Analysis of Structures PowerPoint Presentation

Slide 1 -

Analysis of Structures

Slide 2 -

Structure: a device that is made up of several connected parts and whose function is to support load Ideally structures are designed to be in equilibrium at rest External forces: Forces applied to a structure These external forces have to be transmitted, through the different members that make up the structure, to the supports Internal Forces: process of transmitting the external forces within the members. may be viewed as actions and reactions between the molecules of the structural members (Newton's Third Law)

Slide 3 -

When a structure is in equilibrium, every part of the structure is also in equilibrium. When a free body diagram of the entire structure is drawn, only the external forces (applied forces and support reactions) will act on the structure. If a free body diagram of a part of the structure is drawn, the external forces acting on that part and the internal forces, which represent the effect of the removed part on the part being drawn, will be shown on the free body diagram. Note that the internal force is in effect an external force on the free body diagram since it is exposed.

Slide 4 -

ppt slide no 4 content not found

Slide 5 -

Note: BE is a two force body, hence FBE must be acted along the member and we know its direction (single force of known direction). Member CE is a three force body. Note that FBE, FCE and FCD can only be considered if looking at equilibrium of the dismembered structure. For equilibrium of the entire structure, these internal forces do not come into play. Also observe that the internal forces cancel out if the dismembered structure is reassembled.

Slide 6 -

Analysis of Trusses Trusses: structures that consist of straight pin-jointed members and are designed to support loads over large spans No member is continuous over a joint and all the loads are assumed to be applied only at the joints Individual members of a truss are typically slender and incapable of supporting a lot of load on their own, hence the requirement that the load must be applied to the joints. Application of loads only to the pinned joints reduces every member of a truss to a two-force body and each member is either in tension or compression.

Slide 7 -

Practical Applications Roofs in Buildings Bridges Cranes Transmission Masts / Towers

Slide 8 -

Objective Student shall be able to identify and draw the following types of trusses: Pratt Truss Howe Truss Warren Truss Differentiate between a roof and bridge truss. Fink Truss Baltimore Truss K-Truss

Slide 9 -

Plane Truss: A truss with all the members and applied loads in the plane The basic element of a truss is the triangle as this constitutes a rigid, non-collapsible frame. Simple Truss: A truss that can be built from the basic triangular element is called a simple truss.

Slide 10 -

Truss analysis involves the determination of the magnitude of the forces in the members and whether the member is in tension or compression. A truss is said to be statically determinate if all the unknown forces and reactions can be determined from equilibrium considerations. Otherwise the truss is statically indeterminate (contains additional members or supports that are not necessary to maintain equilibrium) or unstable (contains fewer number of members or support than are required to maintain equilibrium).

Slide 11 -

Note that the total number of unknown forces (forces in the members) and unknown reactions will be equal to b + r, where b = # of members in the truss & r = number of reactions at the supports. Since the truss is in equilibrium, every part of it must be in equilibrium. Considering the equilibrium of a pinned joint (which is in effect a particle), there are two independent equations of equilibrium for the joint. If the total number of joints in the truss is j, then there are a total of 2j independent equations of equilibrium. Hence we have that: b + r - 2j = 0 ==> Truss is statically determinate b + r - 2j > 0 ==> Truss is statically indeterminate b + r - 2j < 0 ==> Truss is unstable

Slide 12 -

Truss Analysis Methods Method of Joints Method of sections

Slide 13 -

Method of Joints involves the consideration of equilibrium of each joint in a truss in a systematic manner and determining the forces in the members of the truss Check if truss is statically determinate, i.e. b + r - 2j =0. Calculate the support reactions by considering equilibrium of the entire truss. Select a joint with not more than two unknowns and determine the forces at the joint by using equilibrium equations: SFx = 0; SFy = 1 Repeat step (c) until all the member forces have been determined. Clearly identify the member forces by using negative sign for compression and the positive sign for tension.

Slide 14 -

Method of Joints Note: that while statically determinate simple trusses can be easily analyzed using the method of joints, compound trusses will require the solutions of large numbers of simultaneous equations before the forces in the members can be determined using the method of joints. Compound trusses should be analyzed using a combination of methods of joints and methods of sections.

Slide 15 -

Example Problem 6.6 (Beer & Johnston Page 283) Using the method of joints, determine the force in each member of the truss shown. State whether each is in tension or compression. Required: Forces in the members using the method of Joints.

Slide 16 -

Example Q 2 Required: Forces in members jointed at H. Forces in members CG and DF.

Slide 17 -

Example Problem 6.12 Using the method of joints, determine the force in each member of the truss shown. State whether each is in tension or compression. Required: Forces in each truss member. Compute Support Reactions Consider equilibrium of the entire truss

Slide 18 -

Joints under Special Loading Conditions Recognition of joints with special features will facilitate and speed-up the analysis of a truss. Special care must however be exercised in using these information. CASE 1: 4 members lying in two intersecting straight lines. No force on joint. FAB = FAD FAC = FAE

Slide 19 -

Case 2: Joint connecting three members, two of which lie along a straight line. Force P acts along the third member. FAB = FAD FAC = P

Slide 20 -

Case 3: Zero force members Joint connecting three members, two of which lie along a straight line. No force on joint. FAB = FAD FAC = 0 Member AC is said to be a zero-force member. Although zero-force members do not carry any load, they help to give the truss its rigidity.

