A cantilever bracket is bolted to a column using three M12 - 1-75 bolts- P- Q and R- The value of maximum shear stress developed in the bolt P in MPa is

Total shear force on P, nbsp; F_s= √((37500)^2+ (3000)^2) =37619.808 N ∴η_max nbsp; = F_sA = 3332.63227619.808π× 0.25 × 12^2 nbsp;= 332.6321 MPa n ...

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Standard XII

Physics

Thermodynamic Processes

A cantilever ...

Question

$A cantilever bracket is bolted to a column using three M 12 \times 1.75 bolts, P, Q and R. The value of maximum shear stress developed in the bolt P (in MPa) is$
332.6321

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Solution

The correct option is A 332.6321

Total shear force on $P$ ,

$F_{s} = \sqrt{(37500)^{2} + (3000)^{2}}$

= $37619.808 N$

$∴ η_{m a x} = \frac{F_{s}}{A} = \frac{3332.63227619.808}{π \times 0.25 \times 12^{2}}$

= $332.6321 M P a$

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A cantilever bracket is bolted to a column using three M12×1.75 bolts, P, Q and R. The value of maximum shear stress developed in the bolt P (in MPa) is332.6321

The correct option is A 332.6321 Total shear force on P, Fs=√(37500)2+(3000)2 =37619.808 N ∴ηmax=FsA=3332.63227619.808π×0.25×122 = 332.6321 MPa

$A cantilever bracket is bolted to a column using three M 12 \times 1.75 bolts, P, Q and R. The value of maximum shear stress developed in the bolt P (in MPa) is$
332.6321

The correct option is A 332.6321

Total shear force on $P$ ,

$F_{s} = \sqrt{(37500)^{2} + (3000)^{2}}$

= $37619.808 N$

$∴ η_{m a x} = \frac{F_{s}}{A} = \frac{3332.63227619.808}{π \times 0.25 \times 12^{2}}$

= $332.6321 M P a$