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CE 102 Statics

CE 102 Statics. Chapter 6 Analysis of Structures. Contents. Introduction Definition of a Truss Simple Trusses Analysis of Trusses by the Method of Joints Joints Under Special Loading Conditions Space Trusses Sample Problem 6.1 Analysis of Trusses by the Method of Sections.

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CE 102 Statics

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  1. CE 102 Statics Chapter 6 Analysis of Structures

  2. Contents Introduction Definition of a Truss Simple Trusses Analysis of Trusses by the Method of Joints Joints Under Special Loading Conditions Space Trusses Sample Problem 6.1 Analysis of Trusses by the Method of Sections Trusses Made of Several Simple Trusses Sample Problem 6.2 Analysis of Frames Frames Which Cease to be Rigid When Detached From Their Supports Sample Problem 6.3 Machines

  3. Introduction • For the equilibrium of structures made of several connected parts, the internal forces as well the external forces are considered. • In the interaction between connected parts, Newton’s 3rd Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense. • Three categories of engineering structures are considered: • Frames: contain at least one one multi-force member, i.e., member acted upon by 3 or more forces. • Trusses: formed from two-force members, i.e., straight members with end point connections • Machines: structures containing moving parts designed to transmit and modify forces.

  4. Definition of a Truss • A truss consists of straight members connected at joints. No member is continuous through a joint. • Most structures are made of several trusses joined together to form a space framework. Each truss carries those loads which act in its plane and may be treated as a two-dimensional structure. • Bolted or welded connections are assumed to be pinned together. Forces acting at the member ends reduce to a single force and no couple. Only two-force members are considered. • When forces tend to pull the member apart, it is in tension. When the forces tend to compress the member, it is in compression.

  5. Definition of a Truss Members of a truss are slender and not capable of supporting large lateral loads. Loads must be applied at the joints.

  6. Definition of a Truss

  7. A rigid truss will not collapse under the application of a load. • A simple truss is constructed by successively adding two members and one connection to the basic triangular truss. Simple Trusses • In a simple truss, m = 2n - 3 where m is the total number of members and n is the number of joints.

  8. Analysis of Trusses by the Method of Joints • Dismember the truss and create a freebody diagram for each member and pin. • The two forces exerted on each member are equal, have the same line of action, and opposite sense. • Forces exerted by a member on the pins or joints at its ends are directed along the member and equal and opposite. • Conditions of equilibrium on the pins provide 2n equations for 2n unknowns. For a simple truss, 2n = m + 3. May solve for m member forces and 3 reaction forces at the supports. • Conditions for equilibrium for the entire truss provide 3 additional equations which are not independent of the pin equations.

  9. Forces in opposite members intersecting in two straight lines at a joint are equal. • The forces in two opposite members are equal when a load is aligned with a third member. The third member force is equal to the load (including zero load). • The forces in two members connected at a joint are equal if the members are aligned and zero otherwise. • Recognition of joints under special loading conditions simplifies a truss analysis. Joints Under Special Loading Conditions

  10. Space Trusses • An elementary space truss consists of 6 members connected at 4 joints to form a tetrahedron. • A simple space truss is formed and can be extended when 3 new members and 1 joint are added at the same time. • In a simple space truss, m = 3n - 6 where m is the number of members and n is the number of joints. • Conditions of equilibrium for the joints provide 3n equations. For a simple truss, 3n = m + 6 and the equations can be solved for m member forces and 6 support reactions. • Equilibrium for the entire truss provides 6 additional equations which are not independent of the joint equations.

  11. Sample Problem 6.1 • SOLUTION: • Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C. • Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements. • In succession, determine unknown member forces at joints D, B, and E from joint equilibrium requirements. Using the method of joints, determine the force in each member of the truss. • All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results.

  12. Sample Problem 6.1 • SOLUTION: • Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C.

  13. Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements. • There are now only two unknown member forces at joint D. Sample Problem 6.1

  14. There are now only two unknown member forces at joint B. Assume both are in tension. • There is one unknown member force at joint E. Assume the member is in tension. Sample Problem 6.1

  15. Sample Problem 6.1 • All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results.

  16. To determine the force in member BD, pass a section through the truss as shown and create a free body diagram for the left side. Analysis of Trusses by the Method of Sections • When the force in only one member or the forces in a very few members are desired, the method of sections works well. • With only three members cut by the section, the equations for static equilibrium may be applied to determine the unknown member forces, including FBD.

