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DYNAMIC FORCE ANALYSIS

DYNAMIC FORCE ANALYSIS. WEEKS-2,3. MASS RELATIONS. Location of Center of Mass : y m 1 m 2 m n O z R x y. m 1 m 2 m n x. O. x 1. G. x 2. x n. R 1. G. R 2. R n. x. dm. G. MASS RELATIONS. Mass Moments and Products of Inertia:

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DYNAMIC FORCE ANALYSIS

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  1. DYNAMIC FORCE ANALYSIS WEEKS-2,3

  2. MASS RELATIONS • Location of Center of Mass: ym1m2 mn O z R x y m1m2mnx O x1 G x2 xn R1 G R2 Rn x dm G

  3. MASS RELATIONS • Mass Moments and Products of Inertia: • Mass Moments of Inertia Mass Products of Inertia • Inertia Tensor If there is symmetry wrt three axes: Ixy= Ixz= Iyz=0 (In this case,Ixx, Iyy, Izz are principal moments of inertia) If there is symmetry wrt two axes: Two of Ixy,Ixz,Iyz=0 If there is symmetry wrt one axis : Oneof Ixy,Ixz,Iyz=0 Unit:massXdistance2

  4. Inertia Forces and D’alembert’s Principle F1 m F3   aG h aG  F G G FG MG F2 Newton’s Law: F=maG(F=FG) MG=Fh=IG (h=IG/maG) h is selected according to the sense of 

  5. Fa F   FGa h aG  aG G G FG (F=FG) TGa MG D’alembert’s Principle: F- maG= 0 F+ FGa = 0 F=0(Inertia Force: Fa=FGa= - maG) (A fictitious force) MG-IG = 0  MG+TGa =0  M=0(Inertia Torque: TGa= -IG) (A fictitious torque)

  6. 3 G3 B 73° FC 3 4 aG3 A 4 18,7° 2 G4 ω2 115° 5,1° C G2,O2 127° O4 aG4 EXAMPLE (Analytical Solution) AO2=60 mm O4O2=100 mm BA=220 mm BO4= 150 mm CO4=CB=120 mm G3A= 90 mm G4O4=90 mm m3=1.5 kg m4=5 kg IG2=0.025 kg.m2 IG3=0.012 kg.m2 IG4=0.054 kg.m2 2=0 3=-119k r/s2 4=-625k r/s2 FC=-0.8j N aG3= 162 aG4= 104 233° m/s2 -73.2 °m/s2 G4O4C = 20.4° G4O4B = 36° G3AB = 30° G

  7. FaG3=-m3aG3 TaG3=-IG33 G3 F43 3 B F23 A F32 A F34 2 FaG4 =-m4aG4 M12 FC F12 G4 G2,O2 C O4 TaG4=-IG44 F14 FREE BODY DIAGRAMS 4 unknowns 3 equations 5 unknowns 3 equations 4 unknowns 3 equations Total: 9 unkowns 9 equations A matrix solution can be performed in the computer. Practically, the most suitable way is to make a coupled solution of links 3 and 4, for F34 and F43 . Therefore, the solution is reduced to 2 equations with 2 unknowns.

  8. F34 FaG4 =-m4aG4 FC G4 C O4 TaG4=-IG44 F14 MOMENT AT LINK 4WRT POINT O4 • ΣMO4=0 • ΣMO4= O4G4xFaG4 + TaG4 + O4C x FC + O4B x F34= 0 • ΣMO4=0.09[cos(20.4+5.1)i + sin25.5j] x 5(104)[cos(233+180)i+sin 53j] + (0.054)(625k) 413°=53° +0.12(cos 5.1i + sin 5.1j) x (-800j) +0.15[cos(36+20.4+5.1)i + sin61.5j] x(F34xi+F34y j)=0 -0.125F34x + 0.083F34y = 37 (1)

  9. FaG3=-m3aG3 TaG3=-IG33 G3 F43 3 B F23 A MOMENT AT LINK 3WRT POINT A • ΣMA=0 • ΣMA= AG3 x FaG3 + TaG3 + AB x F43 =0 • ΣMA= 0.09[cos(30+18.7)i +sin48.7j] x • 1.5 (162) [cos(-73.2+180)i+sin(106.8)j] • +(0.012)(119k) • +0.22(cos18.7i+sin18.7j) x (-F34xi-F34yj)=0 • 0.0705F34x-0.208F34y=-20 (2)

  10. F34 FaG4 =-m4aG4 FC G4 C O4 TaG4=-IG44 F14 Finding Forces F34 and F14 • -0.125F34x + 0.083F34y = 37 • 0.0705F34x-0.208F34y=-20 • F34x=-300 F34y=-5,39 F34=-300i-5.39j N • Force Balance at Link 4 • ΣF=0 F14+F34+FC+FaG4=0 • F14-300i-5.39j-800j+5(104)(cos53i+sin53j)=0 • F14=-13i+390j N

