Bending moments diagram at T intersection

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SUMMARY

The discussion focuses on calculating axial force, shear force, and bending moment diagrams at a T intersection in structural analysis. The user is uncertain about how to approach the intersection at point D, specifically whether to evaluate it from D-E while treating the post from D-F as a pin joint. A solution is provided, indicating that section DE acts as a cantilever, allowing for straightforward calculations of shear and moment at that section using a Free Body Diagram (FBD).

PREREQUISITES
  • Understanding of Free Body Diagrams (FBD)
  • Knowledge of shear force and bending moment calculations
  • Familiarity with cantilever beam behavior
  • Basic principles of structural analysis
NEXT STEPS
  • Study the principles of Free Body Diagrams in structural mechanics
  • Learn how to calculate shear and moment diagrams for cantilever beams
  • Explore the concept of pin joints in structural analysis
  • Review examples of T intersections in beam analysis
USEFUL FOR

Structural engineers, civil engineering students, and anyone involved in analyzing forces and moments in beam structures will benefit from this discussion.

mackhina
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Homework Statement



Draw the axial force, shear force and bending moment diagrams. Show the locations and magnitude of the maximum and minimum values.

Homework Equations



See diagram.

The Attempt at a Solution



I've worked out the values from points A-B and then from points B-D. I don't grasp how I am suppose to approach the T intersection at point D. Should I evaluate it from D-E and treat the post from D-F as a pin joint, with values of 36.13kN and 15.13kN?

Then do the joint D-F after? If the D-E sections wasn't there I would just continue working around from D-F next, but because it is I'm not sure of how to work it out?

Any help would be heaps appreciated. Thanks a lot!

Cheers
 

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Section DE is just a cantilever. If you cut a section thru DE in a FBD that includes the 10 kN force, the shear and moment at that section are easily calculated (no axial force). In DF, your FBD is similar to that of AB ( but with different values).
 

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