ME Statics. Structures. Chapter 4

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1 ME Statics Structures Chapter 4

2 Outline Applications Simple truss Method of joints Method of section

3

4 Germany

5 Tacoma Narrows Bridge UK

6 Applications Trusses are commonly used to support a roof. For a given truss geometry and load, how can we determine the forces in the truss members and select their sizes? A more challenging question is that for a given load, how can we design the trusses geometry to minimize cost?

7 Applications Trusses are also used in a variety of structures like cranes and the frames of aircraft or space stations. How can we design a light weight structure that will meet load, safety, and cost specifications?

8 Simple Trusses A truss is a structure composed of slender members joined together at their end points. Joint connections are formed by bolting or welding the ends of the members to a common plate, called a gusset plate, or by simply passing a large bolt or pin through each of the members

9 Simple Trusses Planar Trusses Planar trusses lie on a single plane and are used to support roofs and bridges The truss ABCD shows a typical roof-supporting truss Roof load is transmitted to the truss at joints by means of a series of purlins, such as DD The analysis of the forces developed in the truss members is 2D

10 Assumptions for Design When designing both the member and the joints of a truss, first it is necessary to determine the forces in each truss member. This is called the force analysis of a truss. When doing this, two assumptions are made: 1. All loadings are applied at the joint Assumption true for most applications of bridge and roof trusses Weight of the members neglected since forces supported by the members are large in comparison If member s weight is considered, apply it as a vertical force, half of the magnitude applied at each end of the member

11 Assumptions for Design 2. The members are joined together by smooth pins Assumption true when bolted or welded joints are used, provided the center lines of the joining members are concurrent

12 Assumptions for Design Each truss member acts as a two force member, therefore the forces at the ends must be directed along the axis of the member If the force tends to elongate the member, it is a tensile force If the force tends to shorten the member, it is a compressive force Important to state the nature of the force in the actual design of a truss tensile or compressive Compression members must be made thicker than tensile member to account for the buckling or column effect during compression

13 Simple Truss To prevent collapse, the form of a truss must be rigid The four bar shape ABCD will collapse unless a diagonal member AC is added for support The simplest form that is rigid or stable is a triangle A simple truss is constructed starting with a basic triangular element such as ABC and connecting two members (AD and BD) to form an additional element. For these trusses, the number of members (M) and the number of joints (J) are related by the equation M = 2 J 3.

14 Method of Joints Procedure for Analysis Truss analysis by the method of joints: 1) draw FBDs for all joints in the structure. 2) apply equations of equilibrium to each joint. 3) solve for unknowns.

15 Method of Joints Before beginning, it is usually necessary to draw a free-body diagram of the entire truss (i.e. treat the truss as a single object) & determine the reactions at its supports: E.g. consider the Warren truss which has members 2 m in length & support loads at B & D

16 Method of Joints From the equilibrium equations: Σ F x = A x = 0 Σ F y = A y + E 400 N 800 N = 0 Σ M point A = (1 m)(400 N) (3 m)(800 N) + (4 m)e = 0 We obtain the reactions: A x = 0, A y = 500 N & E = 700 N The next step is to choose a joint & draw its freebody diagram: Isolate joint A by cutting members AB & AC

17 Method of Joints

18 Method of Joints The terms T AB & T AC are the axial forces in members AB & AC, respectively Although the directions of the arrows representing the unknown axial forces can be chosen arbitrarily, notice that we have chosen them so that a member is in tension if we obtain a positive value for the axial force Consistently choosing the directions in this way helps avoid errors

19 Method of Joints The equilibrium equations for joint A are: Σ F x = T AC + T AB cos 60 = 0 Σ F y = T AB sin N = 0 Solving these equations, we obtain the axial force T AB = 577 N & T AC = 289 N Member AB is in compression & member AC is in tension

20 Method of Joints Next, obtain a free-body diagram of joint B by cutting members AB, BC & BD: From the equilibrium equations for joint B: Σ F x = T BD + T BC cos cos 60 = 0 Σ F y = 400 N sin 60 T BC sin 60 = 0 We obtain T BC = 115 N & T BD = 346 N Member BC is in tension & member BD is in compression

21 Method of Joints By continuing to draw free-body diagrams of the joints, we can determine the axial forces of all the members In 2 dimensions, you can obtain only 2 independent equilibrium equations from the freebody diagram of a joint Summing the moments about a point does not result in an additional independent equation because the forces are concurrent

22 Method of Joints Therefore when applying the method of joints, you should choose joints to analyze that are subjected to no more than 2 unknown forces: In our example, we analyzed joint A first because it was subjected to the known reaction exerted by the pin support & 2 unknown forces, the axial forces T AB & T AC We could then analyze joint B because it was subjected to 2 known forces & 2 unknown forces, T BC & T BD If we had attempted to analyze joint B first, there would have been 3 unknown forces

23 Method of Joints When determining the axial forces in the members of a truss, it will be simpler if you are familiar with 3 particular types of joints: 1.Truss joints with 2 collinear members & no load: the sum of the forces must equal zero, T 1 = T 2. The axial forces are equal.

24 Method of Joints 2.Truss joints with 2 noncollinear members & no load: because the sum of the forces in the x direction must equal zero, T 2 = 0. therefore T 1 must also equal zero. The axial forces are zero.

25 Method of Joints 3.Truss joints with 3 members, 2 of which are collinear & no load: because the sum of the forces in the x direction must equal zero, T 3 = 0. The sum of the forces in the y direction must equal zero, so T 1 = T 2. The axial forces in the collinear members are equal & the axial force in the 3 rd member is zero.

26 Zero force members A member that supports no force is called a zeroforce member. Analysis of trusses is simplified if we can easily identify (such as by inspection) any zero force members. zero force member not a zero force member

27 example: identify any zero force members. F D H B J A L C E G I K

28 example: identify any zero force members. F D H B J A L C E G I K

29 Example 1 Determine the axial forces in the members of the truss in Figure. Strategy 1 st, draw a free-body diagram of the entire truss, treating it as a single object & determine the reactions at the supports. Then apply the method of joints, simplifying the task by identifying any special joints

30 Example 1 Solution Determine the Reactions at the Supports: Draw the free-body diagram of the entire truss:

31 Example 1 Solution From the equilibrium equations: Σ F x = A x + B = 0 Σ F y = A y 2 kn = 0 Σ M point B = (6 m) A x (10 m)(2 kn) = 0 We obtain the reactions A x = 3.33 kn, A y = 2 kn & B = 3.33 kn.

32 Example 1 Solution Identify Special Joints: Because joint C has 3 members, 2 of which are collinear & no load, the axial force in member BC is zero, T BC = 0 & the axial forces in the collinear members AC & CD are equal, T AC = T CD. Draw Free-Body Diagrams of the Joints: We know the reaction exerted on joint A by the support & joint A is subjected to only 2 unknown forces, the axial forces in members AB & AC.

33 Solution Example 1

34 Example 1 Solution The angle = arctan (5/3) = 59.0 The equilibrium equations for joint A are: Σ F x = T AC sin 3.33 kn = 0 Σ F y = 2 kn T AB T AC cos = 0 Solving these equations, we obtain T AB = 0 & T AC = 3.89 kn.

35 Example 1 Solution Now draw the free-body diagram of joint B:

36 Example 1 Solution From the equilibrium equation: Σ F x = T BD kn = 0 We obtain T BD = 3.33 kn. The negative sign indicates that member BD is in compression. The axial forces in the members are: AB: 0 AC: 3.89 kn in tension (T) BC: 0 BD: 3.33 kn in compression (C) CD: 3.89 kn in tension (T)

37 Example 1 Critical Thinking Observe how our solution was simplified by recognizing that joint C is the type of special joint with 3 members, 2 of which are collinear & no load: This allowed us to determine the axial forces in all members of the truss by analyzing only 2 joints

38 Example 2 Each member of the truss in Figure will safely support a tensile force of 10 kn & a compressive force of 2 kn. What is the largest downward load F that the truss will safely support?

39 Example 2 Strategy This truss is identical to the one we analyzed in Example 1. By applying the method of joints in the same way, the axial forces in the members can be determined in terms of the load F. The smallest value of F that will cause a tensile force of 10 kn or a compressive force of 2 kn in any of the members is the largest value of F that the truss will support.

40 Example 2 Solution By using the method of joints in the same way as in Example 1, we obtain the axial forces: AB: 0 AC: 1.94F (T) BC: 0 BD: 1.67F (C) CD: 1.94F (T)

41 Example 2 Solution For a given load F, the largest tensile force is 1.94F (in members AC & CD) & the largest compressive force is 1.67F (in member BD). The largest safe tensile force would occur when 1.94F = 10 kn or when F = 5.14 kn. The largest safe compressive force would occur when 1.67F = 2 kn or when F = 1.20 kn. Therefore, the largest load F that the truss will safely support is 1.20 kn.

