Maximum Load on Beams: Formula & Share/Bend Moment Diagrams

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SUMMARY

The maximum load that can be supported by a beam, whether simply supported, cantilevered, or fixed, can be calculated using established formulas. Key equations involve solving for static equilibrium and determining the maximum moment in the beam, which is then used in the stress bending equation: stress = M*y/I. Recommended resources for further study include "Roark's Equations for Stress and Strain" and various Mechanics & Strengths of Materials textbooks. Additionally, online resources such as efunda provide valuable information on buckling and beam analysis.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Familiarity with stress and strain equations
  • Knowledge of moment of inertia calculations
  • Basic concepts of beam types and loading conditions
NEXT STEPS
  • Study "Roark's Equations for Stress and Strain" for comprehensive formulas
  • Learn about beam deflection calculations for various loading scenarios
  • Research buckling analysis techniques for columns and beams
  • Explore online resources like efunda for practical examples and formulas
USEFUL FOR

Engineers, structural analysts, and students in civil or mechanical engineering who are focused on beam design, load analysis, and structural integrity assessments.

topito2
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Do you know if there is a formula for the maximum load that can be supported by a beam? The beam could be a simply supported beam, cantilevered, or fixed on both ends.
By the way, is finding the equations for share and bending moment diagrams an undetermined problem?
 
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topito2 said:
Do you know if there is a formula for the maximum load that can be supported by a beam? The beam could be a simply supported beam, cantilevered, or fixed on both ends.
Yes there are. Are you talking horizontal or vertical orientation of the principal axis? There are load-deflection formulae for any number of beams, load and load distributions, and end-point or boundary conditions.

By the way, is finding the equations for share and bending moment diagrams an undetermined problem?
There are already standard formulas.
 
Thank you for your answer. Do you recommend any book or website where I could look for further info regarding this topic?
 
"Roarks Equations for Stress and Strain" is the most comprehensive book you'll ever find for stress and strain equations, but the information you seek can be found in any Mechanics & Strengths of Materials textbook. Solve for static equilibrium and find the maximum moment in the beam in terms of force. Plug it into your stress bending equation stress=M*y/I, and solve for force with stress = to yield strength or UTS (depending on which is defined as failure). y=furthest distance from neutral axis of the beam, M is your maximum moment, and I is the moment of inertia of the beam cross section about the neutral axis. It's slightly more difficult if the beam cross section is not uniform throughout but not by much.
 
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topito2 said:
Thank you for your answer. Do you recommend any book or website where I could look for further info regarding this topic?
Roark's is one possiblity.

Then many textbooks and on-line resources have forumulas.

For example - buckling of a column (efunda has a limited number of samples, then one has to register, but there is a lot of valuable information there)
http://www.efunda.com/formulae/solid_mechanics/columns/columns.cfm

http://www.tech.plym.ac.uk/sme/desnotes/buckling.htm

http://www.diracdelta.co.uk/science/source/b/u/buckling load/source.html

Then there are more complicated systems - http://www.ce.washington.edu/em03/proceedings/papers/84.pdf

And of course, for horizontal beams and distributed load, there are formulas for maximum deflection and corresponding stresses in various types of beams and cross-sections. The peak load would be such that some location in the beam would exceed UTS and the beam would progress to failure. However, in practice, given various uncertainties in dimensions and material properties, and tolerances, one normally designs to some safety margin in order to avoid inadvertent failure.
 
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Thanks for all your answers. You have been very helpful.
 

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