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Moments TUTORIAL 4 to answer just click on the button or image related to the answer Distance Force

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Question 1a what do we do first? calculate the clockwise moments a calculate the anti-clockwise moments b either a or b c given a beam loaded as shown, calculate the weight, W required for equilibrium B W1 200N W2 100N W 4m2m4m ACD

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Question 1b about which point do we take the moments? A a B b C or D c if we decide to calculate the clockwise moments first B W1 200N W2 100N W 4m2m4m ACD

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Question 1c what is the total sum of the clockwise moments? 300 N a 1000 N b 1000 Nm c if we decide to calculate the clockwise moments first about B B W1 200N W2 100N W 4m2m4m ACD 1000 N/m d

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Question 1d what do we do next? calculate the anti-clockwise moments a sum the total forces on the beam b B W1 200N W2 100N W 4m2m4m ACD

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Question 1e what is the total sum of the anti-clockwise moments about B? 400 Nm a 4W N/m b 4W Nm c B W1 200N W2 100N W 4m2m4m ACD

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Question 1f sum the clockwise and anti-clockwise moments a equate the clockwise and anti-clockwise moments b take the difference between the clockwise and anti-clockwise moments c B W1 200N W2 100N W 4m2m4m ACD what do we do next?

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Question 1g 4W = 1000 Nm, W = 250 Nm a 4W = 1000 Nm, W = 250 N b B W1 200N W2 100N W 4m2m4m ACD what do we get for W? when we equate the clockwise and anti-clockwise moments

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Question 2a what do we know? this is a statically determinate system and we can use the equations of static equilibrium a this is a statically indeterminate system and we can’t use the equations of static equilibrium b given the beam loaded as shown, calculate the reactions, R L and R R R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R

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Question 2b which equation do we use first? ΣH = 0, sum of all horizontal forces = 0 a ΣV = 0, sum of all vertical forces = 0 b ΣM = 0, sum of all moments = 0 c given that the beam is statically determinate either b or c d R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R

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Question 2c what do we get? R L + R R = 28 kN a R L = R R + 28 kN b R R = R L + 28 kN c if we start with ΣV = 0 R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R

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Question 2d what do we do now? give up a use another one of the equations, ΣH = 0 b we still have two unknowns, R L and R R R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R use another one of the equations, ΣM = 0 c

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Question 2e what do we do first? calculate the clockwise moments a calculate the anti-clockwise moments b given that we are going to use ΣM = 0 select a point about which to take moments c R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R

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Question 2f which point do we select? A or F a B,C, D, or E b given that we select a point about which we will take moments R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R ACDEFB

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Question 2g suppose we select point A R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R ACDEFB what do we do next? calculate the clockwise moments a calculate the anti-clockwise moments b either a or b c

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Question 2h R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R ACDEFB what is the total sum of the clockwise moments about A? R R kN a R R x 6.5 kNm b 105 kNm c if we decide to calculate the clockwise moments about A first 28 kN d

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Question 2i R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R ACDEFB what is the total sum of the anti-clockwise moments about A? R R kN a R R kNm b 6.5 R R kNm c if we calculate the anti-clockwise moments about A 6.5 R L kNm d

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Question 2j R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R ACDEFB what now? equate the clockwise and anti-clockwise moments a use another equation b give up c

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Question 2k R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R ACDEFB 6.5 R R = 28, R R = 4.31 kN a 6.5 R R = 105, R R = 16.15 kN b 6.5 R R = 105, R R = 16.15 kNm c if we equate the clockwise and anti-clockwise moments what do we get?

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Question 2l R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R ACDEFB use the result from equation ΣV = 0 that we calculated previously a take moments about F b now that we’ve got R R we need to get R L how do we do that?

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Question 2m R L 1m2m1.5m 0.5m 4kN8kN 10kN6kN R R ACDEFB R L = 11.85 kN a R L = 44.15 kN b what does using ΣV = 0 give us?

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

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

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

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

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

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Question 4a convert all the UDLs into point loads a take moments b 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 2kN/m 1.5kN/m 2kN R L 4m 6m2m R R what do we do first ?

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Question 4b take moments about A a we now get the loadings as shown what do we do now ? take moments about B b take moments about C c take moments about D d take moments about E e 6kN2kN R L 2m 4m R R 16kN ABCDE 2m6m

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Question 4c clockwise moments a if we take moments about A what moments do we take ? anti-clockwise moments b 6kN2kN R L 2m 4m R R 16kN ABCDE 2m6m

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Question 4d clockwise moments = 24 kNm a if we take clockwise moments about A what do we get ? clockwise moments = 92 kNm b clockwise moments = 52 kNm c 6kN2kN R L 2m 4m R R 16kN ABCDE 2m6m

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Question 4e calculate the anti-clockwise moments about A a what do we do now ? take moments about D b 6kN2kN R L 2m 4m R R 16kN ABCDE 2m6m

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Question 4f anti-clockwise moments = 2 kNm a what do we get ? anti-clockwise moments = 6 x R R kNm b 6kN2kN R L 2m 4m R R 16kN ABCDE 2m6m if we take anti-clockwise moments about A anti-clockwise moments = 8 x R R kNm c

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Question 4g 6kN2kN R L 2m 4m R R 16kN ABCDE 2m6m the anti-clockwise moments about A = 6R R what now? equate the clockwise and anti-clockwise moments a take moments about D b

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Question 4h 6kN2kN R L 2m 4m R R 16kN ABCDE 2m6m R R = 92 / 6 kNm, R R = 15.3 kN a R L = 92 / 6 kNm, R L = 15.3 kN b what do we get? when we equate the clockwise and anti-clockwise moments

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Question 4i 6kN2kN R L 2m 4m R R 16kN ABCDE 2m6m take moments about D a use one of our equations b what now? we have R R, we now want R L

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Question 4j 6kN2kN R L 2m 4m R R 16kN ABCDE 2m6m which one? we use one of our equations ΣH = 0, sum of all horizontal forces = 0 a ΣV = 0, sum of all vertical forces = 0 b ΣM = 0, sum of all moments = 0 c

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Question 4k 6kN2kN R L 2m 4m R R 16kN ABCDE 2m6m what do we get? using ΣV = 0, sum of all vertical forces = 0 R L – R R = 24, R L = 39.3 kN a R L + R R = 24, R L = 8.7 kN b R L + R R = 24, R L = 8.7 kNm c

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Question 5a what is tending to overturn the building? 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 the wind force of 16 kN a the weight of the building, 20 kN b the moment caused by the wind force c 3m 2m 16kN Weight 20kN A

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Question 5b what is the overturning moment? 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 16 kN a 32 kN b 32 kNm c 3m 2m 16kN Weight 20kN A

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Question 5c what is preventing the building from overturning? 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 the weight of the building a the moment caused by the weight of the building b 3m 2m 16kN Weight 20kN A

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Question 5d what is this restraining moment? 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 20 kN a 60 kNm b 3m 2m 16kN Weight 20kN A 30 kNm c

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Question 5e will the building overturn? 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 yes a no b 3m 2m 16kN Weight 20kN A maybe c

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Question 5f how can we make the building safer? 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 increase the weight of the building a lower the height of the building b 3m 2m 16kN Weight 20kN A make the base wider c all of the above d

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

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let me try again let me out of here why did you pick that?

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

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

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next question enough ! (200 N x 2 m) + (100 N x 6 m) = 1000 Nm = 1 kNm

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let me try again let me out of here we are talking about moments not just forces. Remember what a moment is.

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

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next question enough ! now that we’ve got the clockwise moments we want to equate them to the anti-clockwise moments to see what we get.

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let me try again let me out of here would be useful if we were after the reaction force at B

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next question enough ! (W N x 4 m) = 4W Nm

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let me try again let me out of here how did you get that? Remember what a moment is.

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let me try again let me out of here how did you get that? Remember what the units for moments are. M = F x D

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next question enough ! clockwise moments = anti-clockwise moments about a point if the system is in equilibrium

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let me try again let me out of here what would that give us?

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next question enough ! 4 W = 1000, so W = 250 and W is a force So units are Newtons

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let me try again let me out of here is W a moment or a force? So what are its units?

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next question enough ! pin joint = 2 reactions + simple support = 1 reaction gives 3 reactions in total. That means it is statically determinate we can use the 3 equations of static equilibrium to determine the reactions

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let me try again let me out of here how many reactions does a pin joint produce? How many does a simple support? So how many are there in total?

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next question enough ! you can start either with equating the sum of the reactions to the total load or you can find the sum of the moments. This will give you the sum of the reactions However, to actually find one of the reactions you should start by taking moments

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let me try again let me out of here what will that give you? Are there any horizontal forces?

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next question enough ! the sum of the upward forces = the sum of the downward forces total reaction forces = total load forces

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let me try again let me out of here how did you get that? total upward forces = total downward forces

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next question enough ! now we have to actually determine one of the reactions

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let me try again let me out of here c’mon, you can do it

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let me try again let me out of here what will that give you? Are there any horizontal forces?

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next question enough ! before we take moments we need to decide on the right point about which we take the moments

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let me try again let me out of here where are you going to take these moments about?

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next question enough ! we should take moments about a point so as to eliminate one of the unknown reactions. Remember that there is no moment about a point through which the force passes

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let me try again let me out of here you could – but you would still be left with both unknown reactions

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next question enough ! that’s right. It doesn’t matter which we do first

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let me try again let me out of here yes that’s fine but …..

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next question enough ! exactly !!