Slide 21 -

Case 4: Joint connecting two members which lie on a straight line: No force on joint. FAB = FAC

Slide 22 -

Case 4: AB and AC must be zero force members. Joint connecting two members which do not lie on a straight line. No force acting on joint. Joint A can only be in equilibrium if it is acted upon by two forces of equal magnitude, same lines of action and opposite sense. This is not possible the way the members are joined at A; hence, for equilibrium, both members must be zero force members.

Slide 23 -

Example Consider Joint C, FBC = 0; FCA = FCE Consider Joint B, FBE = 0; FBA = FBD Consider Joint E, FED = 0; FEC = FEG Consider Joint K, FKJ = 0; FKI = FKL Consider Joint J, FJI = 0; FJH = FJL Consider Joint I, FIH = 0; FIG = FIK Consider Joint H, FHG = -50kN; FHF = FHJ

Slide 24 -

Note, there is nothing special about joint D:

Slide 25 -

Example Problem 6.8 (Bedford & Fowler Page 286) The load F=10 kN. Determine the axial forces in the members. Required: Axial forces in the members, indicating if tension or compression

Slide 26 -

Example Problem 6.17 (Bedford & Fowler Page 287) Determine the axial forces in the members in terms of the weight W. Required: Axial forces in the members, stating if tension or compression.

Slide 27 -

Analysis of Trusses by Method of Sections The method of joints is used when the forces in every member of the truss is required and involves the consideration of equilibrium of each joint which is treated as a particle. When the forces in only a one, two or three members of the truss are required, the method of sections provide a more efficient way of carrying in the analysis. The method involves the consideration of equilibrium of a part of the truss obtained by passing an imaginary plane through the members we are interested in. This exposes the internal forces in these members and by considering the equilibrium of either part of the truss, the forces in the members can be determined. Recall that there are three independent equilibrium equation for a 2-D rigid body. Hence, the cutting plane should be chosen such that there the number of unknowns do not exceed three.

Slide 28 -

Summary of Procedure Check that the truss is statically determinate Calculate the support reactions using fbd of the entire structure Pass an imaginary cutting plane through not more than three members. Using equilibrium equations, calculate the forces in the members that were cut by the imaginary planes. To eliminate from the equilibrium equation, take moments about a point on its line of action; to eliminate two unknown forces, take moments about the point of intersection of the lines of action of the two forces. Note: The cutting plane can pass through 4 or more members if two or more of the members so cut are collinear (see Problems 6.61-6.64, Page 298, Beer and Johnston). Also keep in mind that compound trusses may best be analysed by combining the method of joints and the method of sections.

Slide 29 -

Example Problem 6.41 Required: Forces in members DE and AE.

Slide 30 -

Example Problem 6.38 Required: Identify the zero force members. Determine the forces in FK and JO.

Slide 31 -

Space Trusses three dimensional structures consisting of pin-jointed members and is loaded only at the joints. Basic building block is a tetrahedron consisting of six bars. Since members are all two-force bodies, they are subjected to only axial forces (tension or compression). The pin-joints are ball-and-socket joints. Method of joints can be used to consider equilibrium of each joint, i.e. SF = 0 (Three equations) ==> joints being analyzed must have no more than three unknown forces. Support reactions must be properly identified for given problems. Method of sections can be used provided that the cutting plane does not pass through more than six members because a total of six independent equations of equilibrium are available (SF = 0 and SM = 0).

Slide 32 -

Example Problem 6.38 (Beer & Johnston Page 288) Given: The truss shown consists of nine members and is supported by a ball and socket at A, two short links at B, and a short link at C. Required: Determine the force in each of the members for the given loading.

Slide 33 -

Analysis of Frames and Machines in 2-D Frames and machines are structures which contain at least one multi-force member (i.e., a member with forces acting at three or more points on it ). Frames are designed to support load and are usually at rest. Machines are generally designed to transmit and modify forces and will usually contain moving parts. The forces on frames and machines can be applied to the members or joints (not just on joints, as is the case for trusses).

Slide 34 -

Analysis of Frames and Machines in 2-D (2) In analyzing a rigid frame, it may be more convenient to dismember it and consider the equilibrium of its constituent members. Recall that if a structure is in equilibrium, then every part of it (every member, every joint) will also be in equilibrium. In general, every multiforce member of a 2-D frame will have 3 independent equations of equilibrium:  Fx = 0  Fy = 0  M = 0 (Moments arise from the application of loads along the members)

Slide 35 -

Analysis of Frames and Machines in 2-D (3) The best way to determine the degree of statical redundancy (indeterminacy) of a frame or machine is to dismember it draw the free body diagram of each constituent multiforce member and count the total number of unknown forces. Since each multiforce member has 3 independent equations of equilibrium (in 2-D), there will be three times as many independent equations of equilibrium as there are multiforce members. The difference between the total number of independent equations of equilibrium and the total number of unknown forces gives the degree of statical redundancy. Recall that for statically determinate structures, the degree of statical redundancy is equal to zero.

