Chapter 4: Force System Resultants Engineering Mechanics: Statics Chapter Objectives To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions. To provide a method for finding the moment of a force about a specified axis. To define the moment of a couple. To present methods for determining the resultants of
non-concurrent force systems. To indicate how to reduce a simple distributed loading to a resultant force having a specified location. Chapter Outline Moment of a Force Scalar Formation Cross Product Moment of Force Vector Formulation Principle of Moments Moment of a Force about a Specified Axis Chapter Outline
Moment of a Couple Equivalent System Resultants of a Force and Couple System Further Reduction of a Force and Couple System Reduction of a Simple Distributed Loading 4.1 Moment of a Force Scalar Formation
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 Consider horizontal force Fx, which acts perpendicular to the handle of the wrench and is located dy from the point O 4.1 Moment of a Force the Scalar Formation
Fx tends to turn pipe about the z axis The larger the force or the distance dy, the greater the turning effect Torque tendency of rotation caused by Fx or simple moment (Mo) z 4.1 Moment of a In General
Force Scalar Formation Consider the force F and the point O which lies in the shaded plane The moment MO about point O, or about an axis passing through O and perpendicular to the plane, is a vector quantity Moment MO has its specified magnitude and direction 4.1 Moment of a
Magnitude Force Scalar Formation For magnitude of MO, MO = Fd where d = moment arm or perpendicular distance from the axis at point O to its line of action of the force Units
for moment is N.m 4.1 Moment of a Direction Direction Force Scalar Formation of MO is specified by using right hand rule - fingers of the right hand are curled to follow the sense of rotation when force rotates
about point O 4.1 Moment of a Force Scalar Formation Resultant Moment of a System of Coplanar Forces Resultant moment, MRo = addition of the moments of all the forces algebraically since all moment forces are collinear MRo = Fd
taking clockwise to be positive 4.1 Moment of a Force Scalar Formation Resultant Moment of a System of Coplanar Forces A clockwise curl is written along the equation to indicate that a positive moment if directed along the + z axis and negative
along the z axis 4.1 Moment of a Force Scalar Formation Moment of a force does not always cause rotation Force F tends to rotate the beam clockwise about A with moment MA = FdA Force F tends to rotate the beam counterclockwise about B with moment
MB = FdB Hence support at A prevents the rotation 4.1 Moment of a Force Scalar Formation Example 4.1 For each case, determine the moment of the
force about point O 4.1 Moment of a Solution Force Scalar Formation Line of action is extended as a dashed line to establish moment arm d Tendency to rotate is indicated and the orbit is shown as a colored curl (a) M o (100 N )(2m) 200 N .m(CW )
(b) M o (50 N )(0.75m) 37.5 N .m(CW ) 4.1 Moment of a Solution Force Scalar Formation (c) M o (40 N )(4m 2 cos 30 m) 229 N .m(CW ) (d ) M o (60 N )(1sin 45 m) 42.4 N .m(CCW ) (e) M o (7kN )(4m 1m) 21.0kN .m(CCW ) 4.1 Moment of a Force Scalar
Formation Example 4.2 Determine the moments of the 800N force acting on the frame about points A, B, C and D. 4.1 Moment of a Force Scalar Formation Solution Scalar Analysis M A (800 N )(2.5m) 2000 N .m(CW ) M B (800 N )(1.5m) 1200 N .m(CW )
M C (800 N )(0m) 0kN .m Line of action of F passes through M C(800 N )(0.5m) 400 N .m(CCW ) D 4.2 Cross Product Cross product of two vectors A and B yields C, which is written as C=AXB Read as C equals A cross B 4.2 Cross Product
Magnitude Magnitude of C is defined as the product of the magnitudes of A and B and the sine of the angle between their tails For angle , 0 180 Therefore, C = AB sin 4.2 Cross Product Direction Vector
C has a direction that is perpendicular to the plane containing A and B such that C is specified by the right hand rule - Curling the fingers of the right hand form vector A (cross) to vector B - Thumb points in the direction of vector C 4.2 Cross Product Expressing vector C when magnitude and direction are known
C = A X B = (AB sin)uC where scalar AB sin defines the magnitude of vector C unit vector uC defines the direction of vector C 4.2 Cross Product Laws of Operations 1. Commutative law is not valid AXBBXA Rather, AXB=-BXA Shown by the right hand rule Cross product A X B yields a vector opposite in direction to C B X A = -C
