ME 201Engineering Mechanics: Statics

Unit 5.3

Reduction of a Simple Distributed

Loading

Distributed Loads

Thus far we’ve been working with loads that

are concentrated at a point:

Many times in engineering we need to be

concerned with another type of loading

referred to as distributed loading:

10 N/m

6 kN/m

Distributed Loads

Instead of being concentrated at a point, a

distributed load is spread out over a distance

It can be thought of as a collection of smaller

loads

To find FR, we need to sum an infinite

number of small forces

Distributed Loads Consider a small differential element, dF

with a width of dx

and a height of w(x) w(x)

wdF

x dx x

The area of the element is

Since infinite number of forces, need to integrate to find FR

dxxwdF )(

AdAdxxwFAl

R )(

Centroids for Simple Shapes

Where is the centroid for these common shapes?

b

h

b

h

2

bxc

2

hyc

3

bxc

3

hyc

Example Problem SolutionGiven:

trapezoid

Find:

FR,

Solution:

FBD

FR

x

50 lb/ft

9 ftA B

100 lb/ft

9 ftA B

FrectFtri

ftftlbFtri 9/)50100(2

1

ftftlbFrect 9/50

lb225

lb450

lbFFF recttriR 675

9 ftA B

FR

x

Example Problem Solution

5.44503225675 x

ftxtri 93

1

ftxrect 92

1

ft3

ft5.4

Given:

trapezoid

Find:

FR,

Solution:

FBD

FR

x

x

ft4

50 lb/ft

9 ftA B

100 lb/ft

9 ftA B

FrectFtri

9 ftA B

FR

x

lbFtri 225 lbFrect 450

lbFR 675

Example Problem Solution

Given:

trapezoid

Find:

FR,

Solution:

FR integral

x

LR dxxwF )(

2

0

260 dxx

N160

|2

0

3

360

x

w=60x2 N/m

2 m

Example Problem Solution

Given:

trapezoid

Find:

FR,

Solution:

FR integral

integral

x

w=60x2 N/m

2 m

L

L

dxxw

dxxxwx

)(

)(

x

2

0

2

2

0

2

60

60

dxx

dxxx

|

|2

0

3

2

0

4

360

460

x

x

160

240 m5.1

Distributed Loads

Thus far we’ve been working with loads that

are concentrated at a point:

Many times in engineering we need to be

concerned with another type of loading

referred to as distributed loading:

10 N/m

6 kN/m

Distributed Loads

Instead of being concentrated at a point, a

distributed load is spread out over a distance

It can be thought of as a collection of smaller

loads

Distributed Loads

To compute the resultant, FR, of a distributed load, consider the following:

To find FR, we need to sum an infinite number of small forces

Distributed Loads

Consider a small differential element, dF

with a width of dx

and a height of w(x)

w(x)

wdF

x dx x

Distributed Loads

The area of the element is

Since infinite number of forces, need to integrate to find FR

dxxwdF )(

AdAdxxwFAl

R )(

w(x)

wdF

x dx x

Distributed Loads

Where does FR act?

Can be determined by equating the moments of

the force resultant and the force distribution

lR dxxwF )(

R

l

F

dxxxwx

)(

This is also the centroid of the area

lR dxxxwFx )(

l

l

dxxw

dxxxw

)(

)(

A

A

dA

xdA

Centroids

For simple shapes, centroid can be found in a table

(see back cover of textbook)

Where is the centroid for these common shapes?

b

h

b

h

2

bxc

2

hyc

3

bxc

3

hyc

Centroids of Simple Shapes

Where does FR act?

A. xc = b/2

B. xc > b/2

C. xc < b/2

D. Cannot tell from information given

b

h1

h2

Example Problem

Given:

w=400 lb/ft

Find:

FR,

w lb/ft

10 ftA B

x

Example Problem SolutionGiven:

w=400 lb/ft

Find:

FR, x

w lb/ft

10 ftA B

Example Problem Solution

Given:

w=400 lb/ft

Find:

FR,

Solution:

FBD

FR

x

w lb/ft

10 ftA B

ftftlbFR 10/400

2

10 ftx

10 ftA B

FR

x

korlb 44000

ft5

x

Example Problem

Given:

w=100x N/m

Find:

FR,

600 N/m

6 mA B

x

Example Problem SolutionGiven:

w=100x N/m

Find:

FR,

600 N/m

6 mA Bx

Example Problem Solution

Given:

w=100x N/m

Find:

FR,

Solution:

FBD

FR

600 N/m

6 mA B

mmNFR 6/6002

1

mx 63

2

x

x

6 mA B

FR

x

kNmorNm 8.11800

m4

Example Problem

Given:

trapezoid

Find:

FR,

50 lb/ft

9 ftA B

100 lb/ft

x

Example Problem SolutionGiven:

trapezoid

Find:

FR, x50 lb/ft

9 ftA B

100 lb/ft

Example Problem SolutionGiven:

trapezoid

Find:

FR,

Solution:

FBD

FR

x

50 lb/ft

9 ftA B

100 lb/ft

9 ftA B

FrectFtri

ftftlbFtri 9/)50100(2

1

ftftlbFrect 9/50

lb225

lb450

lbFFF recttriR 675

9 ftA B

FR

x

Example Problem Solution

5.44503225675 x

ftxtri 93

1

ftxrect 92

1

ft3

ft5.4

Given:

trapezoid

Find:

FR,

Solution:

FBD

FR

x

x

ft4

50 lb/ft

9 ftA B

100 lb/ft

9 ftA B

FrectFtri

9 ftA B

FR

x

lbFtri 225 lbFrect 450

lbFR 675

Example Problem

Given:

function

Find:

FR,

w=60x2 N/m

2 mx

Example Problem SolutionGiven:

function

Find:

FR, x

w=60x2 N/m

2 m

Example Problem Solution

Given:

trapezoid

Find:

FR,

Solution:

FR integral

x

LR dxxwF )(

2

0

260 dxx

N160

|2

0

3

360

x

w=60x2 N/m

2 m

Example Problem Solution

Given:

trapezoid

Find:

FR,

Solution:

FR integral

integral

x

w=60x2 N/m

2 m

L

L

dxxw

dxxxwx

)(

)(

x

2

0

2

2

0

2

60

60

dxx

dxxx

|

|2

0

3

2

0

4

360

460

x

x

160

240 m5.1

In Class Exercise

Solution

In Class Exercise

In Class Exercise