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Understanding Distributed Loads: Uniform and Triangular, Study notes of Motor Vehicle Design

International University Institute of Luxembourg (IUIL)Motor Vehicle Design

An in-depth explanation of distributed loads, specifically uniform and triangular loads. It covers the concept of distributed loads, their representation, and the methods to calculate the equivalent point load for each type. The document also includes an example problem to illustrate the concepts.

Typology: Study notes

2021/2022

Uploaded on 08/05/2022

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char_s67 🇱🇺

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Distributed Loads
Decimals have a point.
Distributed Loads
¢ Up to this point, all the forces we have
considered have been point loads
¢ Single forces which are represented by a vector
¢ Not all loading conditions are of that type
Monday, November 5, 2012 Distrubuted Loads 2
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Download Understanding Distributed Loads: Uniform and Triangular and more Study notes Motor Vehicle Design in PDF only on Docsity!

Distributed Loads

Decimals have a point.

Distributed Loads

¢ Up to this point, all the forces we have

considered have been point loads

¢ Single forces which are represented by a vector

¢ Not all loading conditions are of that type

2 Distrubuted Loads Monday, November 5, 2012

Distributed Loads

¢ Consider how your ears feel as you go deeper

into a swimming pool.

¢ The deeper you go, the greater the pressure on

your ears.

3 Distrubuted Loads Monday, November 5, 2012 Distributed Loads

¢ If we consider how this pressure acts on the

walls of the pool, we would have to consider a

force (generated by the pressure) that was

small at the top and increased as we went

down.

4 Distrubuted Loads Monday, November 5, 2012

Distributed Loads

¢ This load has the same intensity along its

application.

¢ The intensity is given in terms of Force/Length

7 Distrubuted Loads Monday, November 5, 2012 Distributed Loads

¢ The total magnitude of this load is the area

under the loading diagram.

¢ So here it would be the load intensity time the

beam length.

8 Distrubuted Loads Monday, November 5, 2012

Distributed Loads

¢ If, for analysis purposes, we wanted to replace

this distributed load with a point load, the

location of the point load would be in the center

of the rectangle.

9 Distrubuted Loads Monday, November 5, 2012 Distributed Loads

¢ We do this to solve for reactions.

¢ For a uniform load, the magnitude of the

equivalent point load is equal to the area of the

loading diagram and the location of the point

load is at the center of the loading diagram.

10 Distrubuted Loads Monday, November 5, 2012

Distributed Loads

¢ You will often see the intensity represented with

the letter w.

13 Distrubuted Loads Monday, November 5, 2012 Distributed Loads

¢ The magnitude of an equivalent point load will

again be the area under the loading diagram.

14 Distrubuted Loads Monday, November 5, 2012

Distributed Loads

¢ For a triangle, this would be ½ the base times

the maximum intensity.

15 Distrubuted Loads Monday, November 5, 2012 Distributed Loads

¢ The location of the equivalent point load will be

2/3 of the distance from the smallest value in

the loading diagram.

16 Distrubuted Loads Monday, November 5, 2012

Distributed Loads

¢ In this case, we can divide the loading diagram

into two parts, one a rectangular load and the

other a triangular load.

19 Distrubuted Loads Monday, November 5, 2012 Distributed Loads

¢ Now you have two loads that you already have

the rules for.

20 Distrubuted Loads Monday, November 5, 2012

Distributed Loads

¢ Take care to note that the maximum intensity of

the triangular load is now reduced by the

magnitude of the rectangular load.

21 Distrubuted Loads Monday, November 5, 2012 22 Distrubuted Loads Monday, November 5, 2012 Example Problem

¢ Given: The loading and support as shown

¢ Required: Reactions at the supports

5 ft 4 ft

A B

100 lb/ft 200 lb/ft

25 Distrubuted Loads Monday, November 5, 2012 Example Problem

¢ Now we can remove the roller support at B

recognizing that the direction of the reaction is

+y

5 ft 4 ft

A B

100 lb/ft 200 lb/ft Ay Ax By 26 Distrubuted Loads Monday, November 5, 2012 Example Problem

¢ We now have all the reactions identified and can

proceed with the analysis

5 ft 4 ft

A B

100 lb/ft 200 lb/ft Ay Ax By

27 Distrubuted Loads Monday, November 5, 2012 Example Problem

¢ The best idea is to now convert all distributed

loads into point loads

5 ft 4 ft

A B

100 lb/ft 200 lb/ft Ay Ax By 28 Distrubuted Loads Monday, November 5, 2012 Example Problem

¢ Break the load into a rectangular load and a

triangular load

5 ft 4 ft

A B

100 lb/ft 100 lb/ft Ay Ax By

31 Distrubuted Loads Monday, November 5, 2012 Example Problem

¢ We have three unknowns, Ay, Ax, and By

¢ Luckily we have three equilibrium constraints to

solve for them

5 ft 4 ft

A B

900 lb 200 lb Ay Ax By 4.5 ft 7.67 ft 0 0 0 x y F F M = = = ∑ ∑ ∑ 32 Distrubuted Loads Monday, November 5, 2012 Example Problem

¢ Summing the moments about A to solve for By.

5 ft 4 ft

A B

900 lb 200 lb Ay Ax By 4.5 ft 7.67 ft 0 0 0 x y F F M = = = ∑ ∑ ∑

33 Distrubuted Loads Monday, November 5, 2012 Example Problem

¢ Writing the expression for the sum of the

moments around A

5 ft 4 ft

A B

900 lb 200 lb Ay Ax By 4.5 ft 7.67 ft 0 0 0 x y F F M = = =

∑ M^ =^0 =^ −^ (^ 4.5^ ft^ )(^900 lb^^ )^ −^ (^ 7.67^ ft^ )(^200 lb^ )^ +(^9 ft^ )(^ By ) 34 Distrubuted Loads Monday, November 5, 2012 Example Problem

¢ Isolating and solving for By

5 ft 4 ft

A B

900 lb 200 lb Ay Ax By 4.5 ft 7.67 ft 0 0 0 x y F F M = = =

( )( ) ( )( ) ( )

y y

ft lb ft lb

B

ft

lb B

37 Distrubuted Loads Monday, November 5, 2012 Example Problem

¢ Now our final constraint condition

5 ft 4 ft

A B

900 lb 200 lb Ay Ax By 4.5 ft 7.67 ft 0 0 0 x y F F M = = =

x x x x

F A

A

A lb

∑ 38 Distrubuted Loads Monday, November 5, 2012 Example Problem

¢ So our complete solution to the problem is

5 ft 4 ft

A B

900 lb 200 lb Ay Ax By 4.5 ft 7.67 ft 0 0 0 x y F F M = = =

x y y

A lb

A lb

B lb

  • 39 Distrubuted Loads Monday, November 5,
  • 40 Distrubuted Loads Monday, November 5,