PROBLEMS. m s TAC. m = 60 kg/m, determine the tension in the two supporting cables and the reaction at D.

Size: px

Start display at page:

Download "PROBLEMS. m s TAC. m = 60 kg/m, determine the tension in the two supporting cables and the reaction at D."

Phillip Lyons
7 years ago
Views:

2 PRLEMS 1. he uniform I-beam has a mass of 60 kg per meter of its length. Determine the tension in the two supporting cables and the reaction at D. (3/64)

3 PRLEMS A( 500) m (5 23) m m = 60 kg/m determine the tension in the two supporting cables and the reaction at D. (523) m D (000) m G (300) m 2 3k A A k A G 2 3k W D k D Dk kg W 60 8m 9.81 m k k ( ) 2 m s M A 0 rg / AW rd / A D 0 i k i k D D D

4 PRLEMS A( 500) m (5 23) m m = 60 kg/m determine the tension in the two supporting cables and the reaction at D. (523) m D (000) m G (300) m A A k k A W G D F 0 A F 0 2(0.83) A A

5 PRLEMS 2. he light right-angle boom which supports the 400 kg clinder is supported b three cables and a ball and socket oint at attached to the vertical - surface. Determine the reactions at and the cable tensions. (3/73)

6 PRLEMS oom is supported b three cables and a ball and socket oint at. Determine the reactions at and the cable tensions. A( 002) m (1.502) m ( 110) m D E D i 2k i k 2 2k E k k i k W 400(9.81) 3924 ( ) (000) m W D(1.520) m E (1.500) m G (0.7502) m G D E

7 i k E E D D D (2775) PRLEMS D E W G i k i k i E D D D k i D E D D m G m E m D m m m m A (0.7502) (1.500) (1.520) (000) 110) ( (1.502) 002) ( / / / k k k i k i k i k r W r r M E D E D G A

8 PRLEMS oom is supported b three cables and a ball and socket oint at. Determine the reactions at and the cable tensions. F F D 0 F D E D E G W

9 PRLEMS 3. he uniform panel door has a mass of 30 kg and is prevented from opening b the strut which is a light two-force member whose upper end is secured under the door knob and whose lower end is attached to a rubber cup which does not slip on the floor. f the door hinges A and onl can support force in the vertical -direction. alculate the compression in the strut and the horiontal components of the forces supported b hinges A and when a horiontal force P = 200 is applied normal to the plane of the door as shown. (3/89)

10 PRLEMS of the door hinges A and onl can support force in the vertical - direction calculate the compression in the strut and the horiontal components of the forces supported b hinges A and (0 0 0) A ( ) ( ) G ( ) E (0 0 2) D ( ) H ( ) m 0.75 k ndh k 1.25 F F ndh F k A Ai A i k W 30(9.81) k 294.3k P 200 M M M 0 0.6i 0.8k 294.3k i 0.8k F k i k k A i A 0.48F 1.6A 360i 0.8F 1.6A F 240k P E D F G A W H A

11 PRLEMS of the door hinges A and onl can support force in the vertical -direction calculate the compression in the strut and the horiontal components of the forces supported b hinges A and k i 0.6F 0.48F 1.6A A 0.8F A F A A P E D F G A W A F F 0 0 A A F A H

12 PRLEMS 4. onsider the rudder assembl of a radio controlled model airplane. For the 15 position shown in the figure the net pressure acting on the left side of the rectangular rudder area is p = 4(10-5 ) /mm 2. Determine the required force P in the control rod DE and the horiontal components of the reactions at hinge which are parallel to the rudder surface. Assume the aerodnamic pressure to be uniform. (3/92)

13 PRLEMS P Pi A A i A p = 4(10-5 ) /mm 2 determine P and horiontal components of reaction at parallel to rudder surface. A k i k 2 Area of rudder A mm F pa410 F 0.15sin15i 0.15cos i top view of rudder 22 16sin15= F 15 H P H D D A cos15= (3840) (000) A(00 26) mm (0068) mm D( ) mm H (24cos15 24sin15 40) ( ) mm F H P A A A

14 PRLEMS PPi p = 4(10-5 ) /mm 2 determine P and horiontal components of reaction at parallel to rudder surface. A A i A A k i k F i M A M AnA M Ak 0 M A rh / AF r / ArD / AP i k i i k Pi i P Pk 0 M k 0 P0.202 (000) A(00 26) mm (0068) mm D( ) mm H ( ) mm 42k i k i A P F A A H A P

15 PRLEMS 5. Under the action of the 40 m torque applied to the vertical shaft the restraining cable limits the rotation of the arm A and attached shaft to an angle of 60 measured from the ais. he collar D fastened to the shaft prevents downward motion of the shaft in its bearing. alculate the bending moment M the compression P and the shear force V in the shaft at section. (ending moment epressed as a vector is normal to the shaft ais and shear force is also normal to the shaft ais.) (3/94)

16 PRLEMS i k collar D prevents downward motion of the shaft in its bearing calculate the bending moment M the compression P and the shear force V in the shaft at section. (ending moment epressed as a vector is normal to the shaft ais and shear force is also normal to the shaft ais.) A( 0.16sin cos600) ( ) m ( ) m ( ) m M 40k M 0 r / i M 40k M 0.53i k 0.09k i k M M i M M k 0 i M M

17 PRLEMS k i F F collar D prevents downward motion of the shaft in its bearing calculate the bending moment M the compression P and the shear force V in the shaft at section. (ending moment epressed as a vector is normal to the shaft ais and shear force is also normal to the shaft ais.) 0.53i k M m M m M M M V m M M F P320.19

18 PRLEMS 6. he shaft lever and handle are welded together and constitute a single rigid bod. heir combined mass is 28 kg with mass center at G. he assembl is mounted on bearings A and and rotation is prevented b link D. Determine the forces eerted on the shaft b the bearings A and while the 30 m couple is applied to the handle as shown. Would these forces change if the couple were applied to the shaft A rather than to the handle?

19 PRLEMS Determine the forces eerted on the shaft b the bearings A and while the 30 m couple is applied to the handle as shown. Would these forces change if the couple were applied to the shaft A rather than to the handle? A F A W

20 PRLEMS 7. he electric sander has a mass of 3 kg with mass center at G and is held in a slightl tilted position (-ais vertical) so that the sanding disk makes contact at its top with the surface being sanded. he sander is gripped b its handles at and. If the normal force R against the disk is maintained at 20 and is due entirel to the force component (i.e. = 0) and if the friction force F acting on the disk is 60 percent of R determine the components of the couple M which must be applied to the handle at to hold the sander in position. Assume that half of the weight is supported at. (3/95)

21 PRLEMS m =3 kg normal force R =20 due entirel to force component (i.e. = 0) the friction force F acting on the disk 60 percent of R determine the components of the couple M which must be applied to the handle at to hold the sander in position. Assume that half of the weight is supported at. 20 F M M i M 0 r / M k 0 A( ) m ( ) m (00300) m G ( 400 0) m R = 20 i k r W r Ri F 0 G/ G/ A W M F = 0.6R=12 W = k 40i i 14.72k 300k 40i 400k 120i 20i 12 M i M M k k M M

22 PRLEMS m =3 kg normal force R =20 due entirel to force component (i.e. = 0) the friction force F acting on the disk 60 percent of R determine the components of the couple M which must be applied to the handle at to hold the sander in position. Assume that half of the weight is supported at. M M i M M k 0 R = 20 F = 0.6R=12 W = M M M 1856 mm mm 2560 mm 12 M M M

23 PRLEMS 8. ecause of friction a ver large force is required to raise the 10 kg bod when the cable is wound around a fied rough clinder as shown. For the static equilibrium values shown determine all of the reactions at point. he 0.3 m dimension refers to the point A where the cable loses contact with the clinder. eglect the weight of the fied clinder.

24 PRLEMS determine all of the reactions at point neglect the weight of the fied clinder A mm A M M M W=10(9.81)=98.1 = k (0 0 0) A (30sin60 30cos60 800) = ( ) ( ) mm W sin 30 cos k 15 k cos 20 cos30cos 20 sin 20 cos 30sin k 8.89 k

25 PRLEMS i k? M M i M M k? M 0 r / W ra / M M k i k 15i k M M M M M determine all of the reactions at point neglect the weight of the fied clinder 0 M i M M k M i M M k kmm i k kmm 19.4 kmm 12.1 km km km km M km M M M W F F F k 8.89 k k 14.9i k k k

26 PRLEMS 9. he structure is subected to the loadings shown. Member A is supported b a ball and socket oint at A and a smooth collar at. Member D is supported b a pin at which acts like a thrust bearing. Determine the reactions at A and.

27 PRLEMS all and socket oint at A smooth collar at thrust bearing at. Determine reactions at A and.

28 PRLEMS 10. A circular plate with a weight of 900 acting at its center point G is supported b cord DE and a thrust bearing at. Shaft is fied and frictionless. a) Draw the Free od Diagram of the circular plate. b) Determine the cable tension and the reactions acting on the circular plate from the bearing at. G D G

29 PRLEMS W = 900 circular plate supported b cord DE and a thrust bearing at. Shaft is fied and frictionless. Determine the cable tension and the reactions at bearing. G D G

Structural Axial, Shear and Bending Moments

Structural Axial, Shear and Bending Moments Positive Internal Forces Acting Recall from mechanics of materials that the internal forces P (generic axial), V (shear) and M (moment) represent resultants