Shear Force Diagrams and Bending Moment Diagrams

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SUMMARY

This discussion focuses on the calculation of Shear Force Diagrams and Bending Moment Diagrams for a beam supported by pin jointed pivots, specifically with a span of 6 meters. The participant successfully calculated the internal moment as -56 kN/m and the shear force as -6 kN, but encountered difficulties in the subsequent section, where the internal moment was incorrectly calculated as 72 + v(6) kN/m and the shear force as -24 kN. Key steps include identifying resultant forces from uniformly and triangularly distributed loads, summing moments to find reactions, and drawing the shear diagram.

PREREQUISITES
  • Understanding of static equilibrium equations: ƩFy = 0, ƩFx = 0, ƩMP = 0
  • Knowledge of shear and bending moment calculations for beams
  • Familiarity with uniformly and triangularly distributed loads
  • Experience with drawing shear force and bending moment diagrams
NEXT STEPS
  • Learn how to calculate resultant forces from uniformly distributed loads
  • Study the principles of triangularly distributed loads and their effects on beams
  • Practice drawing shear force and bending moment diagrams for various loading conditions
  • Explore the use of software tools for structural analysis, such as SAP2000 or AutoCAD
USEFUL FOR

Students in civil or structural engineering, professionals involved in structural analysis, and anyone seeking to understand the principles of shear and bending moment calculations for beams.

nobodyuknow
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Homework Statement


http://img189.imageshack.us/img189/9131/22465062cfdd4934b51d7f5.png

Hopefully the image isn't too difficult to decipher... Basically a beam balanced on to pin jointed pivots with 6 metres in between each one.

Homework Equations



ƩFy = 0
ƩFx = 0
ƩMP = 0

The Attempt at a Solution



I've solved the first section of the part...
The internal moment comes out to be -56kN/m
Axial Force is 0
And Shear Force is -6kN

Calculating the second section...
I'm stuck here, hence, why I don't think I can proceed with Section 3 either
I calculated the Internal Moment to be... 72 + v(6) kN/m (Which I think is wrong)..
The Shear Force came out to be -24kN
Axial Force is... 0?

Need help with completing the rest
 
Last edited by a moderator:
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You need to be able to identify the resultant force and location of the uniformly distributed load and the resultant force and location of the triangularly distributed load. Then solve for the reactions next by summing moments. Then draw the shear diagram. Practice with a more simple case first, like a uniform load on a simply supported beam. Triangularly distributed loads add a level of difficulty to the problem.
 

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