4.2 Cross Product Laws of Operations 2. Multiplication by a Scalar a( A X B ) = (aA) X B = A X (aB) = ( A X B )a 3. Distributive Law AX(B+D)=(AXB)+(AXD) Proper order of the cross product must be maintained since they are not commutative 4.2 Cross Product Cartesian Vector Formulation Use
C = AB sin on pair of Cartesian unit vectors Example For i X j, (i)(j)(sin90) = (1)(1)(1) = 1 4.2 Cross Product Laws of Operations In a similar manner, i X j = k i X k = -j iXi=0 j X k = i j X i = -k jXj=0
k X i = j k X j = -i kXk=0 Use the circle for the results. Crossing CCW yield positive and CW yields negative results 4.2 Cross Product Laws of Operations Consider cross product of vector A and B A X B = (Axi + Ayj + Azk) X (Bxi + Byj + Bzk) = AxBx (i X i) + AxBy (i X j) + AxBz (i X k) + AyBx (j X i) + AyBy (j X j) + AyBz (j X k) + AzBx (k X i) +AzBy (k X j) +AzBz (k X k) = (AyBz AzBy)i (AxBz - AzBx)j + (AxBy AyBx)k
4.2 Cross Product Laws of Operations In determinant form, i j AXB Ax Ay Bx B y
k Az Bz 4.3 Moment of Force - Vector Formulation Moment of force F about point O can be expressed using cross product MO = r X F where r represents position vector from O to any point lying on the line of action
of F 4.3 Moment of Force - Vector Formulation Magnitude For magnitude of cross product, MO = rF sin where is the angle measured between tails of r and F Treat r as a sliding vector. Since d
= r sin, MO = rF sin = F (rsin) = Fd 4.3 Moment of Force - Vector Formulation Direction Direction and sense of MO are determined by right-hand rule - Extend r to the dashed position - Curl fingers from r towards F - Direction of MO is the same as the direction of the thumb
4.3 Moment of Force - Vector Formulation Direction *Note: - curl of the fingers indicates the sense of rotation - Maintain proper order of r and F since cross product is not commutative 4.3 Moment of Force - Vector Formulation Principle of Transmissibility For
force F applied at any point A, moment created about O is MO = rA x F F has the properties of a sliding vector and therefore act at any point along its line of action and still create the same moment about O 4.3 Moment of Force - Vector Formulation Cartesian Vector Formulation For
force expressed in Cartesian form, i M O r XF rx j ry k rz Fx
Fy Fz where rx, ry, rz represent the x, y, z components of the position vector and Fx, Fy, Fz represent that of the force vector 4.3 Moment of Force - Vector Formulation Cartesian Vector Formulation With
the determinant expended, MO = (ryFz rzFy)i (rxFz - rzFx)j + (rxFy yFx)k MO is always perpendicular to the plane containing r and F Computation of moment by cross product is better than scalar for 3D problems 4.3 Moment of Force - Vector Formulation Cartesian Vector Formulation
Resultant moment of forces about point O can be determined by vector addition MRo = (r x F) 4.3 Moment of Force - Vector Formulation Moment of force F about point A, pulling on cable BC at any point along its line of action, will remain constant
Given the perpendicular distance from A to cable is rd MA = rdF In 3D problems, MA = rBC x F 4.3 Moment of Force - Vector Formulation Example 4.4 The pole is subjected to a 60N force that is directed from C to B. Determine the magnitude of the moment created by this force about the support at A.
4.3 Moment of Force - Vector Formulation Solution Either one of the two position vectors can be used for the solution, since MA = rB x F or MA = rC x F Position vectors are represented as rB = {1i + 3j + 2k} m and rC = {3i + 4j} m Force F has magnitude 60N and is directed from C to B 4.3 Moment of Force - Vector Formulation
Solution F (60 N )u F (1 3)i 93 4) j 92 0)k (60 N ) 2 2 2
( 2) ( 1) (2) 40i 20 j 40k N determinant Substitute into
i j k formulation M r XF 1 3 2 A B 40 20 40
[3(40) 2( 20)]i [1(40) 2( 40)] j [1( 20) 3(40)]k 4.3 Moment of Force - Vector Formulation Solution Or
i M A rC XF 3 j k 4 0
40 20 40 [4(40) 0( 20)]i [3(40) 0( 40)] j [3( 20) 4(40)]k Substitute into determinant formulation
M A 160i 120 j 100k N .m For magnitude, M A (160) 2 (120) 2 (100) 2 224 N .m 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 For F = F1 + F2, MO = r X F1 + r X F2 = r X (F1 + F2) =rXF 4.4 Principles of Moments The guy cable exerts a force F on the pole and creates a moment about
the base at A MA = Fd If the force is replaced by Fx and Fy at point B where the cable acts on the pole, the sum of moment about point A yields the same resultant moment 4.4 Principles of Moments Fy create zero moment about A
MA = Fxh Apply principle of transmissibility and slide the force where line of action intersects the ground at C, Fx create zero moment about A MA = Fyb 4.4 Principles of Moments Example 4.6 A 200N force acts on the bracket.
