w 4 3Lx - x - L

61 7 57. Draw the shear and moment diagrams for the beam. 550 kn/m lb/ft N lb mft 0 6 mft lb N mft + ΣM = 0; 30 (3) + y (6) = 0 30 kn y = 15 kn 50 N m + ΣF x = 0; x = 0 + ΣF y = 0; y = 0 For 0 x 3 m y = 15 kn + ΣF x = 0; 15 5x V = 0 50 N m 15 kn 5x 50 N m 50 N m 15 kn V (kn) 15 6 m 6 m 6 m 5 kn/m 50 N m 15 kn V = 5 (3 x) + ΣM = 0; 15 (x) x + M = 0 M (kn m) m m M = 1 (30x 5x 0.5) M = 15x.5x

62 7 58. Draw the shear and moment diagrams for the compound beam. The beam is pin-connected at E and F. E F C w D SOLUTION Support Reactions: From FD (b), L L 3 L 3 L 3 L a+ M E = 0; + c F y = 0; F y a L 3 b - wl 3 a L 6 b = 0 F y = wl 6 E y + wl 6 - wl 3 = 0 E y = wl 6 From FD (a), a+ M C = 0; D y 1L + wl 6 a L 3 b - 4wL 3 a L 3 b = 0 D y = 7wL 18 From FD (c), a+ M = 0; + c F y = 0; 4wL 3 a L 3 b - wl 6 a L 3 b - y 1L = 0 y = 7wL 18 y + 7wL 18-4wL - wl 3 6 = 0 y = 10wL 9 Shear and Moment Functions: For 0 x<l [FD (d)], + c F y = 0; 7wL 18 - wx - V = 0 V = w 17L - 18x 18 a+ M = 0; M + wxa x b - 7wL 18 x = 0 M = w 18 17Lx - 9x For L x<l [FD (e)], + c F y = 0; 7wL wL - wx - V = 0 9 V = w 13L - x a+ M = 0; M + wxa x b - 7wL 18 x - 10wL 1x - L = 0 9 M = w 18 17Lx - 0L - 9x For L<x 3L [FD (f)], + c F y = 0; V + 7wL 18 - w13l - x = 0 V = w 147L - 18x 18 a+ M = 0; 7wL 18 3L - x - w 3L - x 3L - x - M = 0 M = w Lx - 9x - 60L

63 7 59. Draw the shear and moment diagrams for the beam. 1.5 kn/m 3m SOLUTION + c F y = 0; x (0.5x) - V = 0 V = x V = 0 = x x = 1.73 m a + M = 0; M + a 1 b(0.5 x) (x) a 1 xb x = 0 3 M = 0.75 x x 3 M max = 0.75(1.73) (1.73) 3 = 0.866

64 7 60. The beam will fail when the maximum internal moment is M max. Determine the position x of the concentrated force P and its smallest magnitude that will cause failure. Solution: For < x, M 1 P( L x) L Px L For > x, M L Note that M 1 = M when x = M max M 1 M Px( L x) L P L Lx x d x Lx d x L x x L P Thus, M max L L L L P L P 4M max L

65 7 61. Draw the shear and moment diagrams for the beam. C 3m SOLUTION Support Reactions: From FD (a), a+ M = 0; y 16 = 0 y = 3.00 kn Shear and Moment Functions: For 0 x 6m[FD (b)], 6m 3 kn/m + c F y = 0; x 4 - V = 0 V = b x 4 r kn The maximum moment occurs when V = 0, then 0 = x 4 x = m a+ M = 0; M + x 4 a x 3 b x = 0 M = b 3.00x - x3 1 r kn # m Thus, M max = = 6.93 kn # m

66 7 6. The cantilevered beam is made of material having a specific weight g. Determine the shear and moment in the beam as a function of x. h SOLUTION y similar triangles d x t y x = h d y = h d x W = gv = ga 1 yxtb = gc 1 a h ght xbxt d = d d x + c F y = 0; V - ght d x = 0 V = ght d x a + M = 0; -M- ght d x a x 3 b = 0 M = -ght 6d x3

67 7 63. Draw the shear and moment diagrams for the overhang beam. 8 kn/m C SOLUTION 0 x 6 5 m: + c F y = 0; a + M = 0 ; 5 x 6 10 m: + c F y = 0; a + M = 0;.5 - x - V = 0 V =.5 - x M + xa 1 xb -.5x = 0 M =.5x - x V = 0 V = -7.5 M + 101x x = 0 M = -7.5x m m

