Engineering Mechanics: Statics Chapter 4: Force System Resultants
Engineering Mechanics: StaticsEngineering Mechanics: Statics
Chapter 4: Force System
Resultants
Chapter 4: Force System
Resultants
Chapter ObjectivesChapter 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 OutlineChapter Outline
Moment of a Force – Scalar Formation
Cross ProductMoment of Force – Vector
FormulationPrinciple of MomentsMoment of a Force about a Specified
Axis
Chapter OutlineChapter Outline
Moment of a CoupleEquivalent SystemResultants of a Force and Couple
SystemReduction of a Simple Distributed
Loading
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 axisCase 1Consider 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 – Scalar
Formation
4.1 Moment of a Force – Scalar
Formation
Fx tends to turn the 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 Force
– Scalar Formation
4.1 Moment of a Force
– Scalar Formation
Moment axis (z) is perpendicular to shaded plane (x-y)
Fx and dy lies on the shaded plane (x-y)Moment axis (z) intersects
the plane at point O
4.1 Moment of a Force
– Scalar Formation
4.1 Moment of a Force
– Scalar Formation
Case 2Apply force Fz to the wrenchPipe does not rotate about z axisTendency to rotate about x axisThe pipe may not actually
rotate Fz creates tendency
for rotation so moment (Mo) x is produced
4.1 Moment of a Force
– Scalar Formation
4.1 Moment of a Force
– Scalar Formation
Case 2Moment axis (x) is perpendicular to
shaded plane (y-z)
Fz and dy lies on the shaded plane (y-z)
4.1 Moment of a Force
– Scalar Formation
4.1 Moment of a Force
– Scalar Formation
4.1 Moment of a Force
– Scalar Formation
4.1 Moment of a Force
– Scalar FormationCase 3Apply force Fy to the wrenchNo moment is produced about point OLack of tendency to rotate
as line of action passes through O
4.1 Moment of a Force
– Scalar Formation
4.1 Moment of a Force
– Scalar FormationIn General Consider the force F and the point O which lies
in the shaded plane The moment MO about point O,
or about an axis passingthrough O and perpendicularto the plane, is a vector quantity
Moment MO has its specified
magnitude and direction
4.1 Moment of a Force – Scalar
Formation
4.1 Moment of a Force – Scalar
FormationMagnitudeFor magnitude of MO,
MO = Fdwhere 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 Force – Scalar
Formation
4.1 Moment of a Force – Scalar
FormationDirectionDirection 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
4.1 Moment of a Force – Scalar
FormationDirection
- Thumb points along the moment axis to give the direction and sense of the moment vector- Moment vector is upwards and perpendicular to the shaded plane
4.1 Moment of a Force – Scalar
Formation
4.1 Moment of a Force – Scalar
FormationDirectionMO is shown by a vector arrow
with a curl to distinguish it from force vectorExample (Fig b)MO is represented by the
counterclockwise curl, which indicates the action of F
4.1 Moment of a Force – Scalar
Formation
4.1 Moment of a Force – Scalar
FormationDirectionArrowhead shows the sense of
rotation caused by FUsing the right hand rule, the
direction and sense of the moment vector points out of the page
In 2D problems, moment of the force is found about a point O
4.1 Moment of a Force – Scalar
Formation
4.1 Moment of a Force – Scalar
FormationDirectionMoment acts about an axis
perpendicular to the plane containing F and d
Moment axis intersects the plane at point O
4.1 Moment of a Force – Scalar
Formation
4.1 Moment of a Force – Scalar
FormationResultant Moment of a System of Coplanar ForcesResultant 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
4.1 Moment of a Force – Scalar
FormationResultant Moment of a System of Coplanar ForcesA 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
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
4.1 Moment of a Force – Scalar
FormationExample 4.1For each case, determine the moment of
the force about point O
4.1 Moment of a Force – Scalar
Formation
4.1 Moment of a Force – Scalar
FormationSolution 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
)(.5.37)75.0)(50()(
)(.200)2)(100()(
CWmNmNMb
CWmNmNMa
o
o
4.1 Moment of a Force – Scalar
Formation
4.1 Moment of a Force – Scalar
FormationSolution
)(.0.21)14)(7()(
)(.4.42)45sin1)(60()(
)(.229)30cos24)(40()(
CCWmkNmmkNMe
CCWmNmNMd
CWmNmmNMc
o
o
o
4.1 Moment of a Force – Scalar
Formation
4.1 Moment of a Force – Scalar
FormationExample 4.2Determine 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
4.1 Moment of a Force – Scalar Formation
SolutionScalar Analysis
Line of action of F passes through C )(.400)5.0)(800(
.0)0)(800(
)(.1200)5.1)(800(
)(.2000)5.2)(800(
CCWmNmNM
mkNmNM
CWmNmNM
CWmNmNM
D
C
B
A
4.2 Cross Product4.2 Cross Product
Cross product of two vectors A and B yields C, which is written as
C = A X BRead as “C equals A cross B”
4.2 Cross Product4.2 Cross Product
Magnitude Magnitude of C is defined as the
product of the magnitudes of A and B and the sine of the angle θ between their tails
For angle θ, 0° ≤ θ ≤ 180°Therefore,
C = AB sinθ
4.2 Cross Product4.2 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 righthand form vector A (cross) to vector B- Thumb points in the direction of vector C
4.2 Cross Product4.2 Cross Product
Expressing vector C when magnitude and direction are known
C = A X B = (AB sinθ)uC
where scalar AB sinθ defines the magnitude of vector C unit vector uC
defines the direction of vector C
4.2 Cross Product4.2 Cross Product
Laws of Operations1. Commutative law is not valid
A X B ≠ B X ARather,
A X B = - B X A Shown by the right hand rule Cross product A X B yields a vector opposite
in direction to CB X A = -C
4.2 Cross Product4.2 Cross Product
Laws of Operations2. Multiplication by a Scalar
a( A X B ) = (aA) X B = A X (aB) = ( A X B )a
3. Distributive Law A X ( B + D ) = ( A X B ) + ( A X D )
Proper order of the cross product must be maintained since they are not commutative
4.2 Cross Product4.2 Cross Product
Cartesian Vector FormulationUse C = AB sinθ on pair of
Cartesian unit vectorsExampleFor i X j, (i)(j)(sin90°) = (1)(1)(1) = 1
4.2 Cross Product4.2 Cross Product
Laws of Operations In a similar manner,
i X j = k i X k = -j i X i = 0j X k = i j X i = -k j X j = 0k X i = j k X j = -i k X k = 0
Use the circle for the results. Crossing CCW yield positive and CW yields negative results
4.2 Cross Product4.2 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 Product4.2 Cross Product
Laws of Operations In determinant form,
zyx
zyx
BBB
AAA
kji
BXA
4.3 Moment of Force - Vector Formulation4.3 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 pointlying on the line of action of F
4.3 Moment of Force - Vector Formulation4.3 Moment of Force - Vector Formulation
MagnitudeFor magnitude of cross product,
MO = rF sinθwhere θ is the angle measured between tails of r and F
Treat r as a sliding vector. Since d = r sinθ,
MO = rF sinθ = F (rsinθ) = Fd
4.3 Moment of Force - Vector Formulation4.3 Moment of Force - Vector Formulation
DirectionDirection 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 Formulation4.3 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 Formulation4.3 Moment of Force - Vector Formulation
Principle of TransmissibilityFor 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 Formulation4.3 Moment of Force - Vector Formulation
Cartesian Vector Formulation For force expressed in Cartesian
form,
where rx, ry, rz represent the x, y, zcomponents of the position vectorand Fx, Fy, Fz represent that of the force vector
zyx
zyxO
FFF
rrr
kji
FXrM
4.3 Moment of Force - Vector Formulation4.3 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 Formulation4.3 Moment of Force - Vector Formulation
Cartesian Vector FormulationResultant moment of forces about
point O can be determined by vector addition
MRo = ∑(r x F)
4.3 Moment of Force - Vector Formulation4.3 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 Formulation4.3 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 Formulation4.3 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 Formulation4.3 Moment of Force - Vector Formulation
Solution
Substitute into determinant formulation
kji
kji
FXrM
Nkji
kjiN
uNF
BA
F
)]40(3)20(1[)]40(2)40(1[)]20(2)40(3[
402040
231
402040
)2()1()2(
)092)493)31()60(
)60(
222
4.3 Moment of Force - Vector Formulation4.3 Moment of Force - Vector Formulation
SolutionOr
Substitute into determinant formulation
For magnitude,
mN
M
mNkjiM
kji
kji
FXrM
A
A
CA
.224
)100()120()160(
.100120160
)]40(4)20(3[)]40(0)40(3[)]20(0)40(4[
402040
043
222
4.4 Principles of Moments4.4 Principles of Moments
Also known as Varignon’s 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)
= r X F
4.4 Principles of Moments4.4 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 Moments4.4 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 Moments4.4 Principles of Moments
Example 4.6The force F acts at the end of the angle bracket. Determine the moment of the
force about point O.
4.4 Principles of Moments4.4 Principles of Moments
SolutionMethod 1MO = 400sin30°N(0.2m)-400cos30°N(0.4m)
= -98.6N.m = 98.6N.m (CCW)
As a Cartesian vector, MO = {-98.6k}N.m
4.4 Principles of Moments4.4 Principles of Moments
SolutionMethod 2: Express as Cartesian vector
r = {0.4i – 0.2j}NF = {400sin30°i – 400cos30°j}N = {200.0i – 346.4j}N
For moment,
mNk
kji
FXrMO
.6.98
04.3460.200
02.04.0
4.6 Moment of a Couple4.6 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 Couple4.6 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 = r X Fsince moment of –F about A = 0
4.6 Moment of a Couple4.6 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 Couple4.6 Moment of a Couple
Scalar FormulationMagnitude 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 Couple4.6 Moment of a Couple
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
4.6 Moment of a Couple4.6 Moment of a Couple
Equivalent CouplesTwo couples are equivalent if they
produce the same momentSince 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 Couple4.6 Moment of a Couple
Resultant Couple MomentCouple 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 Couple4.6 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 Couple4.6 Moment of a Couple
Example 4.10A 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 Couple4.6 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 Couple4.6 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 = Fd24N.m = F(0.2m)F = 120N
4.7 Equivalent System4.7 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 System4.7 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 System4.7 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 System4.7 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 System4.7 Equivalent System
Point O is Not on the Line of ActionF is to be moved to point ) without altering
the external effects on the bodyApply equal and opposite forces at point OThe two forces indicated by a slash across
them, form a couple that has a moment perpendicular to F
4.7 Equivalent System4.7 Equivalent System
Point O is Not on the Line of ActionThe moment is defined by cross product
M = r X FCouple moment is free vector and can be
applied to any point P on the body
4.8 Resultants of a Force and Couple
System
4.8 Resultants of a Force and Couple
System Consider 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
4.8 Resultants of a Force and Couple
System
4.8 Resultants of a Force and Couple
System Equivalency 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
4.8 Resultants of a Force and Couple
System
4.8 Resultants of a Force and Couple
System Simplifying any force and couple
system, 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
4.8 Resultants of a Force and Couple
SystemProcedure 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
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
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
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
4.8 Resultants of a Force and Couple System
Solution Force Summation For x and y components of resultant force,
NN
NNF
FF
NN
NNF
FF
Ry
yRy
Rx
xRx
8.8828.882
54sin400600
;
8.3828.382
54cos400100
;
4.8 Resultants of a Force and Couple System
4.8 Resultants of a Force and Couple System
Solution For magnitude of resultant force
For direction of resultant force
6.66
8.382
8.882tantan
962
)8.882()8.382()()(
11
2222
Rx
Ry
RyRxR
F
F
N
FFF
4.8 Resultants of a Force and Couple System
4.8 Resultants of a Force and Couple System
Solution Moment Summation Summation of moments about point A,
When MRA and FR act on point A, they will produce the same external effect or reactions at the support
)(.551.551
)3.0)(54cos400(
)8.0)(54sin400()4.0(600)0(100
;
CWmNmN
mN
mNmNNM
MM
RA
ARA
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingLarge surface area of a body may be
subjected to distributed loadings such as those caused by wind, fluids, or weight of material supported over body’s 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
4.9 Reduction of a Simple Distributed
Loading
4.9 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
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
Loading Loading function p = p(x) 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
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingLoad-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
Magnitude of Resultant Force FR = ∑F Integration is used for infinite number of
parallel forces dF acting along the plate For entire plate length,
Magnitude of resultant force is equal to the total area A under the loading diagram w = w(x)
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
Loading
AdAdxxwFFFL A
RR )(;
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingLocation of Resultant Force MR = ∑MO
Location 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,
x
L
RORo dxxxwFxMM )(;
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingLocation of Resultant Force Solving,
Resultant 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)
A
A
L
L
dA
xdA
dxxw
dxxxw
x)(
)(
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingLocation 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
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
Loading Beam 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
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
Loading Line of action passes through the centroid
or center of the rectangle, = a + b/2 Resultant is equivalent to the distributed
load Both loadings produce same “external”
effects or support reactions on the beam
x
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingExample 4.20Determine the magnitude and location of the equivalent resultant force acting on the shaft
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingSolutionFor the colored differential area element,
For resultant force
N
x
dxxdAF
FF
dxxwdxdA
AR
R
160
30
32
603
60
60
;
60
332
0
3
2
0
2
2
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingSolutionFor location of line of action,
Checking,
mmax
mNmabA
m
xdxxx
dA
xdA
x
A
A
5.1)2(43
43
1603
)/240(23
5.1
160
40
42
60
160
460
160
)60(44
2
0
42
0
2
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingExample 4.21A distributed loading of p = 800x Pa acts over the top surface of the beam. Determine the magnitude and location of the equivalent force.
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingSolution 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 triangleFR = ½(9m)(1440N/m) = 6480 N = 6.48 kN
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingSolutionResultant 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 triangleFR = V = ½(7200N/m2)(0.2m)
= 6.48 kN
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingExample 4.22The granular material exerts the distributed loading on the beam. Determine the magnitude and location of the equivalent resultant of this load
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingSolution Area of loading diagram is trapezoid Magnitude of each force = associated area
F1 = ½(9m)(50kN/m) = 225kN
F2 = ½(9m)(100kN/m) = 450kN Line of these parallel forces act
through the centroid of associated areas and insect beams at
mmxmmx 5.4)9(21
,3)9(31
21
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
LoadingSolution Two parallel Forces F1 and F2 can be
reduced to a single resultant force FR For magnitude of resultant force,
For location of resultant force,
mx
x
MM
kNxF
FF
ORo
R
R
4
)450(5.4)225(3)675(
;
675450225
;
Solution*Note: Trapezoidal area can be divided into two
triangular areas,F1 = ½(9m)(100kN/m) = 450kN
F2 = ½(9m)(50kN/m) = 225kN
4.9 Reduction of a Simple Distributed
Loading
4.9 Reduction of a Simple Distributed
Loading
mmxmmx 3)9(31
,3)9(31
21