STATIC AND FATIGUE BOLT DESIGN

03 May 2015

Dr. Ghassan Mousa

Thread standards & definition• d: major diameter

• p: pitch

• l: lead

• Multiple threads

Fig. 8.1 Terminology of screw threads

http://www.gizmology.net/nutsbolts.htm

• The thread size is specified by the p for metric sizes.

• At is tensile stress area of unthreaded rod.

Table 8-1 Geometrical data for Standard Bolts (SI)

Bolt Types

Three types of threaded fastener. (a) Bolt and nut; (c) Cap screw; (c) stud.

Bolt strength

Bolts in axial loading fail at the:

•Fillet under the head.

•Thread runout.

•1st thread engaged in the nut.

Table 8-11 Material properties of steel bolts

Tension joints• Fi: preload

• Fb = Pb + Fi . Total force in each bolt

• Fm = Pm – Fi. Force Carried by the Joint

• P: external tensile load per bolt

• Pb: portion of P taken by bolt

• Pm: portion of P taken by members

• C: stiffness constant of the joints

(joint constant)

Fig. 8.13 A bolted connection loaded in tension by the forces P

¿𝐹 𝑏=𝐹 𝑖+𝑃𝐶

Static load

Fi can be determined as:where SP is the proof strength and can be found in tables 8 -11

𝑛𝑝=𝑆𝑝 𝐴𝑡

𝐶𝑃+𝐹𝑖

Yield factor of safety guarding against exceeding Sp

𝑛𝐿=𝑆𝑝 𝐴𝑡−𝐹 𝑖

𝐶𝑃Load factor of safety guarding against overloading

𝑛0=𝐹𝑖

¿¿Load factor for joint separation

b: Joint Separation

BOLT FAILURE MODES

0Fm

To separate the joint:

𝑛0=𝑃𝑚𝑎𝑥

𝑃

a: tensile failure

𝜎 𝑏=𝐹 𝑏

𝐴𝑡

𝑛𝑏=𝑆𝑏

𝜎𝑏

𝑛𝑝=𝑆𝑝 𝐴𝑡

𝐶𝑃+𝐹𝑖

𝑛𝐿=𝑆𝑝 𝐴𝑡−𝐹 𝑖

𝐶𝑃

𝑛0=𝐹𝑖

¿¿

An M14x2 grade 5.8 bolt is used in a bolted connection. The joint constant is C = 0.3498. The factors of safety to be applied are 2 against tensile stress failure and 3.5 against joint separation.

Calculate the maximum force (P) that can be applied to this unit for reusable application.

Example of bolt under static load

Solution.

a) Tensile Failure

2t mm115A MPa380S p

𝑛𝐿=𝑆𝑝 𝐴𝑡−𝐹 𝑖

𝐶𝑃

From table 8-1 & 8-11

= 15.6 kN

b) Joint separation

𝑛0=𝐹𝑖

¿¿ = 14.4 kN

𝐹 𝑖=0.75𝑆𝑝 𝐴𝑡

= 32.775 kN

Gasketed joints

𝑛𝑝=𝑆𝑝 𝐴𝑡

𝐶 (𝑃 𝑡𝑜𝑡𝑎𝑙 /𝑁 )+𝐹 𝑖

𝑛𝐿=𝑆𝑝 𝐴𝑡−𝐹 𝑖

𝐶 (𝑃 𝑡𝑜𝑡𝑎𝑙/𝑁 )

𝑛0=𝐹𝑖

¿¿

3≤𝜋 𝐷𝑏

𝑁𝑑≤6

Where:

• N is the no. of bolts

• is the diameter of the bolt circle

• is 0.75 or 0.9

𝑁 𝐴𝑡=𝑛𝐿𝐶𝑃

(1−𝛼)𝑆𝑃

𝑁 𝐴𝑡=𝑛0(1−𝐶 )𝑃

𝛼𝑆𝑃

OR

OR

𝐷𝑏

A cylinder of internal diameter 1000mm and wall thickness 3mm is subjected to an internal pressure equivalent to 1.5 MPa. Assume safety factors of 2.5 against bolt tensile failure, 3.5 against joint separation, and class 8.8 bolts in either fine or coarse threads.

Design a suitable bolt group for the cylinder head for reusable application and an assumed joint constant of 0.45. Give a suitable bolt pitch circle diameter, .

Example of bolt under static load (gasket joint)

Solution

MPaS p 600 From table 8-1 & 8-11

=

𝐶=𝑝𝑖𝐷

f∨a siutable spacingbetween bolts ,𝑁 canbe estimated as 36bolts

Solution (cont.)

, we can select M20x1.5 from Table 8-1

𝐷𝑏canbeestimated as

1000 + 2(3)+2(47) = 1100

𝜋 𝐷𝑏

𝑁𝑑𝜋 110036∗20

=4.8

NUMBER OF BOLTS = 36BOLT DESIGNATION : M20 X 1.5BOLT PITCH CIRCLE DIAMETER = 1100 mm(NOTE: This is only one possible solution)

Load variation Per Bolt

N

CPFi

min

N

CPFi

max

iFt

Dynamic Loading a) Tensile Failure

The alternating and mean forces per bolt are, respectively,

N

PCNCP

FN

CPF

P aii

a

2

minmax

N

PCF

NCP

FN

CPF

P mi

ii

m

2

minmax

Dynamic Loading (cont.)

2

PPP minmaxa

2

PPP minmaxm

where

where

Applying to the external load at both equations and substitute in Goodman line and rearranging:

emuta

puttef SPSP

SS

C

NASn

put

emuta

e

ft SS

SPSP

NS

CnA

where

a) Tensile Failure

Load variation Per Bolt

N

PCFi

min)1(

N

PCFi

max)1(

Fmt

Dynamic Loading (cont.)b) Joint Separation

1)1( max

2

PC

NFn if

Dynamic Loading (cont.)

where

b) Joint Separation

Table 8-17 Material Properties (Endurance Strength) for Standard Bolts (SI)

The fluctuating pressure in the cylinder shown is given by:

Using Class 10.9 M24x2 bolts with a factor of safety of 2.5 against bolt tensile failure and 3 against joint separation.

determine the number of bolts which should be used for the application. Assume that the bolts carry 25% of the external load. Assume a reusable application.

Example of bolt under dynamic load

26m/N)tcos2(10p

Solution.

2384 mmAt MPaS p 830 From table 8-1 & 8-11𝑆𝑢𝑡=1040MPa

𝑆𝑒=162MPa From table 8-17

= 0.75 𝐶=0.25

N4

1*10*3P

26

max

N4

1*10*1P

26

min

N4

1*10*1

2

PPP

26minmax

a

N4

1*10*2

2

PPP

26minmax

m

N4

1*10*3P

26

max

N4

1*10*1P

26

min

N4

1*10*1

2

PPP

26minmax

a

N4

1*10*2

2

PPP

26minmax

m

N4

1*10*3P

26

max

N4

1*10*1P

26

min

N4

1*10*1

2

PPP

26minmax

a

N4

1*10*2

2

PPP

26minmax

m

N4

1*10*3P

26

max

N4

1*10*1P

26

min

N4

1*10*1

2

PPP

26minmax

a

N4

1*10*2

2

PPP

26minmax

m

Solution (cont.)a) Tensile Failure b) Joint separation

1.22

)1( max

tpf AS

PCnN

26

put

emuta

te

f

SS

SPSP

AS

CnN

𝜋 𝐷𝑏

𝑁𝑑𝜋 11602 6∗24

=5 .8

Check the bolt spacing with N is 26:

Check the bolt spacing with N is 30:

𝜋 116030∗24

=5 .06