Bolted joint is a type of joint where minimum two parts are connected to each
other with a bolt. Main functions of a bolted joint are keeping parts together during service life and transmitting torque, moment and force
between parts. Bolt sizing is very critical issue and improper design
of a joint can lead failures.
Metric bolt size calculator roughly determines bolt diameter in a bolted joint
according to guideline VDI 2230 Part-1 , which was prepared by the Association of German
Engineers (VDI), and treats the calculation of concentrically and eccentrically clamped bolted joints.
This calculator can be used for initial sizing of a bolt and preload in a joint
according to metric unit system. The tool is applicable for:
A sample example about
bolt sizing for pressure vessel cap is given in "Examples" section of this page to show usage of
the calculation tool. Reference information on friction coefficients, load introduction factor and
loading types are given in "Supplements" section to assist the usage of the
calculator.
This calculator can be used to roughly estimate bolt diameter
for a joint but more detailed
calculation shall be done according to VDI 2230 Part-1 guideline.
Note 1 : +Minimum values for static friction coefficients between
materials which are used in general engineering application. Reference values
are given in "Supplements" section.
Note 2 : *Graphical representation of axial loading types which is given to
assist selection of loading type among the alternatives is given in
"Supplements" section.
RESULTS |
Parameter |
Value |
Required preload |
FMmax |
----
|
N |
Required bolt size for selected grade |
----
|
--- |
Note 1 : This calculator is applicable for bolt dimension range from M3 to M39
and for strength grades G8.8 , G10.9, G12.9.
Calculation methodology of the calculator is summarized below. [Reference-1]
Load (N) |
Nominal Diameter |
Definition |
Strength Grade |
Step-1:
If only transverse or longitudinal loads exist, then F= FAmax or FQmax
If there are both transverse and longitudinal loads exist, if FAmax< FQmax/ μTmin
Then F= FQmax else F= FAmax
|
12.9 |
10.9 |
8.8 |
250 |
|
|
|
400 |
630 |
Step-2:
Select the closest larger load to F from the column 1 of the table. This load is
new F. F= closest larger load
|
1000 |
M3 |
M3 |
M3 |
1600 |
M3 |
M3 |
M3 |
2500 |
M3 |
M3 |
M4 |
4000 |
M4 |
M4 |
M5 |
Step-3:
If F= FQmax in Step-1 then increase F found in Step-2 by 4 steps
using column-1 of the table and find FMmin.
If F=FAmax in step-1 then increase F found in Step-2 by following
steps using column-1 of the table and find FMmin.
- Increase by 2 steps for dynamically and eccentrically applied axial load
- Increase by 1 step for dynamically and concentrically applied axial load
- Increase by 1 step for statically and eccentrically applied axial load
- No increase for statically and concentrically applied axial load
|
6300 |
M4 |
M5 |
M6 |
10000 |
M5 |
M6 |
M8 |
16000 |
M6 |
M8 |
M10 |
25000 |
M8 |
M10 |
M12 |
Step-4:
Find FMmax by increasing FMin using column-1 of the table with following number
of steps.
- 2 steps for tightening the bolt with a simple tightening spindle which has
been set by the retightening torque
- 1 step for tightening using a torque wrench or precision spindle adjusted by
means of dynamic torque measurement or elongation measurement of the bolt
- No steps for tightening by means of angle control within the plastic range or
by means of computer-controlled yield-point monitoring
|
40000 |
M10 |
M12 |
M14 |
63000 |
M12 |
M14 |
M16 |
100000 |
M16 |
M18 |
M20 |
160000 |
M20 |
M22 |
M24 |
Step-5:
Next to the found value of FMmax, select a bolt size with desired
strength (Column 2 to 4) from the relevant row .
|
250000 |
M24 |
M27 |
M30 |
400000 |
M30 |
M33 |
M36 |
630000 |
M36 |
M39 |
-- |
Note: Increase 1 step means, getting the value which is next row under the
current value. For example if current value is 1000 N, 1 step increase will
result 1600 N.