SOLUTION 6 6. Determine the force in each member of the truss, and state if the members are in tension or compression.

Transcription

1 6 6. etermine the force in each member of the truss, and state if the members are in tension or compression. 600 N 4 m Method of Joints: We will begin by analyzing the equilibrium of joint, and then proceed to analyze joints and E. E 4 m 900 N Joint : From the free-body diagram in Fig. a, : + F x = 0; F E a 3 5 b = 0 6 m F E = 1000 N = 1.00 kn () 1000 a 4 5 b - F = 0 F = 800 N (T) Joint : From the free-body diagram in Fig. b, F E = 0 F E = 900 N () F = 0 F = 800 N (T) Joint E: From the free-body diagram in Fig. c, R+ F x =0; Q+ F y =0; cos F E sin = 0 F E = 750 N (T) F E sin cos = 0 F E = 1750 N = 1.75 kn ()

2 6 7. etermine the force in each member of the ratt truss, and state if the members are in tension or compression. 2 m K J I Joint : 20 - F sin 45 = 0 F = kn () 2 m 2 m 2 m H E F 2 m 2 m 2 m 2 m 2 m 10 kn 10 kn 20 kn G : + F x = 0; F cos 45 = 0 F = 20 kn (T) Joint : : + F x = 0; F - 20 = 0 F = 0 F = 20 kn (T) Joint : R+ F x = 0; +Q F y = 0; F = F K = 0 F K = kn () Joint : : + F x = 0; F - 20 = 0 F = 20 kn (T) F K - 10 = 0 F K = 10 kn (T) Joint K: R+ F x - 0; +Q F y = 0; 10 sin 45 - F K cos ( ) = 0 F K = kn () cos sin ( ) - F KJ = 0 F KJ = kn () Joint J: ue to Symmetry sin 45 - F JI sin 45 = 0 F JI = kn () 2 (23.57 cos 45 ) - F J = 0 F J = 33.3 kn (T) F = F GH = F K = F HI = 28.3 kn () F = F GF = F = F FE = F = F E = 20 kn (T) F = F FH = F = F HE = 0 F K = F EI = 10 kn (T) F KJ = F IJ = 23.6 kn () F K = F I = 7.45 kn ()

3 6 22. etermine the force in each member of the double scissors truss in terms of the load and state if the members are in tension or compression. c+ M = 0; a 3 b + a 2 3 b - ( y)() = 0 y = y = /3 E /3 F /3 /3 Joint F: F F - F FE - F F a 1 22 b = 0 (1) F F - F FE = : + F y = 0; 1 F F a 22 = 0 Similarly, F F = 22 = 1.41 (T) F E = 22 Joint : F a 2 25 b - 22a 1 22 b - F a 1 22 b = 0 +c F y = 0; 2 25 F F = F F F = = 0 = = 1.49 () Joint : F = 22 3 = = () F E a 1 22 b a 2 25 b = 0 F E = 5 = 1.67 (T) 3 Similarly, F F =1.67 (T) From Eq.(1), and Symmetry, F FE = (T) F F = 1.67 (T) F = () F E = 1.67 (T) F = 1.49 () F F = 1.41 (T) F = 1.49 () F E = 1.41 (T) F = ()

4 6 25. etermine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression. u Joint : F sin 2u - = 0 F = csc 2u () csc 2u(cos 2u) - F = 0 F = cot 2 u () Joint : cot 2 u + + F cos 2 u - F cos u = 0 F sin 2 u - F sin u = 0 F = cot 2 u + 1 cos u - sin ucot 2 u F = (cot u cos u - sin u + 2cos u) (T) F = (cot 2 u + 1) () Joint : F - (cot 2 u + 1)(cos 2 u) = 0 F = (cot 2 u + 1)(cos 2 u)() ()

5 6 26. The maximum allowable tensile force in the members of the truss is 1F t 2 max = 2 kn, and the maximum allowable compressive force is 1F c 2 max = 1.2 kn. etermine the maximum magnitude of the two loads that can be applied to the truss.take = 2mand u = 30. u (T t ) max = 2kN (F ) max = 1.2 kn Joint : F cos 30 - = 0 F = = () cos 30 F sin 30 - F = 0 F = tan 30 = () Joint : -F sin 30 + F sin 60 = 0 sin 60 F = F a sin 30 b = F tan F cos 60 - F cos 30 = 0 tan F = a b = () 23 cos 30 - cos 60 F = (T) Joint : F sin 30 = 0 F = () 1) ssume F = 2kN = = N F = 1.577(732.05) = N 6 (F c ) max = 1200 N (O.K.!) Thus, max = 732 N

6 6 31. etermine the force in members, J, KJ,and J of the truss which serves to support the deck of a bridge. State if these members are in tension or compression lb 8000 lb E 5000 lb F G 12 ft a + M = 0; -9500(18) (9) + F KJ (12) = 0 F KJ = lb = 11.2 kip (T) K J I H 9ft 9ft 9ft 9ft 9ft 9ft a + M J = 0; -9500(27) (18) (9) + F (12) = 0 F = 9375 lb = 9.38 kip () F J = 0 Joint : F J = 3125 lb = 3.12 kip () F J = 0

7 *6 32. etermine the force in members EI and JI of the truss which serves to support the deck of a bridge. State if these members are in tension or compression lb 8000 lb E 5000 lb F G 12 ft a+ M E = 0; -5000(9) (18) - F JI (12) = 0 F JI = 7500 lb = 7.50 kip (T) K J I H 9ft 9ft 9ft 9ft 9ft 9ft F EI = 0 F EI = 2500 lb = 2.50 kip ()

8 6 46. etermine the force in members and M of the altimore bridge truss and state if the members are in tension or compression. lso, indicate all zero-force members. M K N O J E F G H 2m 2m I 2kN 5kN 3kN 16m,8@2m 2kN Support Reactions: a+ M I = 0; y 1162 = 0 y = kn x = 0 Method of Joints: y inspection, members N, N, O, O, HJ E and JG are zero force members. Method of Sections: a+ M M = 0; F = 0 F = kn 1T2 a+ M = 0; F M = 0 F M = 2.00 kn T

9 6 47. etermine the force in members EF, E, and K of the altimore bridge truss and state if the members are in tension or compression. lso, indicate all zero-force members. M K N O J E F G H 2m 2m I 2kN 5kN 3kN 16m,8@2m 2kN Support Reactions: a+ M = 0; I y = 0 I y = kn Method of Joints: y inspection, members N, N, O, O, HJ E and JG are zero force members. Method of Sections: a+ M K = 0; F EF 142 = 0 F EF = = 7.88 kn 1T2 a+ M E = 0; F K 142 = 0 F K = 9.25 kn F E sin 45 = 0 F E = 1.94 kn T

10 6 71. etermine the support reactions at,, and E on the compound beam which is pin connected at and. 9 kn 10 kn Equations of Equilibrium: First, we will consider the free-body diagram of segment E in Fig. c. 10 kn m 1.5 m 1.5 m 1.5 m 1.5 m 1.5 m 1.5 m E + M = 0; + M E = 0; N E (3) - 10(1.5) = 0 N E = 5 kn 10(1.5) - y (3) = 0 y = 5 kn x = 0 Subsequently, the free-body diagram of segment in Fig. b will be considered using the results of and obtained above. x y + M = 0; + M = 0; N (1.5) - 5(3) - 10 = 0 N = kn = 16.7 kn y (1.5) - 5(1.5) - 10 = 0 y = kn y = 0 Finally, the free-body diagram of segment in Fig. a will be considered using the results of and obtained above. x y x = y = 0 + M = 0; y = 2.67 kn 11.67(3) - 9(1.5) - M = 0 M = 21.5 kn # m

11 The tractor boom supports the uniform mass of 500 kg in the bucket which has a center of mass at G. etermine the force in each hydraulic cylinder and and the resultant force at pins E and F. The load is supported equally on each side of the tractor by a similar mechanism. G E 0.25 m 0.3 m 0.1 m 1.5 m 0.2 m 1.25 m a+ M E = 0; F = 0 F = 981 N F -E x = 0; E y = 0; F E = = 2.64 kn E x = 981 N E y = N 0.4 m 0.3 m 0.6 m a+ M F = 0; F 1cos F 1sin = 0 F = N = 16.3 kn F x sin 12.2 = 0 F x = 3455 N -F y cos 12.2 = 0 F y = N F F = = 14.0 kn

12 The symmetric coil tong supports the coil which has a mass of 800 kg and center of mass at G. etermine the horizontal and vertical components of force the linkage exerts on plate EIJH at points and E. The coil exerts only vertical reactions at K and. Free-ody iagram: The solution for this problem will be simplified if one realizes that links and F are two-force members. H 300 mm J E I 400 mm mm F mm 100 mm Equations of Equilibrium : From F (a), K G a+ M = 0; 78481x2 - F K 12x2 = 0 F K = 3924 N From F (b), a+ M = 0; F cos F sin = 0 F = N x cos 45 = 0 x = 981 N y sin 45 = 0 y = 4905 N From F (c), a+ M E = 0; 4905 sin sin F F cos = 0 F F = N E x cos 30 = 0 E x = N = 6.79 kn E y sin = 0 E y = N = 1.55 kn t point, x = F cos 45 = cos 45 = 981 N y = F sin 45 = sin 45 = 981 N