68 7 64. Draw the shear and moment diagrams for the beam. w SOLUTION The free-body diagram of the beam s segment sectioned through an arbitrary point shown in Fig. b will be used to write the shear and moment equations. The magnitude of the resultant force of the parabolic distributed loading and the location of its point of application are given in the inside back cover of the book. w w 0 L x C L L w 0 x Referring to Fig. b, we have + c F y = 0; w 0 L a w 0 L x bx - V = 0 V = w 0 1L al3-4x 3 b (1) a M + 1 () 3 w 0 (x) x L x 4 - w 0L 1 x = 0 M = w 0 + M = 0; 1L al3 x - x 4 b The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1) and (), respectively. The location at which the shear is equal to zero can be obtained by setting V = 0 in Eq. (1). 0 = w 0 1L al3-4x 3 b x = 0.630L The value of the moment at x = 0.630L is evaluated using Eq. (). M ƒ x = 0.630L = w 0 1L cl3 (0.630L) - (0.630L) 4 d = w 0 L

69 7 65. Draw the shear and bending-moment diagrams for the beam. 300 N/m C 3m 4m SOLUTION Support Reactions: From FD (a), a+ M = 0; y = 0 y = 650 N Shear and Moment Functions: For 0 x<3 m[fd (b)], + c F y = 0; x - V = 0 V = x 6 N a+ M = 0; M x a x 3 b + 650x = 0 M = 5-650x x 3 6 N # m For 3m<x 7m[FD (c)], + c F y = 0; V x = 0 V = x6 N a+ M = 0; xa 7 - x b - M = 0 M = x 6 N # m

70 7 66. Draw the shear and moment diagrams for the beam. w w SOLUTION L Support Reactions: From FD (a), a+ M = 0; wl 4 a L 3 b + wl a L b - y 1L = 0 y = wl 3 Shear and Moment Functions: For 0 x L [FD (b)], + c F y = 0; wl 3 - w x - 1 a w L xbx - V = 0 V = w 1L 14L - 6Lx - 3x The maximum moment occurs when V = 0, then 0 = 4L - 6Lx - 3x x = 0.575L a+ M = 0; M + 1 a w L xbxa x 3 b + wx a x b - wl 3 1x = 0 M = w 1L 14L x - 3Lx - x 3 Thus, M max = w 1L 34L L - 3L10.575L L 3 4 = wL

71 7 67. Determine the internal normal force, shear force, and moment in the curved rod as a function of u, where 0 u 90. P SOLUTION With reference to Fig. a, a+ M = 0; y (r) - p(r) = 0 y = p> u r Using this result and referring to Fig. b, we have F x = 0; F y = 0; a+ M = 0; p sin u - V = 0 p cos u - N = 0 p 3r (1 - cos u)4 - M = 0 V = p sin u N = p cos u M = pr (1 - cos u)

72 7 68. Determine the normal force, shear force, and moment in the curved rod as a function of Given: c 3 d 4 Solution: For 0 F x = 0; N d c d P cos c c d P sin 0 N P c d d cos c sin F y = 0; V d c d P sin c c d P cos 0 V P c d d sin c cos M = 0; d c d Prrcos c c d Prsin M 0 M Pr c d d dcos c sin

73 7 69. The quarter circular rod lies in the horizontal plane and supports a vertical force P at its end. Determine the magnitudes of the components of the internal shear force, moment, and torque acting in the rod as a function of the angle. Solution: F z = 0; V P M x = 0; M Prcos 0 M Prcos M Prcos M y = 0; T Pr l sin 0 T Pr l sin T Pr 1 sin

74 7 70. Draw the shear and moment diagrams for the beam. kn kn kn kn 4kN 4kN 1.5 m 1 m 1 m 1 m m 1m 0.75 m SOLUTION Support Reactions: a+ M = 0; y = 0 y = 9.50 kn + c F y = 0; y = 0 y = 6.50 kn

75 7 71. Draw the shear and moment diagrams for the beam. 7 kn 1 kn m m m 4 m SOLUTION

76 7 7. Draw the shear and moment diagrams for the beam. 3m SOLUTION 50 N/m Support Reactions: a + M = 0; F C a 3 b = 0 5 F C = 65 N C m m + c F y = 0; y + 65a 3 b = 0 5 y = 65 N 500 N

