Shear stress at centroid vs other point

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SUMMARY

The discussion centers on the calculation of shear stress at the centroid versus point E in a structural analysis problem. The user calculated the shear stress at the centroid as 1.9 x 107 Pa and at point E as 3.6 x 107 Pa, which contradicts the expectation that shear stress at the centroid should be maximum. The discrepancy arises due to the larger thickness at the centroid and a sharp decrease in shear stress at point E caused by the geometry of the structure. This highlights the importance of understanding shear distribution in non-uniform cross-sections.

PREREQUISITES
  • Understanding of shear stress and its calculation in structural engineering
  • Familiarity with the formula τ = VQ / It for shear stress
  • Knowledge of centroid and moment of inertia calculations
  • Basic principles of structural mechanics and material behavior under load
NEXT STEPS
  • Study the effects of cross-sectional geometry on shear stress distribution
  • Learn about shear flow and its implications in beam theory
  • Explore the calculation of shear stress in non-uniform cross-sections
  • Investigate the role of thickness variations in shear stress analysis
USEFUL FOR

Structural engineers, civil engineering students, and anyone involved in analyzing shear stress in beams and other structural elements will benefit from this discussion.

fonseh
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[ Mod Note: moving this to Physics H/W ]

Homework Statement


for this question , I'm having problem with the shear stress at point E and shear stress at centorid.
normally , the shear stress at the centoid will be maximum .

But , in my working , I found that the shear stress at the centroid is smaller than the shear stress at E. What's wrong with the working ?

i get y coordinates of centorid = 66.7mm
For Ixx , i get (5.00x10^-5)(m^4) , For V(shear force ) , I use (437.5x10^3)N

For shear stress at centroid , i use formula of $$\tau = V(Q) / It $$

so at centroid , Q = (66.67x10^-3)(160x10^-3)(66.67x10^-3 / 2 ) = 3.56x10^-4
so $$\tau $$= (437.5x10^3)(3.56x10^-4) / (5.00x10^-5)(160x10^-3) = 1.9x 10^7 Pa

at E , Q = Ay = (40x10^-3)(80x10^-3)(53.33x10^-3)**(2)** = 3.41x10^-4

so , $$\tau $$= (437.5x10^3)(3.41x10^-4) / (5.00x10^-5)(80x10^-3) = 3.6x 10^7 Pa

For Q at E i have labelled it with the orange part ,
for Q at centroid , i have labelled it with the green part ...

Homework Equations

The Attempt at a Solution


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Aything wrong with the question given ?
I'm getting the shear stress at centroid lower than the other point ( point E ) , is it possible ?

Or the question gt error
?
 
Last edited:
Yes it is possible because the thickness at the centroid is large. At point E there is a sharp decrease in shear stress due to the sharp increase in thickness at the webs interface.
 
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