Transcript ENGR-36_Lec-24_Dist_Loads
Engineering 36
Chp09: Distributed Loads
Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]
Engineering-36: Engineering Mechanics - Statics
1 Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Distributed Loads
The Load on an Object may be Spread out, or Distributed over the surface.
Load Profile,
w
(x)
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2 Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Distributed Loads
If the Load Profile,
w
(x), is known then the distributed load can be replaced with at POINT Load at a SPECIFIC Location Magnitude of the Point Load, W,
W
span w
dx
is Determined by Area Under the Profile Curve Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
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Distributed Loads
To Determine the Point Load Location employ Moments (1 st Moment of Force) Recall: Moment = [LeverArm ]•[Intensity] In This Case 4 • LeverArm = The distance from the Baseline Origin, x n • Intensity = The Increment of Load, dW n , which is that load, w(x n ) covering a distance dx located at x n – That is: dW n
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= w(x n )•dx Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Distributed Loads
x
Now Use Centroidal Methodology
span
LeverArm
Intensity
span n
w
n dx
And Recall:
x
x W
x
is the Centroid Location
Equating the Ω Expressions
span
n
w
n dx
find
x
W
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Distributed Loads on Beams
W
0
L w dx
dA
A
• A distributed load is represented by plotting the load per unit length,
w
(
N/m
). The
total load
is equal to the
area under the load curve
.
W
x dW A
0
L
x dA
x A
• A distributed load can be REPLACED by a concentrated load with a magnitude equal to the area under the load curve and a line of action passing through the areal centroid.
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Engineering-36: Engineering Mechanics - Statics
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Example:Trapezoidal Load Profile
A beam supports a distributed load as shown. Determine the equivalent concentrated load and the reactions at the supports.
Solution Plan • The magnitude of the concentrated load is equal to the total load (the area under the curve) • The line of action of the concentrated load passes through the centroid of the area under the Load curve.
• Determine the support reactions by summing moments about the beam ends
Engineering-36: Engineering Mechanics - Statics
8 Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Example:Trapezoidal Load Profile
SOLUTION: • The magnitude of the concentrated load is equal to the total load, or the area under the curve.
F
1500 2 4500 N m 6 m
F
18 .
0 kN • The line of action of the concentrated load passes through the area centroid of the curve.
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X
63 kN m
X
3 .
5 m 18 kN Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Example:Trapezoidal Load Profile
A y B y
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10 Determine the support reactions by summing moments about the beam ends After Replacing the Dist-Load with the Equivalent
M
POINT-Load
A
0 :
B y
18 kN 3 .5
m 0
M B
0 :
A y
6 m
B y
10 .
5 kN 18 kN 6 m 3 .5
m 0
A y
7 .
5 kN Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
3D Distributed Loads
The Previous 2D Dist Load Profile had units of Force per Unit-Length (e.g., lb/in or N/m) If 3D The Force acts over an AREA and the units become Force per Unit Area, or PRESSURE (e.g., psi or Pa) Knowledge of the Pressure Profile allows calculation of an Equivalent Point Load and its Location
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11 Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Pressure Loading
Consider an Area Subject to a Pressure Load The incremental Force, dF mn , Results from pressure p(x m ,y n ) acting on the incremental area dA mn = (dx m ) (dy n )
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Uniform Pressure Profile
Then the Total Force,
F p
F, on the Area
area dF
area p
dA
Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Pressure Loading: Total Force
The Differential Geometry is shown below
dF dA
13 Then the Total Pressure Force
F p
dF mn
area p
x m
,
y n
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area
all x, all y
p x m
,
y n
dA mn
Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Pressure Loading – Pressure Ctr
Use MOMENT Methodology in 2-Dimensions to find the Location for the Point Force F p Then the Moment about the y-axis due to intensity dF mn
d
x
LeverArm x m
x m
p
x m
,
y n
and
dxdy
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x m dF
pdxdy y n
Then the Total y-axis
x
Moment
d
x surface
x surface m p
x m
,
y n
dxdy
Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Pressure Loading – Pressure Ctr
Recall also Ω x = X C •F p Equating the two Ω
X C F p
expressions
x m surface p
x m
,
y n
dxdy
X C
Isolating X C
x m surface p
x m
,
F p y n
dxdy
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15
x m dF
pdxdy y m
The Similar
Y C
Expression for Y C
y surface n
x m
,
n
dxdy F p
Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Pressure Loading Summarized
Given a surface with Pressure Profile 16 The Equivalent Force, F p , Exerted on the Surface due to the Pressure
F p
p
x m
,
y n
all x, all y F p is located at the Center of Pressure
X C
at CoOrds (X C ,Y C )
x surface m p
x m
,
y n
dxdy F p
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Y C
surface y n p
x m
,
y n
dxdy F p
Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
WhiteBoard Work
Lets Work These Nice Problems
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Engineering 36
Appendix
dy dx
sinh
µx T
0
µs T
0 Bruce Mayer, PE Registered Electrical & Mechanical Engineer [email protected]
Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
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Beam Problem
For the Negligible-Wt Beam Find • Equivalent POINT-Load and it’s Location (Point of Application, PoA) • The RCNs at Pt-A
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Pressure Problem
Find the Equivalent POINT-LOAD and its Point of Application (Location) For the Given Pressure Distribution
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Engineering-36: Engineering Mechanics - Statics
22 Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
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Pressure Loading
The Differential Geometry is shown belwo
dF dA
25 Then the Total Pressure Force
F p
area dF mn
area p
Engineering-36: Engineering Mechanics - Statics
all x,
p
all y
x m
,
y n x m
,
y n
dA mn
Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
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26 Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx