Bolted Connection LSM

A bolt is a metal pin with a head formed at one end and shank threaded at the other in order to receive a nut.

Types of bolts

1. Unfinished bolts

2. Finished bolts

3. High Strength Friction Grip (HSFG) bolt

Let's look at the important fomulas for designing a bolted connection based on IS 800:2007

Design Strength of Bolt,
${{V}_{dsb}}=\frac{{{V}_{nsb}}}{{{\gamma }_{mb}}}$

${{V}_{nsb}}=\frac{{{f}_{ub}}}{\sqrt{3}}({{\eta }_{n}}{{A}_{nb}}+{{\eta }_{s}}{{A}_{sb}})$

${{V}_{nsb}}=$ Nominal Shear Capacity of bolt
${{\gamma }_{mb}}=$ Partial factor of safety of materail bolt = 1.25
${{f}_{ub}}=$ Ultimate tensile strength of a bolt
${{\eta }_{n}}=$ Number of shear planes with threads intercepting the shear plane.
${{\eta }_{s}}=$ Number of shear planes without threads intervepting the shear plane.
${{A}_{sb}}=$ Nominal shank area of bolt $=\frac{\pi {{d}^{2}}}{4}$
${{A}_{nb}}=$ Net Shear area of the bolt at threads $=0.78\times \frac{\pi {{d}^{2}}}{4}$
$d=$ Nominal diameter of Bolt

Reduction Factor for Shear Capacity of Bolts

1. Reduction Factor for long joints

If the distance between the first and last bolt in the joint measured in the direction of load exceed 15 d, the shear capacity ${{V}_{db}}$ shall be reduced by the factor,

                             ${{\beta }_{{{l}_{j}}}}=1.075-0.005\frac{{{l}_{j}}}{d}$

                      Subjected to the limits $0.75\le {{\beta }_{lj}}\le 1.0$

2. Reduction Factor if Grip Length is Large

If the total thickness of the connected plate exceeds 5 times the diameter d of the bolts, the shear capacity ${{V}_{db}}$ shall be reduced by the factor,

               ${{\beta }_{{{l}_{g}}}}=\frac{8d}{3d+{{l}_{g}}}$

Subjected to conditions, maximum value $={{\beta }_{{{l}_{j}}}}$ i.e. ${{\beta }_{{{l}_{g}}}}<{{\beta }_{{{l}_{j}}}}$
In no case ${{l}_{g}}$ be greater than $8d$

3. Reduction Factor if Packing Plates are Used

If packing plates of thickness more than 6 mm are used them the shear capacity ${{V}_{db}}$ shall be reduced by the factor,

               ${{\beta }_{pk}}=1-0.0125{{t}_{pk}}$

Thus Shear capacity of bolt

               ${{V}_{dsb}}=\frac{{{f}_{ub}}}{\sqrt{3}{{\gamma }_{mb}}}({{\eta }_{n}}{{A}_{nb}}+{{\eta }_{s}}{{A}_{sb}}){{\beta }_{{{l}_{j}}}}{{\beta }_{{{l}_{g}}}}{{\beta }_{pk}}$

Bearing Capacity of Bolts
               ${{V}_{dpb}}=\frac{{{V}_{npb}}}{{{\gamma }_{mb}}}$

               ${{V}_{npb}}=2.5{{K}_{b}}\times d\times t\times {{f}_{u}}$

${{V}_{dpb}}=$ Design bearing strength
${{V}_{npb}}=$ Nominal bearing srength
${{\gamma }_{mb}}=$ Partial factor of safety of material = 1.25
${{K}_{b}}=$ smaller of

$\frac{e}{3{{d}_{h}}},\left( \frac{p}{3{{d}_{h}}}-0.25 \right),\frac{{{f}_{ub}}}{{{f}_{u}}},1.0$

$e=$ end distance
$p=$ pitch
${{d}_{h}}=$ diameter of hole
${{f}_{u}}=$ Utlimate tensile strength of plate
$t=$ Summation of the thickness of the connected plates experiencing bearing stress in the same direction.

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