SRSD 2093: Engineering Mechanics 2SRRI SECTION 19 ROOM 7, LEVEL 14, MENARA RAZAK

SRSD 2093: Engineering Mechanics 2SRRI SECTION 19 ROOM 7, LEVEL 14, MENARA RAZAK

SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS Today s Objectives: Students will be able to: a) Define a simple truss. b) Determine the forces in members of a simple truss. c) Identify zero-force members. In-Class Activities: Applications Simple Trusses Zero-force Members Group Problem Solving

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

APPLICATIONS Trusses are also used in a variety of structures like cranes and the frames of aircraft or the space station. How can you design a light weight structure satisfying load, safety, cost specifications, is simple to manufacture, and allows easy inspection over its lifetime?

SIMPLE TRUSSES 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. For these trusses, the number of members (M) and the number of joints (J) are related by the equation M = 2 J 3.

ANALYSIS AND DESIGN ASSUMPTIONS When designing the members and 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 the ends together. 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.

STEPS FOR ANALYSIS 1. If the truss s support reactions are not given, draw a FBD of the entire truss and determine the support reactions (typically using scalar 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 on the pin) unless you can determine by inspection that the forces are compression 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.

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 zero-force members. In this example members DE, DC, AF, and AB are zero force members. Zero-force members can be removed (as shown in the figure) when analyzing the truss.

ZERO-FORCE MEMBERS 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 member DA and CA are zero force members. Please note that zero-force members are used to increase stability and rigidity of the truss, and to provide support for various different loading conditions.

EXAMPLE 1 Given: Loads as shown on the truss Find: The forces in each member of the truss. Plan: 1. Check if there are any zero-force members. (Note that member BD is zero-force member. F BD = 0) 2. First analyze pin D and then pin A.

EXAMPLE 1: SOLUTION 45 º F AD D 450 kn 45 º F CD FBD of pin D + F X = 450 + F CD cos 45 F AD cos 45 = 0 + F Y = F CD sin 45 F AD sin 45 = 0 F CD = 318 kn (Tension) or (T) and F AD = 318 kn (Compression) or (C)

EXAMPLE 1: SOLUTION Analyzing pin A: F AD A 45 º F AB A Y FBD of pin A + F X = F AB + ( 318) cos 45 = 0; F AB = 225 kn (T) Could you have analyzed Joint C instead of A?

GROUP PROBLEM SOLVING Given: Loads as shown on the truss 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. Is member CE zero-force member? b) Draw FBDs of pins D, C, and E, and then apply E-of-E at those pins to solve for the unknowns.

GROUP PROBLEM SOLVING FBD of pin D Y 5 F DE 3 4 D 600N X F CD Analyzing pin D: + F X = F DE (3/5) 600 = 0 F DE = 1000 N = 1.00 kn (C) + F Y = 1000 (4/5) F CD = 0 F CD = 800 N = 0.8 kn (T)

GROUP PROBLEM SOLVING FBD of pin C Y F CD = 800 N F CE C 900 N X Analyzing pin C: F BC + F X = F CE 900 = 0 F CE = 900 N = 0.90 kn (C) + F Y = 800 F BC = 0 F BC = 800 N = 0.80 kn (T)

GROUP PROBLEM SOLVING FBD of pin E Y 3 F DE = 1000 N 3 4 5 E F CE = 900 N F AE 4 5 4 5 3 F BE X Analyzing pin E: + F X = F AE (3/5) + F BE (3/5) 1000 (3/5) 900 = 0 + F Y = F AE (4/5) F BE (4/5) 1000 (4/5) = 0 Solving these two equations, we get F AE = 1750 N = 1.75 kn (C) F BE = 750 N = 0.75 kn (T)

THE METHOD OF SECTIONS Today s Objectives: Students will be able to determine: 1. Forces in truss members using the method of sections. In-Class Activities: Applications Method of Sections Group Problem Solving

APPLICATIONS Long trusses are often used to construct large cranes and large electrical transmission towers. The method of joints requires that many joints be analyzed before we can determine the forces in the middle of a large truss. So another method to determine those forces is helpful.

THE METHODS 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 members will also be either tensile or compressive with the same magnitude as the forces at the joint. This result is based on the equilibrium principle and Newton s third law.

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 external reactions). 3. If required, determine any necessary support reactions by drawing the FBD of the entire truss and applying the E-of-E.

STEPS FOR ANALYSIS 4. Draw the FBD of the selected part of the cut truss. You need to indicate the unknown forces at the cut members. Initially, you may assume all the members are in tension, as done when using the method of joints. Upon solving, if the answer is positive, the member is in tension as per the assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either tension or compression by inspection as was done in the figures above.)

STEPS FOR ANALYSIS 5. Apply the scalar equations of equilibrium (E-of-E) to the selected cut section of the truss to solve for the unknown member forces. Please note, in most cases it is possible to write one equation to solve for one unknown directly. So look for it and take advantage of such a shortcut!

EXAMPLE Given: Loads as shown on the truss. Find: Plan: The force in members KJ, KD, and CD. a) Take a cut through members KJ, KD and CD. b) Work with the left part of the cut section. Why? c) Determine the support reactions at A. What are they? d) Apply the E-of-E to find the forces in KJ, KD and CD.

