DYNAMIC FORCE ANALYSIS OF MECHANISMS DYNAMIC FORCE ANALYSIS OF MECHANISMS KINEMATICS α ω DYNAMICS Content Copyright Nanyang Technological University 2 DYNAMIC FORCE ANALYSIS OF MECHANISMS Newton’s Second Law D’Alembert’s Principle Dynamic force analysis - examples NEWTON’S SECOND LAW OF MOTION Translation: a F2 m F1 ∑ F i = ma F3 Two Scalar Equations ∑F i.e. {∑ Content Copyright Nanyang Technological University ix = ma x F iy = ma y 4 NEWTON’S SECOND LAW OF MOTION Rotation about a fixed point: ∑M i = Iα I - moment of inertia about fixed point F1 α - angular acceleration T2 α F3 4-bar Animation T3 T1 F2 Content Copyright Nanyang Technological University Pivot point 5 NEWTON’S SECOND LAW OF MOTION GENERAL PLANAR MOTION Translation of CG ∑ F j = maG G F2 Rotation about CG ∑M jG = I Gα F1 AG α T1 T2 F3 = F m a ∑ j G ∑M jG = I Gα Subscript G is denoted as the center of mass IG is the mass moment of inertia about G Content Copyright Nanyang Technological University 6 End of topic one D’ALEMBERT’S PRINCIPLE – FORCE EQUATION Newton’s law: F1 AG Newton’s law: CG F2 ∑ F j = maG α T1 ∑ M jG = IGα T2 ∑ F j = maG ⇒ ∑ F j − maG = 0 Define inertial force to be: FI = −maG F3 D’Alembert’s Principle: CI = IGα We have (mathematically): F1 ∑ Fi = 0 F2 ∑M G T2 T1 F3 FI = ma G i =0 ∑ F j + FI = 0 ∑ Fi = 0 Imaginary (mathematic) ‘static’ equilibrium state with an additional inertial force FI = −maG Content Copyright Nanyang Technological University 8 D’ALEMBERT’S PRINCIPLE – MOMENT EQUATION Newton’s law: F1 AG Newton’s law: ∑M CG F2 ∑ F j = maG α T1 ∑M T2 jG = I Gα CI = IGα F1 ∑ Fi = 0 F2 ∑M G T2 T1 F3 FI = ma G i =0 = I Gα ⇒ ∑ M jG − I Gα = 0 Define inertial moment to be: C I = − I Gα F3 D’Alembert’s Principle: jG We have (mathematically): ∑M jG ∑M =0 i + CI = 0 Imaginary (mathematic) ‘static’ equilibrium state with an additional inertial moment Content Copyright Nanyang Technological University C I = − I Gα 9 D’ALEMBERT’S PRINCIPLE D’Alembert’s Principle Newton’s law: D’Alembert’s Principle: F1 AG CI = IGα CG F1 F2 F2 α G T1 T2 T2 T1 F3 FI F3 ∑ F j = maG ∑ Fi = 0 ∑M ∑M jG = I Gα Content Copyright Nanyang Technological University i = maG =0 10 D’ALEMBERT’S PRINCIPLE – LINK EXAMPLE D’Alembert’s Principle: Newton’s law: FI FI FA C.G. FA FI = maG A A CI = IGα TI TI AG B B α FB C.G. FB Dynamic Problem ∑ F j = maG ∑M jG Equivalent Static Equilibrium ∑ Fj = 0 ∑M = I Gα Content Copyright Nanyang Technological University j =0 11 D’ALEMBERT’S PRINCIPLE CONVENTION FOR INERTIAL FORCE & MOMENT y G a α x G a x G ma y G ma C.G. Motion(Kinematics) IGα Inertial force & moment Content Copyright Nanyang Technological University 12 D’ALEMBERT’S PRINCIPLE SUMMARY OF THE STEPS FOR DYNAMIC FORCE ANALYSIS 1. Find accel. of CG & angular accel. For each link (or use given ones) 2. Free individual body(s) 3. Draw inertia force at CG & inertia moment about CG 4. Draw applied force/moments 5. Draw assumed constraint forces 6. Write equilibrium equations for each free body (total 3N) 7. Solve the equations for unknowns Content