6.5 Space Trusses A space truss consists of members joined together at their ends to form a stable 3D structure The simplest space truss is a tetrahedron, formed by joined 6 members as shown Any additional members added would be redundant in supporting force P

6.5 Space Trusses Assumptions for Design The members of a space truss may be treated as two force members provided the external loading is applied at the joints and the joints consist of ball and socket connections If the weight of the member is to be considered, apply it as a vertical force, half of its magnitude applied at each end of the member

6.5 Space Trusses Procedure for Analysis Method of Joints To determine the forces in all the members of the truss Solve the three scalar equilibrium ∑Fx = 0, ∑Fy = 0, ∑Fz = 0 at each joint The force analysis begins at a point having at least one unknown force and at most three unknown forces Cartesian vector analysis used for 3D

6.5 Space Trusses Procedure for Analysis Method of Sections Used to determine a few member forces When an imaginary section is passes through a truss and the truss is separated into two parts, the below equations of equilibrium must be satisfied ∑Fx = 0, ∑Fy = 0, ∑Fz = 0 ∑Mx = 0, ∑My = 0, ∑Mz = 0 By proper selection, the unknown forces can be determined using a single equilibrium equation

6.5 Space Trusses Example 6.8 Determine the forces acting in the members of the space truss. Indicate whether the members are in tension or compression.

6.5 Space Trusses View Free Body Diagram Solution Joint A

6.5 Space Trusses Solution Joint A To show

6.6 Frames and Machines Composed of pin-connected multi-force members (subjected to more than two forces) Frames are stationary and are used to support the loads while machines contain moving parts, designated to transmit and alter the effects of forces Apply equations of equilibrium to each member to determine the unknown forces

6.6 Frames and Machines Free-Body Diagram Isolate each part by drawing its outlined shape - show all the forces and the couple moments that act on the part - label or identify each known and unknown force and couple moment with reference to the established x, y and z coordinate system

6.6 Frames and Machines Free-Body Diagram - indicate any dimension used for taking moments - equations of equilibrium are easier to apply when the forces are represented in their rectangular coordinates - sense of any unknown force or moment can be assumed

6.6 Frames and Machines Free-Body Diagram Identify all the two force members in the structure and represent their FBD as having two equal but opposite collinear forces acting at their points of application Forces common to any contracting member act with equal magnitudes but opposite sense on the respective members

6.6 Frames and Machines Free-Body Diagram - treat two members as a system of connected members - these forces are internal and are not shown on the FBD - if the FBD of each member is drawn, the forces are external and must be shown on the FBD

6.6 Frames and Machines Example 6.9 For the frame, draw the free-body diagram of each member, the pin at B and the two members connected together.

6.6 Frames and Machines Solution Part (a) members BA and BC are not two-force members BC is subjected to 3 forces, the resultant force from pins B and C and the external P AB is subjected to the resultant forces from the pins at A and B and the external moment M

6.6 Frames and Machines Solution Part (b) Pin at B is subjected to two forces, force of the member BC on the pin and the force of member AB on the pin For equilibrium, these forces and respective components must be equal but opposite

6.6 Frames and Machines Solution Part (b) But Bx and By shown equal and opposite on members AB ad BC results from the equilibrium analysis of the pin rather from Newton’s third law

6.6 Frames and Machines Solution Part (c) FBD of both connected members without the supporting pins at A and C Bx and By are not shown since they form equal but opposite collinear pairs of internal forces

6.6 Frames and Machines Solution Part (c) To be consistent when applying the equilibrium equations, the unknown force components at A and C must act in the same sense Couple moment M can be applied at any point on the frame to determine reactions at A and C

6.6 Frames and Machines Example 6.10 A constant tension in the conveyor belt is maintained by using the device. Draw the FBD of the frame and the cylinder which supports the belt. The suspended black has a weight of W.

6.6 Frames and Machines Solution Idealized model of the device Angle θ assumed known Tension in the belt is the same on each side of the cylinder since it is free to turn

6.6 Frames and Machines Solution FBD of the cylinder and the frame Bx and By provide equal but opposite couple moments on the cylinder Half of the pin reactions at A act on each side of the frame since pin connections occur on each side

6.6 Frames and Machines Example 6.11 Draw the free-body diagrams of each part of the smooth piston and link mechanism used to crush recycled cans.

