What is the relationship between moment diagrams and the area under the curve?

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SUMMARY

The discussion focuses on the calculation of moment areas in structural analysis, specifically using the area of a moment diagram to determine deflections. The area of a triangle in the moment diagram is calculated as 0.5 times the base (2m) times the height (800Nm), with the centroid located at 4/3 from the left end. The conversation highlights the importance of considering different points for taking moments, such as points A and B, and clarifies why certain moments, like 400Nm, are included or excluded in different solutions. The method emphasizes the need to compute the moment of the area about the reactions for accurate deflection analysis.

PREREQUISITES
  • Understanding of moment diagrams in structural analysis
  • Knowledge of calculating areas of geometric shapes, specifically triangles
  • Familiarity with centroid calculations in engineering mechanics
  • Basic principles of static equilibrium and moment calculations
NEXT STEPS
  • Study the derivation of shear and moment diagrams in structural analysis
  • Learn about calculating deflections using the moment-area method
  • Explore the significance of centroidal moments in engineering applications
  • Review examples of moment calculations involving multiple reactions and their areas
USEFUL FOR

Structural engineers, civil engineering students, and professionals involved in analyzing and designing structures will benefit from this discussion, particularly those focused on deflection calculations and moment diagram interpretations.

chetzread
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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 shouldn't it be (800)(4/3) ??
 

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why there's a need to include 0.5(2) ?
 
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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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.
 
Justus said:
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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why the author didnt take the moment about A into consideration ?
 
He considered that is why there are two solutions (AreaAB)XA and (AreaAB)XB denoting two moment areas
 
Justus said:
He considered that is why there are two solutions (AreaAB)XA and (AreaAB)XB denoting two moment areas
i don't understand , can you explain further?
 
why the author want to 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
and I also noticed that the author didnt include the moment 400Nm in the first solution,is it wrong?
 
  • #11
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
Justus said:
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
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
Justus said:
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
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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