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6. Statics · PDF file 6 / Statics 97 6. Statics Engineering Mechanics includes both Statics and Dynamics. In this chapter we will review Statics, which involves the forces and moments

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Text of 6. Statics · PDF file 6 / Statics 97 6. Statics Engineering Mechanics includes both Statics...

  • 6 / Statics 97

    6. Statics Engineering Mechanics includes both Statics and Dynamics. In this chapter we will review Statics, which involves the forces and moments that act on bodies in static equilibrium: ladders, bridges, levers, beams, and trusses, to name a few. We will also include the properties of areas, volumes, and masses, sometimes referred to as mensuration. We will review the subject of Dynamics in Chapter 7. 6.1 Forces, Moments, and Resultants Forces are classified as body forces or surface forces. A body force is most often due to gravity but can also result from an electric or magnetic field; it is most often positioned at the centroid of the body. A surface force occurs when one body acts on another but can also be caused by friction, such as air moving over an airfoil. If a force is due to a distribution over a small distance or a small area, we assume the force to act at a point resulting in a concentrated force. A vector is used to represent a force, as will be demonstrated by examples. A system of forces can be concurrent, such that all the forces can be positioned at a single point, or non-concurrent, as occurs with forces acting parallel to each other. Most often forces act in a plane (two- dimensional), but they may act spatially (three-dimensional). A force F acting at point B creates a moment M about a point A is defined by the vector relationship =M r×F (6.1) where r is the vector from A to B. If F and r are in the xy-plane, as is often the case, then the moment M is found using the right-hand rule: place r and F end to end, as shown in Fig.6.1, and wrap your fingers such that r is turned into F; your thumb points in the direction of M, perpendicular to the plane of F and r.

    x

    z

    y F

    r r

    F

    M

    Fingers rotate andthumb points indirection of MA

    B Figure 6.1 A moment due to a force.

    The resultant of a system of forces and moments is a force and moment that is equivalent to the total system at any point. A couple is a pair of equal but oppositely directed forces (see Fig.6.2); its resultant is simply a moment equal to the magnitude of one of the forces times the distance between the forces. A couple is often signified by a curved symbol as shown in Fig.6.2.

  • 98 6 / Statics

    Figure 6.2 A couple.

    EXAMPLE 6.1 Find the component of the force 1 15 12 + 9= −F i j k in the direction of the force 2 2 + + 2 .=F i j k Also, find the angle between the two vectors.

    Solution: The unit vector in the direction of F2 is

    22 2

    2 + + 2 3= =

    F i j ki F The component of F1 in the direction of F2 is

    1 2 2 + + 2(15 12 + 9 ) 10 4 + 6 123⋅ = − ⋅ = − = i j kF i i j k

    The angle is found using 1 2 1 cos ,F θ⋅ =F i since 1 2 1 2 cosF F θ⋅ =F F . We have

    2 2 2

    1

    12 12cos 0.5657. 55.55° 15 +12 +9F

    θ θ= = = ∴ =

    EXAMPLE 6.2 At the point (2, 4) find the resultant of 1 15 12= −F i j which acts at (0, 2, 0), 2 2 + =F i j which acts at (2, 2, 0), and the couple M = 20 which acts in the plane of F1 and F2.

    Solution: The resultant force is

    1 2 (15 12 ) (2 ) 17 11= + = − + + = −R F F i j i j i j Next, find the moment about (2, 4) produced by each force:

    1 1 1 (2 2 ) (15 12 ) 24 30 54= × = + × − = − − = −M r F i j i j k k k 2 2 2 (0 2 ) (2 ) 4= × = + × + = −M r F i j i j k

    The total moment is 1 2 20 54 4 20 38= + + = − − + = −M M M k k k k k .

