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工業力学 I(第 1 回) - KIT PML HP...工業力学 I(第 1 回) 本日の予定 Statics I...
Transcript of 工業力学 I(第 1 回) - KIT PML HP...工業力学 I(第 1 回) 本日の予定 Statics I...
工業力学 I(第 1 回)
本日の予定
Statics I(1) Force (2) Moment / Torque
(Vector Operations)
Definition of Force(p.2)
Force is the action of one body on another.
tends to move a body in the direction of its action.
characterized by
✴ its magnitude ✴ the direction of its action ✴ its point of application
Vector quantity
Principle of Transmissibility(p.18)
Fig. 2.2
Internal effect を無視できる場合 力の作用点はどこでもよい
treated as a sliding vector
not as a fixed vector
(注) 数学では free vector として扱う
classification of vectors(p.3)
Fig. 2.3(p.19)
Vector Addition
Parallelogram Lawconcurrent
force
Triangle Law
危険!
作用線が 違う!
Resultant
Fig. 2.3(p.19)
Vector Addition
Parallelogram Lawconcurrent
force
Triangle Law
危険!
作用線が 違う!
Resultant
Fig. 2.4(p.20)
Fig. 2.3 (e)(p.19)
Components & Projections(p.20)
F1, F2: components Fa, Fb: projections
直交する2方向を取ると等しくなる
F = Fx i + Fy j +Fzk( )
i, j, k: x, y, z 方向の単位ベクトル
( )2.2; p.21 2.11; p.37
Fig. 2.3 (e)(p.19)
Components & Projections(p.20)
F1, F2: components Fa, Fb: projections
直交する2方向を取ると等しくなる
F = Fx i + Fy j +Fzk( )
i, j, k: x, y, z 方向の単位ベクトル
( )2.2; p.21 2.11; p.37
F = Fx2+ Fy
2+ Fz2
Fx = F cosθxFy = F cosθyFz = F cosθz
Fig. 2.16(p.37)
各軸の正の部分から測る (正値なので向きはなし)
0≤θ≤π
(2.11; p.37)
Spherical Polar Coordinates(p.38)
Fig. 2.18(p.38)
φ
X
0≤φ≤π (方向なし)
−π≤θ≤π(CCW)
Fx = F cosθsinφ
Fy = F sinθsinφ
Fz = F cosφ
⎫
⎬
⎪⎪⎪⎪
⎭⎪⎪⎪⎪
・polar coordinates(2D) ・cylindrical coordinates(3D)
Dot Product(p.38)
Fig. 2.19 (b)
F の n 方向成分は F と n のスカラー積
Fn =F ⋅n= Fx l+ Fy m+ Fz n
ここでn= l i+m j+ nk
l, m, n: direction cosines (方向余弦; p.5)
l= cosθx m= cosθy n= cosθz
angle between two vectors
θ= cos−1 F ⋅nF⎛
⎝⎜⎜⎜
⎞
⎠⎟⎟⎟⎟ (2.13; p.68)
F·n = 0 → F⊥n
Sample Problem 2.1(p.23)
The forces F1, F2, and F3, all of which act on point A of the bracket, are specified in three different ways. Determine the x and y scalar components of each of the three forces.
✴ 記号を用いること! ✴ 方向余弦で!
θ1x
θ3x
θ2x
F1x = F1 cosθ1x = 600×cos 35!( )= 491.491
F1y = F1 cosθ1y = 600×cos 90!−35!( )= 344.146
Moment / Torque
A force can also tend to rotate a body about an axis.(p.26)
Fig. 2.8(a)(p.26)
Moment / Torque
M = F d
Scalar?
(2.5)
Moment Vector
Fig. 2.8 (b)(p.26)
M= r×F
順番も重要! (交換法則 NG)
M = r×F
= F r sinα= F d
(2.6)
(2.7)
Cross Product(p.41)
Fig. 2.22(p.42)
点 O 回りのモーメント
=
i j krx ry rzFx Fy Fz
=ry rzFy Fz
i+rz ryFz Fy
j+rx ryFx Fy
k
= Mxi+ Myj+ Mzk
MO = r×F (2.14)
(2.15)
= ryFz−rzFy( )i+ rzFx−rxFz( )j+ rxFy−ryFx( )k
Varignon’s Theorem(p.27)
Fig. 2.9(p.28)
MO = r×R & R= P+Q
MO = r× P+Q( )= r×P+r×Q
The moment of a force about any point is equal to the sum of the moments of the components of the force about the same point.
MO =−Rd / MO = pP−qQ
(2.8 Rev.)
Varignon’s Theorem(p.27)
Fig. 2.9(p.28)
MO = r×R & R= P+Q
MO = r× P+Q( )= r×P+r×Q
The moment of a force about any point is equal to the sum of the moments of the components of the force about the same point.
MO =−Rd / MO = pP−qQ
(2.8 Rev.)
Fig. 2.24(p.43)
MO = r×F( )∑ = r×R (2.18)(p.43)
concurrent なら 3D でも成り立つ
Sample Problem 2.5(p.29)
Calculate the magnitude of the moment about the base point O of the 600-N force in five different ways.
F
MO = Fd MO = F1d1 + F2d2
d1
d2MO = F1d1
MO = F2d2
MO = r × F= −2610k [N ⋅m]
Sample Problem 2.5(p.29)
Calculate the magnitude of the moment about the base point O of the 600-N force in five different ways.
次回の予定
Statics II(1) Couple (2) Force - Couple System (3) Resultant