Strength of materials (shear and moment)

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SUMMARY

The discussion focuses on the analysis of shear and moment in structural engineering, specifically addressing the calculation of reactions at supports A and B, which are 3.4375 kN and 5.0625 kN, respectively. The shear diagram was incorrectly drawn by treating concentrated loads as distributed loads, leading to discrepancies in moment calculations at points A and B. The correct shear values at point C should reflect a jump discontinuity due to a 2 kN concentrated load, resulting in shear values of 2.9375 kN from the left and 0.9375 kN from the right. This highlights the importance of accurately representing load types in shear and moment diagrams.

PREREQUISITES
  • Understanding of shear and moment diagrams in structural analysis.
  • Familiarity with concentrated and distributed loads in engineering.
  • Knowledge of basic equilibrium equations in statics.
  • Proficiency in calculating reactions at supports in beam structures.
NEXT STEPS
  • Study the principles of shear and moment diagrams in structural analysis.
  • Learn how to differentiate between concentrated and distributed loads in load analysis.
  • Explore advanced topics in beam deflection and stability analysis.
  • Review case studies involving shear and moment calculations in real-world structures.
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Structural engineers, civil engineering students, and professionals involved in the design and analysis of beam structures will benefit from this discussion.

defdek
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R at B = 5.0625
R at A = 8.5kN-5.0625kN=3.4375kN

x/0.9375=(0.5-x)/4.5625

x=15/176

but moment near x there i get different from both side:
from A = 2.431kNm
from B = 2.346kNm

is there something wrong?
 

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Your shear curve is not drawn correctly. For example, starting at A, the reaction gives a shear of 3.4375 kN. The distributed load between A and C is 1 kN/m over a distance of
0.5 m. At C, there should be a jump discontinuity in the shear. Approaching from the left, the shear is 3.4375 - 0.5 * 1 = 2.9375 kN. The 2 kN concentrated load is applied at point C only. The shear value at C approaching from the right is 0.9375 kN. A similar situation will occur at point E with the 5 kN concentrated load there.

In other words, your shear diagram is incorrect because you have incorporated the concentrated loads as if they were distributed loads.
 
SteamKing said:
Your shear curve is not drawn correctly. For example, starting at A, the reaction gives a shear of 3.4375 kN. The distributed load between A and C is 1 kN/m over a distance of
0.5 m. At C, there should be a jump discontinuity in the shear. Approaching from the left, the shear is 3.4375 - 0.5 * 1 = 2.9375 kN. The 2 kN concentrated load is applied at point C only. The shear value at C approaching from the right is 0.9375 kN. A similar situation will occur at point E with the 5 kN concentrated load there.

In other words, your shear diagram is incorrect because you have incorporated the concentrated loads as if they were distributed loads.

Oh I didnt see that. Thank you so much... :D
 

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