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Structures vary from the mundane. To the iconic And entertaining.

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Presentation on theme: "Structures vary from the mundane. To the iconic And entertaining."— Presentation transcript:

1 structures vary from the mundane

2 To the iconic

3 And entertaining

4 Natural structures can be even more amazing

5 Forces In this course we will deal with several forces acting on objects at one time Because forces have magnitude(size) and direction(angle) they are vectors.

6 See-saw In this see- saw how many forces are acting on the beam?

7 F1 F2 FRFR F 1 +F 2 = FRFR For static equilibrium the sum of the vertical forces must also = 0, F1F1 F2F2 FRFR

8 For static equilibrium the sum of the horizontal forces must also = 0,

9 Moments As well as the sum of vertical and horizontal forces on a body, the moments(or turning effect) of forces must also be zero for equilibrium

10 Moments If the sum of the moments(or turning effect) of forces are not zero you don’t have equilibrium http://www.youtube.com/watch?v=V2- lll4GF5g http://www.youtube.com/watch?v=V2- lll4GF5g

11 Moments

12

13 Now try assignments on moments of forces

14 Moments Conditions of Equilibrium In the force system in this section you shall apply the three condition of equilibrium that you have used in the following order 1. The sum of all the moments = zero. ∑M o = 0 2. The sum of all the vertical forces = zero. ∑ F v = 0 2. The sum of all the horizontal forces = zero. ∑ F h = 0

15 Moments A simply supported beam has no horizontal forces on it

16 Moments A simply supported beam has no horizontal forces on it We have to find R A and R B, the reaction forces supplied by the supports

17 Moments We have to find R A and R B, the reaction forces supplied by the supports

18 Moments We have to find R A and R B, the reaction forces supplied by the supports

19 Moments We have to find R A and R B, the reaction forces supplied by the supports

20 There are no horizontal forces in this simply supported beam ∑F h =0 So we just need to consider the vertical equilibrium 0.88kN ∑ F v = 0 R A + R B = 10 + 8 R A = 17.12kN R A =18 – 0.88 R A + 0.88 = 18 R A = 17.12kN

21 Forces sometimes act at an angle so we need to be able to break them into their vertical and horizontal components to do calculations with them Look at the webpage from a climbing website http://www.ropebook.com/information/vector- forces http://www.ropebook.com/information/vector- forces

22 90N Force at an angle of 30degrees 90 N

23 First To find F v To find F H Now try Resolution of forces questions 1-4


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