A bolt of major diameter 12 mm is required to clamp two steel plates. cross sectional area of the threaded portion of the bolt is $84.3 m m^{2}$ . The length of the threaded portion in grip is 30 mm, while the length of the unthreaded portion in grip is 8 mm. Young's modulus of material is 200 GPa. The effective stiffness (in MN/m) of the bolt in the clamped zone is
468.77

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Solution

The correct option is A 468.77
$k_{t} = \frac{A_{t} E}{L_{t}}$ (Stiffness in threaded portion)

$k_{t} = \frac{84.3 \times 200 \times 10^{3}}{30 \times 10^{- 3}} = 562 \times 10^{6} N / m$

$= 562 M N / m$

$k_{d} = \frac{A_{d} E}{L_{d}}$ (Stiffness in unthreaded region) $A_{d}$ (major diameter c/s area)

$= \frac{π}{4} d^{2} = 0.785 \times 144 = 113.04 m m^{2}$ $A_{d}$ (length of unthreaded portion)

$= 8 m m ∴ k_{d} = \frac{113.04 \times 200 \times 10^{3}}{8 \times 10^{- 3}}$

$= 2826 \times 10^{6} N / m = 2826 M n / m$

$\frac{1}{k} = \frac{1}{k_{t}} + \frac{1}{k_{d}} \Rightarrow k = \frac{k_{d} k_{t}}{k_{d} + k_{t}} = \frac{826 \times 562}{2826 + 562}$

$= 468.77 M n / m$

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