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Chapter Two

Force System

Example 25: Determine the moment MAB produced by force F in Figure which tends to rotate the rod about the AB axis. Solution: Because that F is parallel to the z-axis so it has no moment about z-axis. Its moment about x-axis equals zero because its line action passing the x-axis and so the lever arm (dx = 0) equals zero. Thus it has only moment about y-axis at point A. M y =  Fx d z  Fz d x = 0 + 300 * 0.6 = 180 N .m

∴ M AB

A Mx=180N.m

1 1 = My * = 180 * = 80.5 N .m 5 5

5

2 x

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Chapter Two

Force System

Example 26: Determine the combined moment of P=17.886-lb and Q= 20-lb force about the axis GB. Solution: The Cartesian components of P and Q are: P=17.886 lb

1 Px = P cos α = −17.886 * = −12.647 lb 2 Py = P cos β = 17.886 cos 90 = 0 lb

Q=

1 = 12.647 lb 2 Qx = Q cos α = 20 cos 90 = 0 lb Pz = P cos γ = 17.886 *

3 = −18.97 lb 10 1 Qz = Q cos γ = −20 * = −6.324 lb 10 Q y = Q cos β = −20 *

And their moments about GF and OG are:

2

Qz=18.97 lb

Qz=6.324 lb 10

3

Q= 1

Pz=12.647 lb

Px=12.647 lb

M GF = ( Pz d x  Px d z ) + (Qz d x  Qx d z )

= [12.647 * 4 − 12.647 * 4] + [6.324 * 4 − 0 * 4] = 25.296 lb.in

MOG=75.888 lb.in MGF=25 .296 lb.in

M OG = ( Py d x  Px d y ) + (Qy d x  Qx d y )

1

= [0 * 4 + 12.647 * 12] + [(−18.97) * 4 − 0 * 12] = 75.888 lb.in

10

3

Q=

Finally MGB is: M GB = 25.296 *

3 1 − 75.888 * = 0 lb.in 10 10

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Chapter Two

.University of Qadisiyah\Faculty of Eng.\Civil Dep

Force System

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Mechanics\Statics\1st Class

Chapter Two

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Force System

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Mechanics\Statics\1st Class

Chapter Two

Force System

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: MO = F(a+d) – Fa ∴ M=Fd

…(2-11)

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.

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Chapter Two

Force System

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 1. 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

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Chapter Two

Force System

Example 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 d1 = 4 ft, d2 = 3 ft, and d3 = 5 ft. + MR = ΣM; MR = -F1d1 + F2d2 - F3d3 = (-200)(4) + (450)(3) - (300)(5) = -950 lb.ft = 950 lb.ft

The negative sign indicates that MR has a clockwise rotational sense. Example 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). + MR = ΣMO;

M = (600 cos30o)(0.2) – (600 sin30o)(0.2) = 43.9 N.m

+ MR = ΣMA;

M = (600 cos30o)(0.2) – (600 sin30o)(0.2) = 43.9 N.m

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Chapter Two

Force System

Note: The same result can also be obtained using M = Fd. Example 29: Determine the couple moment acting on the pipe shown in Figure (a). Segment AB is directed 30o below x-y plane.

Solution: The perpendicular distance between the lines of action of the couple forces is d = 6cos30o = 5.196 in

Taking moments of the forces about either point A or point B yields M = Fd = -25(5.196) = -129.9 lb.in

Applying the right hand rule, M acts in the negative y direction M ≅ 130 lb.in

⇐

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Chapter Two

Force System

Example 30: The rigid structural member is subjected to a couple of the two 100 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 =100(0.1) = 10 N.m

The forces P and –P produce a counterclockwise couple M = 400 (0.040) cosθ

Equating the two expressions gives 10 = 400 (0.040) cosθ  10  o  = 51.3  16 

θ = cos −1 

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. Example 31: Determine the magnitude and direction of the couple M which will replace the two given couples and still produce the same external 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

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