Copyright 2010 Pearson Education South Asia Pte Ltd Chapter 4: Force System Resultants 1. Moment of a Force Scalar Formulation 2. Cross Product 3. Moment of Force Vector Formulation 4. Principle of Moments 5. Moment of a Force about a Specified Axis/line 6. Moment of a Couple 7. Simplification of a Force and Couple System 8. Further Simplification of a Force and Couple System 9. Reduction of a Simple Distributed Loading Copyright 2010 Pearson Education South Asia Pte Ltd Chapter Objectives Concept of moment of a force in two and three dimensions Method for finding the moment of a force about a specified axis. Define the moment of a couple. Determine the resultants of non-concurrent force systems Reduce a simple distributed loading to a resultant force having a specified location APPLICATIONS Beams are often used to bridge gaps in walls.We have to know what the effect of the force on the beam will have on the supports of the beam. What do you think is happening at points A and B? APPLICATIONS (continued) Carpenters often use a hammer in this way to pull a stubborn nail. Through what sort of action does the force FH at the handle pull the nail? How can you mathematically model the effect of force FH at point O? Copyright 2010 Pearson Education South Asia Pte Ltd 4.1 Moment of a Force Scalar Formulation Moment of a force about a point or axis a measure of the tendency of the force to cause a body to rotate about the point or axis Torque tendency of rotation caused by F or simple moment (Mo) z

Copyright 2010 Pearson Education South Asia Pte Ltd Magnitude For magnitude of MO, MO = Fd (Nm) where d = perpendicular distancefrom point O to its line of actionof force F Direction Direction using right hand rule Copyright 2010 Pearson Education South Asia Pte Ltd Resultant Moment Resultant moment, MRo = moments of all the forces

MRo = Fd Copyright 2010 Pearson Education South Asia Pte Ltd Example 4.1 For each case, determine the moment of the force about point O. Copyright 2010 Pearson Education South Asia Pte Ltd Copyright 2010 Pearson Education South Asia Pte Ltd 4.2 Cross Product Cross product of two vectors A and B yields C, which is written as C = A X B Magnitude Magnitude of C is the product ofthe magnitudes of A and B andsine of angle For angle , 0 180 C = AB sin Copyright 2010 Pearson Education South Asia Pte Ltd Direction Vector C has a direction that is perpendicular to the plane (x-y plane) containing A and B such that C is specified by the right hand rule To express vector C whenmagnitude and direction are known C = A X B= (AB sin)uC (AB sin) is the magnitude of vector C Copyright 2010 Pearson Education South Asia Pte Ltd Laws of Operations 1. Commutative law is not valid A X B B X A Rather,A X B = - (B X A) Cross product B X A yields avector opposite in direction to C

B X A = -C Copyright 2010 Pearson Education South Asia Pte Ltd 2. Multiplication by a Scalar a( A X B ) = (aA) X B = A X (aB) = ( A X B )a 3. Distributive LawA X ( B + D ) = ( A X B ) + ( A X D ) Proper order of the cross product must be maintained since they are not commutative Copyright 2010 Pearson Education South Asia Pte Ltd Cartesian Vector Formulation A more compact determinant in the form a special utility for a cross-product in Cartesian vector form z y xz y xB B BA A Ak j iB X A CROSS PRODUCTAlso, the cross product can be written as a determinant.Each component can be determined using 2 2 determinants. Copyright 2010 Pearson Education South Asia Pte Ltd 4.3 Moment of Force - Vector Formulation Moment of force F about point O can be expressed using cross product MO = r X F Magnitude For magnitude of cross product,MO = rF sin Copyright 2010 Pearson Education South Asia Pte Ltd Direction Direction and sense of MO are determined by right-hand rule*Note:- curl of the fingers indicates the sense of rotation - Maintain proper order of r and F since cross product is not commutative Copyright 2010 Pearson Education South Asia Pte Ltd Principle of Transmissibility For force F applied at any pointalong the line of action of the force, moment about O is MO = r x F F has the properties of a sliding vector, thus MO = r1 X F = r2 X F = r3 X F Copyright 2010 Pearson Education South Asia Pte Ltd Cartesian Vector Formulation For force expressed in Cartesian form, With the determinant expended,MO = (ryFz rzFy)i (rxFz - rzFx)j + (rxFy ryFx)k z y xz y x OF F Fr r rk j iF X r M Copyright 2010 Pearson Education South Asia Pte Ltd Resultant Moment of a System of Forces Resultant moment of forces about point O can be determined by vector addition MRo = (r x F)Copyright 2010 Pearson Education South Asia Pte Ltd Example 4.4 Two forces act on the rod. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector. Copyright 2010 Pearson Education South Asia Pte Ltd EXAMPLE1) Find F = F1 + F2 and rOA. 2) DetermineMO=rOA F. o Given: F1={100 i - 120 j + 75 k}kN F2={-200 i +250 j + 100 k}kN Find: Resultantmoment by the forces about point O. Plan: Copyright 2010 Pearson Education South Asia Pte Ltd 4.4 Principles of Moments Also known as Varignons Theorem Moment of a force about a point is equal to the sum of the moments of the forces components about the point Since F = F1 + F2, MO = (r X F)= r X (F1 + F2)= r X F1 + r X F2 Copyright 2010 Pearson Education South Asia Pte Ltd Example 4.5 Determine the moment of the force about point O. Copyright 2010 Pearson Education South Asia Pte Ltd GROUP PROBLEM SOLVINGI Since this is a 2-D problem: 1) Resolve the 100 N force along the handles x and y axes. 2) Determine MA using a scalar analysis. Given: A 100 N force is applied to the hammer. Find:The moment of the force at A. Plan: x y GROUP PROBLEM SOLVING II 1) Find F and rAC.

