Example 25: Determine the moment M AB produced by force F in Figure which tends to rotate the rod about the AB axis.

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1 Eample 25: Determine the moment M AB produced by force F in Figure which tends to rotate the rod about the AB ais. Solution: Because that F is parallel to the z-ais so it has no moment about z-ais. Its moment about -ais equals zero because its line action passing the -ais and so the lever arm (d = 0) equals zero. Thus it has only moment about y-ais at point A. M = F d F d = *0.6 = 80N m y z z. M AB = M y * = 80 * = 80.5N. m 5 5 A M =80N.m y 2 5 B.University of Qadisiyah\Faculty of Eng.\Civil Dep 45

2 Eample 26: Determine the combined moment of P=7.886-lb and Q= 20-lb force about the ais GB. Solution: The Cartesian components of P and Q are: P = P cosα = * = lb 2 P = P cos β = cos 90 = 0lb y Pz = P cosγ = * = lb 2 Q = Q cosα = 20 cos 90 = 0lb Q Q y z = Q cos β = 20 * 3 = 8.97 lb 0 = Q cosγ = 20 * = 6.324lb 0 Q= Q z =8.97 lb P=7.886 lb 2 Q z =6.324 lb And their moments about GF and OG are: 0 3 Q= P z =2.647 lb P =2.647 lb M GF = ( Pz d P d z ) + ( Qzd Qd z ) [ * * 4] + [ 6.324* 4 0* 4] = = lb.in M = OG = ( Py d P d y) + ( Qyd Qd y) [ 0 * *2] + [( 8.97) * 4 0*2] = lb.in Finally M GB is: 3 M GB = * * = 0lb. in 0 0 M OG = lb.in M GF = lb.in 0 3 Q=.University of Qadisiyah\Faculty of Eng.\Civil Dep 46

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5 Couples A couple consists of two parallel, noncollinear forces that are equal in magnitude and opposite in direction A couple is a purely rotational effect, it has a moment but no resultant force (resultant equals zero thus it has no tendency to translate the body in any direction). A couple possesses two important characteristics: a) A couple has no resultant force (ΣF = 0), and b) The moment of a couple is the same about any point in the plane of the couple. So it may be considered as a free vector quantity (not localized vector and can be moved to any parallel position) having both magnitude and direction (aspect of plane and sense of rotation). The magnitude of the couple is: M O = F(a+d) Fa M=Fd (2-) where F is the magnitude of one forces, and d is the perpendicular distance or moment arm between the forces. The direction and the sense of the couple moment are determined by the right-hand rule, where the thumb indicates this direction when the fingers are curled with the sense of rotation caused by the couple forces. In all cases, M will act perpendicular to the plane containing these forces..university of Qadisiyah\Faculty of Eng.\Civil Dep 49

6 Equivalent couples: If two couples produce a moment with the same magnitude and direction, then these two couples are equivalent. Figure below illustrates the four operations that may be performed on a couple without changing its moment; all couples shown in the figure are equivalent. The operations are. Changing the magnitude F of each force and the perpendicular distance d while keeping the product Fd constant, 2. Rotating the couple in its plane, 3. Moving the couple to a parallel position in its plane 4. Moving the couple to a parallel plane Resultant of Couples Because couples are vectors, they may be added by the usual rules of vector addition. Being free vectors, the requirement that the couples to be added must have a common point of application does not apply. The resolution of couples is no different than the resolution of moments of forces..university of Qadisiyah\Faculty of Eng.\Civil Dep 50

7 Eample 27: Determine the resultant couple moment of three couples acting on the plate in Figure below. Solution: As shown the perpendicular distances between each pair of couple forces are d = 4 ft, d 2 = 3 ft, and d 3 = 5 ft. + M R = ΣM; M R = -F d + F 2 d 2 - F 3 d 3 = (-200)(4) + (450)(3) - (300)(5) = -950 lb.ft = 950 lb.ft The negative sign indicates that M R has a clockwise rotational sense. Eample 28: Determine the magnitude and direction of the couple moment acting on the gear in Figure (a) Solution: The easiest solution requires resolving each force into its components as shown in Figure (b). + M R = ΣM O ; M = (600 cos30 o )(0.2) (600 sin30 o )(0.2) = 43.9 N.m + M R = ΣM A ; M = (600 cos30 o )(0.2) (600 sin30 o )(0.2) = 43.9 N.m.University of Qadisiyah\Faculty of Eng.\Civil Dep 5

8 Note: The same result can also be obtained using M = Fd. Eample 29: Determine the couple moment acting on the pipe shown in Figure (a). Segment AB is directed 30 o below -y plane. Solution: The perpendicular distance between the lines of action of the couple forces is d = 6cos30 o = 5.96 in Taking moments of the forces about either point A or point B yields M = Fd = -25(5.96) = lb.in Applying the right hand rule, M acts in the negative y direction M 30 lb. in.university of Qadisiyah\Faculty of Eng.\Civil Dep 52

9 Eample 30: The rigid structural member is subjected to a couple of the two 00 N forces. Replace this couple by an equivalent couple consisting of the two forces P and P, each of which has a magnitude of 400 N. Determine the proper angle θ. Solution: The original couple is counterclockwise when the plane is viewed from above, and its magnitude is M = Fd M =00(0.) = 0 N.m The forces P and P produce a counterclockwise couple M = 400 (0.040) cosθ Equating the two epressions gives 0 = 400 (0.040) cosθ 0 cos θ = = 6 o 5.3 Hint: Since the two equal couples are parallel free vectors, the only dimensions which are relevant are those which give the perpendicular distance between the forces of the couples. Eample 3: Determine the magnitude and direction of the couple M which will replace the two given couples and still produce the same eternal effect on the block. Specify the two forces F and F, applied in the two faces of the block parallel to the y-z plane, which may replace the four given forces. The 30 N forces act parallel to the y-z plane. Solution: The couple due to 30 N forces has the magnitude.university of Qadisiyah\Faculty of Eng.\Civil Dep 53