ENGR-36_Lec-24_Dist_Loads

Report
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)
Engineering-36: Engineering Mechanics - Statics
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
W  w x dx
Point Load, W,
span
is Determined
by Area Under
the Profile Curve



Engineering-36: Engineering Mechanics - Statics
3
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Distributed Loads
 To Determine the Point Load Location
employ Moments (1st Moment of Force)
 Recall: Moment = [LeverArm]•[Intensity]
 In This Case
• LeverArm = The distance from
the Baseline Origin, xn
• Intensity = The Increment of Load, dWn,
which is that load, w(xn) covering a
distance dx located at xn
– That is: dWn = w(xn)•dx
Engineering-36: Engineering Mechanics - Statics
4
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Distributed Loads
 Now Use Centroidal Methodology
x 
 LeverArm Intensity    x wx dx 
n
span
n
span
 And Recall:
 x  xW
x is theCentroidLocation
 Equating the
Ω Expressions
x
find
Engineering-36: Engineering Mechanics - Statics
5
 x wx dx
n
n
span
W
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Distributed Loads on Beams
L
W   wdx   dA  A
0
OP W   xdW
L
OP  A   xdA  x A
0
• 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.
• 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.
Engineering-36: Engineering Mechanics - Statics
6
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Engineering-36: Engineering Mechanics - Statics
7
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Example:Trapezoidal Load Profile
 Solution Plan
 A beam supports a
distributed load as
shown. Determine the
equivalent concentrated
load and the reactions
at the supports.
Engineering-36: Engineering Mechanics - Statics
8
• 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
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.
1500  4500 N
F
 6m F  18.0 kN
2
m
• The line of action of the concentrated load
passes through the area centroid of the curve.
63 kN  m
X 
18 kN
Engineering-36: Engineering Mechanics - Statics
9
X  3.5 m
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Example:Trapezoidal Load Profile
 Determine the support reactions
by summing moments about the
beam ends After Replacing the
Dist-Load with the Equivalent
POINT-Load
 M A  0 : By 6 m  18 kN3.5 m  0
By  10.5 kN
 M B  0 :  Ay 6 m  18 kN6 m  3.5 m  0
Ay  7.5 kN
Ay
By
Engineering-36: Engineering Mechanics - Statics
10
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
Engineering-36: Engineering Mechanics - Statics
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, dFmn,
Results from
pressure p(xm,yn)
acting on the
incremental area
dAmn= (dxm) (dyn)
Uniform Pressure Profile
 Then the Total Force,
F, on the Area
Fp 
 dF   px, y dA
area
Engineering-36: Engineering Mechanics - Statics
12
area
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Pressure Loading: Total Force
 The Differential Geometry is shown below
dF
dA
 Then the Total Pressure Force
Fp

 dF   px , y dA
 px , y dydx
mn
area

Engineering-36: Engineering Mechanics - Statics
13
m
mn
area
m
all x, all y
n
n
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 Fp
 Then the Moment
about the y-axis due
to intensity dFmn and
LeverArm xm
dx  xm  pxm , yn dxdy
Engineering-36: Engineering Mechanics - Statics
14
dF 
pdxdy
yn
xm
 Then the Total y-axis
Moment
x 
 d
x
surface

 x p x
m
surface
m
, yn dxdy
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Pressure Loading – Pressure Ctr
 Recall also
Ωx = XC•Fp
dF 
pdxdy
 Equating the two Ω
expressions
X C Fp 
 x p x
m
m
 Isolating XC
m
XC 
m
, yn dxdy
surface
Fp
Engineering-36: Engineering Mechanics - Statics
15
xm
, yn dxdy
surface
 x p x
ym
 The Similar
Expression for YC
 y p x
n
YC 
m
, yn dxdy
surface
Fp
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Pressure Loading Summarized
 Given a surface with Pressure Profile
 The Equivalent Force, Fp, Exerted on
the Surface due to the Pressure
 p x
Fp 
m
, yn dxdy
all x, all y
 Fp is located at the Center of Pressure
at CoOrds (XC,YC)
 x p x
m
XC 
m
, yn dxdy
surface
Fp
Engineering-36: Engineering Mechanics - Statics
16
 y p x
n
YC 
m
, yn dxdy
surface
Fp
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
WhiteBoard Work
Lets Work
These Nice
Problems
Engineering-36: Engineering Mechanics - Statics
17
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Engineering 36
Appendix
dy
µx µs
 sinh

dx
T0 T0
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
18
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Beam Problem
 For the Negligible-Wt Beam Find
• Equivalent POINT-Load and it’s Location
(Point of Application, PoA)
• The RCNs at Pt-A
Engineering-36: Engineering Mechanics - Statics
19
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Pressure Problem
 Find the Equivalent POINT-LOAD and
its Point of Application (Location) For
the Given Pressure Distribution
Engineering-36: Engineering Mechanics - Statics
20
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Engineering-36: Engineering Mechanics - Statics
21
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Engineering-36: Engineering Mechanics - Statics
22
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Engineering-36: Engineering Mechanics - Statics
23
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Engineering-36: Engineering Mechanics - Statics
24
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Pressure Loading
 The Differential Geometry is shown belwo
dF
dA
 Then the Total Pressure Force
Fp

 dF
mn
area

Engineering-36: Engineering Mechanics - Statics
25
 p x

, yn dAmn
area
 p x
all x, all y
m
m
, yn dxdy
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx
Engineering-36: Engineering Mechanics - Statics
26
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx

similar documents