77

78 7 74. Draw the shear and moment diagrams for the simplysupported beam. w 0 w 0 SOLUTION L/ L/

79 7 75. w 0 Draw the shear and moment diagrams for the beam. The support at offers no resistance to vertical load. SOLUTION L

80 7 76. Draw the shear and moment diagrams for the beam. kn/m 8kN SOLUTION Support Reactions: a+ M = 0; y = 0 y = 8.90 kn 5m 3m m 8 + c F y = 0; y = 0 y = 9.10 kn

81 7 77. The shaft is supported by a thrust bearing at and a journal bearing at. Draw the shear and moment diagrams for the shaft. 300 N/m 600 N 300 Nm SOLUTION 1.5 m 0.75 m 0.75 m

82

83 7 79. Draw the shear and moment diagrams for the beam. 0 kn m 8kN 15 kn/m m 1m m 3m SOLUTION

84 7 80. Draw the shear and moment diagrams for the compound supported beam. 3 kn/m 5 kn SOLUTION C D 3 m 3 m 1.5 m 1.5 m

85

86 7 8. Draw the shear and moment diagrams for the overhang beam. 4 kn/m SOLUTION 6 kn m 3 m 3 m m 3 kn

87 7 83. Draw the shear and moment diagrams for the shaft. The support at is a journal bearing and at it is a thrust bearing lb N lb/ft N/m lb N mft 1000 N 000 N/m 0.3 m 1. m 0.3 m N N V (m) N m ftm 1. 4 ft m ftm m x (m) M (N m) x (m)

88 7 84. Draw the shear and moment diagrams for the beam. Units Used: kn 10 3 N Given: w 1 0 kn w m 10 kn a 1.5m m Solution: x a a Vx ( ) w x x w 1 a x 1 kn x Mx ( ) w x x w 1 a x x 3 1 knm 4 Force (kn) Vx ( ) x m Distance (m) Moment (kn-m) Mx ( ) x m Distance (m)

89 7 85. Draw the shear and moment diagrams for the beam. kn/m kn/m 18 knm C 3 m 3 m SOLUTION

90 7 86. Draw the shear and moment diagrams for the beam. w 0 L L SOLUTION Support Reactions: a + M = 0; y (L) - w 0 La L b - w 0L a 4L 3 b = 0 y = 7w 0L 6 + c F y = 0; y + 7w 0L 6 y = w 0L 3 - w 0 L - w 0L = 0

91 7 87. Draw the shear and moment diagrams for the beam. w 0 L L SOLUTION Support Reactions: a + M = 0; + c F y = 0; M - w 0 L a L 4 b - w 0L a L 3 b = 0 y - w 0 L - w 0L M = 7w 0L 1 = 0 y = 3w 0L 1 wl 0 1 wl L wl 0 ( 0 w ) 1 1

92 7 88. Draw the shear and moment diagrams for the beam. w w L/ L/ SOLUTION

93 7 89. The shaft is supported by a smooth thrust bearing at and a smooth journal bearing at. Draw the shear and moment diagrams for the shaft. 300 N 100 N/m 300 Nm SOLUTION 1.5 m 1.5 m 3 m

94 7 90. Draw the shear and moment diagrams for the overhang beam. 3 kn/m 6 kn SOLUTION The maximum span moment occurs at the position where shear is equal to zero within the region 0 x 6 6 m of the beam. This location can be obtained using the method of sections. y setting V = 0, Fig. b, we have 6 m 1.5 m + c F y = 0; a 1 xbx - 1 (6 - x)(x) = 0 x = 1.76 m Using this result, + M = 0; M ƒ x = 1.76 m (6-1.76)(1.76)a b (1.76)(1.76)c (1.76) d - 4.5(1.76) = 0 3 M ƒ x = 1.76 m = 3.73 kn # m

