Question

Determine the bolt value of 20mm diameter bolt connecting 10mm plate in (1) single shear , bolt used are 4.6 grade , plate of 410 grade . Take area of bolt 245mm²

Answer 1

Answer:

F 4g ft rf fcyvyvcfgggvuhiggguggi

Answer 2

Answer:

The bolt value is $45.26 \mathrm{kN}$.

Step-by-step explanation:

GIVEN-

The diameter of the given bolt, $\mathrm{d}=20 \mathrm{~mm}$

The diameter of the hole, $\mathrm{d}_{0}=20+2=22 \mathrm{~mm}$

For bolts of grade 4.6: fub $=4 \times 100=400 \mathrm{MPa}$, fyb $=0.6 \times 400=240 \mathrm{Mpa}$

The thickness of the connecting plate $=10 \mathrm{~mm}$

For Fe 410 grade of steel: fu $= 410 \mathrm{MPa}$

TO FIND-

Strength of the Bolt (Bolt Value).

Note: Bolt value is the smallest of the Design Shear strength of the Bolts and Design Bearing Strength of the Bolt.

The bolts will be in single shear.

The Design Shear strength of bolt:

$\mathbf{V}_{\mathbf{d s b}} V_{s b}=\frac{f_{u}}{\sqrt{3}}\left(n_{n} A_{n b}+n_{s} A_{s b}\right) / \gamma_{m b}$

Here, $n_{n}=1$ and $n_{s}=0$ (Single Shear assuming shear plane to be in threaded portion).

$\mathrm{A}_{\mathrm{sb}}=\Pi \mathrm{d}^{2} / 4=3.14 \times 20^{2} / 4=314\mathrm{~mm}^{2} \\ \mathrm{~A}_{\mathrm{nh}}=0.78 \mathrm{~A}_{\mathrm{sh}}=0.78 \times 314=245 \mathrm{~mm}^{2}$

$V_{d s b}=A_{n b} \frac{f_{u b}}{\sqrt{3} \gamma_{m b}}=245 \times \frac{400}{\sqrt{3} \times 1.25} \times 10^{-3}=45.26 \mathrm{kN}$

Therefore, the bolt value is $45.26 \mathrm{kN}$.