6.6 Frames and Machines Solution Member AB is a two force member FBD of the parts

6.6 Frames and Machines Solution Since the pins at B and D connect only two parts together, the forces are equal but opposite on the separate FBD of their connected members Four components of the force act on the piston: Dx and Dy represent the effects of the pin and Nw is the resultant force of the floor and P is the resultant compressive force caused by can C

6.6 Frames and Machines Example 6.12 For the frame, draw the free-body diagrams of (a) the entire frame including the pulleys and cords, (b) the frame without the pulleys and cords, and (c) each of the pulley.

6.6 Frames and Machines Solution Part (a) Consider the entire frame, interactions at the points where the pulleys and cords are connected to the frame become pairs of internal forces which cancel each other and not shown on the FBD

6.6 Frames and Machines Solution Part (b) and (c) When cords and pulleys are removed, their effect on the frame must be shown

6.6 Frames and Machines Example 6.13 Draw the free-body diagrams of the bucket and the vertical boom of the back hoe. The bucket and its content has a weight W. Neglect the weight of the members.

6.6 Frames and Machines Solution Idealized model of the assembly Members AB, BC, BE and HI are two force members

6.6 Frames and Machines Solution FBD of the bucket and boom Pin C subjected to 2 forces, force of the link BC and force of the boom Pin at B subjected to three forces, force by the hydraulic cylinder and the forces caused by the link These forces are related by equation of force equilibrium

6.6 Frames and Machines Equations of Equilibrium Provided the structure is properly supported and contains no more supports and members than necessary to prevent collapse, the unknown forces at the supports and connections can be determined from the equations of equilibrium The selection of the FBD for analysis are completely arbitrary and may represent each of the members of the structure, a portion or its entirety.

6.6 Frames and Machines Equations of Equilibrium Consider the frame in fig (a) Dismembering the frame in fig (b), equations of equilibrium can be used FBD of the entire frame in fig (c)

6.6 Frames and Machines Procedures for Analysis FBD Draw the FBD of the entire structure, a portion or each of its members Choice is dependent on the most direct solution to the problem When the FBD of a group of members of a structure is drawn, the forces at the connected parts are internal forces and are not shown Forces common to two members which are in contact act with equal magnitude but opposite sense on their respective FBD

6.6 Frames and Machines Procedures for Analysis FBD Two force members, regardless of their shape, have equal but opposite collinear forces acting at the ends of the member In many cases, the proper sense of the unknown force can be determined by inspection Otherwise, assume the sense of the unknowns A couple moment is a free vector and can act on any point of the FBD

6.6 Frames and Machines Procedures for Analysis FBD A force is a sliding vector and can act at any point along its line of action Equations of Equilibrium Count the number of unknowns and compare to the number of equilibrium equations available In 2D, there are 3 equilibrium equations written for each member

6.6 Frames and Machines Procedures for Analysis Equations of Equilibrium Sum moments about a point that lies at the intersection of the lines of action of as many unknown forces as possible If the solution of a force or couple moment magnitude is found to be negative, it means the sense of the force is the reserve of that shown on the FBD

6.6 Frames and Machines Example 6.14 Determine the horizontal and vertical components of the force which the pin C exerts on member CB of the frame.

6.6 Frames and Machines Solution Method 1 Identify member AB as two force member FBD of the members AB and BC

6.6 Frames and Machines Solution

6.6 Frames and Machines Solution Method 2 Fail to identify member AB as two force member

6.6 Frames and Machines Solution Member AB

6.6 Frames and Machines Solution Member BC

6.6 Frames and Machines Example 6.15 The compound beam is pin connected at B. Determine the reactions at its support. Neglect its weight and thickness.

6.6 Frames and Machines Solution FBD of the entire frame Dismember the beam into two segments since there are 4 unknowns but 3 equations of equilibrium

6.6 Frames and Machines Solution Segment BC

6.6 Frames and Machines Solution Member AB

6.6 Frames and Machines Example 6.16 Determine the horizontal and vertical components of the force which the pin at C exerts on member ABCD of the frame.

