What Are the Bending Moment Equations for Sections AB and BC?

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SUMMARY

The discussion focuses on deriving the bending moment equations for sections AB and BC of a beam under specific loading conditions. The reactions at supports A and C were recalculated, resulting in values of 25 kN and 10 kN, respectively. The bending moment equation for the section from 0 to 1 meter is established as 25x - 5x². Participants emphasize the importance of verifying reaction forces and moments about various points to ensure accuracy in calculations.

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  • Understanding of static equilibrium principles, including force and moment balance.
  • Familiarity with bending moment equations and shear force diagrams.
  • Knowledge of beam loading conditions, including point loads and distributed loads.
  • Ability to perform calculations involving reaction forces at supports.
NEXT STEPS
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  • Study the method of sections for calculating internal forces in beams.
  • Explore the effects of different load types on bending moment equations.
  • Investigate the use of software tools like SAP2000 for structural analysis.
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Structural engineers, civil engineering students, and anyone involved in beam analysis and design will benefit from this discussion.

mm391
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Homework Statement



Find the equations for the bending moment along AB and BC?
Reactions at A = -15kN
Reactions at B = 60kN

Homework Equations



F in the Y direction = 0
Moments with clockwise being positive = 0

The Attempt at a Solution



Moments about cross-section from the left before the point load (clockwise is +ve)= 15x-10(x/2)-M=0
M=10x

Moments about cross section-section from the left past the point load
(clockwise is +ve) = 10x-15(x-1)-10((x-1)/2)-M=0 M=-10x-20

Moments about cross-section from RIGHT hand side = M+60(3-x)+10(3-x)((3-x)/2)=0
M= -5(x^2-18x+45)

As I have no answers I am not sure if these are correct or where to go from here?

Any help would be greatly appreciated.

Thanks

Matt
 

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Show your calcs for the reaction forces. They should both be positive upward and sum to 45.
 
In these problems you can check it yourself. Sum of vertical loads = sum of vertical reactions? Checks ok. BUT Sum of moments about any point must be zero. Emphasise any. Do you, for example, have a balance of moments about C? In checking this, make sure that, whatever sign convention you adopt, clockwise moments balance anti-clockwise moments. This should reveal that your reactions are in error. You could have (should have) checked this before attempting any bending moment equation. I suggest you start again by getting reactions right, and check for your self they are correct by checking moments balance about ANY point. Do this for several points to make really sure.
 
My fault. Have just calculated the reaction forces again:

ƩMoments about A (clockwise +ve) = (15*1)+((10*3)/2)+Reactions at C*3)=0

∴ Reactions at C = 10kN

ƩForces in the Y direction = Reactions at A - 15 - 30 + Reactions at C = 0

Reactions at C = 10

∴ Reactions at A = 35kN
 
mm391 said:
My fault. Have just calculated the reaction forces again:

ƩMoments about A (clockwise +ve) = (15*1)+ ((10*3)/2) [/color]+Reactions at C*3)=0

∴ Reactions at C = 10kN
you are incorrecty calculating the moment from the distributed load. When determining force reactions, the distributed load can be represented as a single resultant concentrated load acting at its center of gravity. The uniform load is 10 kN/m over the 3 m length of beam. So its total resultant load is ___?___ and its moment arm is ___?___
ƩForces in the Y direction = Reactions at A - 15 - 30 + Reactions at C = 0

Reactions at C = 10

∴ Reactions at A = 35kN
Your error carries over when doing it this way. Sum moments about C instead, then check your work to be sure ƩForces in y direction = 0.
 
Jay's advice good. Please check your answer by taking moments about any point you haven't used before. This is what engineers have to do in practical cases, because real life examples don't have an answer "in the back of the book". A good way to get reactions in a case like this is to take moments about A to get reaction at C. Then take moments about C to get reaction at A. Then check that vertical forces are balanced.
 
So I find the Reactions to be;

Reactions at A = 25kN
Reactions at B = 20kN

Then the bending moment equation from 0<x<1 meter:

25x-5x^2

Now I am unsure how to carry on, do I make another cut before Reactions at B and take moments then after solving that make another cut and take moments from the left hand side. Sorry I know this may sound vague but I am not sure how to explain my method. I have tried it before but when I look at it none of the bending moments equaitons seem to incude the 15kN load. Should I have three bending moment with one from 0 to 1m, one from 1m to 1<x<3 and then find another eqations.
 
Reactions are at A and C, not A and B.
Your moment equation between x = 0 and x= 1 is OK.
Between x = 1 and x = 3, there are several ways to find the moment. One way is to cut the beam to the right of B and use the left section as your FBD and sum moments about the cut. Note that since the distance from A to the cut is x, then for example the distance from B to the cut is (x - 1), etc.
 

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