Announcements Mountain Dew is an herbal supplement Today s Objectives Define a simple truss Trusses Method of Joints Determine the forces in members of a simple truss Identify zero-force members Class Activities Applications Simple trusses Method of joints Zero-force members Examples Engr221 Chapter 6 1
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? Applications - continued 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? Engr221 Chapter 6 2
Defining a Simple Truss A truss is a structure composed of slender members joined together at their end points. If a truss, along with the imposed load, lies in a single plane (as shown at the top right), then it is called a planar truss. A simple truss is a planar truss which begins with a triangular element and can be expanded by adding two members and a joint. Analysis and Design Assumptions 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 loads are applied at the joints. The weight of the truss members is often neglected as the weight is usually small as compared to the forces supported by the members. 2. The members are joined together by smooth pins. This assumption is satisfied in most practical cases where the joints are formed by bolting or welding. With these two assumptions, the members act as two-force members. They are loaded in either tension or compression. Often compressive members are made thicker to prevent buckling. Engr221 Chapter 6 3
The Method of Joints In this method of solving for the forces in truss members, the equilibrium of a joint (pin) is considered. All forces acting at the joint are shown in a FBD. This includes all external forces (including support reactions) as well as the forces acting in the members. Equations of equilibrium ( F X = 0 and F Y = 0) are used to solve for the unknown forces acting at the joints. Steps for Analysis 1. If the support reactions are not given, draw a FBD of the entire truss and determine all the support reactions using the equations of equilibrium. 2. Draw the free-body diagram of a joint with one or two unknowns. Assume that all unknown member forces act in tension (pulling the pin) unless you can determine by inspection that the forces are compressive loads. 3. Apply the scalar equations of equilibrium, F X = 0 and F Y = 0, to determine the unknown(s). If the answer is positive, then the assumed direction (tension) is correct, otherwise it is in the opposite direction (compression). 4. Repeat steps 2 and 3 at each joint in succession until all the required forces are determined. Engr221 Chapter 6 4
Example A 200 lb 5 4 3 F BA B 500 lb 45 º F BC FBD of pin B Given: P 1 = 200 lb, P 2 = 500 lb Find: Plan: The forces in each member of the truss. First analyze pin B and then pin C + F X = 500 + F BC cos 45 (3 / 5) F BA = 0 + F Y = 200 F BC sin 45 (4 / 5) F BA = 0 + F X = F CA + 525.3 cos 45 = 0 F CA = 371 (T) F BA = 214 lb (T) and F BC = 525.3 lb (C) F CA 45º 525.3 C C Y FBD of pin C Zero-Force Members If a joint has only two non-collinear members and there is no external load or support reaction at that joint, then those two members are zeroforce members. In this example members DE, CD, AF, and AB are zero force members. You can easily prove these results by applying the equations of equilibrium to joints D and A. Zero-force members can be removed (as shown in the figure) when analyzing the truss. Engr221 Chapter 6 5
Zero-Force Members - continued If three members form a truss joint for which two of the members are collinear and there is no external load or reaction at that joint, then the third non-collinear member is a zero force member. Again, this can easily be proven. One can also remove the zero-force members, as shown, on the left, for analyzing the truss further. Note that zero-force members are used to increase stability and rigidity of the truss, and to provide support for various loading conditions. Example B Given: P 1 = 240 lb and P 2 = 100 lb Find: Plan: Determine the force in all the truss members (do not forget to mention whether they are in T or C). a) Check if there are any zero-force members b) Draw FBD s of pins D and B, and then apply the E-of-E at those pins to solve for the unknowns Engr221 Chapter 6 6