95

96 7 9. The beam consists of three segments pin connected at and E. Draw the shear and moment diagrams for the beam. Units Used: kn 10 3 N Given: M 8 knm F 15 kn w 3 kn a 3 m b m m c m d m e m f 4 m Guesses y 1 N y 1 N C y 1 N Given D y 1 N E y 1 N F y 1 N y C y D y F y F wf 0 Fb M y ( a b) 0 w f f F y f 0 M Fa y C y D y E y 0 y ( a b) 0 y c D y d E y ( d e) 0 y y C y D y E y F y Find y y C y D y E y F y y y C y D y E y F y kn x a a 1 V 1 ( x) y kn x a 1.01a a b 1 V ( x) y F kn M 1 ( x) y x M 1 knm M ( x) y x M Fx ( a) 1 knm x 3 a b 1.01( a b) a b c 1 V 3 ( x) y kn M 3 ( x) y ( x a b) 1 knm x 4 a b c 1.01( a b c) a b c d 1 V 4 ( x) y C y kn M 4 ( x) y ( x a b) C y ( x a b c) 1 knm

97 x 5 a b c d 1.01( a b c d) a b c d e 1 V 5 ( x) E y kn M 5 ( x) E y ( a b c d e x) 1 knm x 6 a b c d e 1.01( a b c d e) a b c d e f 1 V 6 ( x) F y wa ( b c d e f x ) kn M 6 ( x) F y ( a b c d e f x) w ( a b c d e f x) 1 knm 10 Force (kn) V 1 x 1 5 V x V 3 x 3 0 V 4 x 4 V 5 x 5 5 V 6 x x 1 x x 3 x 4 x 5 x 6 Distance (m)

98 30 Moment (kn-m) M 1 x 1 M x M 3 x 3 M 4 x 4 M 5 x 5 M 6 x x 1 x x 3 x 4 x 5 x 6 Distance (m)

99 7 93. Draw the shear and moment diagrams for the beam. Units Used: Given: kn 10 3 N w 3 kn a 3 m b 3 m m Solution: y ( a b) wb b 3 wa b a 0 3 y wb 3 wa b a b a 3 w y y ( a b) 0 y w ( a b) y x a a 1 V 1 ( x) y w x a x 1 kn x a 1.01a a b M 1 ( x) y x 1 w x a x x 3 1 knm V ( x) y 1 w a b x b ( a b x) 1 kn M ( x) y ( a b x) 1 w a b x b ( a b x) a b x 3 1 knm 5 Force (kn) V 1 x 1 V x x 1 x Distance (m)

100 10 Moment (kn-m) 5 M 1 x 1 M x x 1 x Distance (m)

101 7 94. Determine the tension in each segment of the cable and the cable s total length. Set P = N. lb. 0.6 ft m ftm C D 50 lbn P ftm 1. 4 ft m ft m From FD (a) 0.9 m 1. m + ΣM = 0; T D cos (0.9) + T D sin (.1) 50 (.1) 400 (0.9) = m T D = N = N + ΣF x = 0; cos x = 0 x = N + ΣF y = 0; y sin = 0 y = N Joint : + ΣF x = 0; T C cos = 0 (1) + ΣF y = 0; T C sin = 0 () Solving Eqs. (1) and () yields : = N x = kn 50 N y = kn T C = N Joint D : + ΣF x = 0; cos T CD cos = 0 (3) T D = kn + ΣF y = 0; sin T CD sin 50 = 0 (4) Solving Eqs. (3) and (4) yields : =.97 T CD = 18.4 N 50 N Total length of the cable: 1.5 l T = sin cos cos 56.8 = m

102 7 95. If each cable segment can support a maximum tension of 375 lb, N, determine the the largest load load P that P that can can be be applied. 0.6 ft m ftm C D 50 lbn P ft m 1. 4 ft m ft m + ΣM = 0; T D (cos ) (0.6) + T D (sin ) (3) 50 (.1) P (0.9) = 0 T D = P ΣF x = 0; x + T D cos = 0 + ΣF y = 0; y P 50 + T D sin = m ssume maximum tension is in cable D, T D = 375 N P = N x = N 0.9 m 1. m 0.6 m Pin : y = N T C = (19.915) + (88.394) = N < 375 N OK 50 N = tan = 56. y = N Joint C : x = N + ΣF x = 0; T CD cos cos 56. = 0 + ΣF y = 0; T CD sin sin = 0 T CD = N < 375 N OK = Thus, P = 360 N T C = N = N

103 7 96. Cable CD supports the lamp of mass M 1 and the lamp of mass M. Determine the maximum tension in the cable and the sag of point. Given: M 1 M 10 kg 15 kg a b c d 1 m 3 m 0.5 m m Solution: Guesses y 1 m T 1 N T C 1 N T CD 1 N Given a b T T a y b C 0 y d y a y y d T b y d T C M 1 g 0 b b y d c T C T c d CD 0 y d b y d T C d c d T CD M g 0 y T T C T CD Find y T T C T CD T T C T CD N T max max T T C T CD T max 157. N y.43 m

