1. Determine the Zero-Force Members in the plane truss.

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1 1. Determine the Zero-orce Members in the plane truss. 1

2 . Determine the force in each member of the loaded truss. Use the Method of Joints.

3 3. Determine the force in member GM by the Method of Section. 3

4 . Determine the forces in members BC and G.

5 Zero-orce Members: EK, E 100 N Cut (Upper Side) Cut BC CJ J G 800 N C CJ J G BC N C M C 0 G G 600 y 0 BC G 0 BC N T 5

6 5. Calculate the forces in members DE, GJ and DG of the simple truss. State whether they are in tension or compression. 6

7 6. Determine the forces in members G, CG, BC, and E for the loaded crane truss. 5 5 E 5 D G C B 5 Zero-orce Members: BG 7

8 1 st Cut G E C B 5 D st Cut (Upper part) x 5 5 G 5 CG 5 E C BC 5 D G 0 CG 0 M 0 50(8) BC() 0 BC y 0 5 G BC 50 0 G C T Joint : 5 5 E C y 0 5 5cos 5 E cos 5 G E 5.7 T 5 0 G 8

9 7. The truss shown consists of 5 triangles. The cross members in the two center panels that do not touch each other are slender bars which are incapable of carrying compressive loads. Determine the forces in members GM and L. Reactions at the supports rom equilibrium of whole truss; x =80 y =60 B y =0 9

10 I. Cut (Left Side) I. Cut G GM L LM x =80 y =60 10

11 8. Determine the force acting in member JI. 0 3 E L D G H K J I 3 C 5 3 m 3 m B 3 m 5 m M m N P 10 m m

12 9. If it is known that the center pin supports one-half of the vertical loading shown, determine the force in member B. 1

13 G y x H y Reactions at the supports y x 0 rom equilibrium of whole truss; 0 Center pin supports one-half of the vertical loading. x 8 10 y 6 Because of symmetry, Gy H y 13 13

14 I. Cut DE D B G y H y =13 DE D B y I. Cut (Right side) There are four unknowns. H y =13 1

15 B Joint 5 o 5 o y =6 0 B cos 5 cos 5 0 x B y 0 Bsin 5 sin B C DE D I. Cut (Right side) M D + 0 D B 5 o H y =13 10(1) 8() 8(36) 13(8) B. T B(16) cos 5(16) sin 5(1)

16 10. Determine the forces acting in members DE, DI, KJ, J. m m m B C D E M I G 3 m 3 m L K J H 37 o 0 6 m 16

17 Zero-orce Members: m m m E, G B T C D E 3 m M I G 3 m L K J H 37 o 0 6 m x Reactions at the supports y rom equilibrium of whole truss; + y 0 y 0sin 37 0 y 1 M 0 T (1) 0cos37(6) 0sin 37(1) 0 T 0 x 0 T x 0cos37 0 x 17

18 Joint : L x = J B T C m m m D E M I G 3 m 3 m x x 0 J J 5 y = C 0 x L K 1 st Cut J H 37 o 0 6 m y DE KJ EI IJ E 1 st Cut (Right side) + M J 0 DE 8 6 0sin 37 0 DE T J J 37 o 0 18

19 nd Cut (Right side) DE E T C m m m D E 3 m K DI KI KJ J I J L 37 o 0 x M K nd Cut I J H 6 m G 3 m 37 o 0 + y M I 0 1 DE 3 KJ 3 J cos373 0cos373 0sin 37 0 KJ T 8 5 M K 0 5 DE 6 J sin 37 0sin 378 DI cos373 DI sin 37 0 DI 7. T

20 m 11. The hinged frames CE and DB are connected by two hinged bars, B and CD, which cross without being connected. Compute the force in B. 3.5 m tana a 3. 5 a 9. 7 o B 0

21 m I. Cut (Left Side) a B CD a 3.5 m tana 3. 5 a 9. 7 o I. Cut (Right Side) a B CD B E x E y + I. Cut (Left Side) 6 Bsin a1.5 CD cosa CDsin a3 M E 0 B cosa 0 CD 3B I. Cut (Right Side) + M B cosa Bsin a3 CD cosa6 CDsin a CD 1.98B 0 B C 60 1

22 1. Determine the forces in members DE, EI, I, and HI of the arched roof truss.

23 Zero-orce Members: BK, H Reactions at the supports rom equilibrium of whole truss; x 0 G 0 Because of symmetry of the truss: y G 150 y x G x M y G G y y y G y (0) 5() 75(10) 100(0) 75(30) 5(36) 0 y G 150 y y 3

24 1 st Cut (Right side) I HI E I H 5 G 1 st Cut + G y =150 y =150 G y =150 M I 0 E E 315. C 1 y 0 E HI HI x E I HI 0 I 05. T T

25 nd Cut (Right side) 75 DE E 5 EI I I IH G nd Cut M I + DE G y =150 y =150 G y =150 6 DE DE C y 0 EI 6. EI T 6 DE HI

26 13. Determine the forces in members ON, NL and DL.

27 x y I y rom equilibrium of whole truss; M x y x y y 6 (18) 6() (15) (9) (6) (3) I y 10 0 I y 6 0 y

28 ON OC BC I.cut I.cut ) ( (3) (3) 6() (6) 0 Compression M ON ON ON y C

29 Joint M ML MN ) ( C ML MN MN y ML MN ML MN x

30 II.cut M y D 0 0 NL y (9) (6) 6() 3 MN MN.5 y ( C) MN DL DL 6 6 NL 0 ( Zero force () 0 member) II.cut MN NL DL DE

31 1. Determine the forces in members HG and IG. 0

32 II.cut I.cut

33 II.cut I.cut CD CD HG HG 0 0 B HI GI GJ I.cut M G =0 0 CD =5.1 (T) II.cut M =0 HG =81.1 (C) I.cut x =0 GI =18.9 (T)

34 15. Determine the forces in members E, NK and LK. 1 C 5 3 D E G m B N M O H m L K J 3 m 3 m 3 m 3 m I

35 3 1 C D E G rom the equilibrium of whole truss I. Cut Top Part B BN N MN M MO O HO H m x, y and I y are determined. x B L K J HI I m I. Cut M H =0 B is determined y 3 m 3 m 3 m 3 m I y

36 3 C D E E G 1 II. Cut Top Part B N M M O H m II. Cut M M =0 B BN MN MO m E and M are determined L K J 3 m 3 m 3 m 3 m I

37 3 C D E E G 1 M m B N M MO O H III. Cut NK L K J LK I m M N =0 LK and NK are determined 3 m 3 m 3 m 3 m III. Cut Left Side

38 16. Determine the forces in members KN and C. G 10 I H P E 1 m m J M K L N C O D 1 m 5 0 m B m 1 m 1 m m

39 G 10 I III. Cut H P I. Cut E 1 m m M N O 1 m 5 J II. Cut K L C 0 D m x B m 1 m 1 m m y B y

1. Determine the Zero-Force Members in the plane truss.

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