Determine the moment of the force about point A. 4.4 Principles of Moments Solution Method 1: From trigonometry using triangle BCD, CB = d = 100cos45 = 70.71mm = 0.07071m Thus, MA = Fd = 200N(0.07071m) = 14.1N.m (CCW) As a Cartesian vector, MA = {14.1k}N.m
4.4 Principles of Moments Solution Method 2: Resolve 200N force into x and y components Principle of Moments MA = Fd MA = (200sin45N)(0.20m) (200cos45)(0.10m) = 14.1 N.m (CCW) Thus, MA = {14.1k}N.m 4.4 Principles of Moments Example 4.7 The force F acts at the end of the angle bracket. Determine the moment of the
force about point O. 4.4 Principles of Moments View Free Body Diagram Solution Method 1 MO = 400sin30N(0.2m)-400cos30N(0.4m) = -98.6N.m = 98.6N.m (CCW) As a Cartesian vector, MO = {-98.6k}N.m 4.4 Principles of Moments
Solution Method 2: Express as Cartesian vector r = {0.4i 0.2j}N F = {400sin30i 400cos30j}N = {200.0i 346.4j}N For moment, i j k
M O r XF 0.4 0.2 0 200.0 346.4 0 98.6k N .m 4.5 Moment of a Force about a Specified Axis For moment of a force about a point, the
moment and its axis is always perpendicular to the plane containing the force and the moment arm A scalar or vector analysis is used to find the component of the moment along a specified axis that passes through the point 4.5 Moment of a Force about a Specified Axis Scalar Analysis Example Consider the pipe assembly that lies in the horizontal plane and is subjected to the vertical
force of F = 20N applied at point A. For magnitude of moment, MO = (20N)(0.5m) = 10N.m For direction of moment, apply right hand rule 4.5 Moment of a Force about a Specified Axis Scalar Analysis Moment tends to turn pipe about the Ob axis Determine the component of MO about the y axis, My since this component tend
to unscrew the pipe from the flange at O For magnitude of My, My = 3/5(10N.m) = 6N.m For direction of My, apply right hand rule 4.5 Moment of a Force about a Specified Axis Scalar Analysis If the line of action of a force F is perpendicular to any specified axis aa, for the magnitude of the moment of F about the axis,
Ma = Fda where da is the perpendicular or shortest distance from the force line of action to the axis A force will not contribute a moment about a specified axis if the force line of action is parallel or passes through the axis 4.5 Moment of a Force about a Specified Axis A
horizontal force F applied to the handle of the flex-headed wrench, tends to turn the socket at A about the z-axis Effect is caused by the moment of F about the z-axis Maximum moment occurs when the wrench is in the horizontal plane so that full leverage from the handle is achieved (MZ)max = Fd 4.5 Moment of a Force about a Specified Axis
For handle not in the horizontal position, (MZ)max = Fd where d is the perpendicular distance from the force line of action to the axis Otherwise, for moment, MA = Fd MZ = MAcos 4.5 Moment of a Force about a Specified Axis
Vector Analysis Example MO = rA X F = (0.3i +0.4j) X (-20k) = {-8i + 6j}N.m Since unit vector for this axis is ua = j, My = MO.ua = (-8i + 6j)j = 6N.m 4.5 Moment of a Force about a Specified Axis Vector Analysis
Consider body subjected to force F acting at point A To determine moment, M , a - For moment of F about any arbitrary point O that lies on the aa axis MO = rXF where r is directed from O to A - MO acts along the moment axis bb, so projected MO on the aa axis is MA
4.5 Moment of a Force about a Specified Axis Vector Analysis - For magnitude of MA, MA = MOcos = MOua where ua is a unit vector that defines the direction of aa axis MA = ua(r X F) i
M a (uaxi uay j uaz k ) rx Fx - In determinant form, j ry Fy k rz Fz
4.5 Moment of a Force about a Specified Axis Vector Analysis - Or expressed as, u u u ax ay az M a uax (r XF ) rx ry rz
Fx Fy Fz where uax, uay, uaz represent the x, y, z components of the unit vector defining the direction of aa axis and rx, ry, rz represent that of the position vector drawn from any point O on the aa axis and Fx, Fy, Fz represent that of the force vector 4.5 Moment of a Force about a Specified Axis
Vector Analysis The sign of the scalar indicates the direction of Ma along the aa axis - If positive, Ma has the same sense as ua - If negative, Ma act opposite to ua Express Ma as a Cartesian vector, Ma = Maua = [ua(r X F)] ua For magnitude of Ma, Ma = [ua(r X F)] = ua(r X F) 4.5 Moment of a Force about a Specified Axis Wind
blowing on the sign causes a resultant force F that tends to tip the sign over due to moment MA created about the aa axis MA = r X F For magnitude of projection of moment along the axis whose direction is defined by unit vector uA is MA = ua(r X F) 4.5 Moment of a Force about a Specified Axis
Example 4.8 The force F = {-40i + 20j + 10k} N acts on the point A. Determine the moments of this force about the x and a axes. 4.5 Moment of a Force about a Specified Axis Solution Method 1 rA { 3i 4 j 6k }m u x i 1
0 0 M x i (rA XF ) 3 4 6 40 20 10 80 N .m Negative sign indicates that sense of Mx is opposite to i 4.5 Moment of a Force about a Specified Axis Solution We can also compute Ma using rA as rA extends
from a a axis to the force point on the u A 3 i 4 j 5 5 3 5 M a u A (rA XF ) 3 40 120 N .m
4 0 5 4 6 20 10 4.5 Moment of a Force about a Specified Axis Solution Method 2 Only 10N and 20N forces contribute moments about the x axis Line of action of the 40N is parallel to this axis
and thus, moment = 0 Using right hand rule Mx = (10N)(4m) (20N)(6m) = -80N.m My = (10N)(3m) (40N)(6m) = -210N.m Mz = (40N)(4m) (20N)(3m) = 100N.m 4.5 Moment of a Force about a Specified Axis Solution 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 4.6 Moment of a Couple Example Position vectors rA and rA are directed from O to A and B, lying on the line of action of F and F Couple moment about O M = rA X (-F) + rA X (F)
Couple moment about A M=rXF since moment of F about A = 0 4.6 Moment of a Couple A couple moment is a free vector - It can act at any point since M depends only on the position vector r directed between forces and not position vectors rA and rB, directed from O to the forces
- Unlike moment of force, it do not require a definite point or axis 4.6 Moment of a Couple Scalar Formulation Magnitude of couple moment M = Fd Direction and sense are determined by right hand rule In all cases, M acts
perpendicular to plane containing the forces 4.6 Moment of a Couple Vector Formulation For couple moment, M=rXF 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 4.6 Moment of a Couple
Equivalent Couples Two couples are equivalent if they produce the same moment Since moment produced by the couple is always perpendicular to the plane containing the forces, forces of equal couples either lie on the same plane or plane parallel to one another 4.6 Moment of a Couple 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) 4.6 Moment of a Couple Frictional forces (floor)
on the blades of the machine creates a moment Mc that tends to turn it An equal and opposite moment must be applied by the operator to prevent turning Couple moment Mc = Fd is applied on the handle 4.6 Moment of a Couple Example 4.10 A couple acts on the gear teeth. Replace it
by an equivalent couple having a pair of forces that cat through points A and B. 4.6 Moment of a Couple Solution Magnitude of couple M = Fd = (40)(0.6) = 24N.m Direction out of the page since forces tend to rotate CW M is a free vector and can be placed anywhere 4.6 Moment of a Couple Solution To preserve CCW motion,
vertical forces acting through points A and B must be directed as shown For magnitude of each force, M = Fd 24N.m = F(0.2m) F = 120N 4.6 Moment of a Couple Example 4.11 Determine the moment of the couple acting on the member. 4.6 Moment of a Couple
Solution Resolve each force into horizontal and vertical components Fx = 4/5(150kN) = 120kN Fy = 3/5(150kN) = 90kN Principle of Moment about point D, M = 120kN(0m) 90kN(2m) + 90kN(5m) + 120kN(1m) = 390kN (CCW) 4.6 Moment of a Couple Solution Principle of Moment about point A, M = 90kN(3m) + 120kN(1m) = 390kN (CCW)
*Note: - Same results if take moment about point B - Couple can be replaced by two couples as seen in figure 4.6 Moment of a Couple Solution - Same results for couple replaced by two couples - M is a free vector and acts on any point on the member -External effects such as support reactions on the member, will be the same if
the member supports the couple or the couple moment 4.7 Equivalent System A force has the effect of both translating and rotating a body The extent of the effect depends on how and where the force is applied We can simplify a system of forces and moments into a single resultant and moment acting at a specified point O A system of forces and moments is then equivalent to the single resultant
force and moment acting at a specified point O 4.7 Equivalent System Point O is on the Line of Action Consider body subjected to force F applied to point A Apply force to point O without altering external effects on body - Apply equal but opposite forces F and F at O 4.7 Equivalent System
Point O is on the Line of Action - Two forces indicated by the slash across them can be cancelled, leaving force at point O - An equivalent system has be maintained between each of the diagrams, shown by the equal signs 4.7 Equivalent System Point O is on the Line of Action - Force has been simply transmitted along its line of action from point A to point O - External effects remain unchanged after force is moved - Internal effects depend on location of F 4.7 Equivalent System
Point O is Not on the Line of Action F is to be moved to point ) without altering the external effects on the body Apply equal and opposite forces at point O The two forces indicated by a slash across them, form a couple that has a moment perpendicular to F 4.7 Equivalent System Point O is Not on the Line of Action The moment is defined by cross product
