The flange and web plates of the doubly symmetric built-up section are connected by continuous 10 mm thick fillet welds as shown in the figure not drawn to the scale- The moment of inertia of the section about its principal axis X-X is 7-73 - 106 mm4- The permissible shear stress in the fillet welds is 100 N-mm2- The design shear strength of the section is governed by the capacity of the fillet welds-The maximum shear forcein kN- round off to one decimal place that can be carried by the section- is

Throat thickness =0.7× 10 =7mm Permissible shear stress =VAyIb Here I = 7.73× 10^6mm^4 b = 4× 7=28mm Ay=100× 10× 55=55000 mm^3 So, permissible shear stress 100 ...

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Standard XII

Physics

Moment of Inertia of a Rod

The flange an...

Question

The flange and web plates of the doubly symmetric built-up section are connected by continuous 10 mm thick fillet welds as shown in the figure (not drawn to the scale). The moment of inertia of the section about its principal axis X-X is 7.73 $\times 10^{6} m m^{4}$ . The permissible shear stress in the fillet welds is $100 N / m m^{2}$ . The design shear strength of the section is governed by the capacity of the fillet welds.
The maximum shear force(in kN, round off to one decimal place) that can be carried by the section, is

393.5

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Solution

The correct option is A 393.5

Throat thickness $= 0.7 \times 10 = 7 m m$
Permissible shear stress $= \frac{V A ¯ ¯ ¯ y}{I b}$
Here $I = 7.73 \times 10^{6} m m^{4}$
$b = 4 \times 7 = 28 m m$
$A ¯ ¯ ¯ y = 100 \times 10 \times 55 = 55000 m m^{3}$
So, permissible shear stress
$100 = \frac{V \times 55000}{7.73 \times 10^{6} \times 28}$
$\Rightarrow V = 393527.27 N$
$\Rightarrow V = 393.5 k N$

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The correct option is A 393.5 Throat thickness =0.7×10=7mm Permissible shear stress =VA¯¯¯yIb Here I=7.73×106mm4 b=4×7=28mm A¯¯¯y=100×10×55=55000mm3 So, permissible shear stress 100=V×550007.73×106×28 ⇒V=393527.27 N ⇒V=393.5kN