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Moments

Force. Distance. Moments. TUTORIAL 4. to answer just click on the button or image related to the answer. let's go !!. W. W1 200N. W2 100N. A. B. C. D. calculate the anti-clockwise moments. calculate the clockwise moments. 4m. 2m. 4m. what do we do first?. either a or b. b. a.

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Moments

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  1. Force Distance Moments TUTORIAL 4 to answer just click on the button or image related to the answer let's go !!

  2. W W1 200N W2 100N A B C D calculate the anti-clockwise moments calculate the clockwise moments 4m 2m 4m what do we do first? either a or b b a c Question 1a given a beam loaded as shown, calculate the weight, W required for equilibrium

  3. W W1 200N W2 100N A B C D B A 4m 2m 4m about which point do we take the moments? C or D b a c Question 1b if we decide to calculate the clockwise moments first

  4. W W1 200N W2 100N A B C D 300 N 1000 N 4m 2m 4m what is the total sum of the clockwise moments? 1000 Nm 1000 N/m d b a c Question 1c if we decide to calculate the clockwise moments first about B

  5. W W1 200N W2 100N A B C D calculate the anti-clockwise moments sum the total forces on the beam 4m 2m 4m a b Question 1d what do we do next?

  6. W W1 200N W2 100N A B C D 400 Nm 4W N/m 4m 2m 4m 4W Nm b a c Question 1e what is the total sum of the anti-clockwise moments about B?

  7. W W1 200N W2 100N A B C D equate the clockwise and anti-clockwise moments sum the clockwise and anti-clockwise moments 4m 2m 4m what do we do next? take the difference between the clockwise and anti-clockwise moments b a c Question 1f

  8. W W1 200N W2 100N A B C D 4W = 1000 Nm, W = 250 N 4W = 1000 Nm, W = 250 Nm 4m 2m 4m a b Question 1g when we equate the clockwise and anti-clockwise moments what do we get for W?

  9. 10kN 6kN 4kN 8kN this is a statically determinate system and we can use the equations of static equilibrium this is a statically indeterminate system and we can’t use the equations of static equilibrium R R L R what do we know? 1m 1.5m 2m 1.5m 0.5m b a Question 2a given the beam loaded as shown, calculate the reactions, RL and RR

  10. 10kN 6kN 4kN 8kN which equation do we use first? ΣH = 0, sum of all horizontal forces = 0 ΣV = 0, sum of all vertical forces = 0 R R L R 1m 1.5m 2m 1.5m 0.5m either b or c ΣM = 0, sum of all moments = 0 b a d c Question 2b given that the beam is statically determinate

  11. 10kN 6kN 4kN 8kN what do we get? RL + RR = 28 kN RL = RR + 28 kN R R L R 1m 1.5m 2m 1.5m 0.5m RR = RL + 28 kN b a c Question 2c if we start with ΣV = 0

  12. 10kN 6kN 4kN 8kN give up use another one of the equations, ΣH = 0 use another one of the equations, ΣM = 0 R R L R what do we do now? 1m 1.5m 2m 1.5m 0.5m b a c Question 2d we still have two unknowns, RL and RR

  13. 10kN 6kN 4kN 8kN calculate the clockwise moments calculate the anti-clockwise moments R R L R what do we do first? 1m 1.5m 2m 1.5m 0.5m select a point about which to take moments b a c Question 2e given that we are going to use ΣM = 0

  14. 10kN 6kN 4kN 8kN R R L R B,C, D, or E A or F 1m 1.5m 2m 1.5m 0.5m which point do we select? b a Question 2f given that we select a point about which we will take moments A B C D E F

  15. 10kN 6kN 4kN 8kN R R L R calculate the clockwise moments calculate the anti-clockwise moments 1m 1.5m 2m 1.5m 0.5m what do we do next? either a or b b a c Question 2g suppose we select point A A B C D E F