M=rXF Couple moment is free vector and can be applied to any point P on the body 4.7 Equivalent System Consider effects when a stick of negligible weight supports a force F at its end When force is applied horizontally, same force Is felt at the grip regardless of where it is applied along line of action This relates the Principle of Transmissibility 4.7 Equivalent System
When force is applied vertically, it cause a downwards force F to be felt at the grip and a clockwise moment of M = Fd Same effects felt when F is applied at the grip and M is applied anywhere on the stick In both cases, the systems are equivalent Consider 4.8 Resultants of a Force and Couple System
a rigid body Since O does not lies on the line of action, an equivalent effect is produced if the forces are moved to point O and the corresponding moments are M1 = r1 X F1 and M2 = r2 X F2 For resultant forces and moments, FR = F1 + F2 and MR = M1 + M2 Equivalency 4.8 Resultants of a Force and Couple System
is maintained thus each force and couple system cause the same external effects Both magnitude and direction of FR do not depend on the location of point O MRo depends on location of point O since M1 and M2 are determined using position vectors r1 and r2 MRo is a free vector and can acts on any point on the body Simplifying
system, 4.8 Resultants of a Force and Couple System any force and couple FR = F MR = MC + MO If the force system lies on the x-y plane and any couple moments are perpendicular to this plane, FRx = Fx
FRy = Fy MRo = MC + MO 4.8 Resultants of a Force and Couple System Procedure for Analysis When applying the following equations, FR = F MR = MC + MO FRx = Fx FRy = Fy
MRo = MC + MO Establish the coordinate axes with the origin located at the point O and the axes having a selected orientation 4.8 Resultants of a Force and Couple System Procedure for Analysis Force Summation For coplanar force system, resolve each force into x and y components If the component is directed along the positive x or y axis, it represent a positive scalar If the component is directed along the negative x or y axis, it represent a negative scalar
In 3D problems, represent forces as Cartesian vector before force summation 4.8 Resultants of a Force and Couple System Procedure for Analysis Moment Summation For moment of coplanar force system about point O, use Principle of Moment Determine the moments of each components rather than of the force itself In 3D problems, use vector cross product to determine moment of each force Position vectors extend from point O to any point on the line of action of each force
4.8 Resultants of a Force and Couple System Example 4.14 Replace the forces acting on the brace by an equivalent resultant force and couple moment acting at point A. 4.8 Resultants of a Force and Couple System Solution Force Summation For x and y components of resultant force, FRx Fx ; FRx 100 N 400 cos 45 N
382.8 N 382.8 N FRy Fy ; FRy 600 N 400 sin 45 N 882.8 N 882.8 N 4.8 Resultants of a Force and Couple System Solution For magnitude of resultant force FR ( FRx ) 2 ( FRy ) 2 (382.8) 2 (882.8) 2 962 N For direction of resultant force
FRy 1 882.8 tan tan 382.8 FRx 66.6 1 4.8 Resultants of a Force and Couple System Solution Moment Summation
Summation of moments about point A, M RA M A ; M RA 100 N (0) 600 N (0.4m) (400 sin 45 N )(0.8m) (400 cos 45 N )(0.3m) 551N .m 551N .m(CW ) When MRA and FR act on point A, they will produce the same external effect or reactions at the support 4.8 Resultants of a Force and Couple System Example 4.15
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. 4.8 Resultants of a Force and Couple System View Free Body Diagram Solution
Express the forces and couple F1 { 800k }N as Cartesian moments vectors rCB F (300 N )u (300 N ) 2 CB
r CB 0.15i 0.1 j 300 { 249 . 6
i 166 . 4 j }N 2 2 (0.15) (0.1) 4 3
M 500 j 500 k { 400 j 300k }N .m 5 5 4.8 Resultants of a Force and Couple System Solution Force Summation FR F ;
FR F1 F2 800k 249.6i 166.4 j { 249.6i 166.4 j 800k }N M Ro M C M O M rC XF1 rB XF2
i j k ( 400 j 300k ) (1k ) X ( 800k ) 0.15 0.1 1 249.6 166.4 0 { 166i 650 j 300k }N .m
4.9 Further Reduction of a Force and Couple System Consider special case where system of forces and moments acting on a rigid body reduces at point O to a resultant force FR = F and MR = MO, which are perpendicular to one another Further simplify the system by moving FR to another point P either on or off the body
4.9 Further Reduction of a Force and Couple System Location of P, measured from point O, can be determined provided FR and MR known P must lie on the axis bb, which is perpendicular to the line of action at FR and the aa axis Distance d satisfies MRo = Frd or d = MRo/Fr 4.9 Further Reduction of a Force and Couple System