104 7 97. The cable supports the three loads shown. Determine the sags y and y D of points and D. Take P 11 = lb, N, P = 150 lb. N ft m y ft m y D E C D P P P mft 0 6 mft ft m ft m t + ΣF x = 0; 6 T C (4. y ) T = 0 y ΣF y = 0; 4. y T C + (4. y ) + 36 y T 150 = 0 y y 15.1 T C = 4500 (1) (4. y ) + 36 y N 4. y 6 t C + ΣF x = 0; 4.5 T CD (4. y D ) T C = 0 (4. y ) ΣF y = 0; 4. yd T CD + (4. y ) D 6 y y D (4. y ) y T C 000 = 0 (4. y ) + 36 T C = 9000 () 4. y 4. y D N 6 y y D (4. y ) D T CD = 1000 (3) t D + ΣF x = 0; 3.6 T Dg (1. + y D ) T CD = 0 (4. y D ) y D y D ΣF y = 0; 1. + yd T Dg (1. + y ) D 4. y D (4. y ) D T CD 150 = N y D (4. y ) D T CD = 4500 (4) Combining Eqs. (1) & () 3.7y + 6y D = Combining Eqs. (3) & (4) 13.5y + 8.8y D = y =.60 m y D =.113 m

105 7 98. The cable supports the three loads shown. Determine the magnitude of of PP 1 1 if if P P= = lb N and y =.4 8 ft. m. lso find the sag yy D D ft m y ft m y D E C D P P P mft 0 6 mft mft mft t + ΣF x = 0; T C 3.6 T = ΣF y = 0; 1.8 T C T = N.4 T 1500 = N T C = N t C + ΣF x = 0; (471.04) T CD = 0 (1) (4. y D ) N y D + ΣF y = 0; (471.04) + 4. y D (4. y ) D T CD P 1 = 0 () t D + ΣF x = 0; 3.6 T Dg (1. + y D ) T CD = 0 (4. y D ) ΣF y = 0; 1. + yd T Dg (1. + y ) D 4. y D (4. y ) D T CD 1500 = 0 T CD = (4. y ) 8.1 y 9.7 D D Substitute into Eq. (1) : y D = m 4. y D y D T CD = N N P 1 = N

106 7 99. If cylinders E and F have a mass of 0 kg and 40 kg, respectively, determine the tension developed in each cable and the sag y C. SOLUTION m 1.5 m m m 1 m y C D First, T will be obtained by considering the equilibrium of the free-body diagram shown in Fig. a. Subsequently, the result of T will be used to analyze the equilibrium of joint followed by joint C. Referring to Fig. a, we have C a + M D = 0; 40(9.81)() + 0(9.81)(4) - T a 3 5 b(1) - T a 4 5 b(4) = 0 E F T = N = 413 N Using the free-body diagram shown in Fig. b, we have : + F x = 0; T C cos u a 3 5 b = 0 + c F y = 0; a 4 5 b - 0(9.81) - T C sin u = 0 Solving, u = 8.44 T C = N = 8 N Using the result of u and the geometry of the cable, y C is given by y C - = tan u = 8.44 y C = m = 3.08 m Using the results of y C, u, and T C and analyzing the equilibrium of joint C, Fig. c, we have : + F x = 0; T CD cos cos 8.44 = 0 + c F y = 0; T CD = N = 358 N sin sin (9.81) = 0 (Check!)

107 If cylinder E has a mass of 0 kg and each cable segment can sustain a maximum tension of 400 N, determine the largest mass of cylinder F that can be supported. lso, what is the sag y C? SOLUTION m 1.5 m m m 1 m y C D We will assume that cable is subjected to the greatest tension, i.e., T = 400 N. ased on this assumption, M F can be obtained by considering the equilibrium of the free-body diagram shown in Fig. a. We have C a+ M D = 0; M F (9.81)() + 0(9.81)(4) - 400a 3 5 b(1) - 400a 4 5 b(4) = 0 E F M F = kg nalyzing the equilibrium of joint and referring to the free-body diagram shown in Fig. b, we have : + F x = 0; T C cos u - 400a 3 5 b = 0 + c F y = 0; 400a 4 5 b - 0(9.81) - T C sin u = 0 Solving, u = 7.9 T C = N Using these results and analyzing the equilibrium of joint C, : + F x = 0; T CD cos f cos 7.9 = 0 + c F y = 0; T CD sin f sin (9.81) = 0 Solving, f = T CD = N y comparing the above results, we realize that cable is indeed subjected to the greatest tension. Thus, M F = 37.5 kg Using the result of either u or f, the geometry of the cable gives y C - = tan u = tan 7.9 y C = 3.03 m or y C - 1 = tan f = tan y C = 3.03 m

