Problem 6.24 The Pratt bridge truss supports five forces ( F D 300 kn). The dimension L D 8m.Determine the axial forces in members BC, BI, and BJ.

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1 Problem 6.24 The Pratt bridge truss supports five forces ( F D 300 kn). The dimension L D 8m.Determine the aial forces in members C, I, and J. L L L L L L C D E G L I J K L M H F F F F F Solution: Find support reactions at and H.From the free bod diagram, F D X D 0, F D Y C H Y D 0, Y G θ I J K L M L H L L L L L L H Y F F F F F L = 8 m F = 300 kn and M D 6 8 H Y C 6 C 24 C 32 C 40 D 0. From these equations, Y D H Y D 750 kn. Joint Joint I From the geometr, the angle D 45 Joint :From the free bod diagram, T θ T I T I I T I T IJ F D X C T cos C T I D 0, Y F F D T sin C Y D 0. From these equations, T D 06 kn and T I D 750 kn. Joint T C θ θ T J T T I Joint I :From the free bod diagram, F D T IJ T I D 0, F D T I 300 D 0. From these equations, T I D 300 kn and T IJ D 750 kn. Joint :From the free bod diagram, F D T C C T J cos T cos D 0, F D T I T J sin T sin D 0. From these equations, T C D 200 kn and T J D 636 kn. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

2 Problem 6.26 The Howe truss helps support aroof. Model the supports at and G as roller supports. Determine the aial forces in members, C, and CD. 400 lb 600 lb C 800 lb D 600 lb E 400 lb F 8 ft H I J K L 4 ft 4 ft 4 ft 4 ft 4 ft 4 ft G Solution: The strateg is to proceed from end,choosing joints with onl one unknown aial force in the -and/or -direction, if possible, and if not, establishsimultaneous conditions in the unknowns. The interior angles HI and HJC differ. The pitch angle is Pitch D tan ( 8 2 ) D The length of the vertical members: ( ) 8 H D 4 D ft, 2 from which the angle ( ) HI D tan D CI D 8 8 D ft, lb α Pitch H 400 lb 600 lb 800 lb 600 lb 400 lb 4 ft 4 ft 4 ft 4 ft 4 ft 4 ft Joint Joint H 600 lb I CI CD α Pitch α Pitch α IJC HI IJ C CI CJ Joint I Joint C Joint H : F D H D 0, or, H D 0. G H 400 lb C α Pitch α Pitch H HI H I Joint from which the angle ( ) IJC D tan D The moment about G : M G D 4 C C 8 C C D 0, from which D D 400 lb. Check: The total load is 2800 lb. 24 From left-right smmetr each support, G supports half the total load. check. F D H C HI D 0, from which HI D 200 lb T Joint : F D cos Pitch C C cos Pitch C I cos Pitch D 0, from which C C I D The method of joints: Denote the aial force in amember joining two points I, K b IK. Joint : F D sin P C 400 D 0, from which D 400 D lb C sin p F D cos Pitch C H D 0, from which H D D 200 lb T c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

3 6.26 (Continued) F D 400 sin Pitch C C sin Pitch I sin Pitch D 0, from which C I D C 400 sin Pitch. Solve the two simultaneous equations in unknowns C, I: I D sin Pitch D lb C, and C D I D lb C Joint I : F D I cos Pitch HI C IJ D 0, from which IJ D 800 lb T F DCI sin Pitch C CI D 0, from which CI D 200 lb (T) Joint C: F D C cos Pitch C CD cos Pitch C CJ cos IJC D 0, from which CD C CJ 0.6 D 800 F D 600 CI C sin Pitch C CD sin Pitch CJ sin IJC D 0, from which CD CJ 0.8 D 400 Solve the two simultaneous equations to obtain CJ D lb C, and CD D lb C c 2008 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the 4

4 Problem 6.42 For the Pratt bridge truss in Problem6.4, use the method of sections to determine the aial force in member EK. Solution: From the solution to Problem 6.4, the support forces are D 0, D H D 850 kn. Method of Sections to find aial force in EK. EK DE θ E G L L L L L L L C D E G I J K L M F F F F F H KL L F F H Y Solution: lso, EK D kn T KL D 360 kn T F : DE EK cos KL D 0 DE D 530 kn C F : H 2 F EK sin D 0 M E : L KL L F C 2 L H D 0 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

