1 / 62

CE 102 Statics

CE 102 Statics. Chapter 4 Equilibrium of Rigid Bodies. Contents. Introduction Free-Body Diagram Reactions at Supports and Connections for a Two-Dimensional Structur e Equilibrium of a Rigid Body in Two Dimensions Statically Indeterminate Reactions Sample Problem 4.1 Sample Problem 4.2

giselle
Download Presentation

CE 102 Statics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CE 102 Statics Chapter 4 Equilibrium of Rigid Bodies

  2. Contents Introduction Free-Body Diagram Reactions at Supports and Connections for a Two-Dimensional Structure Equilibrium of a Rigid Body in Two Dimensions Statically Indeterminate Reactions Sample Problem 4.1 Sample Problem 4.2 Sample Problem 4.3 Equilibrium of a Two-Force Body Equilibrium of a Three-Force Body Sample Problem 4.4 Equilibrium of a Rigid Body in Three Dimensions Reactions at Supports and Connections for a Three-Dimensional Structure Sample Problem 4.5

  3. The necessary and sufficient condition for the static equilibrium of a body are that the resultant force and couple from all external forces form a system equivalent to zero, • Resolving each force and moment into its rectangular components leads to 6 scalar equations which also express the conditions for static equilibrium, Introduction • For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body.

  4. First step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-body diagram. • Select the extent of the free-body and detach it from the ground and all other bodies. Free-Body Diagram • Indicate point of application, magnitude, and direction of external forces, including the rigid body weight. • Indicate point of application and assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body. • Include the dimensions necessary to compute the moments of the forces.

  5. Reactions at Supports and Connections for a Two-Dimensional Structure • Reactions equivalent to a force with known line of action.

  6. Reactions equivalent to a force of unknown direction and magnitude. • Reactions equivalent to a force of unknown direction and magnitude and a couple.of unknown magnitude Reactions at Supports and Connections for a Two-Dimensional Structure

  7. For all forces and moments acting on a two-dimensional structure, • Equations of equilibrium become where A is any point in the plane of the structure. • The 3 equations can not be augmented with additional equations, but they can be replaced Equilibrium of a Rigid Body in Two Dimensions • The 3 equations can be solved for no more than 3 unknowns.

  8. Fewer unknowns than equations, partially constrained • More unknowns than equations • Equal number unknowns and equations but improperly constrained Statically Indeterminate Reactions

  9. Sample Problem 4.1 • SOLUTION: • Create a free-body diagram for the crane. • Determine B by solving the equation for the sum of the moments of all forces about A. Note there will be no contribution from the unknown reactions at A. • Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components. A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B. • Check the values obtained for the reactions by verifying that the sum of the moments about B of all forces is zero.

  10. Determine B by solving the equation for the sum of the moments of all forces about A. • Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces. Sample Problem 4.1 • Create the free-body diagram. • Check the values obtained.

  11. Sample Problem 4.2 • SOLUTION: • Create a free-body diagram for the car with the coordinate system aligned with the track. • Determine the reactions at the wheels by solving equations for the sum of moments about points above each axle. • Determine the cable tension by solving the equation for the sum of force components parallel to the track. A loading car is at rest on an inclined track. The gross weight of the car and its load is 5500 lb, and it is applied at at G. The cart is held in position by the cable. Determine the tension in the cable and the reaction at each pair of wheels. • Check the values obtained by verifying that the sum of force components perpendicular to the track are zero.

  12. Determine the reactions at the wheels. • Determine the cable tension. Sample Problem 4.2 • Create a free-body diagram

  13. Sample Problem 4.3 • SOLUTION: • Create a free-body diagram for the frame and cable. • Solve 3 equilibrium equations for the reaction force components and couple at E. The frame supports part of the roof of a small building. The tension in the cable is 150 kN. Determine the reaction at the fixed end E.

  14. Solve 3 equilibrium equations for the reaction force components and couple. Sample Problem 4.3 • Create a free-body diagram for the frame and cable.

  15. Consider a plate subjected to two forces F1 and F2 • For static equilibrium, the sum of moments about A must be zero. The moment of F2 must be zero. It follows that the line of action of F2 must pass through A. • Similarly, the line of action of F1 must pass through B for the sum of moments about B to be zero. Equilibrium of a Two-Force Body • Requiring that the sum of forces in any direction be zero leads to the conclusion that F1 and F2 must have equal magnitude but opposite sense.

  16. Consider a rigid body subjected to forces acting at only 3 points. • Assuming that their lines of action intersect, the moment of F1 and F2 about the point of intersection represented by D is zero. • Since the rigid body is in equilibrium, the sum of the moments of F1, F2, and F3 about any axis must be zero. It follows that the moment of F3 about D must be zero as well and that the line of action of F3 must pass through D. Equilibrium of a Three-Force Body • The lines of action of the three forces must be concurrent or parallel.

  17. Sample Problem 4.4 • SOLUTION: • Create a free-body diagram of the joist. Note that the joist is a 3 force body acted upon by the rope, its weight, and the reaction at A. • The three forces must be concurrent for static equilibrium. Therefore, the reaction R must pass through the intersection of the lines of action of the weight and rope forces. Determine the direction of the reaction force R. A man raises a 10 kg joist, of length 4 m, by pulling on a rope. Find the tension in the rope and the reaction at A. • Utilize a force triangle to determine the magnitude of the reaction force R.

  18. Create a free-body diagram of the joist. • Determine the direction of the reaction force R. ( ) = = = AF AB cos 45 4 m cos 45 2 . 828 m 1 = = = CD AE AF 1 . 414 m 2 ( ) = + = = BD CD cot( 45 25 ) 1 . 414 m tan 20 0 . 515 m ( ) = - = - = CE BF BD 2 . 828 0 . 515 m 2.313 m CE 2 . 313 a = = = tan 1 . 636 AE 1 . 414 Sample Problem 4.4

  19. Sample Problem 4.4 • Determine the magnitude of the reaction force R.

  20. Six scalar equations are required to express the conditions for the equilibrium of a rigid body in the general three dimensional case. • The scalar equations are conveniently obtained by applying the vector forms of the conditions for equilibrium, Equilibrium of a Rigid Body in Three Dimensions • These equations can be solved for no more than 6 unknowns which generally represent reactions at supports or connections.

  21. Reactions at Supports and Connections for a Three-Dimensional Structure

  22. Reactions at Supports and Connections for a Three-Dimensional Structure

  23. Sample Problem 4.5 • SOLUTION: • Create a free-body diagram for the sign. • Apply the conditions for static equilibrium to develop equations for the unknown reactions. A sign of uniform density weighs 270 lb and is supported by a ball-and-socket joint at A and by two cables. Determine the tension in each cable and the reaction at A.

  24. Sample Problem 4.5 • Create a free-body diagram for the sign. • Since there are only 5 unknowns, the sign is partially constrain. It is free to rotate about the x axis. It is, however, in equilibrium for the given loading.

  25. Sample Problem 4.5 • Apply the conditions for static equilibrium to develop equations for the unknown reactions. Solve the 5 equations for the 5 unknowns,

  26. The semicircular rod ABCD is maintained in equilibrium by the small wheel at D and the rollers at B and C. Knowing that a = 45o, determine the reactions at B , C , and D. a A O D 45o 45o C B Problem 4.6 P

  27. Problem 4.6 Solving Problems on Your Own The semicircular rod ABCD is maintained in equilibrium by the small wheel at D and the rollers at B and C. Knowing that a = 45o, determine the reactions at B , C , and D. P a A O D 45o 45o C B 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.

  28. Problem 4.6 Solving Problems on Your Own a A O D The semicircular rod ABCD is maintained in equilibrium by the small wheel at D and the rollers at B and C. Knowing that a = 45o, determine the reactions at B , C , and D. 45o 45o C B P 2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be: SFx = 0 SFy = 0 SMO = 0 where O is an arbitrary point in the plane of the structure or SFx = 0 SMA = 0 SMB = 0 where point B is such that line AB is not parallel to the y axis orSMA = 0 SMB = 0 SMC = 0 where the points A, B , and C do not lie in a straight line.

