Engineering Mechanics: Statics

Transcription

1 Engineering Mechanics: Statics Chapter 2: Force Systems Part A: Two Dimensional Force Systems Force Force = an action of one body on another Vector quantity External and Internal forces Mechanics of Rigid bodies: Principle of Transmissibility Specify magnitude, direction, line of action No need to specify point of application Concurrent forces Lines of action intersect at a point 1

2 2D Force Systems Rectangular components are convenient for finding the sum or resultant R v of two (or more) forces which are concurrent v v v R = F + F = ( F ˆi + F ˆj ) + ( F ˆi + F ˆj ) 1 2 1x 1y 2x 2y = ( F + F ) ˆi + ( F + F ) ˆj 1x 2x 1y 2y Actual problems do not come with reference axes. Choose the most convenient one! Moment In addition to tendency to move a body in the direction of its application, a force tends to rotate a body about an axis. The axis is any line which neither intersects nor is parallel to the line of action This rotational tendency is known as the moment M of the force Proportional to force F and the perpendicular distance from the axis to the line of action of the force d The magnitude of M is M = Fd 2

3 Moment The moment is a vector M perpendicular to the plane of the body. Sense of M is determined by the right-hand rule Direction of the thumb = arrowhead Fingers curled in the direction of the rotational tendency In a given plane (2D),we may speak of moment about a point which means moment with respect to an axis normal to the plane and passing through the point. +, - signs are used for moment directions must be consistent throughout the problem! Moment A vector approach for moment calculations is proper for 3D problems. Moment of F about point A maybe represented by the cross-product M = r x F where r = a position vector from point A to any point on the line of action of F M = Fr sin α = Fd 3

4 Examples Calculate the magnitude of the moment about the base point O of the 600-N force by using both scalar and vector approaches. Calculate the moment of the 90-N force about point O for the condition θ = 15º. Couple Moment produced by two equal, opposite, and noncollinear forces = couple M = F(a+d) Fa = Fd Moment of a couple has the same value for all moment center Vector approach M = r A x F + r B x (-F) = (r A - r B ) x F = r x F Couple M is a free vector 4

5 Force-Couple Systems Replacement of a force by a force and a couple Force F is replaced by a parallel force F and a counterclockwise couple Fd Example Replace the force by an equivalent system at point O Also, reverse the problem by the replacement of a force and a couple by a single force Problem 2/76 (modified) The device shown is a part of an automobile seat-back-release mechanism. The part is subjected to the 4-N force exerted at A and a 300-N-mm restoring moment exerted by a hidden torsional spring. Find an equivalent force-couple system at point O of the 4-N force 5

6 Resultants The simplest force combination which can replace the original forces without changing the external effect on the rigid body Resultant = a force-couple system v v v v v R= F + F + F + K=ΣF x =Σ x, y =Σ y, = ( Σ x) + ( Σ y) R F R F R F F θ = tan -1 R R y x Resultants Choose a reference point (point O) and move all forces to that point Add all forces at O to form the resultant force R and add all moment to form the resultant couple M O Find the line of action of R by requiring R to have a moment of M O M =Σ M= Σ( Fd) O v v R=ΣF Rd = M O 6

7 Problem 2/87 Replace the three forces acting on the bent pipe by a single equivalent force R. Specify the distance x from point O to the point on the x-axis through which the line of action of R passes. Problem 2/76 The device shown is a part of an automobile seat-back-release mechanism. The part is subjected to the 4-N force exerted at A and a 300-N-mm restoring moment exerted by a hidden torsional spring. Determine the y-intercept of the line of action of the single equivalent force. 7

8 Force Systems Part B: Three Dimensional Force Systems Three-Dimensional Force System Rectangular components in 3D Express in terms of unit vectors î, ĵ, ˆk v F= F ˆi + F ˆj + F kˆ x y z F = Fcos θ, F = Fcos θ, F = Fcosθ x x y y z z F= F + F + F x y z cosθ x, cosθ y, cosθ z are the direction cosines cosθ x = l, cosθ y = m, cosθ z = n v F= F( li ˆ + mj ˆ + nkˆ ) 8

9 Three-Dimensional Force System Rectangular components in 3D If the coordinates of points A and B on the line of action are known, v v v AB ( x ˆ ˆ ˆ 2 x1) i + ( y2 y1) j + ( z2 z1) k F = FnF = F = F AB ( x x ) + ( y y ) + ( z z ) If two angles θ and φ which orient the line of action of the force are known, F = Fcos φ, F = F sinφ xy F = Fcosφ cos θ, F = Fcosφ sinθ x z y Problem 2/98 The cable exerts a tension of 2 kn on the fixed bracket at A. Write the vector expression for the tension T. 9

10 Moment and Couple Moment of force F about the axis through point O is M O = r x F r runs from O to any point on the line of action of F Point O and force F establish a plane A The vector M o is normal to the plane in the direction established by the right-hand rule Evaluating the cross product ˆi ˆj kˆ M = r r r O x y z F F F x y z Sample Problem 2/10 A tension T of magniture 10 kn is applied to the cable attached to the top A of the rigid mast and secured to the ground at B. Determine the moment M z of T about the z-axis passing through the base O. 10

11 Resultants A force system can be reduced to a resultant force and a resultant couple v v v v v R= F1 + F2 + F3 L= F v v v v v v M= M1 + M2 + M3 + L= ( r F) Problem 2/154 The motor mounted on the bracket is acted on by its 160-N weight, and its shaft resists the 120-N thrust and 25-N.m couple applied to it. Determine the resultant of the force system shown in terms of a force R at A and a couple M. 11