  11. FaG3=-m3aG3 TaG3=-IG33 G3 F43 3 B F23 A Finding Force F23 • Force Balance at Link 3 • ΣF=0 F23+F43+FaG3=0 • F23+300i+5.39j+1.5(162)(cos106.8i+sin106.8j)=0 • F23=-230i-238j N

  12. F32 A 2 M12 F12 G2,O2 Finding Force F12 and MomentM12 • Force Balance at Link 2 • ΣF=0 F12+F32=0 • F12=-F32=F23=-230i-238j N • Moment Balance at Link 2 • ΣMO2=0 O2A x F32 + M12 + TaG2=0 • 0.06(cos115i+sin115j)x(230i+238j)+M12k+0=0 • M12=18.6 N.m M12=18.6 k N.m

  13. B F34O4B FaG4 =-m4aG4 FaG3=-m3aG3 TaG3=-IG33 G3 FC F43 G4 3 B C O4 TaG4=-IG44 F23 A F14 EXAMPLE (Graphical Solution) MO4=0 -(F34O4B)O4B-FC(O4C)x- (FG4)ax (O4G4)y+(FG4)ay(O4G4)x+(TG4)a =0 -(F34O4B)(0.15)- (800)(0.12 cos5.1)- (5)(104cos53)(0.09sin25.5)+ (5)(104sin53)(0.09cos25.5)+(0.054)(625)=0 (F34O4B)=-268.4 MA=0 -(F43AB)AB+(FG3)ax (AG3)y+(FG3)ay(AG3)x+(TG3)a=0 -(F43AB)(0.22)+(1.5)(-162cos106.8)(0.09sin48.7)+ (1.5)(162sin106.8)(0.09cos48.7)+(0.012)(119)=0 (F43AB)= 90.9 (F34AB)= 90.9 F43AB

  14. B F34O4B FaG4 =-m4aG4 FC G4 C O4 TaG4=-IG44 181 F14 Link 4: F=0 !!! (F34O4B) is negative Force scale (hf) : 1cm=100N Of Force scale (hf) : 1cm=50N F14 F34AB 92 F34O4B F34 F34//O4B FC F34//AB F34 (FG4)a F34= (6cm)(50 N/cm)=300 N F14= (3.9cm)(100 N/cm)=390 N F34= 300 /181 N F14= 390 /92 N

  15. FaG3=-m3aG3 TaG3=-IG33 G3 F43 3 B F23 A F32 A 2 M12 F12 G2,O2 Link 3: F=0 Force scale (hf) : 1cm=50N Link 2: F=0 F12=-F32=F23= 330 /226 N 226 F43 A F32 M12 FG3a F23 F12 h Of O2 MO2=0 M12 – F32 h =0 M12 =18.6 N.m M12 =18.6k N.m F23= (6.6cm)(50 N/cm)=330 N F23= 330 /226 N

  16. Example y O2A=3 in. W2=18 lb BA=12 in. W3=3.5 lb G2O2=1.25in. W4=2.5 lb G3A=3.5 in IG2=0.00369in.lb.s2 IG3=0.110in.lb.s2 B 4 FB O2 G4  30 2 2 3 G2 G3 A 2=0 3=-3090k r/s2 FB=800 180 °N aG2= 2640 150  ft/s2 aG3= 6130 158.3° ft/s2aG4= 6280 180° ft/s2 O2A sin30 = BA sin =7.18 ° Problem will be solved by including gravity forces.

  17. W4 aG4 B All forces pass through mass center G4 for the moment balance 4// G4 FB FG4a Sense is assumed F34 F14 W4 MA=0 AB x(FB+ F14+W4+FG4a)+ + AG3X(W3+FG3a) + TG3a=0 12(cos7.18i+sin7.18j)X[-800i+F14j-2.5j+ +(2.5/386)(6280)(12)i] + 3.5(cos7.18i+sin7.18j)X[-3.5j+ +(3.5/386)(6130)(12)(-cos158.3i-sin158.3j)]+ +(0.110)(3090)k=0  F14=30.374 j lb 3+4// FG4a FB B TG3a aG3 F14 G3 A FG3a W3 F23

  18. 3+4// F=0 F23+ FG3a + W3+FB+F14+W4+FG4a=0 F23+(3.5/386)(6130)(12)(-cos158.3i-sin158.3j)]-3.5j-800i+30.374j-2.5j+ (2.5/386)(6280)(12)i=0 F23=-307.8i+222.245j lb 4// F=0 F34+FB+F14+W4+FG4a=0 F34-800i+30.374j-2.5j+(2.5/386)(6280)(12)i=0  F34=311.92i-27.874j lb F12 2// F=0 F32+F12+W2+FG2a=0 307.8i-222.245j+F12-0.95j+(0.95/386)(2640)(12)(-cos150i-sin150j)=0 F12=-375.32i+256.58j lb MO2=0 O2G2X FG2a+ O2G2X W2+O2AX F32+M12=0 0+1.25(cos30i-sin30j)X(-0.95j)+3(cos30i-sin30j)X(307.8i- 222.245j)+M12k =0 O2 G2 FG2a M12 F32 W2 A M12=104.67 k in.lb

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