Ksenia P. Frolova (student of SPbSTU, Department for Theoretical Mechanics) Оlga S. Loboda (PhD SPbSTU, Department for Theoretical Mechanics) Plan of the presentation Brief review Problem definition Results Experiment Conclusion 2 Principle of action Limbs bending 3 Classification of the bow Types of the bow 4 Compound bow Limbs bending 5 Problem definition Modeling: 1. Straight rigid beams jointed with a spiral spring 2. Thin flexible beams; deformations are small Function F(𝑥) dynamic curve Energy Initial velocity of arrow Compound bows: Comparison with classical bows 6 Model №1 Limbs are absolutely rigid rods Spiral spring placed between the rods Bow string is inextensible 1 – spring, 2 – limbs, 3 – bow string 7 Problem definition №1 M= 𝑐𝜑 Tℎ = M F = 2Tcos𝛽 𝑐𝜑 F=2 cos𝛽 ℎ M – spring moment с – spring stiffness 𝜑/2 – angle of limb deviation T – bow string tension force F – bow tension force 𝛽 – angle between the spring and the line of it’s removal 8 Function F(x) 𝐹 ′ (0) 𝐹 ′′ (0) 2 𝐹 ′′′ (0) 3 𝐹(𝑥)=F(0)+ 𝑥+ 𝑥 + 𝑥 +… 1! 2! 3! 2𝑐(𝑙 2 − 2𝑥02 ) 3 𝐹(𝑥)= 2 2 2 𝑥 2 𝑙 𝑥0 (𝑙 − 𝑥0 ) cubic dependence 9 Energy, Initial velocity of arrow Energy 𝑐 (𝑙 2 − 2𝑥02 ) 4 U= 𝑥 2𝑥02 𝑙 2 (𝑙 2 − 𝑥02 ) Initial velocity of arrow 𝑉0 = 2 𝑐(𝑙 2 − 2𝑥02 ) 𝑙𝑥0 𝑚(𝑙 2 − 𝑥02 ) 𝑥2 10 Model №2 Linear theory Static task Limbs - Bernoulli-Euler beams 11 Problem definition №2 𝐍′ = 0 M’ + 𝝉 × 𝐍= 0 ε = u’ + 𝝉 × 𝝍 = 0 𝟏 Ф=(с 𝒌𝒌 1 + 𝟏 𝒏𝒏 с2 + M = Mk Ф= 𝝍′ 𝟏 с3 𝝉 𝝉 )∙M N – force in cross-section of the limb M – moment in crosssection of the limb 𝝉 – vector of limb tangent ε – deformation shift vector u –removal vector 𝝍 – turn vector Ф – deformation vector с1, с2, с3 – stiffness tensor components 12 Solution boundary conditions: u|s=0 = 0 𝜓|s=0 = 0 N|s=l = N0 M|s=l = 0 u – movement vector 𝜓 - turn vector N – force in cross-section of the limb M – moment in crosssection of the limb s – coordinate along the limb l – length of the limb u|s=l = 𝑙3 N0 3𝑐1 u – movement vector N0 – force in cross-section of the limb 13 Decision 𝑥 F = 𝑝2 − (𝑙𝑠𝑖𝑛𝛼 − 𝑙3 F𝑐𝑡𝑔𝛼)2 6с1 + 𝑙𝑐𝑜𝑠𝛼+ 𝑙3 6с1 F − 𝑥0 3с1 𝑠𝑖𝑛𝛼( 𝑥 + 𝑥0 𝑠𝑖𝑛𝛼 + 𝑝2 − (𝑙 − 𝑥 + 𝑥0 𝑐𝑜𝑠𝛼)2 ) 𝑉0 = 2 𝑥 3 𝑚𝑙 14 Modeling of compound bow BC F T AB F – cable tension force , T – string tension force 15 Experiment №1 b o w Fixing of the handle loads Measurement of spring removal 16 Experiment №2 compound bow «Ястреб» block 17 List of parameters №1 Parameter Designation Value Length of the limb l 0.650 m Considering length of the string p 0.625 m Initial removal x0 0.180 m Young modulus E 21 GPa Cross-section height h 0.009 m Cross-section width b = b(z), z=[0..l], (м) b = 0.0316z+0.01, m Fmax 11.5 kgf Maximum effort 18 List of parameters №2 Parameter Designation Parameter Designation Length of the limb 0.410 m Radius of the big block 0.027 м Length of the string 1. 032 m Radius of the small block 0.020 м Length of the cable 1.040 m Initial removal 0.190 м 20 kgf Height width of cross-section 0.008 м 0.05 0.08 м Maximum effort 19 Bow stiffness in model №1 The Method of Least Squares 𝑁 S= Fэксп 𝑖=1 𝑙2 2 2𝑐 − 2𝑥0 3 𝑥𝑖 − 2 2 2 𝑥 𝑙 𝑥0 𝑙 − 𝑥02 𝑖 2 𝜕𝑆 =0 𝜕𝑐 𝑐 = 23.169 N ∙ m 20 Bow stiffness in model №2 с = 𝐸𝐼 bℎ3 I= 12 𝑏 = 0.0316𝑧 + 0.01 𝑐 = 24.852 Н ∙ м2 E – Young modulus, I – moment of inertia of cross-section of the limb 21 Dynamic curve 22 Bow energy 23 Initial velocity of arrow 24 Dependence of effort from removal 25 Conclusion 1. 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