Understanding Moment in Beam: Analysis and Solution | Beam Homework

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SUMMARY

This discussion focuses on the analysis of bending moments in beams, specifically addressing the moment function EIy" as it relates to a distributed load of 300 N/m. The author’s approach is validated, demonstrating that the distributed load begins at x = 0 and extends to x = 2, followed by an equal and opposite load beyond this point, resulting in a net load of zero. Participants express confusion regarding the formulation of the bending moment, particularly the terms -0.5(300)(x^2) and +0.5(300)[(x-2)^2], which are crucial for accurate calculations in beam analysis.

PREREQUISITES
  • Understanding of beam mechanics and bending moment theory
  • Familiarity with distributed loads and their effects on structures
  • Knowledge of the moment function EIy" in structural analysis
  • Proficiency in algebraic manipulation of equations related to beam loading
NEXT STEPS
  • Study the derivation of bending moment equations in beam theory
  • Learn about the principles of static equilibrium in structural analysis
  • Explore the application of the superposition principle in beam loading scenarios
  • Investigate the use of software tools for beam analysis, such as SAP2000 or ANSYS
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Structural engineers, civil engineering students, and professionals involved in beam design and analysis will benefit from this discussion. It provides insights into the complexities of bending moment calculations and the importance of understanding distributed loads.

chetzread
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Homework Statement


refer to the circled part , i don't understand the author's working .
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Homework Equations

The Attempt at a Solution


I think it should be 450x - 150(4-x) -30092)(x-1) , correct me if I'm wrong . (taking moment about x ) ...
 
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is the author wrong , coz the distributed load start from R1 to 2m from R1 ...
 
chetzread said:
is the author wrong , coz the distributed load start from R1 to 2m from R1 ...
No, the author is not wrong.

The bending moment function EIy" is carefully constructed to show that the distributed load starts at x = 0 and runs to x = 2, after which an equal and opposite distributed load is applied to the beam at x > 2, so that the net distributed load is zero.
 
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SteamKing said:
No, the author is not wrong.

The bending moment function EIy" is carefully constructed to show that the distributed load starts at x = 0 and runs to x = 2, after which an equal and opposite distributed load is applied to the beam at x > 2, so that the net distributed load is zero.
why ? i still don't understand , i can only understand the 450x , why should be -0.5(300)(x^2) +0.5(300)[(x-2)^2 ]
 
SteamKing said:
No, the author is not wrong.

The bending moment function EIy" is carefully constructed to show that the distributed load starts at x = 0 and runs to x = 2, after which an equal and opposite distributed load is applied to the beam at x > 2, so that the net distributed load is zero.
why shouldn't it be 450x - 150(4-x) -300(0.5)(2)(x-1) ??
 
chetzread said:
why shouldn't it be 450x - 150(4-x) -300(0.5)(2)(x-1) ??
I'm not going to speculate on that.

Look, the author has a method of analysis of this beam which he is trying to teach. If you want to develop your own methods, write your own textbook.
 
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SteamKing said:
I'm not going to speculate on that.

Look, the author has a method of analysis of this beam which he is trying to teach. If you want to develop your own methods, write your own textbook.
i don't understand why it is -0.5(300)(x^2) +0.5(300)[(x-2)^2 ] , can you explain ?
 
SteamKing said:
No, the author is not wrong.

The bending moment function EIy" is carefully constructed to show that the distributed load starts at x = 0 and runs to x = 2, after which an equal and opposite distributed load is applied to the beam at x > 2, so that the net distributed load is zero.
why it's 0.5(300)(x^2) ? the 300N/m only 'defined ' from R1 to 2m from R1, am i right ? why shouldn't it be 0.5(300)(2)(x-2) instead? as you said , after 2m , net distributed load is zero
 
SteamKing said:
I'm not going to speculate on that.

Look, the author has a method of analysis of this beam which he is trying to teach. If you want to develop your own methods, write your own textbook.
the location of 300n/m is at R1 and extend to 2m from R1, so ,the EIy" should be = 450x-300(2)(x-1) ?
why there's -0.5(300)(x^2) +0.5(300)( (x-2)^2 )?
I'm confused

Or do you mean the 300(X) force still exist at X >2?
 
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