ENGR-36_Lec-24_Dist_Loads

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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)

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 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

Engineering-36: Engineering Mechanics - Statics

3

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

Engineering-36: Engineering Mechanics - Statics

= 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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5 Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx

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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7 Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx

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

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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

Engineering-36: Engineering Mechanics - Statics

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

Engineering-36: Engineering Mechanics - Statics

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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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

Engineering-36: Engineering Mechanics - Statics

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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20 Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx

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24 Bruce Mayer, PE [email protected] • ENGR-36_Lec-24_Dist_Loads.pptx

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