Example B - continued FBD of pin D Analyzing pin D: + F Y = 100 (5 / 13) F DB = 0 F DB = 260 lb (C) + F X = 240 F DC (12 / 13) ( 260) = 0 F DC 13 5 12 F DB Y D 240 lb X 100 lb F DC = 480 lb (T) F BC Y 260 lb Analyzing pin B: + F Y = F BC (5 / 13) 260 = 0 B X B 13 5 12 X F BC = 100 lb (T) FBD of pin B Questions 1. Using this FBD, you find that F BC = 500 N. Member BC must be in F BC A) Tension B) Compression B F BD C) Can not be determined B Y 2. For the same magnitude of force to be carried, truss members in compression are generally made as compared to members in tension. A) Thicker B) Thinner C) The same size Engr221 Chapter 6 7
Questions 1. One of the assumptions used when analyzing a simple truss is that the members are joined together by A) welding B) bolting C) riveting D) smooth pins E) super glue 2. In the method of joints, typically equations of equilibrium are applied at every joint. A) two B) three C) four D) six Textbook Problem 6-7 Determine the force in each member of the truss and state if the members are in tension or compression. F BC = 3 kn (C) F BA = 8 kn (C) F AC = 1.46 kn (C) F AF = 4.17 kn (T) F CD = 4.17 kn (C) F CF = 3.12 kn (C) F EF = 0 kn F ED = 13.1 kn (C) F DF = 5.21 kn (T) Engr221 Chapter 6 8
Textbook Problem 6-8 Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression. F BC = 1.89P (C) F CD = 1.37P (T) F AB = 0.471P (C) F BD = 1.67P (T) F AD = 1.37P (T) Summary Define a simple truss Determine the forces in members of a simple truss Identify zero-force members Engr221 Chapter 6 9
Announcements Today s Objective Trusses Method of Sections Determine forces in truss members using the method of sections Class Activities Applications Method of sections Examples Engr221 Chapter 6 10
Applications Long trusses are often used to construct bridges. The method of joints requires that many joints be analyzed before we can determine the forces in the middle part of the truss. There is another method to determine these forces directly. The Method of Sections In the method of sections, a truss is divided into two parts by taking an imaginary cut (shown here as a-a) through the truss. Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut member will also be either tensile or compressive with the same magnitude. This result is based on the equilibrium principle and Newton s third law. Engr221 Chapter 6 11
Steps for Analysis 1. Decide how you need to cut the truss. This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general). 2. Decide which side of the cut truss will be easier to work with (minimize the number of forces you have to find). 3. If required, determine the necessary support reactions by drawing the FBD of the entire truss and applying the E-of-E. Steps for Analysis - continued 4. Draw the FBD of the selected part of the cut truss. You need to indicate the unknown forces at the cut members. Initially we assume all the members are in tension, as we did when using the method of joints. Upon solving, if the answer is positive, the member is in tension as per your assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either in tension or compression by inspection as was done in the figures above.) Engr221 Chapter 6 12
Steps for Analysis - continued 5. Apply the E-of-E to the selected cut section of the truss to solve for the unknown member forces. Note that in most cases it is possible to write one equation to solve for one unknown directly. Textbook Example 6.7 1. Can you determine the force in member ED by making the cut at section a-a? Explain your answer. A) No, there are 4 unknowns B) Yes, using Σ M D = 0 C) Yes, using Σ M E = 0 D) Yes, using Σ M B = 0 Engr221 Chapter 6 13
