Pin-Jointed rame Structures (rameworks) 1
Pin Jointed rame Structures (rameworks) A pin-jointed frame is a structure constructed from a number of straight members connected together at their ends by frictionless pinned joints. Since the joints are free to rotate they cannot transmit torque from one to another, they are therefore only subjected to axial loading, which may be tensile or compressive. A member in tension is called a tie, while one in compression is a strut. Tie in Tension Strut in Compression The arrow head within the members, represent the equal and opposite forces acting in the member. An example of a typical framework is given below: Typical framework with loading and supports Such a frame is said to be statically determinate, that is, capable of being solved using equations of equilibrium. When this is not the case it would be referred to as statically indeterminate. We shall only be considering the former. Problem of this type can be solved either graphically or analytically. We will consider both methods. raphical Method Using this method of solution, we make use of Bow s notation which is use to describe the framework and the forces within each member. We first place capital letter between all of the external forces and then internally between the member which make up the framework. You need to work around the frame in a clockwise direction and from left to right across it, similar to that shown below. Where you start is unimportant. A D E B C The load on the top right-hand corner of the framework would be referred to as force AB and its sense which is downwards would be ab as we go clockwise from A to B. Commencing at a joint at which there are no more than two unknown forces, we construct a force polygon for all internal and external forces acting at that joint. Proceed systematically though the frame, constructing a force polygon at each joint, to produce one composite force diagram. The magnitude and direction of each unknown force can then be scaled of the diagram. 2
This method has largely been superseded by modern computer methods but is still useful with complex frames. (See later for a worked example.) Analytical Methods We will consider two analytical methods, namely; Method of Resolution at Joint and Method of Section. Method of Resolution at Joint This method is best suited when forces in all members are required. At any joint in a plane frame at which there are no more than two unknown forces, write down two equations of equilibrium by resolving forces at the joint in two mutually perpendicular directions. Solve these equations for the two unknown forces. Proceed systematically through the frame until all forces are known. External support reactions are usually determined first by the use of moment equilibrium. (See worked example later.) Method of Section This method is useful if you only need to determine the forces in one or a few members of the framework. Cut the frame, as shown in the example below by a section through the member under consideration and no more than two other members in which the member forces are unknown. Both parts of the structure can then be treated as structures in equilibrium and either part can then be solved by resolving forces or by taking moments of force about a suitable point. The choice as to whether to resolve or to take moments will depend upon the geometry of the frame and will be illustrated in a worked example that follows shortly. rame cut for solution using Method of Section Let s use the following example to illustrate all of the above three methods. Note that Bow s notation has been used to label the frame. 50kN A 30 kn D E B All angles are 60 o and sides are 3m C R 1 R 2 raphical Solution 1. Calculate the support reactions Taking moments about R 1 and assuming clockwise moments to be positive M = 0 = (50 x 1.5) + (30 x 4.5) 6R 2 3
This gives R 2 = 35 kn Equating vertical forces will give R 1 (assume forces up to be positive) V = 0 = R 1 +35-50 30 This gives R 1 = 45 kn 2. Draw the force diagram to scale. Start with the external forces which should all be in a straight line. d e g c f a orce Diagram (NTS) b 3. Scale off the magnitude of all unknown forces. B = -40 = +5.7 E = -5.7 DE = -52 CE = +26 C = +20 A = -23 Accuracy of answers depends upon scale of diagram and how well you draw it. Note the + and signs have come from the next stage. 4. Determine whether forces are compressive or tensile using Bow s notation will be explained in class. 50kN A 30 kn D E B Diagram shows nature of forces in each member. C R 1 R 2 4
Method of Resolution at Joint We already have values for the support reactions otherwise calculate them at this point. Next consider each joint in turn starting with one where at least one force is known. We will begin by considering joint BC C B Notice that we are assuming that all internal forces are tensile. Actual values are then substituted. A negative result implies a force is compressive. R 2 Resolving vertical forces V = 0 = R 2 + BCos30 (We are assuming forces up are positive) Therefore 0 = 35 +BCos30 This gives B = -40.41 kn Resolving Horizontal forces (We are assuming that forces left to right are positive) H = 0 = -C BSin30 Therefore 0 = -C (-40.41Sin30) Hence C = 40.41Sin30 = 20.21 kn Consider next joint AB AB A B Resolving vertical forces V = 0 = -AB BCos30 Sin60 0 = -30 (-40.41Cos30) Sin60 This gives = 5.77 kn Resolving horizontal forces H = 0 = -A Cos60 + BSin30 5