EXAMPLE: SOLUTION 56.7 kn Analyzing the entire truss for the reactions at A, we get F X = A X = 0. A moment equation about G to find A Y results in: M G = A Y (18) 20 (15) 30 (12) 40 (9) = 0; A Y = 56.7 kn Now take moments about point D. Why do this? + M D = 56.7 (9) + 20 (6) + 30 (3) F KJ (4) = 0 F KJ = 75.1 kn or 75.1 kn ( C )

EXAMPLE: SOLUTION 56.7 kn Now use the x and y-directions equations of equilibrium. + F Y = 56.7 20 30 (4/5) F KD = 0; F KD = 8.38 kn (T) + F X = ( 75.1) + (3/5) (8.38) + F CD = 0; F CD = 70.1 kn (T)

GROUP PROBLEM SOLVING Given: Loads as shown on the truss. Find: The force in members GB and GF. Plan: a) Take the cut through members GF, GB, and AB. b) Analyze the left section. Determine the support reactions at A. Why? c) Draw the FBD of the left section. d) Apply the equations of equilibrium (if possible, try to do it so that every equation yields an answer to one unknown.

GROUP PROBLEM SOLVING: SOLUTION 1) Determine the support reactions at A by drawing the FBD of the entire truss. + F X = A X = 0 + M D = A Y (28) + 600 (18) + 800 (10) = 0; A Y = 671.4 kn Why is A x equal zero by inspection?

GROUP PROBLEM SOLVING: SOLUTION 2) Analyze the left section. + M B = 671.4 (10) + F GF (10) = 0; F GF = 671 kn (C) + F Y = 671.4 F GB = 0; F GB = 671 kn (T)

FRAMES AND MACHINES Today s Objectives: Students will be able to: a) Draw the free-body diagram of a frame or machine and its members. b) Determine the forces acting at the joints and supports of a frame or machine. In-Class Activities: Applications Analysis of a Frame/Machine Group Problem Solving

APPLICATIONS Frames are commonly used to support various external loads. How is a frame different than a truss? To be able to design a frame, you need to determine the forces at the joints and supports.

APPLICATIONS Machines, like those 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? These forces and moments are required when designing the machine s members.

DEFINITION: FRAMES AND MACHINES Frame Machine 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 1. Draw a FBD of the frame or machine and its members, as necessary. Hints: a) Identify any two-force members, F AB b) Note that forces on contacting surfaces (usually between a pin and a member) are equal and opposite, and c) For a joint with more than two members or an external force, it is advisable to draw a FBD of the pin.

STEPS FOR ANALYZING A FRAME OR MACHINE 2. Develop a strategy to apply the equations of equilibrium to solve for the unknowns. Look for ways to form single equations and single unknowns. F AB Problems are going to be challenging since there are usually several unknowns. A lot of practice is needed to develop good strategies and ease of solving these problems.

EXAMPLE Given: The frame supports an external load and moment as shown. Find: The horizontal and vertical components of the pin reactions at C and the magnitude of reaction at B. Plan: a) Draw FBDs of the frame member BC. Why pick this part of the frame? b) Apply the equations of equilibrium and solve for the unknowns at C and B.

EXAMPLE: SOLUTION 800 N m 400 N C X 1 m F AB 45 FBD of member BC (Note AB is a 2-force member!) Please note that member AB is a two-force member. Equations of Equilibrium: Start with B 1 m 2 m M C since it yields one unknown.. + M C = F AB sin45 (1) F AB cos45 (3) + 800 N m + 400 (2) = 0 F AB = 1131 N C Y

EXAMPLE: SOLUTION. 800 N m 400 N C X 1 m B 1 m 2 m C Y F AB 45 FBD of member BC + F X = C X + 1131 sin 45 = 0 C X = 800 N + F Y = C Y + 1131 cos 45 400 = 0 C Y = 400 N

GROUP PROBLEM SOLVING Given: A frame supports a 50-kN load as shown. Find: Plan: The reactions exerted by the pins on the frame members at B and C. a) Draw a FBD of member BC and another one for AC. b) Apply the equations of equilibrium to each FBD to solve for the four unknowns. Think about a strategy to easily solve for the unknowns.

GROUP PROBLEM SOLVING: SOLUTION FBDs of members BC and AC 3.5 m 50 kn C Y 6 m C X A X 8 m Applying E-of-E to member AC: A Y + M A = C Y (8) + C X (6) + 50 (3.5) = 0 (1) + F X = C X A X = 0 + F Y = 50 A Y C Y = 0

GROUP PROBLEM SOLVING: SOLUTION FBDs of members BC and AC 50 kn 3.5 m C Y 6 m C X A X 8 m Applying E-of-E to member BC: A Y + M B = 50 (2) 50 (3.5) + C Y (8) = 0 ; C Y = 34.38 = 34.4 kn From Eq (1), C X can be determined; C X = 16.67 = 16.7 kn + F X = 16.67 + 50 B X = 0 ; B X = 66.7 kn + F Y = B Y 50 + 34.38 = 0 ; B Y = 15.6 kn

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