Copyright Nanyang Technological University 13 End of topic 2 MASS CENTRE, MOMENT OF INERTIA FOR RIGID BODY The mass moment of inertia of a rigid body is calculated in the following way: I G = ∫ r 2 ρdV Where: ρ isi the density of rigid body . r dV G r is the distance from the centre of mass to an arbitrary point inside the body The mass moment of inertia is taken about the centre of mass point G. Content Copyright Nanyang Technological University 15 MASS CENTRE, MOMENT OF INERTIA FOR RIGID BODY For a homogenous body of regular geometry, the mass moment of inertia can be calculated from equation directly. The following table gives some mass moment of inertia of a body with simple geometry L 1 mL2 12 1 I O = mL2 3 IG = Slender rod G O r IG = Disk/Cylinder 1 mr 2 2 a Rectangular Plate b Content Copyright Nanyang Technological University IG = 1 m(a 2 + b 2 ) 12 16 MASS CENTRE, MOMENT OF INERTIA FOR RIGID BODY r2 r1 Ring r Thin ring IG = 1 m ( r12 + r22 ) 2 I G = mr 2 IG = G Semi-circular plate r h O Content Copyright Nanyang Technological University 1 mr 2 2 1 I O = mr 2 − mh 2 2 4r h= 3π 17 PARALLEL AXIS THEOREM For mass moment of inertia being taken about a point A other than the centre of mass G, we can derive the mass moment of inertia using the following parallel axis theorem: I A = I G + m | r AG | 2 Where: is the position vector from rA IA IG G point A to the centre of mass G r AG A The mass moment of inertia is taken about an arbitrary point A. Content Copyright Nanyang Technological University 18 EXAMPLE FOR MASS CENTRE, MOMENT OF INERTIA An assembly of a uniform slender and disk A uniform slender rod of mass ( and length ( ) ) and a uniform disk of mass ( ) and radius ( ) are welded to form an assembly as shown in the figure. Determine the mass moment of inertia of the assembly about its centre of mass G and point C. The moments of inertia of the disk and the rod about their mass centres A and B are known to be ( ) and ( ) respectively. Content Copyright Nanyang Technological University 19 EXAMPLE FOR MASS CENTRE, MOMENT OF INERTIA An assembly of a uniform slender and disk Solution: • Center of Mass: Using point A as the origin, the centre of mass can located as Note rAA=0 ⇒ (m) Content Copyright Nanyang Technological University 20 EXAMPLES FOR MASS CENTRE, MOMENT OF INERTIA An assembly of a uniform slender and disk Solution (continued): • Mass Moment of Inertia about G: Using the parallel axis theorem, the moment of inertia can be calculated as follows: the disk: (kg m2) the rod: (kg m2) and • (kg m2) Mass Moment of Inertia about C: Similarly, we have (kg m2) Content Copyright Nanyang Technological University 21 End of topic 3 EXAMPLES FOR DYNAMIC FORCE ANALYSIS Rotation of a uniform slender rod hinged at point A A G α g L 4 Example 1: The uniform slender rod of mass m and length L is pivoted at A in the position as shown. The rod is released from rest. Determine the initial angular acceleration of the rod and the constraint force at A. (Given IG = 1 mL2 12 ) Content