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Engineering 36 Chp 6: Frames Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Engineering-36: Engineering Mechanics - Statics 1 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Introduction: MultiPiece Structures • For the equilibrium of structures made of several connected parts, the internal forces as well the external forces are considered. • In the interaction between connected parts, Newton’s 3rd Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense. • Three categories of engineering structures are considered: • Frames: contain at least one multi-force member, i.e., a member acted upon by 3 or more forces. • Trusses: formed from two-force members, i.e., straight members with end point connections • Machines: structures containing moving parts designed to transmit and modify forces. Engineering-36: Engineering Mechanics - Statics 2 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Analysis of Frames • Frames and machines are structures with at least one multiforce member. Frames are designed to support loads and are usually stationary. Machines contain moving parts and are designed to transmit & modify forces. • A free body diagram of the complete frame is used to determine the external forces acting on the frame. • Internal forces are determined by dismembering the frame and creating free-body diagrams for each component. • Forces on two force members have known lines of action but unknown magnitude and sense. • Forces on multiforce members have unknown magnitude and line of action. They must be represented with two unknown components. • Forces between connected components are equal, have the same line of action, and opposite sense. Engineering-36: Engineering Mechanics - Statics 3 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Not Fully Rigid Frames • Some frames may collapse if removed from their supports. Such frames can NOT be treated as rigid bodies. • A free-body diagram of the complete frame indicates four unknown force components which can not be determined from the three equilibrium conditions. • The frame must be considered as two distinct, but related, rigid bodies. • With equal and opposite reactions at the contact point between members, the two free-body diagrams indicate 6 unknown force components. • Equilibrium requirements for the two rigid bodies yield 6 independent equations. Engineering-36: Engineering Mechanics - Statics 4 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Example: Pin & Roller Frame SOLUTION PLAN • Create a free-body diagram for the complete frame and solve for the support reactions. (won’t collapse) • Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown magnitude (2-frc member); determined by summing moments about C. Members ACE and BCD are • With the force on the link DE known, connected by a pin at C and the sum of forces in the x and y by the link DE. For the directions may be used to find the force loading shown, determine components at C. the force(s) in link DE and • With member ACE as a free-body, the components of the check the solution by summing force exerted by the Pin at moments about A C on member BCD. Engineering-36: Engineering Mechanics - Statics 5 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Example: Pin & Roller Frame SOLUTION • Create a free-body diagram for the COMPLETE FRAME and solve for the support reactions. F M 0 A y 480 N y A y 480 N 0 480 N 100 mm B 160 mm A B 300 N F x A x 300 N 0 B Ax Note tan Engineering-36: Engineering Mechanics - Statics 6 1 80 150 28 . 07 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Example: Pin & Roller Frame • Define a free-body diagram for member BCD. The force exerted by the 2-force link DE has a known line of action but unknown magnitude. It is determined by summing moments about C. M C 0 F DE sin 250 mm 300 N 8 0 mm 480 N 100 mm F DE 561 N (DE in Compression) F DE 561 N C • Use the Sum of forces in the x and y directions to find the force components at C. F x 0 C x F DE cos 300 N 0 C x 561 N cos 300 N F y 0 C y F DE sin 480 N 0 C y 561 N sin 480 N Engineering-36: Engineering Mechanics - Statics 7 C x 795 N C y 216 N Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Example: Pin & Roller Frame With member ACE as a freebody, check the solution by summing moments about A M A F DE cos 300 mm F DE sin 100 mm C x 220 mm 561 cos 300 mm 561 sin 100 mm 795 220 Engineering-36: Engineering Mechanics - Statics 8 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx mm 0 Example: Pin & Pin Frame For the Frame Shown Below Find • The RCNs at points A&C Draw FBDs noting that the forces at B are Equal & Opp • FBD for AB • The SHEAR FORCES acting on the PINS at A & C Engineering-36: Engineering Mechanics - Statics 9 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Example: Pin & Pin Frame • FBD for Member BC For AB take ΣMB = 0 M B 0 150 lb 3 ft F Ay 6 ft F Ay 450 ft lb 6 ft 75 . 0 lb F Ay 75 . 0 lb Engineering-36: Engineering Mechanics - Statics 10 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Example: Pin & Pin Frame For AB take ΣFy = 0 • Recall FAy = 75 lbs For AB take ΣFx = 0 F x 0 F Ax F Bx F Bx F Ax • Will use Later: FBx = FAx F y 0 F Ay F By 150 lb F By 150 lb F Ay F By 150 lb 75 lb 75 lb F By 75 . 0 lb Engineering-36: Engineering Mechanics - Statics 11 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Example: Pin & Pin Frame For BC take ΣMC = 0 M C 0 0 150 ft lb F By 2 ft F Bx 4 ft sin 60 Recall : F By 75 . 0 lb F Bx 150 ft lb 150 ft lb 3 . 46 ft F Bx 86 . 6 lb F Bx 86 . 6 lb For BC take ΣFx = 0 F x 0 F Bx FCx FCx F Bx 86 . 6 lb FCx 86 . 6 lb Engineering-36: Engineering Mechanics - Statics 12 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Example: Pin & Pin Frame For BC take ΣFy = 0 F y 0 FCy F By FCy F By 75 lb FCy 75 . 0 lb Recall from Before: FAx = FBx, and FBx = 86.6 lbs • Thus F Ax 86 . 6 lb Engineering-36: Engineering Mechanics - Statics 13 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Example: Pin & Pin Frame Now Find SHEAR FORCE on Pins at A&C Find the Shear Force Magnitude by F F x F 2 2 y In this Case The Forces at A & C F A F Ax iˆ F Ay ˆj F C FCx iˆ FCy ˆj Engineering-36: Engineering Mechanics - Statics 14 FA 86 . 6 lb 2 7 5 lb 2 115 lb FC 86 . 6 lb 2 7 5 lb 2 115 lb Thus the connecting pins must resist a shear force of 115 lbs Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx WhiteBoard Work Let’s Work This Nice Problem 30cm Find the Forces Acting on Each of the Members, and on the Frame at Pts A & D Engineering-36: Engineering Mechanics - Statics 15 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Engineering-36: Engineering Mechanics - Statics 16 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx Engineering 36 Appendix Bruce Mayer, PE Registered Electrical & Mechanical Engineer [email protected] Engineering-36: Engineering Mechanics - Statics 17 Bruce Mayer, PE [email protected] • ENGR-36_Lec-16_Frames.pptx