Shear Flow in Beams

Example: (continued)

Recall that I is the moment of inertia of the entire x-section about the neutral axis, z.

So first find the location of the neutral axis, z.

(A₁ + A₂) y_sect = A₁ y₁ + A₂ y₂

(200)(50) + (200)(50)] y_sect = (200)(50)(100) + (50)(200)(200 + 25)

y_sect = 162.5 mm

Now Q = A’y_bar here A’ = (200)(50) = 10,000 mm²

and y_bar = 225 – 162.5 = 62.5 mm

So Q = A’y_bar = (200)(50)(62.5) = 0.625 x 10⁶ mm³

Now use the parallel axis theorem. I = I₁ + A₁d₁² + I₂ + A₂d₂² where

I₁ = (1/12)(50)(200)³ = 33.33 x 10⁶ mm⁴ , A₁ = (50)(200) = 10000 mm², d₁ = 62.5 mm

I₂ = (1/12)(200)(50)³ = 2.083 x 10⁶ mm⁴ , A₂ = (50)(200) = 10000 mm², d₂ = 62.5 mm

I = 113.5 x 10⁶ mm⁴ and t = 50 mm

Now stress τ = VQ/It = (1600) ( 0.625 x 10⁶ ) / (113.5 x 10⁶ )(50)

The resulting shear stress in the glue is τ = 0.176 N/mm²

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