0 = -A 5.77Cos60 + (-40.41Sin30) This gives A = -23.09 kn The remainder of the problem is carried out in a similar fashion. You can complete the rest as a tutorial problem. Method of Section 1. Calculate the support reactions already done. 2. Let s assume we require the forces in members A,, and C. We therefore cut the frame as shown below. A D E B C 3. Draw the BD A 30 kn B Again the cut members are assumed to be in tension. 4. Calculate the forces C 35 kn It should be obvious from the free body diagram that we can determine by equating vertical forces. V = 0 = -30 + 35 Sin60 (taking forces up as positive) This gives = 5.77 kn Again by careful examination of the BD we see that by taking moments about AB we should be able to determine C. M AB = 0 = (-35 x 1.5) + C x 3Cos30 (taking clockwise moments as positive) This gives C = 20.21 kn Next by equating horizontal forces we should be able to determine A H = 0 = -A Cos60 C 0 = -A 5.77Cos60 20.21 6
This gives A = - 23.09 kn Summary of the Procedure of Method of Section 1. Calculate the reactions of the frame using a BD of the whole frame. 2. Isolate a portion of the frame by passing a cutting plane through the frame, cutting no more than three members in which the forces are unknown (unless all but one of the lines of action of the cut member intersect at a point). 3. Apply the three equations of equilibrium ( M = 0, H = 0, and V = 0) to the isolated portion of the frame and solve for the unknowns. 7
Tutorial Problems 1. or the framework given in below determine the magnitude of the force in each member, using the graphical method, and state whether each is a tie or a strut. (Note that it is not usual to label the framework for you.) 2kN A 3kN E H B All members are 1m D C R1 1kN R2 [R 1= 2.75kN; R 2= 3.25kN - All of the following have been found by measurement - members: hb = 3.75kN (strut), ch = 1.9kN (tie), fd = 1.6kN (tie), ef = 3.2kN (strut), ag = 2kN (strut), fg = 0.85kN (tie), gh = 0.3kN (tie)] 2. Use the method of resolution at joints to verify the forces in the members that make up the top right hand joint of the framework shown on question1 above. (hb = -3.75kN hg = 0.286 kn ga = -2.018 kn) 3. or the pin jointed frame shown below solve all the forces and reactions in the frame stating which members are tie and which are struts. The length of all vertical and horizontal members are 3m. A E H D C B 15kN 30kN [Reactions: R 1 = 20kN; R 2 = 25kN; members: ha = 35.35kN (strut), bh = 25kN (tie), ed = 20kN (tie), ea = 29kN (strut), cg = 25kN (tie), gf = 7.07kN (strut), fa = 20kN (strut), ef = 20kN (tie), gh = 30kN (tie)] 4. The Warren truss shown below is composed of similar members all of which are 3 m long. Determine the forces in all members due to a vertical load of 90 kn at. Use the method of resolution at joints to solve this problem. 8
(Reactions: R E = 30; R A = 60 kn; members: DE = -34.64; E = 17.32; D = 34.64; CD = - 34.64; C = -34.64; = 51.96; C = 34.64; BC = -69.28; B = 69.28; A = 34.64; AB = - 69.28 kn) 5. Structure shown, is a truss which is pinned to the floor at point A, and supported by a roller at point D. Determine the forces in all members of the truss. It is suggested that you use the method of joints to solve the problem. Note: Start by calculating reaction at D which you should find is 12 kn. ( AB = 8.73 kn tensile, A = 21.82 kn compressive, BC = 15.71 kn tensile, B = 8.73 kn compressive, B = 8.73 kn tensile, CD = 5.24 kn tensile, CE =13.09 kn tensile, C = 13.09 kn compressive, DE = 13.9 kn compressive, E = 0.48 kn compressive, = 12.22 kn compressive) 6. Using the method of section, determine the forces in members BD, CD, and CE of the roof truss shown below. (BD = -160 kn; CD = -200 kn; CE = 320 kn) Note: You need to determine reaction at A which you should fine is 120 kn 9
7. Determine the forces, and state whether tensile of compressive, in all members of the framework shown below. It is suggested you use Method of Joints. ( AB = 5.56 kn tensile, AE = 75.56 kn tensile, BC = 4.45 kn tensile, BE = 3.34 kn compressive, CD = 88.87 kn compressive, CE = 5.57 kn tensile, C = 50 kn tensile, D = 71.11 kn tensile, E = 71.11 kn tensile) 8. The framework shown below is supported by a roller at C and hinge at. By the method of section determine the forces in BC, D, and CE. ( BC=44.72 kn compressive, CE=10 kn tensile, D=80 kn compressive) 9. Determine the force in members D, D, and E of the Howe truss shown below. (D = -280 kn, D = -150 kn, E = 400 kn) 10. or the framework shown below determine the forces in members BC, C and E, stating whether they are tensile or compressive. 10
E 1.2 m A B C D 1.2 m 1.2 m 1.2 m 2.7 kn 2.7 kn 3.6 kn [ BC = 9.9 kn (C), C =8.91 kn (T), E = 3.6 kn (T)] 11
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