Lecture 0. Statics Module 1. Overview of Mechanics Analysis
Overview of Mechanics Analysis Procedure of Solving Mechanics Problems Objective : Estimate the force required in the flexor muscle Crandall, An Introduction to the Mechanics of solid, Mc Graw-Hill, 1999 1
Overview of Mechanics Analysis Procedure of Solving Mechanics Problems Step 1) of interest - forearm + hand Step 2) = Idealization and simplification. Neglect the friction at the joint Flexor muscle attached to the forearm at one point etc. Crandall, An Introduction to the Mechanics of solid, Mc Graw-Hill, 1999 Step 3) : Apply the principle of Mechanics to the idealized system to find force, stress and deformation in the system Step 4) : Compare the predicted results with the behavior of the actual system or similar system. ( Test / experiment ) 2
Overview of Mechanics Analysis Procedure of Solving Mechanics Problems Remarks : Modeling (Step 2): most difficult in general Analysis (Step 3) Professional Analysis Software Available such as - NASTRAN, ANSYS, COMSOL, HYPERWORKS, ABAQUS,etc BUT, Understanding of Mechanics is Crucial! For Verification (step 4), Experimental analysis needs to be studied. 3
Overview of Mechanics Analysis Statics Fundaentals first before Solid Mecanics Statics Statics -Force [equilibrium] 14
Module 2. Equilibrium Condition Force and Moment Equilibrium for Bodies
Equilibrium Condition Moment of force Moment of Force cause rotational tendency of the body. A In components of vectors, o e e e x y z M r F x y z F F F x y z Vectors are very useful in dealing with force, moments, etc. yf zf e zf xf e xf yf e z y x x z y y x z 1
Equilibrium Condition Couple Couple : A system of two forces F and -F having the same magnitude, parallel line of action, and opposite sense is said to form a couple cause a rotation of a rigid body. z r B r A F y d F M r F r F A B M F d A B r r F x In the limit of d 0 with M=fixed:. 2
Equilibrium Condition Equilibrium conditions (1) For a particle F 3 F 1 F 2 Fi i 0 Newton s Law F i (2) For a system of particles or bodies. System F 2 F i r i N particles 3
Equilibrium Condition Example Q. Determine the pull P on the rope exerted by the man to hold the box in the position shown. Also find the tension T in the upper rope. Free-Body Diagram (FBD, 자유물체도 ) P Crandall, An Introduction to the Mechanics of solid, Mc Graw-Hill, 1999 y y' T P 200 9 81. N 30 x' x 1 15. sin 17 45. 5 cos 0 954. 4
Equilibrium Condition Example (continued) Analysis Using x-y axes F x 0 P 871 N, T 2513 N F y Or Analysis Using x -y axes 0 y' T y P 30 x' x 200 9 81. N 5
Equilibrium Condition Remark We considered equilibrium conditions around a point (Thus only O F 0 used; i.e. force eqm. Only!) * For a 2-D Body* 3 Conditions A A x y z F 0, F 0, M 0 B OR (See page Module 7 of Lecture 0 to see the equivalence of these conditions.) *Body under forces: a collection of particles (such as a continuum) where applied forces may be applied at different locations. 6
Equilibrium Condition Special case 1: Two-force system * Consider Two Forces acting on a 2-D body at different locations Eqm conditions: 7
Equilibrium Condition Special case 2: Three-force system The lines of action of the three forces ( i.e. pass through the same point ) Limiting case of three-force system 8
Module 3. Free Body Diagram
Free Body Diagram Free-Body Diagram FBD is the most important concept in mechanics (require some experience of proper modeling) sketch of an isolated system of interest with all forces (moment) acting on it shown. The system should be disconnected from boundaries or other systems that are not of interest Example F 1
Free Body Diagram Mechanical action at supports things to know for FBD Analysis 1) Flexible cable support Flexible: no shear, no compression, no bending.. Tension only! System of interest Flexible cable support 2) surface Discussion on friction will be given later Contact surface System of interest System of interest Smooth surface Rough surface System of interest 2
Free Body Diagram Mechanical action at supports 3) Roller support Roller Roller Bull Rocker 4) Pinned support 5) Built-in, clamped, or fixed Free to turn (usual case) Not free to turn (unusual case) Built-in Weld (or fixed) 3
Module 4. Use of Free-Body Diagram (FBD) for Force Analysis in Statics
Use of Free-body diagram Example 1 2m B C Ex 1. A uniform 8kg bar with small end rollers is supported by the horizontal and vertical surfaces and by wire AC. Calculate the tension T in the wire and the reaction forces at A and B. Solve by using two moment equations and one force equation. (use g=10m/sec 2 here for simple calculation) 2m A 3m 1
Use of Free-body diagram Example 1 - solution System of interest: bar+two end rollers 2 1 2 tan tan 33.7 3 3 M M F y A E 0: 0: 0: T R y 2