With FR located, it will produce the same resultant effects on the body If the system of forces are concurrent, coplanar or parallel, it can be reduced to a single resultant force acting For simplified system, in each case, FR and MRo will always be perpendicular to each other 4.9 Further Reduction of a Force and Couple System Concurrent Force Systems All
the forces act at a point for which there is no resultant couple moment, so that point P is automatically specified 4.9 Further Reduction of a Force and Couple System Coplanar Force Systems May include couple moments directed perpendicular to the plane of forces Can be reduced to a single resultant force
When each force is moved to any point O in the x-y plane, it produces a couple moment perpendicular to the plane 4.9 Further Reduction of a Force and Couple System Coplanar Force Systems For resultant moment, MRo = M + (r X F) Resultant
moment is perpendicular to resultant force FR can be positioned a distance d from O to create this same moment MRo about O 4.9 Further Reduction of a Force and Couple System Parallel Force Systems Include couple moments that are perpendicular to the forces Can be reduced to a single resultant force When each force is moved to any point O in the
x-y plane, it produces a couple moment components only x and y axes 4.9 Further Reduction of a Force and Couple System Parallel Force Systems For resultant moment, MRo = MO + (r X F) Resultant moment is perpendicular to resultant force FR can be positioned a distance d from O to create this same moment MRo about O
Three 4.9 Further Reduction of a Force and Couple System parallel forces acting on the stick can be replaced a single resultant force FR acting at a distance d from the grip To be equivalent, FR = F1 + F2 + F3 To find distance d, FRd = F1d1 + F2d2 +F3d3
4.9 Further Reduction of a Force and Couple System Procedure for Analysis Establish the x, y, z axes and locate the resultant force an arbitrary distance away from the origin of the coordinates Force Summation For coplanar force system, resolve each force into x and y components If the component is directed along the positive x or y axis, it represent a positive scalar If the component is directed along the
negative x or y axis, it represent a negative scalar 4.9 Further Reduction of a Force and Couple System Procedure for Analysis Force Summation Resultant force = sum of all the forces in the system Moment Summation Moment of the resultant moment about
point O = sum of all the couple moment in the system plus the moments about point O of all the forces in the system Moment condition is used to find location of resultant force from point O 4.9 Further Reduction of a Force and Couple System Reduction to a Wrench In general, the force and couple moment system acting on a body will reduce to a single resultant force and a couple moment at
o that are not perpendicular FR will act at an angle from MRo MRo can be resolved into one perpendicular M and the other M parallel to line of action of FR 4.9 Further Reduction of a Force and Couple System Reduction to a Wrench M can be eliminated by moving FR to point P that lies on the axis bb, which is perpendicular to both MRo and FR
To maintain equivalency of loading, for distance from O to P, d = M/FR When F is applied at P, moment of F tends R R to cause rotation in the same direction as M 4.9 Further Reduction of a Force and Couple System Reduction to a Wrench
Since M is a free vector, it may be moved to P so that it is collinear to FR Combination of collinear force and couple moment is called a wrench or screw Axis of wrench has the same line of action as the force Wrench tends to cause a translation and rotation about this axis 4.9 Further Reduction of a Force and Couple System
Reduction to a Wrench A general force and couple moment system acting on a body can be reduced to a wrench Axis of the wrench and the point through which this axis passes can always be determined 4.9 Further Reduction of a Force and Couple System Example 4.16 The beam AE is subjected to a system of coplanar
forces. Determine the magnitude, direction and location on the beam of a resultant force which is equivalent to the given system of forces measured from E 4.9 Further Reduction of a Force and Couple System Solution Force Summation FRx Fx ; FRx 500 cos 60 N 100 N
350.0 N 350.0 N FRy Fy ; FRy 500 sin 60 N 200 N 233.0 N 233.0 N 4.9 Further Reduction of a Force and Couple System Solution For magnitude of resultant force FR ( FRx ) 2 ( FRy ) 2 (350.0) 2 (233.0) 2 420.5 N For
direction of resultant force FRy 233.0 tan 1 tan F 350 . 0 Rx 33.7
1 Solution 4.9 Further Reduction of a Force and Couple System - Moment Summation Summation of moments about point E, M RE M E ; 233.0 N (d ) 350N (0) (500 sin 60 N )(4m) (500 cos 60 N )(0m) (100N )(0.5m) (200N )(2.5m) 1182.1 d
5.07m 233.0 4.9 Further Reduction of a Force and Couple System Example 4.17 The jib crane is subjected to three coplanar forces. Replace this loading by an equivalent resultant force and specify where the resultants line of action intersects the column AB and boom BC.
4.9 Further Reduction of a Force and Couple System Solution Force Summation FRx Fx ; 3 FRx 2.5kN 1.75kN 5 3.25kN 3.25kN FRy Fy ;
4 FRy 2.5 N 0.6kN 5 2.60kN 2.60 N 4.9 Further Reduction of a Force and Couple System Solution For magnitude of resultant force FR ( FRx )2 ( FRy )2 (3.25)2 (2.60)2 4.16kN For
direction of resultant force 1 FRy 1 2.60 tan tan 3.25 FRx 38.7 4.9 Further Reduction of a Force and Couple System
Solution Moment Summation Method 1 Summation of moments about point M A, M ; RA A 3.25kN ( y ) 2.60kN (0) 1.75kn(1m) 0.6kN (0.6m) 3
4 2.50kN (2.2m) 2.50kN (1.6m) 5 5 y 0.458m 4.9 Further Reduction of a Force and Couple System Solution Principle of Transmissibility M RA M A ;
3.25kN (2.2m) 2.60kN ( x) 1.75kn(1m) 0.6kN (0.6m) 3 4 2.50kN (2.2m) 2.50kN (1.6m) 5 5 x 2.177 m 4.9 Further Reduction of a Force and Couple System Solution
Method 2 Take moments about point A, M RA M A ; 3.25kN ( y ) 2.60kN ( x) 1.75kn(1m) 0.6kN (0.6m) 3 4 2.50kN (2.2m) 2.50kN (1.6m) 5 5 3.25 y 2.60 x 1.49 4.9 Further Reduction of a Force and Couple
System Solution To find points of intersection, let x = 0 then y = 0.458m Along BC, set y = 2.2m then x = 2.177m Large 4.10 Reduction of a Simple Distributed Loading surface area of a body may be
subjected to distributed loadings such as those caused by wind, fluids, or weight of material supported over bodys surface Intensity of these loadings at each point on the surface is defined as the pressure p Pressure is measured in pascals (Pa) 1 Pa = 1N/m2 4.10 Reduction of a Simple Distributed Loading Most common case of distributed pressure
loading is uniform loading along one axis of a flat rectangular body Direction of the intensity of the pressure load is indicated by arrows shown on the load-intensity diagram Entire loading on the plate is a system of parallel forces, infinite in number, each acting on a separate differential area of the plate 4.10 Reduction of a Simple Distributed Loading function p = p(x) Loading Pa, is a function of
x since pressure is uniform along the y axis Multiply the loading function by the width w = p(x)N/m2]a m = w(x) N/m Loading function is a measure of load distribution along line y = 0, which is in the symmetry of the loading Measured as force per unit length rather than per unit area 4.10 Reduction of a Simple Distributed Loading Load-intensity
diagram for w = w(x) can be represented by a system of coplanar parallel This system of forces can be simplified into a single resultant force FR and its location can be specified 4.10 Reduction of a Simple Distributed Loading Magnitude of Resultant Force FR
= F Integration is used for infinite number of parallel forces dF acting along the plate For entire plate length, FR F ; FR w( x)dx dA A Magnitude L A of resultant force is equal to the
total area A under the loading diagram w = w(x) 4.10 Reduction of a Simple Distributed Loading Location of Resultant Force MR = MO Location x of the line of action of FR can be determined by equating the moments of the
force resultant and the force distribution about point O dF produces a moment of xdF = x w(x) dx about O For the entire plate, M Ro M O ; x FR xw( x)dx L 4.10 Reduction of a Simple Distributed Loading Location of Resultant Force Solving,
xw( x)dx xdA x L w( x)dx L Resultant A dA A
force has a line of action which passes through the centroid C (geometric center) of the area defined by the distributed loading diagram w(x) 4.10 Reduction of a Simple Distributed Loading Location of Resultant Force Consider 3D pressure loading p(x), the resultant force has a magnitude equal to the volume under the distributed-loading curve p
= p(x) and a line of action which passes through the centroid (geometric center) of this volume Distribution diagram can be in any form of shapes such as rectangle, triangle etc Beam 4.10 Reduction of a Simple Distributed Loading supporting this stack of lumber is subjected to a uniform distributed loading,
and so the load-intensity diagram has a rectangular shape If the load-intensity is wo, resultant is determined from the are of the rectangle FR = wob Line 4.10 Reduction of a Simple Distributed Loading of action passes through the centroid or center of the rectangle, = a + b/2
x Resultant is equivalent to the distributed load Both loadings produce same external effects or support reactions on the beam 4.10 Reduction of a Simple Distributed Loading Example 4.20 Determine the magnitude and location of the equivalent resultant force acting on the shaft 4.10 Reduction of a
Simple Distributed Loading Solution For the colored differential area element, dA wdx 60 x 2 dx For resultant force FR F ; 2 FR dA 60 x 2dx A 0
3 2 x 23 03 60 60 3 0 3 3 160 N 4.10 Reduction of a Simple Distributed Loading Solution
For location of line of action, 2 2 x4 24 04 2 xdA x(60 x )dx 60 4 60 4 4 0 x A 0 160
160 160 dA A 1.5m Checking, ab 2m(240 N / m) 160 3 3
3 3 x a (2m) 1.5m 4 4 A 4.10 Reduction of a Simple Distributed Loading Example 4.21 A distributed loading of p = 800x Pa acts over the top surface of the beam. Determine the magnitude and location of the equivalent
force. 4.10 Reduction of a Simple Distributed Loading Solution Loading function of p = 800x Pa indicates that the load intensity varies uniformly from p = 0 at x = 0 to p = 7200Pa at x = 9m For loading, w = (800x N/m2)(0.2m) = (160x) N/m Magnitude of resultant force = area under the triangle FR = (9m)(1440N/m)
= 6480 N = 6.48 kN 4.10 Reduction of a Simple Distributed Loading Solution Resultant force acts through the centroid of the volume of the loading diagram p = p(x) FR intersects the x-y plane at point (6m, 0) Magnitude of resultant force = volume under the triangle FR = V = (7200N/m2)(0.2m)
= 6.48 kN Chapter Summary Moment of a Force A force produces a turning effect about the point O that does not lie on its line of action In scalar form, moment magnitude, M = Fd, O where d is the moment arm or perpendicular distance from point O to its line of action of the force Direction of the moment is defined by right hand rule
For easy solving, - resolve the force components into x and y components Chapter Summary Moment of a Force - determine moment of each component about the point - sum the results Vector cross product are used in 3D problems MO = r X F where r is a position vector that extends from point O to any point on the line of action of F
Chapter Summary Moment about a Specified Axis Projection of the moment onto the axis is obtained to determine the moment of a force about an arbitrary axis provided that the distance perpendicular to both its line of action and the axis can be determined If distance is unknown, use vector triple product Ma = uar X F where ua is a unit vector that specifies the direction of the axis and r is the position vector that is directed from any point on the
axis to any point on its line of action Chapter Summary Couple Moment A couple consists of two equal but opposite forces that act a perpendicular distance d apart Couple tend to produce rotation without translation Moment of a couple is determined from M = Fd and direction is established using the righthand rule If vector cross product is used to determine the couple moment, M = r X F, r extends from any
point on the line of action of one of the forces to any point on the line of action of the force F Chapter Summary Reduction of a Force and Couple System Any system of forces and couples can be reduced to a single resultant force and a single resultant couple moment acting at a point Resultant force = sum of all the forces in the system Resultant couple moment = sum of all the forces and the couple moments about the point Only concurrent, coplanar or parallel force
system can be simplified into a single resultant force Chapter Summary Reduction of a Force and Couple System For concurrent, coplanar or parallel force systems, - find the location of the resultant force about a point - equate the moment of the resultant force about the point to moment of the forces and couples in the system about the same point Repeating the above steps for other force
system will yield a wrench, which consists of resultant force and a resultant collinear moment Chapter Summary Distributed Loading A simple distributed loading can be replaced by a resultant force, which is equivalent to the area under the loading curve Resultant has a line of action that passes through the centroid or geometric center of the are or volume
under the loading diagram Chapter Review Chapter Review Chapter Review Chapter Review Chapter Review Chapter Review
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