108 If h = 5 m, determine the maximum tension developed in the chain and its length. The chain has a mass per unit length of 8 kg>m. h 5 m 50 m

109

110 7 10. The cable supports the uniform distributed load of w 0 = 1 600kN/m. lb>ft. Determine the tension in the cable at each support and ft m 10 3 mft Use the equations of Example 7.1. y = 4.5 = 3 = w 0 x 1 x 1 (7.5 x) ft m w 0 1 (4.5) x = 1 (7.5 x) (3) 3 m 4.5 m x = 1.5 ( x + x ) 0.5x.5x = x Choose root < 7.5 m x = m = w 0 x = y 1 (4.5) (4.188) =.79 kn t : y = w 0 x = 1 (.79) x dy dx = tan = x x = =.180 = T = cos =.79 cos = 54.5 kn t : y = w 0 x = 1 (.79) x dy dx = tan = x x = ( ) = = T = cos =.79 cos = kn

111 Determine the maximum uniform distributed load w 0 the cable can support if the maximum tension the cable can sustain is kn. lb ft m 10 3 mft w 0 Use the equations of Example 7.1. y = w 0 x ft m 4.5 = 3 = w w 0 0 x (7.5 x) 3 m 4.5 m x 4.5 = 1 (7.5 x) x x = 1.5 ( x + x ) 0.5x.5x = 0 Choose root < 7.5 m x = m = w 0 x = y w 0 (4.5) (4.188) = w 0 Maximum tension occurs at since the slope y of the cable is greatest there. w0 y = x dy = tan max = w x 0 dx x = = w 0 (4.188) max = T max = cos max 0 = cos w 0 = 4.40 kn/m

112 The cable is subjected to a uniform loading of 00 N/m. If the weight of the cable is neglected and the slope angles at points and are 30 and 60, respectively, determine the curve that defines the cable shape and the maximum tension developed in the cable. y 60 SOLUTION y = 1 L 00 dx dx L y = x + C 1 x + C dy dx = 1 100x + C m x 00 N/m t x = 0, y = 0; C = 0 t x = 0, dy dx = tan 30 ; C 1 = tan 30 y = x + tan 30 x t x = 15 m, dy dx = tan 60 ; = 598 N y = 138.5x + 577x110-3 m T max = u max = 60 cos u max = 598 cos 60 = 5196 N T max = 5.0 kn

113 The bridge deck has a weight per unit length of 80 kn>m. It is supported on each side by a cable. Determine the tension in each cable at the piers and. 150 m 1000 m 75 m

114

115 If each of the two side cables that support the bridge deck can sustain a maximum tension of 50 MN, determine the allowable uniform distributed load w 0 caused by the weight of the bridge deck. 150 m 1000 m 75 m

116

117 Cylinders C and D are attached to the end of the cable. If D has a mass of 600 kg, determine the required mass of C, the maximum sag h of the cable, and the length of the cable between the pulleys and. The beam has a mass per unit length of 50 kg> m. 3 m 1 m h SOLUTION From the free-body diagram shown in Fig. a, we can write a+ M = 0; 600(9.81) sin u (1) - 600(9.81) cos u (3) - 50(1)(9.81)(6) = 0 u = : + F x = 0; 600(9.81) cos m C (9.81) cos u = 0 C D + c F y = 0; Solving, m C (9.81) sin u + 600(9.81) sin (1)(9.81) = 0 m C = kg = 478 kg u = 3.47 Thus, = T cos u = N. s shown in Fig. a, the origin of the x - y coordinate system is set at the lowest point of the cable. Using Eq. (1) of Example 7 1, y = w 0 x = 50(9.81) ( ) x y = x Using Eq. (4) and applying two other boundary conditions y = (h + 3) m at x = x 0 and y = h at x = -(1 - x 0 ), we have h + 3 = x 0 h = [-(1 - x 0 )] Solving these equations yields h = m = 0.87 m x 0 = 8.19 m The differential length of the cable is ds = dx + dy = 1 + a dy dx b dx = x dx Thus, the total length of the cable is 8.19 m L = ds = x L L m 8.19 m = x dx L m = e 1 cx x ln ax x 8.19 m bdf` m = 13. m

118 .

119 If the pipe has a mass per unit length of 1500 kg> m, 30 m determine the maximum tension developed in the cable. 3 m SOLUTION s shown in Fig. a, the origin of the x, y coordinate system is set at the lowest point of the cable. Here, w(x) = w 0 = 1500(9.81) = (10 3 ) N>m. Using Eq. 7 1, we can write y = 1 L a L w 0dxbdx = 1 a (103 ) x + c 1 x + c b dy pplying the boundary condition at x = 0, results in c 1 = 0. dx = 0 pplying the boundary condition y = 0 at x = 0 results in c = 0. Thus, y = (103 ) x pplying the boundary condition y = 3 m at x = 15 m, we have 3 = (103 ) (15) = (10 3 ) N Substituting this result into Eq. (1), we have dy dx = x The maximum tension occurs at either points at or where the cable has the greatest angle with the horizontal. Here, Thus, u max = tan - 1 a dy dx ` b = tan -1 [0.0667(15)] = m T max = = 551.8(103 ) cos u max cos 1.80 = 594.3(103 ) N = 594 kn

120 If the pipe has a mass per unit length of 1500 kg> m, 30 m determine the minimum tension developed in the cable. 3 m SOLUTION s shown in Fig. a, the origin of the x, y coordinate system is set at the lowest point of the cable. Here, w(x) = w 0 = 1500(9.81) = (10 3 ) N>m. Using Eq. 7 1, we can write y = 1 L a L w 0dxbdx = 1 a (103 ) x + c 1 x + c b dy pplying the boundary condition at x = 0, results in c 1 = 0. dx = 0 pplying the boundary condition y = 0 at x = 0 results in c = 0. Thus, y = (103 ) x pplying the boundary condition y = 3 m at x = 15 m, we have 3 = (103 ) (15) = (10 3 ) N Substituting this result into Eq. (1), we have dy dx = x The minimum tension occurs at the lowest point of the cable, where u = 0. Thus, T min = = (10 3 ) N = 55 kn

121 If the slope of the cable at support is zero, determine the deflection curve y = f(x) of the cable and the maximum tension developed in the cable. y 1 m SOLUTION 4.5 m x Using Eq. 7 1, y = 1 L a L w(x)dxbdx y = 1 L a L 4 cos p 4 * dxbdx 4 kn/m p w 4 cos x 4 y = 1 4 L p c4(103 ) d sin p 4 x + C 1 y = - 4 p c 96(103 ) p cos p 4 x d + C 1x + C dy pplying the boundary condition at x = 0 results in C 1 = 0. dx = 0 pplying the boundary condition y = 0 at x = 0, we have 0 = - 4 p c 96(103 ) p cos 0 d + C Thus, C = 304(103 ) p y = 304(103 ) p c1 - cos p 4 x d pplying the boundary condition y = 4.5 m at x = 1 m, we have 4.5 = 304(103 ) p c1 - cos p 4 (1) d = (10 3 ) N Substituting this result into Eqs. (1) and (), we obtain dy dx = 96(10 3 ) p(51.876)(10 3 ) sin p 4 x = sin p 4 x and y = 304(10 3 ) p (51.876)(10 3 ) c1 - cos p 4 x d = 4.5a1 - cos p 4 xb m The maximum tension occurs at point where the cable makes the greatest angle with the horizontal. Here, u max = tan -1 a dy dx ` x = 1 m b = tan-1 c sina p (1)b d = Thus, T max = = (103 ) cos u max cos = 60.07(103 ) N = 60. kn

122 7 11. Determine the maximum tension developed in the cable if it is subjected to a uniform load of 600 N/m. y 10 0 m x SOLUTION The Equation of The Cable: 100 m 600 N/m y = 1 L 1 1w1xdx dx = 1 w 0 x + C 1 x + C dy dx = 1 1w 0 x + C 1 (1) () oundary Conditions: y = 0 at x = 0, then from Eq. (1) 0 = 1 1C F C = 0 H dy dx = tan 10 at x = 0, then from Eq. () tan 10 = 1 1C F 1 C 1 = tan 10 H Thus, y = w 0 x + tan 10 x (3) y = 0 m at x = 100 m, then from Eq. (3) 0 = tan = N and dy dx = w 0 x + tan 10 u = u max at x = 100 m and the maximum tension occurs when u = u max. tan u max = dy dx ` = tan 10 x = 100 m The maximum tension in the cable is = 600 x + tan = x + tan 10 u max = 1.61 T max = = = N = 1.30 MN cos u max cos 1.61

123 Determine the maximum uniform distributed loading w 0 N/m that the cable can support if it is capable of sustaining a maximum tension of 60 kn. 60 m 7 m w 0.

124 cable has a weight density and is supported at points that are a distance d apart and at the same elevation. If it has a length L, determine the sag. Given: 30 N cm d 5m L 6m Solution: Guess 1000N Given L sinh d = 0 Find 770 N h cosh 1 d 1 F h m H

125 wire has a weight denisty. If it can span a distance L and has a sag h determine the length of the cable. The ends of the cable are supported from the same elevation. Given : N d 10m h 1.m m Solution : From Eq. (5) of Example 7-15 : d 1 h = F H 8 d F h H 0.83 N From Eq. (3) of Example 7-15 : L sinh d = L sinh 1 d F L 10.39m H

126 The 10 kg>m cable is suspended between the supports and. If the cable can sustain a maximum tension of 1.5 kn and the maximum sag is 3 m, determine the maximum distance L between the supports 3 m L SOLUTION The origin of the x, y coordinate system is set at the lowest point of the cable. Here w 0 = 10(9.81) N>m = 98.1 N>m. Using Eq. (4) of Example 7 13, y = w 0 cosh w 0 x - 1R y = x cosh - 1R pplying the boundary equation y = 3 m at x = L we have, 3 = L cosh - 1R The maximum tension occurs at either points or where the cable makes the greatest angle with the horizontal. From Eq. (1), tan u max = sinh 49.05L y referring to the geometry shown in Fig. b, we have cos u max = sinh 49.05L = 1 cosh 49.05L Thus, T max = cos u max 1500 = cosh 49.05L (3) Solving Eqs. () and (3) yields L = 16.8 m = N

127 Show that the deflection curve of the cable discussed in Example 7 13 reduces to Eq. 4 in Example 7 1 when the hyperbolic cosine function is expanded in terms of a series and only the first two terms are retained. (The answer indicates that the catenary may be replaced by a parabola in the analysis of problems in which the sag is small. In this case, the cable weight is assumed to be uniformly distributed along the horizontal.) SOLUTION Substituting into cosh x = 1 + x 1 + Á y = w 0 cosh w 0 x - 1R = w w 0x F H = w 0x Using Eq. (3) in Example 7 1, + Á - 1R We get = w 0L 8h y = 4h L x QED

128 If the horizontal towing force is T = 0 kn and the chain has a mass per unit length of 15 kg>m, determine the maximum sag h. Neglect the buoyancy effect of the water on the chain. The boats are stationary. T 40 m h T

129 The cable has a mass of 0.5 kg>m, and is 5 m long. Determine the vertical and horizontal components of force it exerts on the top of the tower. SOLUTION x = L ds b F (w 0 ds) r H Performing the integration yields: m rom Eq x = dy dx = 1 L w 0ds b sin h c (4.905s + C 1 ) d + C r (1) dy dx = 1 (4.905s + C 1 ) t s = 0 dy ; = tan 30. Hence C 1 = tan 30 dx dy dx = 4.905s + tan 30 () pplying boundary conditions at x = 0; s = 0 to Eq.(1) and using the result C yields C = -sin h = tan 30 (tan 30 ). Hence x = b sin h c (4.905s+F tan 30 ) d - sin h - 1 (tan 30 )r H (3) t x = 15 m; s = 5 m. From Eq.(3) 15 = b sin h c (4.905(5) + F tan 30 )R - sin h - 1 (tan 30 ) r H y trial and error = N t point, s = 5 m From Eq.() tan u = dy dx = 4.905(5) s = 5 m tan 30 u = (F v ) = tan u = tan = 165 N ( ) = = 73.9 N

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