Area of moment diagram

  • Thread starter chetzread
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  • #1
801
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Homework Statement


can someone explain about the area of moment diagram ? taking the circled part as example , why it's 0.5(2)(800)(4/3) ?why shouldnt it be (800)(4/3) ??
 

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  • #2
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why there's a need to include 0.5(2) ???
 
  • #3
PhanthomJay
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I haven't done this in a long long time, but this is part of a method to determine deflections. The method involves computing the moment of the area of the moment diagram about the reactions. The area of the triangle is one half the base (one half of 2) times the height (800). It's centroid is 4/3 from left end. Result is in Nm^3.
 
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  • #4
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We are looking for the moment area so the formula would be (area of triangle)(centroidal moment of area) and that would be (0.5)(2)(800) as triangle area and (4/3) as the centroid.
 
  • #5
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We are looking for the moment area so the formula would be (area of triangle)(centroidal moment of area) and that would be (0.5)(2)(800) as triangle area and (4/3) as the centroid.
why the author taking the moment about B and 2m from A ? (in the first solution , the author taking moment about the point which 2m from A )( in the second solution , the author taking moment about B , am i right ?
why the author didnt take the moment about A ?
 

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  • #6
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why the author didnt take the moment about A into consideration ???
 
  • #7
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He considered that is why there are two solutions (AreaAB)XA and (AreaAB)XB denoting two moment areas
 
  • #8
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He considered that is why there are two solutions (AreaAB)XA and (AreaAB)XB denoting two moment areas
i dont understand , can you explain further?
 
  • #9
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why the author wanna calculate the area moment by 2 method which is (area AB) x A and (area AB) x B ???
What's the purpose of doing so ?
 
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  • #10
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and I also noticed that the author didnt include the moment 400Nm in the first solution,is it wrong?
 
  • #11
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You should review how to draw moment diagrams. It is already known that when we derive from shear to moment diagrams , any given moment is considered a straight line thus no area and that is why it isn't included in computing the moment area.
 
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  • #12
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You should review how to draw moment diagrams. It is already known that when we derive from shear to moment diagrams , any given moment is considered a straight line thus no area and that is why it isn't included in computing the moment area.
then, why the author include the moment 400Nm in the second solution ????
 
  • #13
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Again , you should read on how to solve moment areas. The 2nd solution is the separate areas for every reaction or simply it is another way to solve the problem. Also you could see that he took the moment area about A to solve the problem and that is why there is no moment diagram for reaction A.
 
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  • #14
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Again , you should read on how to solve moment areas. The 2nd solution is the separate areas for every reaction or simply it is another way to solve the problem. Also you could see that he took the moment area about A to solve the problem and that is why there is no moment diagram for reaction A.
what do you mean by separate areas for every reaction???
 
  • #15
801
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the second solution is taking moment about B ? so the 400Nm is included?
While the first solution is taking moment about the point which has 400Nm moment , that's why the moment is not included in the first solution , since dM/ dx = 0 ?
 

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