  17. Compound trusses are statically determinant, rigid, and completely constrained. • Truss contains a redundant member and is statically indeterminate. • Additional reaction forces may be necessary for a rigid truss. • Necessary but insufficient condition for a compound truss to be statically determinant, rigid, and completely constrained, non-rigid rigid Trusses Made of Several Simple Trusses

  18. Sample Problem 6.2 • SOLUTION: • Take the entire truss as a free body. Apply the conditions for static equilib-rium to solve for the reactions at A and L. • Pass a section through members FH, GH, and GI and take the right-hand section as a free body. • Apply the conditions for static equilibrium to determine the desired member forces. Determine the force in members FH, GH, and GI.

  19. Sample Problem 6.2 • SOLUTION: • Take the entire truss as a free body. Apply the conditions for static equilib-rium to solve for the reactions at A and L.

  20. Pass a section through members FH, GH, and GI and take the right-hand section as a free body. • Apply the conditions for static equilibrium to determine the desired member forces. Sample Problem 6.2

  21. Sample Problem 6.2

  22. Frames and machines are structures with at least one multiforce member. Frames are designed to support loads and are usually stationary. Machines contain moving parts and are designed to transmit and modify forces. • A free body diagram of the complete frame is used to determine the external forces acting on the frame. • Internal forces are determined by dismembering the frame and creating free-body diagrams for each component. Analysis of Frames • Forces on two force members have known lines of action but unknown magnitude and sense. • Forces on multiforce members have unknown magnitude and line of action. They must be represented with two unknown components. • Forces between connected components are equal, have the same line of action, and opposite sense.

  23. Some frames may collapse if removed from their supports. Such frames can not be treated as rigid bodies. • A free-body diagram of the complete frame indicates four unknown force components which can not be determined from the three equilibrium conditions. • The frame must be considered as two distinct, but related, rigid bodies. Frames Which Cease To Be Rigid When Detached From Their Supports • With equal and opposite reactions at the contact point between members, the two free-body diagrams indicate 6 unknown force components. • Equilibrium requirements for the two rigid bodies yield 6 independent equations.

  24. Sample Problem 6.3 • SOLUTION: • Create a free-body diagram for the complete frame and solve for the support reactions. • Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown magnitude. It is determined by summing moments about C. Members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the force in link DE and the components of the force exerted at C on member BCD. • With the force on the link DE known, the sum of forces in the x and y directions may be used to find the force components at C. • With member ACE as a free-body, check the solution by summing moments about A.

  25. Sample Problem 6.3 • SOLUTION: • Create a free-body diagram for the complete frame and solve for the support reactions. Note:

  26. Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown magnitude. It is determined by summing moments about C. • Sum of forces in the x and y directions may be used to find the force components at C. Sample Problem 6.3

  27. Sample Problem 6.3 • With member ACE as a free-body, check the solution by summing moments about A. (checks)

  28. Machines are structures designed to transmit and modify forces. Their main purpose is to transform input forces into output forces. • Create a free-body diagram of the complete machine, including the reaction that the wire exerts. • The machine is a nonrigid structure. Use one of the components as a free-body. • Taking moments about A, Machines • Given the magnitude of P, determine the magnitude of Q.

  29. Problem 6.4 15 in 15 in 15 in The pin at B is attached to member ABCD and can slide along a slot cut in member BE. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium. 80 lb 40 lb A B C D 36 in M E

  30. Problem 6.4 Solving Problems on Your Own 15 in 15 in 15 in The pin at B is attached to member ABCD and can slide along a slot cut in member BE. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium. 80 lb 40 lb A B C D 36 in For this problem we note that there are no two-force members. In solving this problem, we M E 1. Dismember the frame, and draw a free-body diagram for each member. 2. To simplify the solution, seek a way to write an equation involving a single unknown.

