CE Statics Lecture 15

Moment of a Force on a Rigid Body If a force is to be moved from one point to another, then the external effects of the force should remain the same. The external effects are the translation and rotation of the body by the force.

Case 1: If point O is located on the line of action of the force The rigid body is subjected to force F at A It is required to move F to point O without altering the external effects Apply equal but opposite forces F at point O Forces F at A and –F at O can be cancelled Force F at O remains It is noted that force F was transmitted along its line of action from A to O The force can act at any point along its line of action, that is why it is a sliding vector. The external effects remain unchanged if a force was transmitted along its line of action. A O F A O F F -F A O F

Case 2: If point O is not located on the line of action of the force Force F is applied at A and needed to be moved to point O without changing the external effects Apply F and –F at O It can be noticed that F at A and F at O form a couple which has moment (M = r  F) Couple moment is a free vector, so it can be applied at any point on the body In addition to couple moment, F acts at point O A O F A O F F -F r A O P F M = r  F

Example 1 The effect of F on the hand will not change whether F is applied at A or O. A O F d A O F d

Example 2 F has moment about A (figure 1). F has no moment about A (figure 2). So, moment has to be added when F is moved from A to O so that the external effects will not change. A O F d A F d

Resultant of a Force and Couple System The rigid body is subjected to couple moments and a system of forces. r2 F2 r1 F1 O M

The forces and moments can be simplified by moving them to point O. Since O is not on the lines of action of F1 and F2, couple moments must be applied at O. M2 = r2  F2 F2 M1 = r1  F1 F1 O M

The resultant F R can be found by : F R = F1 + F2 And the resultant couple moment can be found by: M R = M + M1 + M2 F R is independent of the location of O, while M R depends on the location of O since M is a function of the position vector (r). FRFR  O MRMR

Procedure for Analysis A. Three-dimensional System Use Cartesian vector analysis F R =  F M RO =  M +  M O B. Coplanar Force System Use scalar analysis F R = F Rx + F Ry F Rx =  Fx F Ry =  Fy The resultant moment M R is perpendicular to the x-y plane containing the forces (M RO =  M +  M O ).

Reduction of a Simple Distributed Loading In many cases, bodies or structures are subjected to distributed loading. The loading intensity at each point is measured in lb/ ft 2 or N/m 2 which are units of pressure. Uniform distributed loading will be discussed. If we have the distributed loading shown: L a y x

The loading is a system of parallel forces. The number of forces is infinite and separated by differential distances. The loading function is a function of x: p = p (x) If [ p = p (x) ] was multiplied by the width ( a ), the loading along y will be obtained: w = [ p (x) N/m 2 ] [a m] = w (x) N/m L a y x

The loading intensity is represented by a coplannar system of forces. L x dx dA dF w = w(x)

Magnitude of Resultant Force To find the resultant force ( F R =  F ), integration must be followed since infinite number of forces (dF) are acting on the body. dF = w(x) dx = dA (since w(x) is the force per unit length, x) for the entire length, + F R =  F =  w (x) dx =  dA = A Therefore, the magnitude of the resultant force is equal to the entire area under the loading diagram [ w= w(x) ].

Location of Resultant Force Using M R =  M The location of F R (D) is found by equating the moments of F R and distributed forces about point (O) D F R =  x [w(x)] dx Then, D = {  x [w(x)] dx} / F R Substituting for F R D = {  x [w(x)] dx} / {  w(x) dx} = {  x dA} / {  dA} which is the equation of the geometric center (centroid). Therefore, F R acts at the centroid of the distributed loading diagram L D FRFR w = w(x)