Distributed load

Reduction of a Simple Distributed Load

Wednesday, September 28, 2011

Reduction of a Simple Distributed Load


Distributed LoadingIn many situations, a surface area of a body is subjected to a distributed load or pressure. Such forces are caused by winds, fluids, or the weight of items on the body’s surface.

We will worry about loads that are uniform along one axis, w is a function of x and has units of force per length.


Concept of a Distributed Load

In many situations, a surface area of a body is subjected to a distributed load or pressure. Such forces are caused by winds, fluids, or the weight of items on the body’s surface.

We analyze simple cases where a pressure is a uniform load along one axis of a flat rectangular body.

In such cases, the distributed load is only x and has units of force per length.


Magnitude of the Resultant (Equivalent Force)

Consider an element of length dx.

Note that w(x) is the local height of the loading curve.

Locally, the force magnitude dF acting on it is given as

dF = w(x) dx


Magnitude of the Resultant (Equivalent Force)

Consider an element of length dx.

Note that w(x) is the local height of the loading curve.

Locally, the force magnitude dF acting on it is given as

dF = w(x) dx

So the net force on the beam is given by

FR = ∫L dF = ∫L w(x) dx = A

Here A is the area under the loading curve


Location of the Resultant Force

The force dF will produce a moment of x(dF) about point O.

The total moment about point O is given as + MRO = ∫L x dF = ∫L x w(x) dx



The force dF will produce a moment of x(dF) about point O.

The total moment about point O is given as + MRO = ∫L x dF = ∫L x w(x) dx

Assuming that FR acts at x , it will produce the moment about point O as

+ MRO = ( x ) (FR) = x ∫L w(x) dx



Comparing the last two equations, we get ….

So point ‘C’ is just the …


Important Examples

RectangleFeq = ∫Lwdx = wLMagnitude:

Location: x = ∫Lwxdx = wL2/2 = L/2∫Lwdx wL


Important Examples

RectangleFeq = ∫Lwdx = wLMagnitude:

Location: x = ∫Lwxdx = wL2/2 = L/2∫Lwdx wL

?


Important Examples

Triangle

Feq = ∫Lwdx = HL/2Magnitude:

Location: x = ∫LH/Lx2dx = H/L L3/3 = 2L/3∫Lwdx HL/2

w = 100x = H/L x


Important Examples

Triangle



?

w = 100x = H/L x


Important Examples

Triangle



?

w = 100x = H/L x

?


In General

Magnitude of the resultant is the area under the loading curve

Location is the centroid of the loading curve

Back of the book has these and other examples tabulated - when in doubt, integrate


Do it in your head

What is the location of FR, i.e., the distance d?

A) 2 m B) 3 m C) 4 m D) 5 m E) 6 m

B A

3 m 3 m

F R

B A d


Do it in your head

1. FR = ____________

A) 12 N B) 100 N

C) 600 N D) 1200 N

2. x = __________.

A) 3 m B) 4 m

C) 6 m D) 8 m

F R 100 N/m

12 m x


Do it in your head

If F1 = 3 N, x1 = 1 m, F2 = 3 N and x2 = 2 m, what is the location of FR, i.e., the distance x.

A) 1 m B) 1.33 m C) 1.5 m

D) 3 m E) 2 m

F 2 F 1

x 1

x 2


ExampleFind the equivalent force to replace the loading and identify its location from the point A.

Plan:



Plan:



1) The distributed loading can be divided into three parts. (one rectangular loading and two triangular loadings).

2) Find FR and its location for each of these three distributed loads.3) Determine the overall FR of the three point loadings and its location.

Plan:


For the left triangular loading of height 8 kN/m and width 3 m,FR1 = (0.5) 8 kN/m 3 m = 12 kN

x1 = (2/3)(3m) = 2 m from A



x1 = (2/3)(3m) = 2 m from A

x2

For the top right triangular loading of height 4 kN/m and width 3 m, FR2 = (0.5) (4 kN/m) (3 m) = 6 kNand its line of action is at = (1/3)(3m) + 3 = 4 m from A



x1 = (2/3)(3m) = 2 m from A

x2

For the top right triangular loading of height 4 kN/m and width 3 m, FR2 = (0.5) (4 kN/m) (3 m) = 6 kNand its line of action is at = (1/3)(3m) + 3 = 4 m from A

For the rectangular loading of height 4 kN/m and width 3 m, FR3 = (4 kN/m) (3 m) = 12 kNand its line of action is at = (1/2)(3m) + 3 = 4.5 m from Ax 3


For the combined loading of the three forces ... FR = 12 kN + 6 kN + 12 kN = 30 kN + MRA = (2) (12) + 4 (6) + (4.5) 12 = 102 kN • m


For the combined loading of the three forces ... FR = 12 kN + 6 kN + 12 kN = 30 kN + MRA = (2) (12) + 4 (6) + (4.5) 12 = 102 kN • m

Now, (FR• x) has to equal MRA = 102 kN • mSo solve for x to find the equivalent force’s location.Hence, x = (102 kN • m) / (30 kN) = 3.4 m from A.


Example


Example


Example

To the board …


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