  31. 13 39 12 36 5 15 + Problem 6.4 Solution B 13 Dismember the frame, and draw a free-body diagram for each member. 5 12 Free-Body: Member BE BE = (152 + 362)1/2 = 39 in 36 in M Ex E Ey To simplify the solution, seek a way to write an equation involving a single unknown. 15 in S ME = 0: M - B (39 in) = 0 M = B (39 in) (1)

  32. + Problem 6.4 Solution Dismember the frame, and draw a free-body diagram for each member. 15 in 15 in 15 in Free-Body: Member BE 80 lb 40 lb Dx To simplify the solution, seek a way to write an equation involving a single unknown. A C D 13 5 B Dy 12 5 13 S MD = 0: (40 lb)(45 in) + (80 lb)(15 in) - B(30 in) = 0 B = 260 lb

  33. Problem 6.4 Solution B 13 15 in 15 in 15 in 5 80 lb 12 40 lb Dx A C D 13 36 in 5 B Dy 12 M B = 260 lb From EQ (1) Ex E M = B (39 in) = (260 lb) (39 in) = 10,140 lb-in Ey 15 in M = B (39 in) M = 10.14 kip-in

  34. Problem 6.5 For the frame shown and neglecting the effect of friction at the horizontal and vertical surfaces, determine the forces exerted at B and C on member BCE. 50 lb 6 in 12 in 6 in E 4 in D 2 in C 6 in A B

  35. Problem 6.5 Solving Problems on Your Own 50 lb 6 in For the frame shown and neglecting the effect of friction at the horizontal and vertical surfaces, determine the forces exerted at B and C on member BCE. 12 in 6 in E 4 in D 2 in C 6 in A B For this problem we note that there are no two-force members. In solving this problem, we 1. Dismember the frame, and draw a free-body diagram for each member. 2. To simplify the solution, seek a way to write an equation involving a single unknown.

  36. Problem 6.5 Solution Dismember the frame, and draw a free-body diagram for each member. 50 lb Cy Cy D D H C 2 in Cx Cx 6 in A A Bx 18 in By 6 in 6 in

  37. + Problem 6.5 Solution To simplify the solution, seek a way to write an equation involving a single unknown. 50 lb Cy Cy D D H C 2 in Cx Cx 6 in A A Bx 18 in 6 in 6 in By Free-Body: Member ACD S MH = 0: Cx(2 in) - Cy(18 in) = 0 Cx = 9Cy

  38. + Problem 6.5 Solution To simplify the solution, seek a way to write an equation involving a single unknown. 50 lb Cy Cy D D H C 2 in Cx = 9Cy Cx Cx 6 in A A Bx 18 in 6 in 6 in By Free-Body: Member BCE S MB = 0: Cx(6 in) + Cy(6 in) - (50 lb)(12 in) = 0 9Cy(6 in) + Cy(6 in) - 600 = 0 Substitute Cx = 9Cy: Cy = + 10 lb; Cx = 9Cy = 9(10) = 90 lb

  39. Problem 6.5 Solution Free-Body: Member BCE C 50 lb C = 90.6 lb 10 lb Cy 6.3o C C 90 lb Cx + S Fx = 0: Bx - 90 lb = 0 Bx = 90 lb S Fy = 0: By + 10 lb - 50 lb = 0 By = 40 lb Bx + 6 in 6 in By B 24.0o 40 lb B = 98.5 lb B 90 lb

  40. Problem 6.6 12.5 kN 12.5 kN 12.5 kN 12.5 kN Using the method of joints, determine the force in each member of the truss shown. 2 m 2 m 2 m A C B D 2.5 m G F E

  41. Problem 6.6 Solving Problems on Your Own 12.5 kN 12.5 kN Using the method of joints, determine the force in each member of the truss shown. 12.5 kN 12.5 kN 2 m 2 m 2 m A C B D 1. Draw the free-body diagram of the entire truss, and use this diagram to determine the reactions at the supports. 2.5 m G F E 2. Locate a joint connecting only two members, and draw the free-body diagram for its pin. Use this free-body diagram to determine the unknown forces in each of the two members. Assuming all members are in tension, if the answer obtained from SFx = 0 and SFy = 0 is positive, the member is in tension. A negative answer means the member is in compression.

  42. Problem 6.6 Solving Problems on Your Own 12.5 kN 12.5 kN Using the method of joints, determine the force in each member of the truss shown. 12.5 kN 12.5 kN 2 m 2 m 2 m A C B D 2.5 m G F E 3. Next, locate a joint where the forces in only two of the connected members are still unknown. Draw the free-body diagram of the pin and use it as indicated in step 2 to determine the two unknown forces. 4. Repeat this procedure until the forces in all the members of the truss have been determined.

  43. + Problem 6.6 Solution 12.5 kN 12.5 kN 12.5 kN 12.5 kN Draw the free-body diagram of the entire truss, and use it to determine reactions at the supports. 2 m 2 m 2 m A Ax C B D Ay 2.5 m G F E E S MA = 0: E(2.5 m) - (12.5 kN)(2 m) - (12.5 kN)(4 m) - (12.5 kN)(6 m) = 0 E = 60 kN S Fy = 0: Ay - (4)(12.5 kN) = 0 Ay = 50 kN S Fx = 0: Ax - E = 0 Ax= 60 kN + +

  44. 12.5 kN 12.5 kN 12.5 kN 12.5 kN 2 m 2 m 2 m 60 kN A B C D 50 kN 2.5 m G F 60 kN E 2.5 6.5 + Problem 6.6 Solution Locate a joint connecting only two members, and draw the free-body diagram for its pin. Use the free-body diagram to determine the unknown forces in each of the two members. Joint D 12.5 kN S Fy = 0: FGD - 12.5 kN = 0 FCD FGD = 32.5 kN C 6.5 2.5 6 6.5 6 S Fx = 0: FGD- FCD = 0 + FGD FCD = 30 kN T

  45. 12.5 kN 12.5 kN 12.5 kN 12.5 kN 2 m 2 m 2 m 60 kN A B C D 50 kN 2.5 m G F 60 kN E S F= 0: FCG = 0 Problem 6.6 Solution Next, locate a joint where the forces in only two of the connected members are still unknown. Draw the free-body diagram of the pin and use it to determine the two unknown forces. Joint G FCG 32.5 kN S F = 0: FFG - 32.5 kN = 0 FFG = 32.5 kN C FFG

  46. 12.5 kN 12.5 kN 12.5 kN 12.5 kN 2 m 2 m 2 m 60 kN A B C D 50 kN 2.5 m G F 60 kN E Problem 6.6 Solution Repeat this procedure until the forces in all the members of the truss have been determined. Joint C 2 3 BF = (2.5 m) = 1.6667 m 12.5 kN BF 2 b = BCF = tan-1 = 39.81o FBC FCD = 30 kN b S Fy = 0: - 12.5 kN - FCF sin b = 0 - 12.5 kN - FCF sin39.81o= 0 FCF = 19.53 kN C + FCF S Fx = 0: 30 kN - FCF cos b -FBC = 0 30 kN - (-19.53) cos39.81o-FBC = 0 FBC = 45.0 kN T +

  47. 12.5 kN 12.5 kN 12.5 kN 12.5 kN 2 m 2 m 2 m 60 kN A B C D 50 kN 2.5 m G F 60 kN E 6 6.5 6 6.5 6.5 6 2.5 6.5 2.5 6.5 2.5 6.5 Problem 6.6 Solution FCF = 19.53kN Joint F FBF b=39.81o 6.5 FFG = 32.5 kN FEF 2.5 6 + S Fx = 0: - FEF - (32.5 kN) - FCF cos b = 0 FEF = -32.5 kN - ( ) (19.53) cos39.81oFEF = 48.8 kN C S Fy = 0: FBF - FEF - (32.5 kN) - (19.53) sin b = 0 + FBF - (-48.8 kN)- 12.5 kN - 12.5 kN = 0 FBF = 6.25 kN T

  48. 12.5 kN 12.5 kN 12.5 kN 12.5 kN 2 m 2 m 2 m 60 kN A B C D 50 kN 2.5 m G F 60 kN E Problem 6.6 Solution 12.5 kN Joint B FAB FBC = 45.0 kN g FBF = 6.25kN FBE 2.5 m 2 m tan g = ; g = 51.34o S Fy = 0: -12.5 kN -6.25 kN - FBE sin 51.34o = 0 FBE = -24.0 kN + FBE = 24.0 kN C + S Fx = 0: 45.0 kN - FAB + (24.0 kN) cos 51.34o = 0 FAB = 60.0 kN FAB= 60.0 kN T

  49. 12.5 kN 12.5 kN 12.5 kN 12.5 kN 2 m 2 m 2 m 60 kN A B C D 50 kN 2.5 m G F 60 kN E 2.5 6.5 Problem 6.6 Solution Joint E FBE = 24 kN FAE g FEF = 48.75 kN 6.5 2.5 60 kN 6 g = 51.34o S Fy = 0: FAE - (24 kN) sin 51.34o - (48.75 kN) = 0 + FAE = 37.5 kN FAE = 37.5 kN T

  50. Problem 6.7 3 kN A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH , FI, and GI. 3 kN 3 kN F 3 kN 3 kN D H 6.75 m 1.5 kN 1.5 kN B J L A I C E G K 3 m 3 m 3 m 3 m 3 m 3 m

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