Distributed Loads - Engineering Mechanics Statics - Lecture Slides, Slides of Mechanical Engineering

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Engineering 36

Chp09:

Distributed Loads

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,

is Determined

by Area Under

the Profile Curve

    span

W w x dx

Distributed Loads

• Now Use Centroidal Methodology

           

span

n n span

x LeverArm Intensity x w x dx

 And Recall:

x  xW ^ x is theCentroid Location

 Equating the

Ω Expressions

find

    

W

x w x dx

x span

 n n 

Distributed Loads on Beams

• 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  (^)  wdx dA A

L 0

  OP  A xdA xA

OP W xdW L  







0

• 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.

Example:Trapezoidal Load Profile

SOLUTION:

• The magnitude of the concentrated load is equal to the total load, or the area under the curve. F  18. 0 kN
• The line of action of the concentrated load passes through the area centroid of the curve.

18 kN

63 kN m

X  X ^3.^5 m

F ^1500  24500 mN 6 m

Example:Trapezoidal Load Profile

 MA ^0 : By^6 m^ ^18 kN^3 .5m^ ^0

By  10. 5 kN

 MB ^0 : Ay^6 m^ ^18 kN^6 m^3 .5m^ ^0 Ay  7. 5 kN

 Determine the support reactions by summing moments about the beam ends After Replacing the Dist-Load with the Equivalent POINT-Load

Ay By

Pressure Loading

• Consider an Area Subject to a Pressure Load

Uniform Pressure Profile

 The incremental Force, dFmn, Results from pressure p(xm,yn) acting on the incremental area dAmn= (dxm) (dyn)

 Then the Total Force, F, on the Area     ^  area area

Fp dF p x , y dA

Pressure Loading: Total Force

• The Differential Geometry is shown below

 Then the Total Pressure Force

dA

dF

 

 ^ ^ ^ 

 



 

allx,all y

,

,

p x y dy dx

F dF p x y dA

m n

area

m n mn area

p mn

Pressure Loading – Pressure Ctr

• Recall also Ωx = XC•Fp  Equating the two Ω expressions

 The Similar Expression for YC

pdxdy

dF 

xm m y

    surface

X (^) C Fp xmp xm , yn dxdy

 Isolating XC

 

p

surface

m m n C (^) F

x p x y dxdy X

 

,  

p

surface

n m n C (^) F

y p x y dxdy Y

 

,

Pressure Loading Summarized

• Given a surface with Pressure Profile
• The Equivalent Force, Fp, Exerted on the

Surface due to the Pressure

      allx,all y

Fp p xm , yn dx dy

 Fp is located at the Center of Pressure

at CoOrds (XC,YC)

 

p

surface

m m n C (^) F

x p x y dxdy

X

 

,  

p

surface

n m n C (^) F

y p x y dxdy Y

 

,

Pressure Loading

• The Differential Geometry is shown belwo

 Then the Total Pressure Force

dA

dF

 

 ^ ^ ^ 

 



 

allx,all y

p x y dx dy

F dF p x y dA

m n

area

m n mn area

p mn

,

,