2) Determine MA = rAC FGiven: The force and geometry shown. Find: Moment of F about point A Plan: Copyright 2010 Pearson Education South Asia Pte Ltd 4.5 Moment of a Force about a Specified Axis For moment of a force about a point, the moment vector and its axis is always perpendicular to the plane A scalar or vector analysis is used to find the component of the moment along a specified axis that passes through the point Copyright 2010 Pearson Education South Asia Pte Ltd Vector Analysis For magnitude of Ma, Ma = MOua

where ua = unit vector In determinant form, z y xz y xaz ay axax aF F Fr r ru u uF X r u M ) ( Copyright 2010 Pearson Education South Asia Pte Ltd Example 4.8 Determine the moment produced by the force F which tends to rotate the rod about the AB axis. Copyright 2010 Pearson Education South Asia Pte Ltd EXAMPLE 1) Use Mz = u (r F). 2) First, find F in Cartesian vector form. 3) Note that u= 1 i in this case. 4)The vector r is the position vector from O to A. A B Given: A force is applied to the tool as shown. Find: The magnitude of the moment of this force about the x axis of the value. Plan:Copyright 2010 Pearson Education South Asia Pte Ltd 4.6 Moment of a Couple Couple two parallel forces same magnitude but opposite direction separated by perpendicular distance d Resultant force = 0 Tendency to rotate in specified direction Couple moment = sum of moments of both couple forces about any arbitrary point Page148 Slide 85 Copyright 2010 Pearson Education South Asia Pte Ltd Scalar Formulation Magnitude of couple momentM = Fd Direction and sense are determined by right hand rule M acts perpendicular to plane containing the forces Copyright 2010 Pearson Education South Asia Pte Ltd Vector Formulation For couple moment,M = r X F If moments are taken about point A, moment of F is zero about this point r is crossed with the force to which it is directed(according to the arrowhead) Copyright 2010 Pearson Education South Asia Pte Ltd Equivalent Couples 2 couples are equivalent if they produce the same moment Forces of equal couples lie on the same plane orparallel plane Copyright 2010 Pearson Education South Asia Pte Ltd Resultant Couple Moment Couple moments are free vectors and may be applied to any point P and added vectorially For resultant moment of two couples at point P, MR = M1 + M2 For more than 2 moments, MR = (r X F) Copyright 2010 Pearson Education South Asia Pte Ltd Example 4.12 Determine the couple moment acting on the pipe. Segment AB is directed 30 below the xy plane. EXAMPLE : SCALAR APPROACH 1)Add the two couples to find the resultant couple. 2)Equate the net moment to 1.5 kNm clockwise to find F. Given: Two couples act on the beam with the geometry shown. Find: The magnitude of F so that the resultant couple moment is 1.5 kNm clockwise. Plan: EXAMPLE: VECTOR APPROACH 1) UseM = r Fto find the couple moment. 2) Setr=rABand F = FB. 3) Calculate the cross product to find M. Given:A 450 N force couple acting on the pipe assembly. Find: The couple moment in Cartesian vector notation. Plan: 1) UseM = r Fto find the couple moment. 2) Set r=rABand F = {15 k} N. 3) Calculate the cross product to find M. Given: F= {15 k} N and F = { 15 k} NFind:The couple moment acting on the pipe assembly using Cartesian vector notation. Plan: GROUP PROBLEM SOLVINGII 4.7 SIMPLIFICATION OF FORCE AND COUPLE SYSTEMWhen a number of forces and couple moments are acting on a body, it is easier to understand their overall effect on the body if they are combined into a single force and couple moment having the same external effect. The two force and couple systems are called equivalent systems since they have the same external effect on the body. MOVING A FORCE ON ITS LINE OF ACTION Moving a force from A to O, when both points are on the vectors line of action, does not change the external effect.Hence, a force vector is called a sliding vector. MOVING A FORCE ON ITS LINE OF ACTION Moving a force from A to B, when both points are on the vectors line of action, does not change the external effect.Hence, a force vector is called a sliding vector. (But the internal effect of the force on the body does depend on where the force is applied). MOVING A FORCE OFF OF ITS LINE OF ACTION Moving a force from point A to O (as shown above when both points are not on the vectors line of action) requires creating an additional couple moment.Since this new couple moment is a free vector, it can be applied at any point P on the body. MOVING A FORCE OFF OF ITS LINE OF ACTION When a force is moved, but not along its line of action, there is a change in its external effect! Essentially, moving a force from point A to B (as shown above) requires creating an additional couple moment. So moving a force means you have to add a new couple. Since this new couple moment is a free vector, it can be applied at any point on the body. B FINDING THE RESULTANT OF A FORCEANDCOUPLE SYSTEM (Section 4.8) When several forces and couple moments act on a body, you can move each force and its