Copyright Nanyang Technological University 23 EXAMPLES FOR DYNAMIC FORCE ANALYSIS Acceleration analysis of a uniform slender rod hinged at point A Rotation of a uniform slender rod hinged at point A G A α y g G α A x r AG L 4 (Given IG = 1 mL2 12 L 4 ) aG Solution: • Kinematics Analysis: Letting α be the angular acceleration of the rod, we have a =0 x G L a =− α 4 y G Content Copyright Nanyang Technological University 24 EXAMPLE FOR DYNAMIC FORCE ANALYSIS • Solution (use D’Alembert’s Principle): ∑M A =0 α= ∑F x y FAy L L L − mg + m α × + I Gα = 0 4 4 4 ⇒ FI = maG = m . L α 4 C = I α I G G 12 g 7 L A FAx mg =0 FAx = 0 ∑F y G α A =0 x r AG FAy + m ⇒ y FAy = L α − mg = 0 4 L 4 4 mg 7 Content Copyright Nanyang Technological University aG a =0 x G L a =− α 4 y G 25 EXAMPLES FOR DYNAMIC FORCE ANALYSIS Example 2: Analyse the planar in-line slider-crank mechanism for the given dimensions as shown in diagram. Given the input ω1 and α1, determine the required torque T1 and the bearing forces at joints. An in-line slider-crank mechanism Content Copyright Nanyang Technological University 26 EXAMPLES FOR DYNAMIC FORCE ANALYSIS Solution: • Kinematics Analysis (velocity): y Using the relative velocity, we can find v C = v B + v C / B = ω 1 × rO1B + ω 2 × rBC ⇒ ω 2 = −2.1k̂ (rad/s)(CW) and vC = −14.85iˆ (in/s) vB B 66.4 G1 1" T1 O1 1.88" G2 2" 70 vC / B 3.76" 30 Content Copyright Nanyang Technological University C G3 vC 27 EXAMPLES FOR DYNAMIC FORCE ANALYSIS • Kinematics Analysis (acceleration): Using the relative acceleration, we can find: n + t + n + t aC = a B + aC / B = a B a B aC / B aC / B ⇒ α 2 = 46kˆ (rad/s2)(CCW) and aC = −70iˆ (in/s2). With the angular velocities and accelerations, the following quantities can be obtained: aG1 x = −71.6iˆ (in/s2), aG1 y = −79.5j (in/s2), aG 2 x = −106iˆ (in/s2), aG 2 y = −80.0j (in/s2). Content Copyright Nanyang Technological University 28 EXAMPLES FOR DYNAMIC FORCE ANALYSIS Free body diagram of the slider-crank using D’alembert Principle FBy FBx m 2 aG 2 y B G2 m 2 aG 2 x I G 2α 2 FCy 30 C FBx FCy B m1aG1 y FO1 y FCx FBy G1 FCx m 3 aG 3 m1aG1 x T1 I G1α1 70 O1 FN FO1 x Content Copyright Nanyang Technological University 29 EXAMPLES FOR DYNAMIC FORCE ANALYSIS Solution (continued): • Dynamic Analysis of Link 3 (Slider): From free body diagram, we have ⇒ ∑ F =0 ⇒ ∑F ∑F FCy FCx m 3 aG 3 FN Free body diagram of the slider using D’Alembert Principle ⇒ x = 0 : FCx + m3 aG 3 x = 0 y = 0 : FN − FCy = 0 FCx = −7.776 ×10 −3 × 70 = −0.543 (lb) F = F Cy N Content Copyright Nanyang Technological University 30 EXAMPLES FOR DYNAMIC FORCE ANALYSIS • Dynamic Analysis of Link 2: From free body diagram, we have ∑MB = 0 rBC × ( − FCx iˆ + FCy ˆj ) − I G 2α 2 + rBG 2 × m 2 (aG 2 x iˆ + aG 2 y ˆj ) = 0 ⇒ − rBC y × FCx + rBC x × FCy − I G 2α 2 + rBG 2 y × m 2 aG 2 x + rBG 2 x × m 2 aG 2 y = 0 ⇒ ⇒ FCy = −0.396 ∑F ∑F x = 0 : FBx − FCx + m2 aG 2 x = 0 y = 0 : FBy + FCy + m2 aG 2 y = 0 F = −1.09 ⇒ Bx FBy = −0.018 FBy FCy FBx (lb) m 2 aG 2 y B G2 m 2 aG 2 x I G 2α 2 FCy 30 C FCx m 3 aG 3 FCx FN Free body diagram of the slider-crank using D’Alembert Principle Content Copyright Nanyang Technological University 31 EXAMPLES FOR DYNAMIC FORCE ANALYSIS • Dynamic Analysis of Link 1: From free body diagram, we have ∑ M O1 = 0 ⇒ ˆ ˆ ˆ T1 k + rO 1 B × ( − FBx i − FBy j ) − I G 1α 1 + rO 1G 1 × m1 (aG 1 x iˆ + aG 1 y ˆj ) = 0 T1 + rO 1 By × FBx − rO 1 Bx × FBy − I G 1α 1 − rO 1G 1 y × m1aG 1 x + rO 1G 1 x × m1aG 1 y = 0 ⇒ T1 = 2.54 (lb-in) FBy FBx m 2 aG 2 y ∑F = 0 ⇒ ∑F ∑F x y B = 0 : FO1x − FBx − m1aG1x = 0 = 0 : FO1 y − FBy − m1aG1 y = 0 FO1 x = −1.28 ⇒ FO1 y = −0.224 G2 FBx I G 2α 2 m1aG1 y FO1 y (lb) B FBy G1 m 2 aG 2 x FCy 30 C FCx m1aG1 x T1 I G1α1 70 O1 FO1 x Free body diagram of the slider-crank using D’Alembert Principle Content Copyright Nanyang Technological University 32 End of topic 4 Example 3 (exam AY00/01 S2): Figure 1 shows a mechanism driven by link 1. At the position shown, link 1 rotates with an angular acceleration α1. At the output end F, the driven link 5 is pushing a mechanical part, which is idealized as a spring of stiffness Ks. The mass of link 3 and the mass of the roller link 4 are considered to be negligible. The dimensions of the links are given in Figure 1. Given that the spring is compressed by a displacement ∆s at this instant, and the following: Mass Location Moment of of CG inertia about CG Acceleration of CG Angular acceleration Link 1 m1 G1 at O1 I1 0 α1 (cw) Link 2 m2 G2 I2 a2xi+a2yj α2 (cw) Link 3 massless Link 4 massless G5 at O5 I5 0 α5 (ccw) Link 5 m5 y Neglecting the gravitational force and friction forces, 1) Draw the free-body diagram of each link (except the ground link) for dynamic force analysis; x 2) Derive the expressions for the constraint force at D between links 4 and 5, the constraint force at bearing O5, the constraint force at B (between links 2 and 3), the constraint force at A (between links 1 and 2), as well as the required input driving torque T. D E α5 Ks Link 5 F Link 4 C B O5 G5 massless G2 G1 O1 A T α1 ο 30 α2 Link 3 Link 2 O3 45 ο Link 1 34 Solution Free Body Diagrams 35 Solution: From free body diagrams, we have • For Free-body link 5: ∑ TO5 = 0 ⇒ ⇒ r5 2 − I 5α 5 − K s ∆ ⋅ r5 = 0 2 I 5α 5 F45 = + 2 K s ∆ = 23 I 5α 5 + 2 K s ∆ r5 F45 ⋅ ∑ Fx = 0 ⇒ F05x = 0 ∑ Fy = 0 ⇒ F05y + F45 − K s ∆ = 0 ⇒ F05y 2 I 5α 5 = K s ∆ − F45 = − − Ks∆ r5 36 • For Free-body link 2+4: ∑ TA = 0 ⇒ I 2α 2 + F32 (r2 cos 30o − r3 cos 45o ) sin 45o + m2 a2 x − m2 a2 y ⇒ r2 2 cos 30o − F45 r2 2 r2 2 sin 30 o cos 30o = 0 F32 = (3 3m2 a2 y + 2 3I 5α 5 + 6 3K s ∆ − 1.5m2 a2 x − I 2α 2 ) / 2 2 ∑ Fx = 0 ⇒ ⇒ F12x − m2 a2 x + F32 cos 45o = 0 F12x = 85 m2 a2 x − 14 (3 3m2 a2 y + 2 3I 5α 5 + 6 3K s ∆ − I 2α 2 ) ∑ Fy = 0 ⇒ ⇒ F12y − m2 a2 y − F45 + F32 sin 45o = 0 B 4−3 3 4−3 3 m2 a2 y + I 5α 5 4 6 + ( 2 − 1.5 3 )K s ∆ + 14 ( 1.5m2 a2 x + I 2α 2 ) F12y = A 37 •For Free-body link 1: ∑ TO1 = 0 ⇒ T + I1α1 + F12y ⋅ 1 = 0 ⇒ 3 3−4 3 3−4 m2 a2 y + I 5α 5 + (1.5 3 − 2) K s ∆ − 14 (1.5m2 a2 x + I 2α 2 ) T = − I1α1 + 4 6 38 End of Chapter