Transcript Простейшая механическая модель лука

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
Initial velocity
of arrow
𝑐 (𝑙 2 − 2𝑥02 )
4
U=
𝑥
2𝑥02 𝑙 2 (𝑙 2 − 𝑥02 )
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.
Mathematical description of bow with straight rigid limbs
jointed with a spiral spring
2. Mathematical description of bow with thin flexible limbs
 Bows with flexible beams more powerful then bows with straight rigid
beams
 Model of bow with flexible beams describe the real construction more
correctly. It is also more effective.
 System of blocks can be used in more powerful bows
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REFERENCES
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
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