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)
𝐹𝑏 = 𝐹𝑖 + 𝑃𝐶
𝐹𝑚 = −𝐹𝑖 + 𝑃(1 − 𝐶
Fig. 8.13 A bolted connection loaded in
tension by the forces P
Static load
Fi can be determined as:
where SP is the proof
strength and can be
found in tables 8 -11
Yield factor of safety guarding against exceeding Sp
𝑆𝑝 𝐴𝑡
𝑛𝑝 =
𝐶𝑃 + 𝐹𝑖
Load factor of safety guarding against overloading
𝑛𝐿 =
𝑆𝑝 𝐴𝑡 − 𝐹𝑖
𝐶𝑃
𝑛0 =
𝐹𝑖
𝑃(1 − 𝐶
Load factor for joint separation
BOLT FAILURE MODES
a: tensile failure
𝐹𝑏
𝜎𝑏 =
𝐴𝑡
𝑛𝑏 =
𝑆𝑏
𝜎𝑏
𝑆𝑝 𝐴𝑡
𝑛𝑝 =
𝐶𝑃 + 𝐹𝑖
𝑛𝐿 =
𝑆𝑝 𝐴𝑡 − 𝐹𝑖
𝐶𝑃
𝑛0 =
𝐹𝑖
𝑃(1 − 𝐶
b: Joint Separation
To separate the joint:
Fm  0
𝑃𝑚𝑎𝑥
𝑛0 =
𝑃
Example of bolt under static load
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.
Solution.
At  115 mm
2
S p  380 MPa
From table 8-1 & 8-11
𝐹𝑖 = 0.75 𝑆𝑝 𝐴𝑡
𝐹𝑖 = 32.775 kN
a) Tensile Failure
b) Joint separation
𝐹𝑖
𝑃(1 − 𝐶
𝑆𝑝 𝐴𝑡 − 𝐹𝑖
𝑛𝐿 =
𝐶𝑃
𝑛0 =
𝑃 = 15.6 kN
𝑃 = 14.4 kN
𝐷𝑏
Gasketed joints
𝑛𝑝 =
𝑆𝑝 𝐴𝑡
𝐶(𝑃𝑡𝑜𝑡𝑎𝑙 /𝑁 + 𝐹𝑖
𝑆𝑝 𝐴𝑡 − 𝐹𝑖
𝑛𝐿 =
𝐶(𝑃𝑡𝑜𝑡𝑎𝑙 /𝑁
𝑛0 =
𝐹𝑖
(𝑃𝑡𝑜𝑡𝑎𝑙 /𝑁 (1 − 𝐶
𝜋𝐷𝑏
3≤
≤6
𝑁𝑑
OR
𝑁𝐴𝑡 =
𝑛𝐿 𝐶𝑃
(1 − 𝛼 𝑆𝑃
OR
𝑁𝐴𝑡 =
𝑛0 (1 − 𝐶 𝑃
𝛼𝑆𝑃
Where:
• N is the no. of bolts
• 𝐷𝑏 is the diameter of the bolt circle
• 𝛼 is 0.75 or 0.9
Example of bolt under static load (gasket joint)
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, 𝐷𝑏 .
Solution
S p  600 MPa
𝑁𝐴𝑡 =
𝑛𝐿 𝐶𝑃
(1−𝛼 𝑆𝑃
From table 8-1 & 8-11
= 8836 𝑚𝑚2
𝐴𝑡 can be estimated by calculating the cerconferance of the circle
𝐶 = 𝑝𝑖 𝐷
for a siutable spacing between bolts, 𝑁 can be estimated as 36 bolts
𝐴𝑡 = 245.4 𝑚𝑚2
Solution (cont.)
for 𝑁 = 36 and𝐴𝑡 = 245.4 𝑚𝑚2 , we can select M20x1.5 from Table 8-1
𝐷𝑏 can be estimated as
𝐷𝑏 =1000 + 2(3)+2(47) = 1100
𝜋𝐷𝑏
𝑁𝑑
𝜋1100
= 4.8
36 ∗ 20
NUMBER OF BOLTS = 36
BOLT DESIGNATION : M20 X 1.5
BOLT PITCH CIRCLE DIAMETER = 1100 mm
(NOTE: This is only one possible solution)
Dynamic Loading
a) Tensile Failure
CPmax
Fi 
N
CPmin
Fi 
N
Fi
t
Load variation Per Bolt
Dynamic Loading (cont.)
a) Tensile Failure
The alternating and mean forces per bolt are, respectively,
CP  
CP

 Fi  max    Fi  min
N  
N
Pa  
2


  CPa
N
where
Pa 
Pmax  Pmin
2
CP  
CP

 Fi  max    Fi  min
N  
N
Pm  
2


  F  CPm
i
N
where
Pm 
Pmax  Pmin
2
Applying 𝑛𝐿 to the external load at both equations and substitute in Goodman
line and rearranging:
nf 
S e NAt
C
 Sut  S p 


 Pa Sut  Pm S e 
At 
n f C  Pa Sut  Pm S e 



S e N  Sut  S p 
where 𝑛𝑓 & N are the fatigue factor of safety and no. of bolts respectively
Dynamic Loading (cont.)
b) Joint Separation
(1  C ) Pmax
 Fi 
N
(1  C ) Pmin
 Fi 
N
Fm
t
Load variation Per Bolt
Dynamic Loading (cont.)
nf 2 
b) Joint Separation
NFi
1
(1  C ) Pmax
where 𝑛𝑓2 is the seperation factor of safety
Table 8-17 Material Properties (Endurance Strength) for
Standard Bolts (SI)
Example of bolt under dynamic load
The fluctuating pressure in the cylinder
shown is given by:
6
p  10 (2  cos t ) N / m
2
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.
Solution.
At  384 mm 2
S p  830 MPa
𝑆𝑢𝑡 = 1040 MPa
𝑆𝑒 = 162 MPa
2 2
6 6*
1*1 
Pmax 3*310
*10
𝐶 =0.25  N N
max
𝛼 =P
0.75
  4 4  
2
2 2



*
1
6
1
6 6*

*
1
Pmax Pmin
3P*10
 1*110
*410  N  N N
min

  4 4  
From table 8-1 & 8-11
From table 8-17
2
6   *1 
Pmax  3 *10 
 N
 4 
2
6   *1 
Pmin  1 *10 
 N
 4 
2
2


2
2
P

P

*
1
6
6  *1
Pmax
P
6 6*
Pmax
min
Pmin  N
1*1  Pa  max min  1 *10 
 N
Pmin
1
*10
Pa Pa 

1
*
10
N



1
*
10
N


2
4 
 4 
2 2 
4
  4  
2
  *12 
P

P
6
Pmax  Pmin
6   *1 
min
2 2 P  max
 2 *10 

 N


Pa 

1
*
10
N


m
Pmax
P
1*1
6 6 *
P

P

min
max
min
2
4*10 
 4 
PmPm 2
 2*210
 N N
2
4
2
P
P
  4  
2
6   *1
Solution (cont.)
a) Tensile Failure
n f C  Pa Sut  Pm S e 


N
S e At  Sut  S p 
 26
Check the bolt spacing with N is 26:
𝜋𝐷𝑏
𝑁𝑑
𝜋1160
= 5.8
26 ∗ 24
Check the bolt spacing with N is 30:
𝜋1160
= 5.06
30 ∗ 24
b) Joint separation
(1  C ) Pmax
N  nf
S p At
 22.1