Slide 36 -

Analysis of Frames and Machines in 2-D (4) For frames that have an external load applied at a joint which connects multiple members, the entire external load may be applied to only one of the members at the joint and must be treated separately from the forces that the members exert at one another at the joint. Recall that the forces that members exert on each other are always equal in magnitude, have the same line of action but opposite sense (Newton's 3rd Law). Regardless of the member on which the entire external load at a joint is placed, the total load at each point of the frame will be the same.

Slide 37 -

Example Problem 6.56 Required: Determine the components of all forces acting on member ABCD when q = 0.

Slide 38 -

Example Problem 6.64 Required: Determine all forces exerted on member AI if the frame is loaded by clockwise couple of magnitude 230 Nm applied (a) at point D, (b) at point E.

Slide 39 -

Problem 6.85 (B&F Page 321) Required: Determine the forces on AD.

Slide 40 -

Problem 6.93 (B&F Page 323) Required: Determine the force exerted on the bolt by the bolt cutters and the magnitude of the force the members exert on each other at the pin connection A.

Slide 41 -

Problem 6.76 (Beer & Johnston Page 310) Required: For the frame and loading shown, determine the force acting on memberABC (a) at B, (b) at C.

Slide 42 -

Problem 6.100 (Beer & Johnston Page 313) Required: For the frame and loading shown, determine the components of all forces acting on member ABE.

Slide 43 -

Example Problem 6.58 Given: The low-bed trailer shown is designed so that the rear end of the bed can be lowered to ground level in order to facilitate the loading of equipment or wrecked vehicles. A 14000-kg vehicle has been hauled to the position shown by a winch; the trailer is the returned to a traveling position where a = 0 and both AB and BE are horizontal. Required: Considering only the weight of the disabled automobile, determine the force which must be exerted by the hydraulic cylinder to maintain a position with a = 0.

Slide 44 -

Problem 6.59 Required: For the marine crane shown, which is used in offshore drilling operations, determine (a) the force in link CD, (b) the force in the brace AC, (c) the force exerted at A on the boom AB.

Slide 45 -

Problem 6.70 (B&F Page 318) Required: Determine the forces on member ABC. Obtain the answers in two ways: (a) draw the free-body diagrams of the individual members, place the 400 N load on the free-body diagram of member ABC. (b) draw the free-body diagrams of the individual members, place the 400 N load on the free-body diagram of member CD.

Slide 46 -

Analysis of Machines Machines are designed to transmit and modify forces and usually have moving members. The analysis procedure is similar to that for frames as the machine needs to be dismembered and the equilibrium of several parts considered by using their free-body diagrams. The free-body diagrams should be chosen to include the input forces and the reactions to the output forces.

Slide 47 -

Consider a pair of pliers that is used to cut a wire. Typically a>>b Given that P = 5 N, a= 250 mm and b = 25 mm, calculate the force Q that cuts the wire and the reaction force at the pivot pin A. P is the input force Q is the output force R is the reaction to the output force . Note that Q = R

Slide 48 -

Free-body diagrams of the two parts that make up the plier

Slide 49 -

Example 6.10 (B&F Page 314) Required: What forces are exerted on the bolt at E in Fig. 6.40 as a result of the 150 N forces on the pliers?

Slide 50 -

Problem 6.104 Given: A 3600-kg steel ingot is lifted by a pair of tongs as shown. Required: Determine the forces exerted at C and E on tong BCE.

Slide 51 -

Example Problem 6.62 (B&F Page 316) Required: For the frame shown, determine the reactions at the built-in support A and the force exerted on the structure by the smooth floor at C.

Slide 52 -

Example Problem 6.68 (B&F Page 317) Given: A football player works out with a 'squat thrust' machine. To rotate the bar ABD, he must exert a vertical force at A such that the axial force in member BC is 1100 N. Required: When the bar is on the verge of rotating, what force does he exert at A and what forces are exerted on member CDE at D?

Slide 53 -

Problem 6.88 (B&F Page 322) Given: The weight W = 300kN. Required: What is the magnitude of the force the members exert on each other at D?

Slide 54 -

Problem 6.92 (B&F Page 323) Given: The scoop C of the front-end loader is supported by two identical arms, one on each side of the loader. One of the two arms (ABC) is visible in the figure. It is supported by a pin supported at A and the hydraulic actuator BD. The sum of the other loads exerted on the arm, including its own weight, is F = 1.6 kN. Required: Determine the axial force in the actuator BD and magnitude of the reaction at A.

Slide 55 -

Example Problem 6.98 (Beer & Johnston Page 262) Given: The shear shown is used to cut and trim electronic-circuit-board laminates. Required: For the position shown, determine (a) the vertical component of force exerted on the shearing blade at D, (b) the reaction at C.

Slide 56 -

Example Problem 6.99 (Beer & Johnston Page 262) Given: The control rod CE passes through a horizontal hole in the body of the toggle clamp shown. Required: Determine (a) the force Q required to hold the clamp in equilibrium, (b) the corresponding force in link BD.