  16. 10kN 6kN 4kN 8kN R R L R RR x 6.5 kNm RR kN 1m 1.5m 2m 1.5m 0.5m what is the total sum of the clockwise moments about A? 28 kN 105 kNm b a d c Question 2h if we decide to calculate the clockwise moments about A first A B C D E F

  17. 10kN 6kN 4kN 8kN R R L R RR kNm RR kN 1m 1.5m 2m 1.5m 0.5m what is the total sum of the anti-clockwise moments about A? 6.5 RL kNm 6.5 RR kNm b a d c Question 2i if we calculate the anti-clockwise moments about A A B C D E F

  18. 10kN 6kN 4kN 8kN what now? R R L R equate the clockwise and anti-clockwise moments 1m 1.5m 2m 1.5m 0.5m give up use another equation b a c Question 2j A B C D E F

  19. 10kN 6kN 4kN 8kN R R L R what do we get? 6.5 RR = 105, RR = 16.15 kN 6.5 RR = 28, RR = 4.31 kN 1m 1.5m 2m 1.5m 0.5m 6.5 RR = 105, RR = 16.15 kNm b a c Question 2k if we equate the clockwise and anti-clockwise moments A B C D E F

  20. 10kN 6kN 4kN 8kN R R L R take moments about F use the result from equation ΣV = 0 that we calculated previously 1m 1.5m 2m 1.5m 0.5m how do we do that? a b Question 2l now that we’ve got RR we need to get RL A B C D E F

  21. 10kN 6kN 4kN 8kN R R L R RL = 44.15 kN RL = 11.85 kN 1m 1.5m 2m 1.5m 0.5m what does using ΣV = 0 give us? a b Question 2m A B C D E F

  22. 3 kN/m UDL M 8m this is a statically determinate system and we can use the equations of static equilibrium this is a statically indeterminate system and we can’t use the equations of static equilibrium what do we know? a b Question 3a given the cantilever beam loaded with a Uniformly Distributed Load (UDL) of 3 kN/m as shown, calculate the moment reaction, M

  23. 3 kN/m UDL M 8m calculate the total load on the beam calculate the total load on the beam and make it into an equivalent point load take moments what do we do first ? a b c Question 3b given that the beam is statically determinate

  24. M 12 kN 24 kN 24 kNm a b c Question 3c 3 kN/m UDL if we make the UDL into an equivalent point load 8m what is the total load ?

  25. M at the wall, point A at the end of the beam, point C at the middle of the beam, point B a b c Question 3d 3 kN/m UDL if we make the UDL into an equivalent point load 8m 4m 4m where does this point load act ? A B C

  26. M 24 kNm 96 KNm 96 kN so what is the moment M at A ? a b c Question 3e 3 kN/m UDL 8m 4m 4m A B C

  27. 2kN 1.5kN/m 2kN/m R R L R 4m 4m 6m 2m convert all the UDLs into point loads take moments what do we do first ? b a Question 4a beam loadings in real structures are often complex. consider the beam as shown with a point load of 2 kN at its end and a UDL of 1.5 kN/m over part of its main span. The beam also carries its own weight of 2 kN/m. given that the beam is statically determinate

  28. 6kN 16kN 2kN R R A B C D E L R take moments about A take moments about E take moments about D take moments about C take moments about B what do we do now ? 2m 2m 4m 6m 2m e d b a c Question 4b we now get the loadings as shown

  29. 6kN 16kN 2kN R R A B C D E L R anti-clockwise moments clockwise moments 2m 2m 4m 6m 2m b a Question 4c if we take moments about A what moments do we take ?

  30. 6kN 16kN 2kN R R A B C D E L R clockwise moments = 52 kNm clockwise moments = 92 kNm clockwise moments = 24 kNm 2m 2m 4m 6m 2m b a c Question 4d if we take clockwise moments about A what do we get ?

  31. 6kN 16kN 2kN R R A B C D E L R take moments about D calculate the anti-clockwise moments about A 2m 2m 4m 6m 2m b a Question 4e what do we do now ?