5 Problem 6.38 The Pratt bridge truss is loaded as shown. Use themethod of sections to determine theaial forces in members D, E, and CE. D F C E G 0 kip 30 kip 20 kip 7 ft 7 ft 7 ft 7 ft H 8 ft Solution: Use the whole structure to find the reaction at. M H : 20 kip 7 ft C 30 kip 34 ft C 0 kip 5 ft 68 ft D 0 ) D 27. 5kip 0 kip 30 kip 20 kip H Now cut through D, E, CE and use the left section F D M : 7 ft C F CE 8ft D 0 ) F CE D 58. 4kip 8 7 M E : 0 kip 7 ft 34 ft F D 8ft D 0 F E ) F D D 95. 6kip C F CE F : 0 kip 8 p 353 F E D 0 ) F E D 4. kip In Summar F CE D 58. 4kip T, F D D 95. 6kip C, F E D 4. kip T 0 kip c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

Problem 6.33 Determine the reactions on member C at and C. 4 kn 0.2 m D 2 kn-m 0.2 m C E 0.2 m 0.2 m Solution: We draw free-bod diagrams for the entire structure, and for members D and C.

6 Problem 6.33 Determine the reactions on member C at and C. 4 kn 0.2 m D 2 kn-m 0.2 m C E 0.2 m 0.2 m Solution: We draw free-bod diagrams for the entire structure, and for members D and C. From the entire structure: 6F : C C 4kND0 6M E : C 0.4 m 4 kn 0.4 m 2 kn-m D 0 From bod C 6M : 0.2 m C C 0.4 m D 0 nd from bod D 6M D : 0.2 m 2 kn-m D 0 Solving these four equations ields D 8kN, D 0 kn C D 4kN, C D 9kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the 507

7 Problem 6.79 The frame supports a 6-kN load at C. Determine the reactions on the frame at and D. 0.4 m.0 m 6 kn C 0.5 m Solution: Notethat members E and CF are twoforce members. Consider the 6kNload as being applied to member C. D 0.8 m E 0.4 m F.0 m 6 kn C 0.4 m F E F CF θ φ tan D tan D D D Member DEF F E F CF D θ E F φ 0.8 m 0.4 m D Equations of equilibrium: Member C: F : C F E cos F CF cos D 0 F : F E sin F CF sin 6 D 0 C M : 0. 4 F E sin. 4 F CF sin. 4 6 D 0 Member DEF: F : D F E cos C F CF cos D 0 F : D C F E sin C F CF sin D 0 C M D : 0. 8 F E sin C. 2 F CF sin D 0 Unknowns,, D, D, F E, F CF we have 6 eqns in 6 unknowns. Solving, we get D 6. 8kN D. 25 kn D D 6. 3kN D D kn lso, F E D 20. 2kN T F CF D. 3kN C c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

8 Problem 7.30 Determine the coordinates of the centroids. 0 in 20 in Solution: The strateg is to find the centroid for the half circle area, and use the result in the composite algorithm. The area: The element of area is avertical strip high and d wide. From theequation of the circle, Dš p R 2 2.The height of the strip will be twice the positive value, so that d D 2 p R 2 2 d, from which R D d D 2 R 2 2 / 2 d 0 D 2 The -coordinate: [ p R 2 2 R d D 2 R 2 2 d 0 2 [ D 2 R2 2 3 / 2 ] R 3 0 Divide b : D 4 R 3 C R 2 ( 2 sin ) ] R D R2 R 2 0 D 2 R 3 3. The -coordinate: From smmetr, the -coordinate is zero. The composite: For acomplete half circle D 4 20 D in. For 3 the inner half circle 2 D in. The areas are D in 2 and 2 D in 2. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

9 Problem 7.36 Determine the coordinates of the centroids. 5 mm 5 mm 50 mm 5 mm 5 mm 5 mm 5 mm 0 mm mm 5 mm 5 mm 0 Solution: Comparison of the solution to Problem 7.29 and our areas, 2, and 3, we see that in order to use the solution of Problem 7.29, we must set a D 25 mm, R D 5 mm, and r D 5mm. If we do this, wefind that for this shape, measuring from the ais, D mm. The corresponding areas for regions, 2, and 3is 025 mm 2.The centroids of the rectangular areas are at their geometric centers. inspection, we how have the following information for the five areas mm 5 mm 50 mm rea : rea D 025 mm 2, D mm, and D 50 mm. rea 2: rea 2 D 025 mm 2, 2 D mm, and 2 D 0mm. rea 3: rea 3 D 025 mm 2, 3 D mm, and 3 D 0mm. rea 4: rea 4 D 600 mm 2, 4 D 0mm, and 4 D 25 mm. 5 mm mm 5 mm 5 mm 0 mm mm 5 mm 5 mm 0 rea 5: rea 5 D 450 mm 2, 5 D 7. 5mm, and 5 D 50 mm. Combining the properties of the five areas, we can calculate the centroid of the composite area made up of the five regions shown. rea TOTL D rea C rea 2 C rea 3 C rea 4 C rea 5 D 425 mm 2. Then, D rea C 2 rea 2 C 3 rea 3 C 4 rea 4 C 5 rea 5 /rea TOTL D mm, and D rea C 2 rea 2 C 3 rea 3 C 4 rea 4 C 5 rea 5 /rea TOTL D mm. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

10 Problem 8.50 Determine I and k. 20 mm 20 mm 80 mm 40 mm 80 mm Solution: We must first find thelocation of thecentroid of thetotal area. Let us use the coordinates XY to do this. Let be the rectangle and 2 be the circular cutout. Note that b smmetr X c D 40 mm Y 80 mm rea X c Y c Rectangle 9600 mm 2 40 mm 60 mm Circle mm 2 40 mm 80 mm D 9600 mm 2 R = 20 mm 20 mm 2 D 257 mm 2 For the composite, 80 mm X c D X c 2 X c 2 2 D 40 mm X Y c D Y c 2 Y c 2 2 D 57. 0mm Now let us determine I and k about the centroid of the composite bod. Rectangle about its centroid (40, 60) mm I D 2 bh3 D I D. 52 ð 0 7 mm 3, Now toc I c D I C 60 Y c 2 I c D. 6 ð 0 7 mm 4 Circular cut out about its centroid 2 D R 2 D 20 2 D 257 mm 2 40 mm 40 mm Now to C! d 2 D D 23 mm I c2 D I 2 C d I c2 D 7. 9 ð 0 5 mm 4 For the composite about the centroid I D I c I c2 I D. 08 ð 0 7 mm 4 The composite rea D mm 2 k D D 8343 mm 2 I D 36. 0mm I 2 D 4 R4 D 20 4 / 4 I 2 D. 26 ð 0 5 mm 4 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

11 Problem 8.53 Determine I and k. 2 in 20 in Solution: Treat the area as a circular area with a half-circular cutout: From ppendi, I D in 4 and I 2 D in 4, 2 in. 20 in. 20 in. 2 2 in. so I D D. 8 ð 0 5 in 4. The area is D D 030 in 2 so, k D I D. 8 ð ð 0 3 D 0. 7in c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

12 Problem 9. In Problem 9.0, bo weighs 00 lb, and the coefficients of friction between bo and the ramp are s D 0.30 and k D For what range of the weights of the bo will the sstem remain stationar? Solution: The upper and lower limits on the range are determined b the weight required to move the bo up the ramp, and the weight that will allow the bo to slip down the ramp. ssume impending slip. The friction force opposes the impending motion. For impending motion up the ramp the sum of forces parallel to the ramp are MX α F D sin MX C S cos D 0, µ s N N from which MIN MX D sin C s cos D 00 sin 30 C 0.3 cos 30 D lb N µ s N α For impending motion down the ramp: F D sin MIN s cos D 0, from which D sin s cos D 00 sin cos 30 D lb Problem 9.2 The mass of the bo on the left is 30 kg, and the mass of the bo on the right is 40 kg. The coefficient of static friction between each bo and the inclined surface is s D 0.2. Determine the minimum angle for with the boes will remain stationar. a 30 Solution: If the boes slip when is decreased, the will slip toward the right. ssume that slip toward the right impends, the free bod diagrams are as shown. The equilibrium equations are F D T 0.2 N sin D 0, () T T 30 α (30)(9.8) (40)(9.8) 0.2 N 0.2 N N N F D N cos D 0, (2) F D T 0.2 N C sin 30 D 0, (3) F D N cos 30 D 0, (4) Summing Equations () and (3), we obtain 0.2 N 0.2 N sin C sin 30 D 0. Solving Equation (2) for N and Equation (4) for N and substituting the results into Equation (5) gives 5 sin C 3 cos D 0 4 cos 30. (6) Using the identit cos D sin 2 and solving Equation (6) for sin, we obtain sin D 0.242, so D 4.0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the 69

13 Problem 9.20 The masses of the boes are m D 5 kg and m D 60 kg. The coefficient of static friction between boes and and between bo and the inclined surface is 0.2. What is the largest force F for which the boes will not slip? F 20 Solution: We have 6 unknowns, 4 equilibrium equations and 2 friction equations. F% : T F 47.5 N sin 20 C f 2 D N T F- : N N cos 20 D 0 F% : T N sin 20 f f 2 D 0 F f 2 T F- : N N N cos 20 D 0 f D 0.2N,f 2 D 0.2N 2 Solving we find F D 267 N 20 N N f N Problem 9.2 In Problem 9.20, what is the smallest force F for which the boes will not slip? Solution: See the solution for 9.20 change the directions of all of the friction forces. F% : T F 47.5 N sin 20 f 2 D 0 F- : N N cos 20 D 0 F% : T N sin 20 C f C f 2 D 0 ) F D 34.8 N F- : N N N cos 20 D 0 f D 0.2N,f 2 D 0.2N 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the 697

14 Problem 0.6 Determine the internal forces and moment at for each loading. 2 m 8 kn m 4 m (a) 2 kn/m m 4 m (b) Solution: (a) Denote the reaction at the pinned left end b R, and the reaction at the roller support b.the reaction at : (a) R = 4 kn M P M D 2 8 C 4 D 0, m V from which D 4kN. The reaction at R : (b) 2 kn/m M P F D R 8 C D 0, 4 kn V from which R D 4kN. F D R D 0. Make acut at :Isolate the left hand part. The sum ofmoments: M D M 4 D 0, from which M D R D 3kNm. V D R 2 d D 4 2 D 2kN 0 P D 0. from which M D 4kN-m V D 4kN P D 0 (b) Determine the reaction at :The sum ofthe moments about R : 4 M R D 2 d C 4 D 0, 0 from which ( D )[2 2 ] 4 D D 4kN. The reaction at R : 4 F D R 2 d C D 0, 0 from which R D 8 4 D 4kN, F D R D 0. Make acut at :Isolate the left hand part. The sum ofmoments: M D M R C 2 d D 0, 0 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

15 Problem 0.3 Determine the internal forces and moment at. 200 lb/ ft 300 lb/ ft 6 ft 8 ft 4 ft Solution: Use the whole bod to find the reactions M C : 8 ft C 600 lb 4 ft C 400 lb ft 600 lb. 33 ft D 0 ) D 833 lb 400 lb 600 lb 600 lb C Now eamine the section to the left of the cut F : P D 0 F : 200 lb 225 lb V D 0 M : 6 ft C 200 lb 3 ft C 225 lb 2 ft C M D 0 Solving P D 0, V D 592 lb, M D 950 ft-lb 225 lb 200 lb 275 lb/ ft 200 lb/ ft M 6 ft P V c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

16 Problem.20 If the load on the car jack is L D 6. 5kN, what is the tension in the threaded shaft between and? L 65 mm 65 mm 20 mm Solution: We have the constraint ( ) 2 ( ) 2 C D b 2 ) υ C υ D b υ D ( ) υ From virtual work we know see ( υu D Pυ Lυ D P C L ) υ D 0 P D L ( ) 240 mm D 6. 5kN D 2. 0kN 30 mm c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

Problem.22 This device raises aload W b etending the hdraulic actuator DE. The bars D and C are each 2mlong, and the distances b D.

b W h C D E Solution: Perform a virtual vertical displacement of the load. Denote the distance CD b.

The distances are related: h D p L 2 2,where L is the length of bar C, from which υh υ D p L 2 D 2 h. Thus D D W p L h D W 2 h 2.

17 Problem.22 This device raises aload W b etending the hdraulic actuator DE. The bars D and C are each 2mlong, and the distances b D. 4mand h D 0. 8m.If W D 4kN, what force must the actuator eert tohold the load in equilibrium? b W h C D E Solution: Perform a virtual vertical displacement of the load. Denote the distance CD b.the virtual work is υu D Wυh C Dυ D 0, from which D D W υh υ. The distances are related: h D p L 2 2,where L is the length of bar C, from which υh υ D p L 2 D 2 h. Thus D D W p L h D W 2 h 2. h Substitute numerical values: p D D 4 D D 9. 7 kn, 0. 8 where the negative sign implies that the force is directed parallel to the negative ais. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

0.3 m. 0.4 m. 0.3 m. A 2 kn. 0.4 m. tan γ = 7. (BC = kn) γ = Fx = BC cos θ + AC cos γ =0

0.3 m. 0.4 m. 0.3 m. A 2 kn. 0.4 m. tan γ = 7. (BC = kn) γ = Fx = BC cos θ + AC cos γ =0 Problem 6.4 etermine the aial forces in the members of the truss. kn 0.3 m 0.4 m 0.6 m 1. m Solution: irst, solve for the support reactions at and, and then use the method of joints to solve for the forces