6.6 Frames and Machines Solution Member BC is a two force member FBD of the entire frame FBD of each member

6.6 Frames and Machines Solution Entire Frame

6.6 Frames and Machines Solution Member CEF

6.6 Frames and Machines Example 6.17 The smooth disk is pinned at D and has a weight of 20N. Neglect the weights of others member, determine the horizontal and vertical components of the reaction at pins B and D

6.6 Frames and Machines Solution FBD of the entire frame FBD of the members

6.6 Frames and Machines Solution Entire Frame

6.6 Frames and Machines Solution Member AB

6.6 Frames and Machines Solution Disk

6.6 Frames and Machines Example 6.18 Determine the tension in the cables and also the force P required to support the 600N force using the frictionless pulley system.

6.6 Frames and Machines Solution FBD of each pulley Continuous cable and frictionless pulley = constant tension P Link connection between pulleys B and C is a two force member

6.6 Frames and Machines Solution Pulley A Pulley B Pulley C

6.6 Frames and Machines Example 6.19 A man having a weight of 750N supports himself by means of the cable and pulley system. If the seat has a weight of 75N, determine the force he must exert on the cable at A and the force he exerts on the seat. Neglect the weight of the cables and pulleys.

6.6 Frames and Machines Solution Method 1 FBD of the man, seat and pulley C

6.6 Frames and Machines Solution Man Seat Pulley C

6.6 Frames and Machines Solution Method 2 FBD of the man, seat and pulley C as a single system

6.6 Frames and Machines Solution

6.6 Frames and Machines Example 6.20 The hand exerts a force of 35N on the grip of the spring compressor. Determine the force in the spring needed to maintain equilibrium of the mechanism.

6.6 Frames and Machines Solution FBD for parts DC and ABG

6.6 Frames and Machines Solution Lever ABG Pin E

6.6 Frames and Machines Solution Arm DC

6.6 Frames and Machines Example 6.21 The 100kg block is held in equilibrium by means of the pulley and the continuous cable system. If the cable is attached to the pin at B, compute the forces which this pin exerts on each of its connecting members

6.6 Frames and Machines Solution FBD of each member of the frame View Free Body Diagram Solution FBD of each member of the frame Ad and CB are two force members

6.6 Frames and Machines Solution Pulley B

6.6 Frames and Machines Solution Pin E Two force member BC subjected to bending as caused by FBC Better to make this member straight so that the force would only cause tension in the member

Chapter Summary Truss Analysis A simple truss consists of triangular elements connected by pin joints The force within determined by assuming all the members to be two force member, connected concurrently at each joint Method of Joints For the truss in equilibrium, each of its joint is also in equilibrium

Chapter Summary Method of Joints For a coplanar truss, the concurrent force at each joint must satisfy force equilibrium For numerical solution of the forces in the members, select a joint that has FBD with at most 2 unknown and 1 known forces Once a member force is determined, use its value and apply it to an adjacent joint Forces that pull on the joint are in tension

Chapter Summary Method of Joints Forces that push on the joint are in compression To avoid simultaneous solution of two equations, sum the force in a direction that is perpendicular to one of the unknown To simplify problem-solving, first identify all the zero-force members

Chapter Summary Method of Sections For the truss in equilibrium, each section is also in equilibrium Pass a section through the member whose force is to be determined Draw the FBD of the sectioned part having the least forces on it Forces that pull on the section are in tension

Chapter Summary Method of Sections Forces that push on the section are in compression For a coplanar force system, use the three equations of equilibrium for solving If possible, sum the force in a direction that is perpendicular to two of the three unknown forces Sum the moment about a point that passes through the line of action of two of the three unknown forces

Chapter Summary Frames and Machines The forces acting at the joints of a frame or machine can be determined by drawing the FBD of each of its members or parts Principle of action-reaction should be observed when drawing the forces For coplanar force system, there are three equilibrium equations for each member

Chapter Review

Chapter Review

Chapter Review

Chapter Review

Chapter Review