Shearing Force Direction and Sign Convention

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Discussion Overview

The discussion revolves around the concepts of shearing force direction and sign convention in the context of beam analysis, particularly in relation to a homework problem involving forces and moments. Participants explore the implications of different assumptions regarding force directions and the calculations of shear forces and moments at specified points along the beam.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions why a 10kN force is ignored for the range 0 < x < 3, suggesting that its position should affect the analysis.
  • Another participant inquires whether the downward force is assumed to be positive, based on the author's treatment of the shearing force.
  • A participant states that for 0 < x < 3, the shearing force V is 4kN, indicating that since the support force at A is upwards, the shearing force should be downwards, raising questions about the sign convention used.
  • There is a question regarding the moment calculation for 3 < x < 5, specifically asking at which point the moment is being taken.
  • One participant explains that making an imaginary cut in the beam allows for the determination of shear force direction, stating that if the shear force is upwards to the left of the cut, it is positive by convention.
  • Another participant reiterates the importance of equilibrium in beam analysis, emphasizing that the shear force and bending moment on either side of the cut must be equal and opposite.
  • A further response highlights that if internal forces are not equal and opposite, it could lead to beam failure, reinforcing the concept of equilibrium in the analysis.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the sign convention for forces and the implications of ignoring certain forces in the analysis. There are multiple viewpoints on the treatment of forces and moments, and the discussion remains unresolved with respect to these conventions.

Contextual Notes

Participants have not fully clarified the assumptions underlying the sign conventions or the specific points at which moments are calculated, leaving some mathematical steps and definitions unresolved.

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Homework Statement


why for 0 <x <3 , why the author ignore the 10kN ? as we can see , the 10kN is located 3m from A ...
2nd question is , the author assume downwards force is positive , am i right ?
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Homework Equations

The Attempt at a Solution


for 0 <x <3 , the shearing force V = 4kN , as we can can see , at A the support force (4kN) is pointing upwards , so the shearing force 4kN should be pointing downwards , am i right ? so , the author assume downwards force is positive??
if so , why in the shearing force diagram (SFD) , the 4kN is at positive x direction , shouldn't it at 4kN bottom of the x-axis ?
 
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one more question , why the moment at 3 <x<5 is 4x - 10(x-3) , the author taking moment at which point ?
 
Make an imaginary cut in the beam. If the shear force is pointing up to the left of the cut then shear force is positive by convention.

Everything to the right of the cut we ignore (for purposes of calculating shear force) because we know beam is in equilibrium. That means you can automatically assume whatever shear force (or bending moment) exists on the left face of the cut, it is equal & opposite to that which exists on the right face.
 
Last edited:
David Lewis said:
Make an imaginary cut in the beam. If the shear force is pointing up to the left of the cut then shear force is positive by convention.

Everything to the right of the cut we ignore (for purposes of calculating shear force) because we know beam is in equilibrium. That means you can automatically assume whatever shear force (or bending moment) exists on the left face of the cut, it is equal & opposite to that which exists on the right face.
Why??
 
Because if internal forces are not equal & opposite, the beam will fail. Any imaginary slice will render mirror image shear, torsion, tension and compression forces between the left and right faces of that section.
 

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