[email protected] ENGR-36_Lec-24_Dist_Loads.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed Electrical

[email protected] • ENGR-36_Lec-24_Dist_Loads.pptx1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp09: Distributed

Loads



Distributed Loads

The Load on an Object may be Spread out, or Distributed over the surface.

Load Profile, w(x)



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 thePoint Load, W, is Determined by Area Under the Profile Curve

span

dxxwW



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



Distributed Loads

Now Use Centroidal Methodology

span

nn

span

x dxxwxIntensityLeverArm

And Recall:

Location Centroid theis xWxx Equating the

Ω Expressionsfind

W

dxxwx

x span

nn



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.

AdAdxwWL

0

AxdAxAOP

dWxWOP

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

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



Example:Trapezoidal Load ProfileSOLUTION:

• The magnitude of the concentrated load is equal to the total load, or the area under the curve.

kN 0.18F

• The line of action of the concentrated load passes through the area centroid of the curve.

kN 18

mkN 63 X m5.3X

m6m

N

2

45001500

F



Example:Trapezoidal Load Profile

0m .53kN 18m 6:0 yA BM

kN 5.10yB

0m .53m 6kN 18m 6:0 yB AM

kN 5.7yA

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

ByAy



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



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

areaarea

p dAyxpdFF ,



Pressure Loading: Total Force

The Differential Geometry is shown below

Then the Total Pressure Force

dA

dF

y all x,all

,

,

dxdyyxp

dAyxpdFF

nm

area

mnnm

area

mnp



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

Then the Total y-axis Moment

dxdyyxpxd nmmx ,

pdxdy

dF

mx ny

surface

nmm

surface

xx

dxdyyxpx

d

,



Pressure Loading – Pressure Ctr

Recall also Ωx = XC•Fp

Equating the two Ω expressions

The Similar Expression for YC

pdxdy

dF

mx my

surface

nmmpC dxdyyxpx FX ,

Isolating XC

p

surface

nmm

C F

dxdyyxpx

X

,

p

surface

nmn

C F

dxdyyxpy

Y

,



Pressure Loading Summarized

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

the Surface due to the Pressure

yall x, all

dydxyxpF nmp ,

Fp is located at the Center of Pressure at CoOrds (XC,YC)

p

surface

nmm

C F

dxdyyxpx

X

,

p

surface

nmn

C F

dxdyyxpy

Y

,



WhiteBoard Work

Lets WorkThese NiceProblems



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix 00

sinhT

µs

T

µx

dx

dy



Beam Problem

For the Negligible-Wt Beam Find• Equivalent POINT-Load and it’s Location

(Point of Application, PoA)• The RCNs at Pt-A



Pressure Problem

Find the Equivalent POINT-LOAD and its Point of Application (Location) For the Given Pressure Distribution











Pressure Loading

The Differential Geometry is shown belwo

Then the Total Pressure Force

dA

dF

yall x, all

dydxyxp

dAyxpdFF

nm

area

mnnm

area

mnp

,

,



Documents

[email protected] ENGR-36_Lec-24_Dist_Loads.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed Electrical