  29. Problem 4.6 Solution a A O D 45o 45o C B P sina D A O D P cosa 45o 45o C B R C/ 2 B/ 2 B C B/ 2 C/ 2 P Draw a free-body diagram of the body. P

  30. P sina D A O D P cosa 45o 45o C B R C/ 2 B/ 2 Problem 4.6 Solution B C B/ 2 C/ 2 P Write three equilibrium equations and solve for the unknowns. + S MO = 0: (P sina) R_D (R) = 0 D = P sina (1) + S Fx = 0: P cosa + B/ 2 _C / 2 = 0 (2) + S Fy = 0: _P sina + B/ 2 + C / 2 _P sina = 0 _2P sina + B/ 2 + C / 2 = 0 (3)

  31. P sina Problem 4.6 Solution D A O D P cosa 45o 45o C B R C/ 2 B/ 2 B C B/ 2 C/ 2 2 2 2 2 P (2) + (3) P(cosa_ 2sina) + 2 B/ 2 = 0 B = (2sina_ cosa) P (4) (2) _ (3) P(cosa + 2sina) _ 2 C/ 2 = 0 C = (2sina + cosa) P (5)

  32. P sina Problem 4.6 Solution D A O D P cosa 45o 45o For a = 45o sina = cosa = 1/ 2 C B R C/ 2 B/ 2 B C B/ 2 C/ 2 1 2 2 1 1 2 2 2 2 2 2 1 2 3 3 2 2 2 2 2 P EQ. (4) : B = ( _ ) P = P ; B = P 45o EQ. (5) : C = ( _ ) P = P ; C = P 45o EQ. (1) : D = P/ 2 D = P/ 2

  33. Problem 4.7 4 in 4 in 20 lb 40 lb The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. A B 2 in C 3 in D 3 in E q

  34. Problem 4.7 4 in 4 in Solving Problems on Your Own The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. 20 lb 40 lb A B 2 in C 3 in D 3 in E q 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.

  35. Problem 4.7 4 in 4 in Solving Problems on Your Own The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. 20 lb 40 lb A B 2 in C 3 in D 3 in E q 2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be: SFx = 0 SFy = 0 SMO = 0 where O is an arbitrary point in the plane of the structure or SFx = 0 SMA = 0 SMB = 0 where point B is such that line AB is not parallel to the y axis orSMA = 0 SMB = 0 SMC = 0 where the points A, B , and C do not lie in a straight line.

  36. 4 in 4 in Problem 4.7 Solution 20 lb 40 lb A B Draw a free-body diagram of the body. 2 in C 3 in D 4 in 4 in 3 in E 20 lb 40 lb q A B 2 in C C 3 in D D 3 in E E 30o

  37. 4 in 4 in Problem 4.7 Solution 60 lb cos 30o 20 lb 40 lb A B 2 in C Write equilibrium equations and solve for the unknowns. C 3 in D D 3 in E E 30o + S Fy = 0: E cos 30o_ 20 _ 40 = 0 E = = 69.28 lb E = 69.3 lb 60o

  38. 4 in 4 in Problem 4.7 Solution + S MD = 0: (20 lb)( 4 in) _ ( 40 lb)( 4 in) _C ( 3 in) + E sin 30o ( 3 in) = 0 _ 80 _ 3C + 69.28 ( 0.5 )( 3 ) = 0 C = 7.974 lb C = 7.97 lb 20 lb 40 lb A B 2 in C C 3 in D D 3 in E E 30o

  39. 4 in 4 in Problem 4.7 Solution 20 lb 40 lb A B 2 in C C + S Fx= 0: E sin 30o + C_D = 0 ( 69.28 lb )( 0.5 ) + 7.974 lb _D = 0 D = 42.6 lb 3 in D D 3 in E E 30o

  40. y 1.2 m E 1.2 m D C x 1.5 m A B 5 kN z f 1 m 2 m Problem 4.8 A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the line of action of the 5-kN force forms an angle f=30o with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A.

  41. Problem 4.8 y 1.2 m E 1.2 m D C x 1.5 m A B 5 kN z f 1 m 2 m Solving Problems on Your Own A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the line of action of the 5-kN force forms an angle f=30o with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A. 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.

  42. y Problem 4.8 1.2 m E 1.2 m D C x 1.5 m A B 5 kN z f 1 m 2 m Solving Problems on Your Own A 3-m pole is supported by a ball- and-socket joint at A and by the cables CD and CE. Knowing that the line of actionof the 5-kN force forms an angle f=30o with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A. 2. Write equilibrium equations and solve for the unknowns. For three-dimensional body the six scalar equations SFx = 0 SFy = 0 SFz = 0 SMx = 0 SMy = 0 SMz = 0 should be used and solved for six unknowns. These equations can also be written as SF = 0 SMO = S (rxF ) = 0 where F are the forces and r are position vectors.

  43. Problem 4.8 Solution y 1.2 m E 1.2 m D C x 1.5 m A B y 1.2 m 5 kN z f E 1.2 m 1 m 2 m D TCE TCD C Ax i x 1.5 m B A 30o Az k Ay j z 5 kN 1 m 2 m Draw a free-body diagram of the body.

  44. y Problem 4.8 Solution 1.2 m E 1.2 m D TCE TCD C Ax i x 1.5 m B A 30o Az k rB/A = 2 i rC/A = 3 i Ay j z 5 kN 1 m 2 m TCD CD CD 3.562 TCD CE CE 3.562 Write equilibrium equations and solve for the unknowns. 5 unknowns and 6 equations of equilibrium, but equilibrium is maintained, S MAC = 0 . Load at B, FB = _ ( 5 cos 30o ) j + ( 5 sin 30o ) k = _ 4.33 j + 2.5 k CD = _ 3 i+ 1.5 j + 1.2 kCD = 3.562 m TCD = TCD = (_ 3 i + 1.5 j + 1.2 k) TCE = TCE = (_ 3 i + 1.5 j_ 1.2 k)

  45. y Problem 4.8 Solution 1.2 m E 1.2 m D TCE TCD C Ax i x 1.5 m B A 30o Az k Ay j z 5 kN 1 m 2 m TCD TCE 3.562 3.562 SMA = 0: rC/A x TCD + rC/A x TCE + rB/A x FB = 0 i j k 3 0 0 _3 1.5 1.2 i j k 3 0 0 _3 1.5 _1.2 i j k 2 0 0 0 _4.33 2.5 + + = 0

  46. y Problem 4.8 Solution 1.2 m E 1.2 m TCD TCE D TCE j: _3.6 + 3.6 _ 5 = 0 _3.6 TCD+3.6 TCE_17.81 = 0 (1) TCD 3.562 3.562 C Ax i x 1.5 m B A 30o Az k TCD TCE Ay j z 5 kN k: 4.5 + 4.5 _ 8.66 = 0 4.5 TCD+4.5 TCE = 30.85 (2) 3.562 3.562 1 m 2 m Equate coefficients of unit vectors to zero. (2) + 1.25 (1): 9TCE_ 53.11 = 0 ; TCE = 5.90 kN Eq. (1): _3.6TCD + 3.6 (5.902) _ 17.81 = 0 TCD = 0.954 kN

  47. y Problem 4.8 Solution 1.2 m E 1.2 m 0.954 5.902 D i: Ax + (_3) + (_3) = 0 Ax = 5.77 kN TCE 3.562 3.562 TCD C Ax i x 1.5 m B A 30o Az k Ay j z 5 kN 1 m 2 m 0.954 5.902 j: Ay + (1.5) + (1.5) _ 4.33 = 0 Ay = 1.443 kN 3.562 3.562 0.954 5.902 3.562 3.562 SF = 0: A + TCD + TCE + FB = 0 k: Az + (1.2) + (_1.2) + 2.5 = 0 Az = _ 0.833 kN A = ( 5.77 kN) i + ( 1.443 kN ) j - ( 0.833 kN ) k

  48. A a 20 in B C 60 lb 10 in Problem 4.9 Rod AC is supported by a pin and bracket at A and rests against a peg at B. Neglecting the effect of friction, determine (a) the reactions at A and B when a = 8 in., (b) the distance a for which the reaction at A is horizontal and the corresponding magnitudes of the reactions at A and B.

  49. Problem 4.9 A a 20 in B C 60 lb 10 in Solving Problems on Your Own Rod AC is supported by a pin and bracket at A and rests against a peg at B. Neglecting the effect of friction, determine (a) the reactions at A and B when a = 8 in., (b) the distance a for which the reaction at A is horizontal and the corresponding magnitudes of the reactions at A and B. 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.

  50. Problem 4.9 A a 20 in B C 60 lb 10 in Solving Problems on Your Own Rod AC is supported by a pin and bracket at A and rests against a peg at B. Neglecting the effect of friction, determine (a) the reactions at A and B when a = 8 in., (b) the distance a for which the reaction at A is horizontal and the corresponding magnitudes of the reactions at A and B. 2. For a three-force body, solution can be obtained by constructing a force triangle. The resultants of the three forces must be concurrent or parallel. To solve a problem involving a three-force body with concurrent forces, draw the free-body diagram showing that the three forces pass through the same point. Complete the solution by using a force triangle.

More Related