Lecture 5, Sections ST4.5 ST4.10 ST4.5) Moment of a Force about a Specified Axis ST4.6) Moment of a Couple ST4.7) Equivalent System ST4.8) Resultant of a Force and a Couple System ST4.9) Further Reduction of a Force and Couple ST4.10) Reduction of Simple Distributed Loading
ST4.5) Moment of Force about a Specified Axis Moment and its axis are always to the plane containing the force and moment arm. In some problems we need to find the component of moment along a specified axis that passes through a point (2 methods are available). 1
Scalar Analysis: 2-Steps: a) M O = ( 20N)(0.5m) = 10N. m But on Ob axis, so 3 b) M y = (10N. m) = 6N. m 5 1- Step M y = ( 20N)(0.3m) = 6N. m But need distance to y axis If the line of action of F is to an axis aa, the magnitude of moment around aa is given by 2
Vector Analysis: 2-Steps: a) M O = r A F = ( 0.3i + 0.4 j) ( 20k) = { 8i + 6 j} N. m But on Ob axis, so b) M y = M u = ( 8i + 6 j) j = 6N O a m 1- Step M = u ( r F) y a But need distance to y axis 3
Vector Analysis (General): The magnitude of a moment around an arbitrary axis, aa`: M u ( r F) a = a This is called Triple Scalar Product, given by: ua ua u x x ax M a = rx ry rz F F F x For the moment vector M a = M u = [ u ( r F)] u a a y a z a The resultant moment of a series of forces: M a = [ ua ( r F)] = ua ( r F) 4
Problem ST4.58: Determine the resultant moment of the two forces about the Oa axis. Express the result as a Cartesian Vector. 5
Problem ST4.58: The hood of the automobile is supported by the strut AB, which exerts a force of F = 100 N on that hood. Determine the moment of this force about the hinged axis y. 6
Problem ST4.65: If a torque of 80 N.m is required to loosen that bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench. 7
ST4.6) Moment of a Couple Couple: two parallel forces that have the same magnitude, have opposite directions, and are separated by a distance d. The only effect of a couple is to produce a rotation or tendency of rotation in a specified direction (resultant force = 0) Moment produced by a couple can be determined by finding the sum of the moments of both couple forces about any arbitrary point, O. To simplify the problem, take moments about a point lying on the line of action of one of the forces (force at chosen point produce zero moment) 8
If point A is chosen M = r F -A couple moment is a free vector that can act at any point since M depends only upon the position vector r directed between the forces and not the position vectors r A and r B directed from the arbitrary point O to the forces. - Two couples are said to be equivalent if they produce the same moment. Thus, their forces must lie either same plane or parallel planes. 9
Scalar Analysis: Magnitude: M = Fd Direction: Right-Hand Rule Vector Analysis: M = r F Resultant M = r F 10
Problem ST4.74: The resultant couple moment created by two couples acting on the disk is M R = {10k} kn.cm. Determine the magnitude of force T. 11
Problem ST4.78: Two couples act on the beam. Determine the magnitude of F so that the resultant couple moment is 450 N.m, counterclockwise. Where on the beam does the resultant couple act on? 12
Problem ST4.80: If the couple moment acting on the pipe has a magnitude of 400 N.m, determine the magnitude F of the vertical force applied to each wrench. 13
Illustration: Horizontal force on a stick Illustration: Vertical force on a stick 14
Illustration: Reducing multiple forces to single force with moment Illustration: Reducing multiple force to single force without moment 15
Problem ST4-116: Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from B. 16
Problem ST4-121: Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD, measured from end C. 17
Reduction of Distributed Load: - In many situations a very large surface of a body may be subjected to distributed loadings (e.g. wind, fluids, weight of material). - Intensity of these loads at each point on the surface is defined as pressure p. - pressure = p = force per unit area = N/m2 = Pascal = Pa. -1 Pa = 1 N/m2. - In this section most common case will be considered - Most common case uniform along one axis of a flat rectangular body upon which the loading is applied. 18
Example of Uniform Loading: - Direction and intensity of the pressure load is indicated by arrows (Load-intensity diagram). - The entire loading on the plate is therefore a system of parallel forces, infinite in number and each acting on a separate differential area of the plate. - The loading function, p = p (x) Pa, is only a function of x since the pressure is uniform along the y axis. -If we multiply p = p(x) by the width = a meters of the plate we obtain w = [p(x) N/m2 ] a m = w (x)n/m. - It is measured as force per unit length rather than force per unit area. 19
2-D Representation of Uniform Load: -Load intensity can be represented by a system of coplanar parallel forces in 2D. -Using integration, this system of forces can be simplified to a single resultant force F R and its location can be specified. w( x dx = da = F = ) R x = L L L xw( x) dx = w( x) dx L L L xda da A (Area Under Curve) (Centroid of Area Under Curve) 20
Problem ST4-139: The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent resultant location, measured from point O. 21
Problem ST4-141: Replace the loading by an equivalent force and couple moment acting at point O. 22
Problem ST4-148: Replace the distributed loading by an equivalent resultant force and specify its location, measured from point A. 23
Problem ST4-153: Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member BC, measured from C. 24
Problem ST4-154: Replace the distributed loading by an equivalent resultant force and specify its line of action intersects the beam measured from point O. 25