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
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
Load factor of safety guarding against overloading
Load factor for joint separation
b: Joint Separation
BOLT FAILURE MODES
0 Fm
To separate the joint:
a: tensile failure
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
2
t 115 mm
A Sp 380 MPa
From table 8-1 & 8-11
b) Joint separation
Gasketed joints
OR
OR
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.
Example of bolt under static load (gasket
joint)
Solution
MPa
S
p 600
From table 8-1 & 8-11
Solution (cont.)
NUMBER OF BOLTS = 36
BOLT DESIGNATION : M20 X 1.5 BOLT PITCH CIRCLE DIAMETER = 1100 mm
(NOTE: This is only one possible solution)
Load variation Per Bolt
N F
i CP
minN F
i CP
maxF
it
Dynamic Loading
a) Tensile FailureThe alternating and mean forces per bolt are, respectively,
N P N C
F CP N
F CP
P a
i i
a
2
min max
N P F C
N F CP N
F CP
P i m
i i
m
2
min max
Dynamic Loading (cont.)
2 P Pa Pmax min
2 P Pm Pmax min where
where
e m ut
a
p t ut
e
f P S P S
S S
C NA
n S
p ut
e m ut
a e
f
t S S
S P S
P N S
C A n
a) Tensile Failure
Load variation Per Bolt
N P F
i ( 1 C )
min
N P F
i ( 1 C )
max
Fm
t
Dynamic Loading (cont.)
b) Joint Separation) 1 1
( max
2
P C nf NFi
Dynamic Loading (cont.)
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
2
6
( 2 cos t ) N / m 10
p
Solution.
384 mm2
At Sp 830 MPa From table 8-1 & 8-11
From table 8-17
4 N 1 10*
* 3 P
6 2
max
N
4 1 10*
* 1 P
6 2
min
4 N 1 10 *
* 2 1
P P P
6 2 min a max
4 N 1 10 *
* 2 2
P P P
6 2 min mmax
Solution (cont.)
a) Tensile Failure b) Joint separation
1 . 22
) 1
(
max
t p
f
S A
P n C
N
26
p ut
e m ut
a t
e f
S S
S P S
P A
S C N n
Check the bolt spacing with N is 26:
Check the bolt spacing with N is 30:
