ME 201 Engineering Mechanics: Statics 201/Supplemental Material/In Clas… · Example Problem...
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ME 201 Engineering Mechanics: Statics Unit 5.3 Reduction of a Simple Distributed Loading
Transcript of ME 201 Engineering Mechanics: Statics 201/Supplemental Material/In Clas… · Example Problem...
ME 201Engineering Mechanics: Statics
Unit 5.3
Reduction of a Simple Distributed
Loading
Distributed Loads
Thus far we’ve been working with loads that
are concentrated at a point:
Many times in engineering we need to be
concerned with another type of loading
referred to as distributed loading:
10 N/m
6 kN/m
Distributed Loads
Instead of being concentrated at a point, a
distributed load is spread out over a distance
It can be thought of as a collection of smaller
loads
To find FR, we need to sum an infinite
number of small forces
Distributed Loads Consider a small differential element, dF
with a width of dx
and a height of w(x) w(x)
wdF
x dx x
The area of the element is
Since infinite number of forces, need to integrate to find FR
dxxwdF )(
AdAdxxwFAl
R )(
Centroids for Simple Shapes
Where is the centroid for these common shapes?
b
h
b
h
2
bxc
2
hyc
3
bxc
3
hyc
Example Problem SolutionGiven:
trapezoid
Find:
FR,
Solution:
FBD
FR
x
50 lb/ft
9 ftA B
100 lb/ft
9 ftA B
FrectFtri
ftftlbFtri 9/)50100(2
1
ftftlbFrect 9/50
lb225
lb450
lbFFF recttriR 675
9 ftA B
FR
x
Example Problem Solution
5.44503225675 x
ftxtri 93
1
ftxrect 92
1
ft3
ft5.4
Given:
trapezoid
Find:
FR,
Solution:
FBD
FR
x
x
ft4
50 lb/ft
9 ftA B
100 lb/ft
9 ftA B
FrectFtri
9 ftA B
FR
x
lbFtri 225 lbFrect 450
lbFR 675
Example Problem Solution
Given:
trapezoid
Find:
FR,
Solution:
FR integral
x
LR dxxwF )(
2
0
260 dxx
N160
|2
0
3
360
x
w=60x2 N/m
2 m
Example Problem Solution
Given:
trapezoid
Find:
FR,
Solution:
FR integral
integral
x
w=60x2 N/m
2 m
L
L
dxxw
dxxxwx
)(
)(
x
2
0
2
2
0
2
60
60
dxx
dxxx
|
|2
0
3
2
0
4
360
460
x
x
160
240 m5.1
Distributed Loads
Thus far we’ve been working with loads that
are concentrated at a point:
Many times in engineering we need to be
concerned with another type of loading
referred to as distributed loading:
10 N/m
6 kN/m
Distributed Loads
Instead of being concentrated at a point, a
distributed load is spread out over a distance
It can be thought of as a collection of smaller
loads
Distributed Loads
To compute the resultant, FR, of a distributed load, consider the following:
To find FR, we need to sum an infinite number of small forces
Distributed Loads
Consider a small differential element, dF
with a width of dx
and a height of w(x)
w(x)
wdF
x dx x
Distributed Loads
The area of the element is
Since infinite number of forces, need to integrate to find FR
dxxwdF )(
AdAdxxwFAl
R )(
w(x)
wdF
x dx x
Distributed Loads
Where does FR act?
Can be determined by equating the moments of
the force resultant and the force distribution
lR dxxwF )(
R
l
F
dxxxwx
)(
This is also the centroid of the area
lR dxxxwFx )(
l
l
dxxw
dxxxw
)(
)(
A
A
dA
xdA
Centroids
For simple shapes, centroid can be found in a table
(see back cover of textbook)
Where is the centroid for these common shapes?
b
h
b
h
2
bxc
2
hyc
3
bxc
3
hyc
Centroids of Simple Shapes
Where does FR act?
A. xc = b/2
B. xc > b/2
C. xc < b/2
D. Cannot tell from information given
b
h1
h2
Example Problem
Given:
w=400 lb/ft
Find:
FR,
w lb/ft
10 ftA B
x
Example Problem SolutionGiven:
w=400 lb/ft
Find:
FR, x
w lb/ft
10 ftA B
Example Problem Solution
Given:
w=400 lb/ft
Find:
FR,
Solution:
FBD
FR
x
w lb/ft
10 ftA B
ftftlbFR 10/400
2
10 ftx
10 ftA B
FR
x
korlb 44000
ft5
x
Example Problem
Given:
w=100x N/m
Find:
FR,
600 N/m
6 mA B
x
Example Problem SolutionGiven:
w=100x N/m
Find:
FR,
600 N/m
6 mA Bx
Example Problem Solution
Given:
w=100x N/m
Find:
FR,
Solution:
FBD
FR
600 N/m
6 mA B
mmNFR 6/6002
1
mx 63
2
x
x
6 mA B
FR
x
kNmorNm 8.11800
m4
Example Problem
Given:
trapezoid
Find:
FR,
50 lb/ft
9 ftA B
100 lb/ft
x
Example Problem SolutionGiven:
trapezoid
Find:
FR, x50 lb/ft
9 ftA B
100 lb/ft
Example Problem SolutionGiven:
trapezoid
Find:
FR,
Solution:
FBD
FR
x
50 lb/ft
9 ftA B
100 lb/ft
9 ftA B
FrectFtri
ftftlbFtri 9/)50100(2
1
ftftlbFrect 9/50
lb225
lb450
lbFFF recttriR 675
9 ftA B
FR
x
Example Problem Solution
5.44503225675 x
ftxtri 93
1
ftxrect 92
1
ft3
ft5.4
Given:
trapezoid
Find:
FR,
Solution:
FBD
FR
x
x
ft4
50 lb/ft
9 ftA B
100 lb/ft
9 ftA B
FrectFtri
9 ftA B
FR
x
lbFtri 225 lbFrect 450
lbFR 675
Example Problem
Given:
function
Find:
FR,
w=60x2 N/m
2 mx
Example Problem SolutionGiven:
function
Find:
FR, x
w=60x2 N/m
2 m
Example Problem Solution
Given:
trapezoid
Find:
FR,
Solution:
FR integral
x
LR dxxwF )(
2
0
260 dxx
N160
|2
0
3
360
x
w=60x2 N/m
2 m
Example Problem Solution
Given:
trapezoid
Find:
FR,
Solution:
FR integral
integral
x
w=60x2 N/m
2 m
L
L
dxxw
dxxxwx
)(
)(
x
2
0
2
2
0
2
60
60
dxx
dxxx
|
|2
0
3
2
0
4
360
460
x
x
160
240 m5.1
In Class Exercise
Solution
In Class Exercise
In Class Exercise