42 Example 2 Critical Thinking This example demonstrates why engineers analyze structures: By doing so, they can determine the loads that an existing structure will support or design a structure to support given loads In this example, the tensile & compressive loads the members of the truss will support are given

43 Example 2 Critical Thinking Information of that kind must be obtained by applying the methods of mechanics of materials to the individual members Then statics can be used, as we have done in this examples, to determine the axial loads in the members in terms of the external loads on the structure

44 Example 3 Determine the force in each member of the truss and indicate whether the members are in tension or compression.

45 Example 4 Determine the force in each member of the truss and indicate whether the members are in tension or compression.

46 Example 5

47 Example

48 The Method of Sections When we need to know the axial forces only in certain members of a truss, we often can determine them more quickly using the method of sections than the method of joints E.g. consider the Warren truss we used for the method of joints: It supports loads at B & D & each member is 2 m in length Suppose we need to determine only the axial force in member BC

49 The Method of Sections Just as in the method of joints, we begin by drawing a free-body diagram of the entire truss & determining the reactions at the supports: The next step is to cut the members AC, BC & BD to obtain a free-body diagram of a part, or a section, of the truss:

50 The Method of Sections Summing moments about point B, the equilibrium equations for the section are: Σ F x = T AC + T BD + T BC cos 60 = 0 Σ F y = 500 N 400 N T BC sin 60 = 0 Σ M point B = (2 sin 60 m)t AC (2 cos 60 m)(500 N) = 0 Solving them, we obtain T AC = 289 N, T BC = 115 N & T BD = 346 N.

51 The Method of Sections Notice how similar this method is to the method of joints: Both methods involve cutting members to obtain free-body diagrams of parts of a truss In the method of joints, we move from joint to joint, drawing free-body diagrams of the joints & determining the axial forces in the members as we go In the method of sections, we try to obtain a single free-body diagram that allows us to determine the axial forces in specific members

52 The Method of Sections In our example, we obtained a free-body diagram by cutting 3 members, including the 1 (member BC) whose axial force we wanted to determine In contrast to the method of joints, the forces on the free-body diagrams used in the method of sections are not usually concurrent: As in our example, we can obtain 3 independent equilibrium equations Although there are exceptions, it is usually necessary to choose a section that requires cutting no more than 3 members, or there will be more unknown axial forces than equilibrium equations

53 Example The truss in Figure supports a 100-kN load. The horizontal members are each 1 m in length. Determine the axial force in member CJ & state whether it is in tension or compression.

54 Example Strategy We need to obtain a section by cutting members that include member CJ. By cutting members CD, CJ & IJ, we will obtain a free-body diagram with 3 unknown axial forces. Solution To obtain a section, we cut members CD, CJ & IJ & draw the free-body diagram of the part of the truss on the right side of the truss

55 Example Solution From the equilibrium equation: Σ F y = T CJ sin kn = 0 We obtain T CJ = kn. The axial force in member CJ is kn (T).

56 Example Critical Thinking We designed this example to demonstrate that the method of sections can be very advantageous when you only need to determine the axial forces in particular members of a truss Imagine calculating the axial force in member CJ using the method of joints But in engineering applications it is usually necessary to know the axial forces in all the members of a truss & in that case the 2 methods are comparable

57 Example Determine the axial forces in members DG & BE & CG of the truss in Figure.

58 Example Strategy We can t obtain a section that involves cutting members DG & BE without cutting more than 3 members. However, cutting members DG, BE, CD & BC results in a section with which we can determine the axial forces in members DG & BE.

59 Example Solution Determine the Reactions at the Supports: Draw the free-body diagram of the entire truss:

60 Example Solution From the equilibrium equations: Σ F x = A x = 0 Σ F y = A y + K F 2F F = 0 Σ M point A = LF (2L)(2F) (3L)F + (4L)K = 0 We obtain the reactions A x = 0, A y = 2F & K = 2F.

61 Example Solution Choose a Section: We obtain a section by cutting members DG, CD, BC & BE. Because the lines of action of T BE, T BC & T CD pass through point B, we can determine T DG by summing moments about B: Σ M point B = L(2F) (2L)T DG = 0

62 Example Solution The axial force T DG = F. Then from the equilibrium equation: Σ F x = T DG + T BE = 0 We see that T BE = T DG = F. Member DG is in compression & member BE is in tension.

63 Example Critical Thinking This is a clever example but not 1 that is typical of problems faced in practice: The section used to solve it might not be obvious even to a person with experience analyzing structures Notice that the free-body diagram of the section of the truss is statically indeterminate, although it can be sued to determine the axial forces in members DG & BE

64 Example Problem For the truss shown, determine the forces in members DC and FG.

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