Distrubuted Loads. Decimals have a point. Distributed Loads

Transcription

1 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 b a vector Not all loading conditions are of that tpe 2 1

2 Consider how our ears feel as ou go deeper into a swimming pool. The deeper ou go, the greater the pressure on our ears. 3 Distributed Loads If we consider how this pressure acts on the walls of the pool, we would have to consider a force (generated b the pressure) that was small at the top and increased as we went down. 4 2

3 This is known as a distributed force or a distributed load. It is represented b a series of vectors which are connected at their tails. 5 Distributed Loads One tpe of distributed load is a uniforml distributed load 6 3

4 This load has the same intensit along its application. The intensit is given in terms of Force/Length 7 Distributed Loads The total magnitude of this load is the area under the loading diagram. So here it would be the load intensit time the beam length. 8 4

5 If, for analsis 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 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 5

6 second tpe of loading we often encounter is a triangular load 11 Distributed Loads triangular load has an intensit of 0 at one end and increases to some maimum at the other end. 12 6

7 You will often see the intensit represented with the letter w. 13 Distributed Loads The magnitude of an equivalent point load will again be the area under the loading diagram. 14 7

8 For a triangle, this would be ½ the base times the maimum intensit. 15 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 8

9 There are other tpes of loading diagrams but these will be sufficient for now. 17 Distributed Loads You ma see a diagram that appears to be a trapezoidal loading. 18 9

10 In this case, we can divide the loading diagram into two parts, one a rectangular load and the other a triangular load. 19 Distributed Loads Now ou have two loads that ou alread have the rules for

11 Take care to note that the maimum intensit of the triangular load is now reduced b the magnitude of the rectangular load. 21 Eample Problem Given: The loading and support as shown Required: Reactions at the supports 100 lb/ft 200 lb/ft 5 ft 4 ft 22 11

12 Eample Problem Isolate the selected sstem from all connections Start with the pin at 100 lb/ft 200 lb/ft 5 ft 4 ft 23 Eample Problem We have a pin, so we have an and a component of the reaction and we will assume that both of them are lb/ft 200 lb/ft 5 ft 4 ft 24 12

13 Eample Problem Now we can remove the roller support at recognizing that the direction of the reaction is lb/ft 200 lb/ft 5 ft 4 ft 25 Eample Problem We now have all the reactions identified and can proceed with the analsis 100 lb/ft 200 lb/ft 5 ft 4 ft 26 13

14 Eample Problem The best idea is to now convert all distributed loads into point loads 100 lb/ft 200 lb/ft 5 ft 4 ft 27 Eample Problem reak the load into a rectangular load and a triangular load 100 lb/ft 100 lb/ft 5 ft 4 ft 28 14

15 Eample Problem For the rectangular load 900 lb 100 lb/ft 4.5 ft 5 ft 4 ft 29 Eample Problem For the triangular load 7.67 ft 4.5 ft 900 lb 200 lb 5 ft 4 ft 30 15

16 Eample Problem We have three unknowns,,, and Luckil we have three equilibrium constraints to solve for them 31 F F M = 0 = 0 = ft 4.5 ft 5 ft 4 ft 900 lb 200 lb Eample Problem Summing the moments about to solve for. 32 F F M = 0 = 0 = 0 5 ft 4 ft 7.67 ft 4.5 ft 900 lb 200 lb 16

17 Eample Problem Writing the epression for the sum of the moments around ( )( ) ( )( ) ( )( ) M = 0= 4.5ft 900lb 7.67ft 200lb + 9ft 33 5 ft 4 ft 7.67 ft 4.5 ft 900 lb 200 lb M = 0 Eample Problem Isolating and solving for 4.5 ft 900lb ft 200lb 34 ( )( ) ( )( ) ( 9 ft) lb = 5 ft 4 ft 7.67 ft 4.5 ft = 900 lb 200 lb M = 0 17

18 Eample Problem Sum of the forces in the -direction = 900lb 200lb + 0= 900lb 200lb lb =+ 900lb + 200lb lb = lb 7.67 ft 4.5 ft 900 lb 200 lb M = ft 4 ft Eample Problem Since the magnitude of our solution came out positive, we assumed the correct direction = 900lb 200lb lb 0= 900lb 200lb lb 200 lb =+ 900lb + 200lb lb 36 = lb 7.67 ft 4.5 ft M = 0 5 ft 4 ft 18

19 Eample Problem Now our final constraint condition F 0 = = 0 = = 0lb 7.67 ft 4.5 ft M = lb 200 lb 37 5 ft 4 ft Eample Problem So our complete solution to the problem is = 0lb = lb = lb 7.67 ft 4.5 ft M = lb 200 lb 38 5 ft 4 ft 19

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21 41 Homework Problem Problem Problem