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let me try again let me out of here this is just a force

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let me try again let me out of here we did say clockwise moments

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let me try again let me out of here we did say clockwise moments – not sum the loads

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next question enough ! exactly !! The moment is F (R R kN) x D (6.5 m)

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let me try again let me out of here this is just a force

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let me try again let me out of here the units are ok but this is just a force Think! what is a moment

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let me try again let me out of here But R L goes through point A So what does that mean about the moment of R L about A?

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next question enough !

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let me try again let me out of here we’ve got everything we need for the moment!

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let me try again let me out of here you’re nearly there

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next question enough ! we’re nearly at the end

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let me try again let me out of here Where did you get 28 from? What was the sum of the clockwise moments?

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let me try again let me out of here R R is a force. What are the units of a force?

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next question enough ! We could take moments about RR (and maybe you should for practice’s sake) But this is simpler

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let me try again let me out of here but there’s an easier way (Actually you should for practice’s sake)

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next question enough ! 28.00 = 16.15 + R L R L = 11.85

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let me try again let me out of here How did you get that? ΣV = 28 So how can R L be more than 28?

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next question enough ! 3 possible reactions means that the system is statically determinate

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let me try again let me out of here I would have thought you would have this pat by now How many possible reactions are there?

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next question enough ! good one !!

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let me try again let me out of here but then what ?

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let me try again let me out of here but we need to do something first ?

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next question enough ! yes ! 3 kN per m times 8 m = 24 kN

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let me try again let me out of here how did you arrive at that ? The UDL is 3 kN/m and the length is 8 m

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let me try again let me out of here I thought we had this straight We are after a force !

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next question enough ! yes ! Uniformly distributed means that we have symmetry. So it is equally balanced

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let me try again let me out of here how did you arrive at that ? It is a uniformly distributed load, i.e. UDL over the beam. So ?

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next question enough ! yes ! Force (24 kN) x Distance (4 m) = 96 kNm 8m M 4m ABC 24 kN

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let me try again let me out of here how did you arrive at that ? M = F x D What is the force? What is the distance?

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let me try again let me out of here I thought we had this straight. What are the units for moments M = F x (times) D

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next question enough !

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let me try again let me out of here how would we do that?

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next question enough ! yes ! This way we eliminate one unknown force

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let me try again let me out of here what would that give us? Wouldn’t we still have two unknown forces?

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next question enough ! either will do

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next question enough ! (6 x 2) kNm + (16 x 4) kNm + (2 x 8) kNm = 92 kNm

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let me try again let me out of here you can’t just sum the forces. Try summing the moments

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let me try again let me out of here How did you get that What are the forces? How far is each force from A?

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next question enough ! Doing well!

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let me try again let me out of here We haven’t finished with A yet

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next question enough ! Force = R R kN and distance = 6 m

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let me try again let me out of here What’s the force? What’s the distance of this force from A? THINK

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let me try again let me out of here How did you get that? How far is R R from A?

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next question enough ! well on the way !

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let me try again let me out of here we haven’t finished getting a value for R R

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next question enough ! Doing well!

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let me try again let me out of here Didn’t we say that the anti-clockwise moments = 6 x R R kNm

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next question enough ! that’s why we have the equations

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let me try again let me out of here We could … but there is a simpler way ( actually you should for practice)

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next question enough ! makes life simple

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let me try again let me out of here do we have any horizontal forces?

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let me try again let me out of here we’ve already done that

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next question enough ! another question bites the dust

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let me try again let me out of here what is the total load? How can the reaction exceed the total load?

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let me try again let me out of here Haven’t we got this straight yet? Isn’t the reaction a force? What are the units for force?

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next question enough ! Moments turn. The wind force produces a moment which tends to overturn the building

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let me try again let me out of here forces don’t overturn things What does overturn things?

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let me try again let me out of here how exactly is the weight going to overturn the building? Actually, the weight is our friend

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next question enough ! OTM = F x D = 16 kN x 2 m = 32 kNm

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let me try again let me out of here this just the force. What happened to the moment?

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let me try again let me out of here By now you should know the units for moments. You should go back to the fundamental concepts M = F x D

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next question enough ! yes! Moments turn things !

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let me try again let me out of here Repeat after me forces do not turn things !! …..

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next question enough ! RM = F (20 kN) x D (1.5 m) = 30 kNm

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let me try again let me out of here This is just the weight (force). We are talking about moments

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let me try again let me out of here Where is the weight acting? How far is this from A?

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next question enough ! OTM = 32, RM = 30, OTM > RM

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let me try again let me out of here what is the OTM? what is the RM? which is greater?

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let me try again let me out of here there’s no maybe about this ! Think again

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the end ! well done ! (i) increase the weight of the building to at least 32/1.5 = 21.4kN (ii) lower the height of the building so that the action of the wind is lowered by at least 125 mm (to 1.875m above ground) (iii) make the building base wider - at least 3.2 m wide (iv) fix it to the ground. It will act as a cantilever

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let me try again let me out of here is that it?

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