Example 1 Remarks Equivalence MA 0 (1) MA 0 (1) ME 0 (2) Fx 0 (4) Fy 0 (3) Fy 0 (3) ME rei Fi 0 (2) i M A rai Fi 0 (1) i Here, Fi (F xi, Fyi ) : vector (2)- (1) Thus, M M r r F E A Ei Ai i i r F 0 (5) EA M M r F 0 (6) E A EA i i Because M 0 (Eq. (1)), A Eq. (6) becomes : AE yi y AE xi x r F 0 i.e., r ( F e F e ) 0 (7) AE i AE xi x yi y Because r / / F e, r F e 0 Thus, F xi 0 (i.e., F 0) x y B R x r EA C T E 80 N R y A x Use of Free-body diagram 3
Use of Free-body diagram Example 2 Q. Determine the axial force in member BH of the truss 6 m 20N 40N 6 m 6 m 12 m Roller Crandall, An Introduction to the Mechanics of solid, Mc Graw-Hill, 1999 4
Use of Free-body diagram Example 2 - Solution Solution Procedure ( Point : how to use FBD ) Step 1 : Reaction First at A and B -> FBD of whole system F x R Ax 0 F R R 20 40 0 y Ay Ey M A 24 R 20 6 4012 0 Ey R R R Ax Ay Ey 0 35 N 25 N 5
Use of Free-body diagram Example 2 Solution (continued) Solution Procedure ( Point : how to use FBD ) Step 2-(a) : Equilibrium around point A F F F cos 45 0 F x AB AG y 35 F sin 45 0 AG F 35 N, F 49.5 N AB AG 6
Use of Free-body diagram Example 2 Solution (continued) Solution Procedure ( Point : how to use FBD ) Step 2-(b) : Equilibrium around point G F F F sin 45 0 x GH AG F F F cos 45 0 y BG AG F 35 N, F 35 N BG GH 7
Use of Free-body diagram Example 2 Solution (continued) Solution Procedure ( Point : how to use FBD ) Step 2-(c) : Equilibrium around point B G H F BG F BH A F AB B F BG FAB B B B F BH F BC B F BC C F BG F F F F sin 45 20 0 y BG BH 35 N ( known) BH 21.2 N 20 N 8
Module 5. More Examples Using FBD in Statics
More examples Example 3 Q. Determine the reactions on its members. (note that Member a-b-c is a single rigid member.) All joints are pined joints except at E. 2 g 9.81m/s Roller support Crandall, An Introduction to the Mechanics of solid, Mc Graw-Hill, 1999 1
More examples Example 3 - solution Solution Procedure ( Point : how to use FBD ) Step 1 : FBD of the whole system to find reactions at A and E F R R 0 x Ax Ex F R 20g 0 y Ay M 0.6 R 0.6 20g 0 A Ex R R R 20g 196.2 N Ax Ay Ex 2
More examples Example 3 - solution (continued) Solution Procedure ( Point : how to use FBD ) Step 2-(a) : FBD of each member 3
More examples Example 3 - solution (continued) Solution Procedure ( Point : how to use FBD ) Step 2-(b) : For member ABC F R R 0 x Ax Cx F R R F 196.2 0 y Ay Cy BD M 0.3 F 0.6 R 0.6196.2 0 A BD Cy R 196.2 N, R F 392.4 N Cx Cy BD From eqm of member CDE, the same results are found. 4
Module 6. Two Kinds of Mechanical Systems from Mechanics Viewpoint
Two Kinds of Mechanical Systems Two kinds of mechanical systems 1. Statically Determinate System [SDS] ( 靜定系, 정정계 ) - - - Solved within Statics. The exampled we solved so far! 2. Statically Indeterminate System [SIS] ( 不靜定系, 부정정계 ) - cannot determine forces/moments of interest only by equilibrium conditions - 1
Two Kinds of Mechanical Systems SDS (Statically Determinate System) SDS (Statically Determinate System) y A x L P B Reaction force at A =? F x y H A A A 0 F V P A 0 M M PL 0 2
Two Kinds of Mechanical Systems SIS (Statically Indeterminate System) y A x L 1 P L 2 B Reaction force at A and B=? Fx HA 0 Fy VA VB P 0 A B M M V L L PL A 3 equations for 4 unknowns 1 2 1 0 cannot solve them without additional equations. 3
Two Kinds of Mechanical Systems SIS (continued) Additional Equations SIS (Statically Indeterminate System)? L 1 L 2 y A x P B y x w L 1 4
Module 7. Appendix
Appendix Appendix Derivation of F 0, r F 0 ext ext for a body System - For the i th particle, F i i f ij f ji j i +1 N F f 0 for i 1,2,, N i j1 i j ij (assuming single external force is applied to each particle) F i f ij N particles : external force : Internal force - Summing up the equations from i =1 to i =N N N F f 0, f f ( i j) i ij ij ji i1 j1 i j N i1 F i 0 1 Only External forces matter!!!
Appendix Appendix Consider F 0 but NOT in equilibrium ext By summing up the eqm. equations for particles, we did not consider along which lines the forces are applied. We also need to consider the moment equilibrium. - For each particle, N ri Fi fij 0 j1 i j for i 1,2,, N By equilibrium condition for a particle O System r i r j F j i f ij f ji j F i i +1 Fi 1 N particles 2
Appendix Appendix Summing up the equations from i =1 to i =N N N r F r f 0 i i i ij i1 j1 i j The result is N i 1 r F 0 i i External forces only N r N i i1 j1 i j f ij r1 f12 r1 f13 r2 f21 r2 f23 i F i r f r f 1 12 2 21 r f r d f 1 12 1 12 21 r f f d f 1 12 21 12 21 d f 0 d / / f 12 21 12 21 O r i r j F j d ij j f ij f ji 3
Appendix Appendix Thus we find for a system of particles that the equilibrium condition is dictated by (*) N i1 N i1 F r i i 0 F i 0 F i r i : External forces only : position of the loaded point in the body. Remarks: If (*) is satisfied, each particle of the system is also in equilibrium. 4
Appendix Appendix F i F j i j i +1 Fi 1 i F i Isolated i th particle : equilibrium Extension to a continuous body (Rigid or Deformable) (*) N i1 N i1 Fi 0 F 0 rifi 0 rf 0 F 1 F 2 F 3 F 1 F 2 Any possible sub-body must be also in equilibrium. F 3 5