Bending shear stress distribution

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SUMMARY

The discussion centers on the shear stress distribution in a cantilever rectangular beam modeled in Patran using Hexa8 elements and analyzed with Nastran. The vertical shear stress distribution was found to be parabolic, with values varying from 5.15 at the center to 8.07 at the ends, contrary to the expected uniform distribution. The discrepancy was attributed to the effects of Poisson's ratio, as simulations with a Poisson ratio of zero resulted in a uniform shear stress distribution. The calculated maximum shear stress at the center was 6.25, closely aligning with the simulation results.

PREREQUISITES
  • Understanding of shear stress distribution in beams
  • Familiarity with finite element analysis (FEA) using Patran and Nastran
  • Knowledge of Poisson's ratio and its effects on stress distribution
  • Basic principles of cantilever beam mechanics
NEXT STEPS
  • Explore the effects of Poisson's ratio on shear stress in beam analysis
  • Learn about plane strain versus plane stress conditions in FEA
  • Investigate shear locking effects in higher-order elements like Hexa20
  • Study the implications of boundary conditions on stress distributions in cantilever beams
USEFUL FOR

Structural engineers, finite element analysts, and students studying beam mechanics and shear stress distribution in cantilever structures.

piygar
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I modeled a simple cantilever rectangular beam in Patran of (L,W,H = 100, 6, 4) with Hexa8 elements and applied a vertical shear load (upwards) of some magnitude at the free end.
Ran in Nastran and plotted the vertical shear stress distribution on a cross-section in middle of length (in order to avoid any boundary condition effects). which is parabolic as expected being zero at top and bottom edges and maximum at center.
But shear stress distribution is not uniform across width. I was expecting rectangular bands parallel to horizontal edges but i get curved bands across widths. what could be the reason of that?
Have attached the fringe plot of vertical shear stress distribution.
 

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  • Capture.PNG
    Capture.PNG
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The parabolic shear stress formula VQ/It assumes a uniform shear stress distribution along the width. The computer shows otherwise. What sort of a range of shear stresses do you get along the width at say the neutral axis? How do the values compare to the hand calculation?
 
Shear stress at neutral axis along the width varies from 5.15 at the center to 8.07 at the either end of the width.
Have inserted fringe plot with spectrum.
I applied 100 unit shear load at the cantilever tip so according to formula max shear stress at center should be 1.5* 100/(6*4) = 6.25

Capture.PNG
 
piygar said:
Shear stress at neutral axis along the width varies from 5.15 at the center to 8.07 at the either end of the width.
Have inserted fringe plot with spectrum.
I applied 100 unit shear load at the cantilever tip so according to formula max shear stress at center should be 1.5* 100/(6*4) = 6.25

View attachment 98605
Thanks for the data. Funny things happen at edges and corners and when beams are not narrow. Anyway, the 6.25 seems like a good weighted average. With the addition of a good catch-all safety factor, of course! Unless the cantilever is of short length, shear doesn't add much in comparison to bending stresses, and I've always ignored it in long cantilever designs using steel.
 
Thanks for reply. Wanted to understand the corners/edges effect clearly.
Also plotted the horizontal shear stress (due to vertical shear load!) on the same cross-section and got this.

Capture1.PNG

Horizontal shear is small but not numeric zero.
Discussed with a colleague and suspicion pointed to Poisson's ratio. Ran another simulation with Poisson ratio = zero and voila, Vertical shear stress is constant across width, and zero horizontal shear stress due to vertical shear load.
Since width of beam is not small,came to understanding that cross-section behaving like a plane strain condition instead of plane stress.
Though plane strain condition should produce secondary normal stress, not shear stress.
 
When you set Poisson's ratio equal to 0 and got uniform stress distribution across the width, how did the result compare to the VQ/It shear stress 6.25 value at the neutral axis?
 
It comes 6.12 at the neutral axis with zero Poisson ratio.
(Tried with higher order Hexa, Hexa20 elements and got same result, as i had shear locking effect in mind)
 

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