An I-beam section is of 100 mm width flange and 20 mm thick web. Under a given shear force, the increase in the shear stress at the junction of flange and web would be

80 times 120 times 2000 times 5 times

Concept:

Shear stress at the Junction of flange and web is given by,

τ = $\frac{F A Y ?}{I b}$

where,

F = Force applied

A = Area of the flange, I = Moment of Inertia of I section, be t= width of the web.

Y? = Distance of centroid of the flange from N.A.

Calculation:

Let shear stress at flange at the junction of flange and web = τ₁

width of flange, b₁ = 100 mm

width of the web, b₂ = 20 mm

Let shear stress at the web at the junction of flange and web = τ₂

Then,

τ₁ = $\frac{F A Y ?}{I b_{1}}$ ------- 1

τ₂ = $\frac{F A Y ?}{I b_{2}}$ --------2

$\frac{τ_{1}}{τ_{2}} = \frac{b_{2}}{b_{1}}$

$\frac{τ_{1}}{τ_{2}} = \frac{20}{100}$

τ₂ = 5 τ₁

Hint Answer

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An I-beam section is of 100 mm width flange and 20 mm thick web. Under a given shear force, the increase in the shear stress at the junction of flange and web would be

5 times

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