    6.2 Equilibrium A body is in equilibrium if the resultant of all forces acting on the body is zero and the resultant moment about every point is zero. This is stated as 0 and 0A= =∑ ∑F M (6.2)

    d

    F

    F

    M = Fd

  • 6 / Statics 99

    where A is some arbitrary point. These two vector equations can be written as the six scalar equations

    0 0 0 0 0 0

    x x A

    y y A

    z z A

    F M F M F M

    = =

    = =

    = =

    ∑ ∑ ∑ ∑ ∑ ∑

    ,

    ,

    ,

    (6.3)

    Most often, the problems you will face on the exam involve forces acting in a single plane, selected as the xy-plane. Then, only three of the six equations are useful, namely 0 0 0x y zF F M= = =∑ ∑ ∑ (6.4) When solving problems involving forces and moments, it is extremely important to draw a free-body diagram, a sketch showing all forces and moments and where they act, and all dimensions. Any body that has only two forces acting on it is a two-force member and the two forces must be equal, opposite, and co-linear, as shown in Fig.6.3a. If three non-parallel forces act on a body, they must be coplanar and intersect at a point, as on the three-force member in Fig.6.3b. If they are parallel forces, such as the three forces of Fig.6.3c, they could act as shown.

    a) two-force member

    b) three-force member c) parallel forces

    Figure 6.3 Plane force systems. When drawing free-body diagrams, it is important to realize that a force acts in tension in the direction of a rope or cable; it acts normal to a roller or ball, and at a wall there exists a force and a moment. EXAMPLE 6.3 Find the reaction at the wall of the beam loaded as shown.

    Solution: The reaction at the wall will consist of a force and a moment. The upward force is found by summing forces in the vertical direction:

    wall 100 50 40 4 310 NF = + + × =

    The moment is found by taking moments about the end at the wall:

    20 N

    20 N

    F1

    F2 F3 F2

    F1 F3

    ������������� ������������� ������������� ������������� ������������� �������������

    2 m 2 m 4 m

    40 N/m 100 N 50 N

  • 100 6 / Statics

    wall 100 2 50 8 160 6 1560 N mM = × + × + × = ⋅

    Note: the force due to the distributed load is positioned at the center of the distribution. EXAMPLE 6.4 A weight is suspended by two ropes, as shown. Determine the tension in each rope.

    Solution: Let T1 be the tension in rope 1 and T2 the tension in rope 2. Summing forces in the x- and y-directions gives

    x-dir: 1 2 1 2cos45 cos30 or 0.707 0.866T T T T° = ° = y-dir: 1 2 1 2400 sin 45 sin30 or 400 0.707 0.5T T T T= ° + ° = +

    These two equations are solved simultaneously to provide

    1 2359 N and 293 NT T= = 6.3 Trusses and Frames A truss is an assembly of two-force members coupled together with pins, as in some bridge structures; forces are located only at the pin connections and a member can be either in tension or compression. A frame member may have forces acting on it at any point on the frame; it is typically much heavier than a truss member. Examples illustrate. EXAMPLE 6.5 Find the force in link CE, CD, and EF in the truss shown on the left. All links are of length L except CD which has length L/2.

    Solution: Cut the truss so that the unknown forces are exposed, as shown on the right above. Sum forces in the vertical direction:

    400 N

    45o 30o �������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������� ��������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������

    Rope 1 Rope 2

    4000 N E

    B C D

    A F

    2000 N

    4000 N

    B C

    A F

    2000 N

    FCD

    FCE

    FEF

  • 6 / Statics 101

    cos30 4000 0. 4620 NCE CEF F° + = ∴ = −

    Moments about point E gives

    4000 2 2000 cos30 cos30 . 11240 NCD CDL L F L F× + × ° = × ° ∴ =

    Sum forces in the horizontal direction to obtain

    11240 4620cos60 2000 0. 11550 NEF EFF F+ + ° − = ∴ = −

    The minus sign means members CE and EF are in compression.

    EXAMPLE 6.6 What is the magnitude of the force acting on the pin at A?

    Solution: Sketch a free-body diagram (it’s not done here, but it’s very helpful). There will be unknown force components Fx and Fy at A. There will be only a vertical force FC at C since BC is a two-force member (it could actually be a rope). So,

    400cos45 283 NxF = ° =

    Sum moments about point A and find (assume FC in tension acting downward)

    20 400 0.707 40 400 0.707 20 . 848 NC CF F× × + × × = × ∴ =

    Sum vertical comp

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