Engineering Mechanics 정역학 (Statics) 4장힘계의합력 1 GeoPave Lab. 4.1 힘의모멘트 스칼라공식 1
4.1 힘의모멘트 스칼라공식 4.1 힘의모멘트 스칼라공식 모멘트크기 (resultant moment) 2
4.1 힘의모멘트 The moment does not always cause r otation. The actual rotation would occur if the support at B were removed. Condition to remove the Nail 4.1 힘의모멘트 스칼라공식 For each case, determine the moment of the force about point O. Å, M R = F d - F d + F d 0 1 1 2 2 3 3 3
4.1 힘의모멘트 예제 4.1 Line of action is extended as a dashed line to establish moment arm d. Tendency to rotate is indicated and the orbit is shown as a colored curl. 4.1 힘의모멘트 예제 4.1 4
4.1 힘의모멘트 예제 4.1 4.1 힘의모멘트 예제 4.2 5
4.1 힘의모멘트 예제 4.3 Resultant moment of the four forces about point O. 4.1 힘의모멘트 예제 4.3 6
4.2 벡터외적 (Cross Product) Cross product of two vectors A and B yields C, which is written as Magnitude Magnitude of C is the product of the magnitudes of A and B For angle θ, 4.2 벡터외적 (Cross Product) Direction Vector C has a direction that is perpendicular to the plane containing A and B such that C is specified by the right hand rule Expressing vector C when magnitude and direction are known 7
4.2 벡터외적 (Cross Product) Laws of Operations 1. Commutative law is not valid Rather, Cross product B A yields a vector opposite in direction to C 4.2 벡터외적 (Cross Product) 2. Multiplication by a Scalar a( A X B ) = (aa) X B = A X (ab) = ( A X B )a 3. Distributive Law A X ( B + D ) = ( A X B ) + ( A X D ) Proper order of the cross product must be maintained since they are not commutative 8
4.2 벡터외적 (Cross Product) 직교벡터공식 Use C = AB sinθ on pair of Cartesian unit vectors A more compact determinant in the form as i r k r Å r j 9
4.2 벡터외적 (Cross Product) In determinant form, 4.3 힘의모멘트 벡터공식 Moment of force F about point O can be expressed using cross product & position vector, r à r : O 에서부터 F 의작용선위의 A 점까지의위치벡터 Magnitude For magnitude of cross product, Treat r as a sliding vector. Since d = r sinθ, 10
4.3 힘의모멘트 벡터공식 Direction Direction M O : by right-hand rule *Note: - 4.3 힘의모멘트 벡터공식 Principle of Transmissibility( 전달성원리 ) For force F applied at any point A, moment created about O is M O = r A x F F has the properties of a sliding vector, thus 11
4.3 힘의모멘트 벡터공식 Cartesian Vector Formulation For force expressed in Cartesian form, r r r i j k r r r M = F = r r r O With the determinant expended, M O = (r y F z r z F y )i (r x F z - r z F x )j + (r x F y y F x )k x F x y F y z F z 4.3 힘의모멘트 - 벡터공식 O 점에대한힘계의합모멘트 12
4.3 힘의모멘트 - 벡터공식 힘 F 로케이블 BC 위의한점에서잡아당길때전신주의기초 A 에대한모멘트는항상같음 A 점에서케이블에대한모멘트팔길이는 r d 3 차원에서벡터의외적을이용하여모멘트팔길이결정 4.3 힘의모멘트 벡터공식 : 예제 4.4 The pole is subjected to a 60N force that is directed from C to B. Determine the magnitude of the moment created by this force about the support at A. 13
4.3 힘의모멘트 벡터공식 : 예제 4.4 4.3 힘의모멘트 벡터공식 : 예제 4.4 14
4.3 힘의모멘트 벡터공식 : 예제 4.4 4.3 힘의모멘트 벡터공식 : 예제 4.5 Two forces act on the rod. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector. 15
4.3 힘의모멘트 벡터공식 : 예제 4.5 4.3 힘의모멘트 벡터공식 : 예제 4.5 16
4.4 모멘트의원리 Since F = F 1 + F 2, 4.4 모멘트의원리 고정용케이블이힘 F 를받을때, 이힘은기초 A 에대하여모멘트를발생 이힘을 B 점에작용하는 F x 및 F y 를이용하여모멘트를결정 F y 의값은 F x 힘은 17
4.4 모멘트의원리 : 예제 4.6 A 200N force acts on the bracket. Determine the moment of the force about point A. 4.4 모멘트의원리 : 예제 4.6 18
4.4 모멘트의원리 : 예제 4.6 4.4 모멘트의원리 : 예제 4.7 The force F acts at the end of the angle bracket. Determine the moment of the force about point O. 19
4.4 모멘트의원리 : 예제 4.7 4.4 모멘트의원리 : 예제 4.7 20
4.5 측정축에대한힘의모멘트 한점에대한힘모멘트계산시모멘트와모멘트축은힘과모멘트팔을포함하는평면에항상수직 필요시그점을통과하는특정축에방향에대한모멘트성분계산이필요함. 스칼라해석또는벡터해석을이용함. 4.5 측정축에대한힘의모멘트 1) 스칼라해석 xy 평면에 A 점에수직으로작용하는힘 F = 20N O 점에대한모멘트크기 : 방향 : right hand rule Ob axis 의모멘트 M O 의 y axis, M y : O 점에서플랜지로부터파이프를풀려고하는성분 For magnitude of M y, For direction of M y, apply right hand rule F 작용선으로부터 y 축까지의수직거리 : 0.3m 모멘트 M y, 힘 F 의작용선이특정축 aa 와수직할때 : M a =Fd a 21
4.5 측정축에대한힘의모멘트 Z 축선상의 A 점에힘 F 작용시 최대모멘트는수평면과평행한힘을작용시킬때, where d is the perpendicular distance from the force line of action to the axis 그외의경우 4.5 측정축에대한힘의모멘트 2) 벡터해석 M O = r A X F = (0.3i +0.4j) X (-20k) = {-8i + 6j}N.m Since unit vector for this axis is u a = j, M y = M O.u a = (-8i + 6j) j = 6N.m 22
4.5 측정축에대한힘의모멘트 A 점에작용하는힘 F 모멘트 M a, - bb 축선상의 O 점에대한모멘트 : where r is directed from O to A - M O 는 bb 축선에작용하는모멘트를 aa 축에투영한크기 = M A where u a : aa 축선상의단위벡터 4.5 측정축에대한힘의모멘트 - determinant form, r M a r r r = ( u i + u j + u k ) ax ay az r i r x F x r j r F y y r k r z F z r r r r M = u ( F) = a ax u r ax x F x u r ay y F y u r az z F z u ax, u ay, u az : x, y, z 성분의단위벡터 r x, r y, r z : O 점으로부터의위치벡터 F x, F y, F z : 힘벡터 23
4.5 측정축에대한힘의모멘트 The sign of the scalar indicates the direction of M a along the aa axis - If positive, M a has the same sense as u a - If negative, M a act opposite to u a Express M a as a Cartesian vector, For magnitude of M a, 4.5 측정축에대한힘의모멘트 교통표지판에바람에의한힘 F 에의한모멘트 M A 단위벡터 u a 로정의되는축에대한모멘트의투영시 : 24
4.5 측정축에대한힘의모멘트 : 예제 4.8 The force F = {-40i + 20j + 10k} N acts on the point A. Determine the moments of this force about the x and a axes. 4.5 측정축에대한힘의모멘트 : 예제 4.8 Solution Method 1 Negative sign indicates that sense of M x is opposite to i 25
4.5 측정축에대한힘의모멘트 : 예제 4.8 We can also compute M a using r A as r A extends from a point on the a axis to the force 4.5 측정축에대한힘의모멘트 : 예제 4.8 Solution : Method 2 Only 10N and 20N forces contribute moments about the x axis Line of action of the 40N is parallel to this axis and thus, moment = 0 Using right hand rule 26
4.5 측정축에대한힘의모멘트 : 예제 4.9 Determine the moment produced by the force F which tends to rotate the rod about the AB axis. 4.5 측정축에대한힘의모멘트 : 예제 4.9 27
4.5 측정축에대한힘의모멘트 : 예제 4.8 4.5 측정축에대한힘의모멘트 : 예제 4.8 28