  32. 6kN 16kN 2kN R R A B C D E L R anti-clockwise moments = 2 kNm anti-clockwise moments = 6 x RR kNm anti-clockwise moments = 8 x RR kNm 2m 2m 4m 6m 2m a b c Question 4f if we take anti-clockwise moments about A what do we get ?

  33. 6kN 16kN 2kN what now? R R A B C D E L R equate the clockwise and anti-clockwise moments 2m 2m 4m 6m 2m take moments about D a b Question 4g the anti-clockwise moments about A = 6RR

  34. 6kN 16kN 2kN R R A B C D E L R RL = 92 / 6 kNm, RL = 15.3 kN RR = 92 / 6 kNm, RR = 15.3 kN what do we get? 2m 2m 4m 6m 2m a b Question 4h when we equate the clockwise and anti-clockwise moments

  35. 6kN 16kN 2kN R R A B C D E L R take moments about D use one of our equations 2m 2m 4m 6m 2m a b Question 4i we have RR, we now want RL what now?

  36. 6kN 16kN 2kN R R A B C D E L R ΣV = 0, sum of all vertical forces = 0 ΣH = 0, sum of all horizontal forces = 0 which one? 2m 2m 4m 6m 2m ΣM = 0, sum of all moments = 0 a b c Question 4j we use one of our equations

  37. 6kN 16kN 2kN R R A B C D E L R RL + RR= 24, RL = 8.7 kN RL – RR = 24, RL = 39.3 kN what do we get? 2m 2m 4m 6m 2m RL + RR= 24, RL = 8.7 kNm a b c Question 4k using ΣV = 0, sum of all vertical forces = 0

  38. Weight 20kN 16kN 2m A 3m the wind force of 16 kN the weight of the building, 20 kN what is tending to overturn the building? the moment caused by the wind force a b c Question 5a a lightweight prefabricated building has a total (empty) weight of 20 kN. a wind load of 16 kN acts at 2 m above the ground

  39. Weight 20kN 16kN 2m A 3m 16 kN 32 kN what is the overturning moment? 32 kNm a b c Question 5b a lightweight prefabricated building has a total (empty) weight of 20 kN. a wind load of 16 kN acts at 2 m above the ground

  40. Weight 20kN 16kN 2m A 3m the weight of the building what is preventing the building from overturning? the moment caused by the weight of the building a b Question 5c a lightweight prefabricated building has a total (empty) weight of 20 kN. a wind load of 16 kN acts at 2 m above the ground

  41. Weight 20kN 16kN 2m A 3m 20 kN what is this restraining moment? 60 kNm 30 kNm a b c Question 5d a lightweight prefabricated building has a total (empty) weight of 20 kN. a wind load of 16 kN acts at 2 m above the ground

  42. Weight 20kN 16kN 2m A 3m yes no maybe a b c Question 5e a lightweight prefabricated building has a total (empty) weight of 20 kN. a wind load of 16 kN acts at 2 m above the ground will the building overturn?

  43. Weight 20kN 16kN 2m A 3m increase the weight of the building all of the above lower the height of the building make the base wider d a b c Question 5f a lightweight prefabricated building has a total (empty) weight of 20 kN. a wind load of 16 kN acts at 2 m above the ground how can we make the building safer?

  44. enough ! yes, yes, yes !! you can start from either. It makes no difference next question

  45. let me try again let me out of here Nearly ! why did you pick that?

  46. enough ! Grrr...eat !! taking moments about a point so as to eliminate an unknown force (the reaction) is always good next question

  47. let me try again let me out of here Possibly, .... but .... when you next calculate the anti-clockwise moments you will have the unknown reaction force at B

  48. enough ! You've got it !! (200 N x 2 m) + (100 N x 6 m) = 1000 Nm = 1 kNm next question

  49. let me try again let me out of here A devil of an answer !! we are talking about moments not just forces. Remember what a moment is.

  50. let me try again let me out of here A devil of an answer !! we are talking about moments not just forces. Remember what a moment is and what are the units for moments

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