Textbook Example 6.7 - continued 2. If you know F ED, how will you determine F EB? A) By taking section b-b and using ΣM E = 0 B) By taking section b-b, and using ΣF X = 0 and ΣF Y = 0 C) By taking section a-a and using ΣM B = 0 D) By taking section a-a and using ΣM D = 0 Textbook Problem 6.36 Given: Loading on the truss as shown Find: The force in members BE, EF, and CB Plan: a) Take a cut through members BE, EF, and CB b) Analyze the top section (no support reactions!) c) Draw the FBD of the top section d) Apply the E-of-E such that every equation yields an answer to one unknown Engr221 Chapter 6 14
Textbook Problem 6.36 - continued + ΣF X = 5 + 10 F BE cos 45º = 0 F BE = 21.2 kn (T) + Σ M E = 5(4) + F CB (4) = 0 F CB = 5 kn (T) + Σ M B = 5 (8) 10 (4) 5 (4) F EF (4) = 0 F EF = 25 kn or 25 kn (C) Textbook Problem 6.53 Given: Loads as shown on the roof truss Find: The force in members DE, DL, and ML Plan: a) Take a cut through the members DE, DL, and ML. b) Work with the left part of the cut section. Why? c) Determine the support reactions at A. d) Apply the E-of-E to find the forces in DE, DL, and ML. Engr221 Chapter 6 15
Problem 6.53 - continued Analyzing the entire truss, we get Σ F X = A X = 0. By symmetry, the vertical support reactions are A Y = I Y = 36 kn + M D = 36 (8) + 6 (8) + 12 (4) + F ML (5) = 0 F ML = 38.4 kn (T) Problem 6.53 - continued +Σ M L = 36 (12) + 6 (12) + 12 (8) + 12 (4) F DE ( 4/ 17)(6) = 0 F DE = 37.11 kn or 37.1 kn (C) + Σ F X = 38.4 + (4/ 17) ( 37.11) + (4/ 41) F DL = 0 F DL = 3.84 kn or 3.84 kn (C) Engr221 Chapter 6 16
Questions 1. In the method of sections, generally a cut passes through no more than members in which the forces are unknown. A) 1 B) 2 C) 3 D) 4 2. If a simple truss member carries a tensile force of T along its length, then the internal force in the member is A) Tensile with magnitude of T/2 B) Compressive with magnitude of T/2 C) Compressive with magnitude of T D) Tensile with magnitude of T Question As shown, a cut is made through members GH, BG, and BC to determine the forces in them. Which section will you choose for analysis and why? A) Right, fewer calculations B) Left, fewer calculations C) Either right or left, same amount of work D) None of the above, too many unknowns Engr221 Chapter 6 17
Question When determining the force in member HG in the previous question, which one equation of equilibrium is best to use? A) ΣM H = 0 B) ΣM G = 0 C) ΣM B = 0 D) ΣM C = 0 Summary Determine forces in truss members using the method of sections Engr221 Chapter 6 18
Announcements Frames and Machines Today s Objectives Draw the free body diagram of a frame or machine and its members Determine the forces acting at the joints and supports of a frame or machine Class Activities Applications Frame/machine analysis Examples Engr221 Chapter 6 19
Applications Frames are commonly used to support various external loads. How is a frame different than a truss? How can you determine the forces at the joints and supports of a frame? Applications - continued Machines, like these above, are used in a variety of applications. How are they different from trusses and frames? How can you determine the loads at the joints and supports? It is required to know these loads when designing the machine members. Engr221 Chapter 6 20
Frames and Machines - Definitions Frames and machines are two common types of structures that have at least one multi-force member. (Recall that trusses have nothing but two-force members). Frames are generally stationary and support external loads. Machines contain moving parts and are designed to alter the effect of forces. Steps for Analyzing a Frame or Machine F AB 1. Draw the FBD of the frame or machine and its members, as necessary. Hints: a) Identify any two-force members b) Forces on contacting surfaces (usually between a pin and a member) are equal and opposite c) For a joint with more than two members or an external force, it is advisable to draw a FBD of the pin 2. Develop a strategy to apply the E-of-E to solve for the unknowns. Problems are going to be challenging since there are usually several unknowns. Practice is needed to develop good strategies. Engr221 Chapter 6 21
Textbook Problem 6.72 Given: A frame and loads as shown Find: The reactions that the pins exert on the frame at A, B and C Plan: a) Draw a FBD of members AB and BC b) Apply the E-of-E to each FBD to solve for the six unknowns. Think about a strategy to easily solve for the unknowns. Textbook Problem 6.72 - continued FBD s of members AB and BC B Y BX B Y B B X B A X 1000N A 45º 0.2m 0.2m 0.4m 500N 0.2m 0.4m C C Y A Y Writing moment equations at A and C, we get: + M A = B X (0.4) + B Y (0.4) 1000 (0.2) = 0 + M C = - B X (0.4) + B Y (0.6) + 500 (0.4) = 0 B Y = 0 and B X = 500 N Engr221 Chapter 6 22
Textbook Problem 6.72 - continued FBD s of members AB and BC B B Y BX B X B Y B Applying E-of-E to bar AB: A X 1000N A 45º 0.2m 0.2m A Y 0.4m 500N 0.2m 0.4m C C Y + F X = A X 500 = 0 ; A X = 500 N + F Y = A Y 1000 = 0 ; A Y = 1000 N Applying E-of-E to bar BC: + F X = 500 C X = 0 ; C X = 500 N + F Y = C Y 500 = 0 ; C Y = 500 N Textbook Problem 6.83 Given: The wall crane supports an external load of 700 lb Find: The force in the cable at the winch motor W and the horizontal and vertical components of the pin reactions at A, B, C, and D Plan: a) Draw FBD s of the frame s members and pulleys b) Apply the E-of-E and solve for the unknowns Engr221 Chapter 6 23
Textbook Problem 6.83 - continued FBD of the Pulley E T T 700 lb E Necessary Equations of Equilibrium: + F Y = 2 T 700 = 0 T = 350 lb 350 lb Textbook Problem 6.83 - continued C Y C X C 350 lb + F X = C X 350 = 0 + F Y = C Y 350 = 0 C X = 350 lb C Y = 350 lb 30 A FBD of pulley C B X B Y 350 lb B 350 lb A FBD of pulley B + F X = B X + 350 350 sin 30 = 0 B X = 175 lb + F Y = B Y 350 cos 30 = 0 B Y = 303.1 lb Engr221 Chapter 6 24
Textbook Problem 6.83 - continued Note that member BD is a two-force member. A X A FBD of member ABC + M A = T BD sin 45 (4) 303.1 (4) 700 (8) = 0 T BD = 2409 lb + F Y = A Y + 2409 sin 45 303.1 700 = 0 A Y = 700 lb + F X = A X 2409 cos 45 + 175 350 = 0 A X = 1880 lb A Y A 45 T BD B 175 lb 303.11 lb 4 ft 4 ft 350 lb 700 lb Textbook Problem 6.83 - continued A FBD of member BD 2409 lb D 45 B 2409 lb At D, the x and y components are: + D X = 2409 cos 45 = 1700 lb + D Y = 2409 sin 45 = 1700 lb Engr221 Chapter 6 25
Questions 1. Frames and machines are different as compared to trusses since they have A) Only two-force members B) Only multiforce members C) At least one multiforce member D) At least one two-force member 2. Forces common to any two contacting members act with on the other member. A) Equal magnitudes but opposite sense B) Equal magnitudes and the same sense C) Different magnitudes but opposite sense D) Different magnitudes but the same sense Questions 1. When determining the reactions at joints A, B, and C, what is the minimum number of unknowns for solving this problem? A) 3 B) 4 C) 5 D) 6 2. For the above problem, imagine that you have drawn a FBD of member AB. What will be the easiest way to directly solve for the first unknown? A) M C = 0 B) M B = 0 C) M A = 0 D) F X = 0 Engr221 Chapter 6 26
Example Problem The compound beam is fixed at A and supported by a rockers at B and C. There are hinges (pins) at D and E. Determine the reactions at the supports. C y B y = 0N = 7.5 KN A x = 0N A y = 7.5 KN M A = 45.0 KN-m Example Problem The pumping unit is used to recover oil. When the walking beam ABC is horizontal, the force acting in the wire-line at the well head is 250 lb. Determine the torque M which must be exerted by the motor in order to overcome this load. The horse-head C weighs 60 lb and has a center of gravity at G C. The walking beam ABC has a weight of 130 lb and a center of gravity at G B, and the counterweight has a weight of 200 lb and a center of gravity at G W. The pitman, AD, is pin-connected at its ends and has negligible weight. F AD = 449 lb M = 313 ft-lb Engr221 Chapter 6 27
Summary Draw the free body diagram of a frame or machine and its members Determine the forces acting at the joints and supports of a frame or machine Engr221 Chapter 6 28