associated couple moment to a common point O. Now you can add all the forces and couple moments together and find one resultant force-couple moment pair. FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM (Section 4.8) = = In three special cases, concurrent, coplanar, and parallel systems of forces, the system can always be reduced to a single force. If FR and MRO are perpendicular to each other, then the system can be further reduced to a single force, FR , by simply moving FR from O to P.Copyright 2010 Pearson Education South Asia Pte Ltd Example 4.16 A structural member is subjected to a couple moment M and forces F1 and F2. Replace this system with an equivalent resultant force and couple moment acting at its base, point O. Copyright 2010 Pearson Education South Asia Pte Ltd EXAMPLE I 1) Sum all the x and y components of the forces to find FRA. 2) Find and sum all the moments resulting from moving each force component to A. 3) Shift FRA to a distance d such that d = MRA/FRy Given: A 2-D force systemwith geometry as shown.Find:The equivalent resultant force and couple moment acting at A and then the equivalent single force location measured from A. Plan: EXAMPLEII 1) Find FRO=Fi= FRzo k 2) Find MRO = (ri Fi) = MRxO i+MRyO j 3) The location of the single equivalent resultant force is given as x = MRyO/FRzOand y=MRxO/FRzO Given: The slab is subjected to three parallel forces. Find: The equivalent resultant force and couple moment at the origin O. Also find the location (x, y) of the single equivalent resultant force. Plan: GROUP PROBLEM SOLVING I 1) Sum all the x and y components of the two forces to find FRA. 2) Find and sum all the moments resulting from moving each force to A and add them to the 1500 Nm free moment to find the resultant MRA . Given: A 2-D force and couple system as shown. Find:The equivalent resultant force and couple moment acting at A. Plan: GROUP PROBLEM SOLVING II Given: Forces F1 and F2 are applied to the pipe.Find: An equivalent resultant force and couple moment at point O. Plan:a) Find FRO = Fi = F1 + F2 b) Find MRO = MC + ( ri Fi ) where,MC are any free couple moments (none in this example). ri are the position vectors from the point O to any point on the line of action of Fi . APPLICATIONS To analyze the loads effect on the steel beams, it is often helpful to reduce this distributed load to a single force. How would you do this? There is a bundle (called a bunk) of 50 mm x 100 mm boards stored on a storage rack.This lumber places a distributed load (due to the weight of the wood) on the beams holding the bunk. APPLICATIONS (continued) The uniform wind pressure is acting on a triangular sign (shown in light brown). To be able to design the joint between the sign and the sign post, we need to determine a single equivalent resultant force and its location. Copyright 2010 Pearson Education South Asia Pte Ltd 4.9 Reduction of a Simple Distributed Loading Large surface area of a body may be subjected to distributed loadings Loadings on the surface is defined as pressure Pressure is measured in Pascal (Pa): 1 Pa = 1N/m2 Uniform Loading Along a Single Axis Most common type of distributedloading is uniform along asingle axis Copyright 2010 Pearson Education South Asia Pte Ltd 4.9 Reduction of a Simple Distributed Loading Magnitude of Resultant Force Magnitude of dF is determined from differential area dA under the loading curve. For length L, Magnitude of the resultant force is equal to the total area A under the loading diagram. A dA dx x w FA LR Copyright 2010 Pearson Education South Asia Pte Ltd 4.9 Reduction of a Simple Distributed Loading Location of Resultant Force MRo = MO dF produces a moment of x[dF] = x [w(x) dx] about O For the entire distributed load on plate, Solving for LR O Rodx x xw F x M M ) ( x AALLdAxdAdx x wdx x xwx) () (Copyright 2010 Pearson Education South Asia Pte Ltd Example 4.21 Determine the magnitude and location of the equivalent resultant force acting on the shaft. EXAMPLES Until you learn more about centroids, we will consider only rectangular and triangular loading diagrams. Finding the area of a rectangle and its centroid is easy! Note that triangle presents a bit of a challenge but still is pretty straightforward. EXAMPLES The rectangular load: FR=10 5 = 50 kN and = 2.5 m. x The triangular loading: FR = (0.5) (600) (6) = 1,800 Nand= 6 (1/3) 6=4 m.Please note that the centroid of a right triangle is at a distance one third the width of the triangle as measured from itsbase. x Now lets complete the calculations to find the concentrated loads (which is a common name for the resultant of the distributed load).

GROUP PROBLEM SOLVING 1) The distributed loading can be divided into three parts. (one rectangular loading and two triangular loadings). 2) Find FR and its location for each of these three distributed loads. 3) Determine the overall FR of the three point loadings and its location. Given: The loading on thebeam as shown. Find: The equivalent force and its locationfrom point A. Plan: