1. Static and Fatigue Bolt Design
03 May 2015
Dr. Ghassan Mousa

2. Thread standards & definition
d: major diameter
p: pitch
l: lead
Multiple threads
Fig. 8.1 Terminology of screw threads

3. 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)

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

5. 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

6. Fig. 8.13 A bolted connection loaded in tension by the forces P
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)
𝐹 𝑏 = 𝐹 𝑖 +𝑃𝐶
𝐹 𝑚 = − 𝐹 𝑖 +𝑃(1−𝐶
Fig A bolted connection loaded in tension by the forces P

7. Static load Fi can be determined as:
where SP is the proof strength and can be found in tables 8 -11
𝑛 𝑝 = 𝑆 𝑝 𝐴 𝑡 𝐶𝑃+ 𝐹 𝑖
Yield factor of safety guarding against exceeding Sp
𝑛 𝐿 = 𝑆 𝑝 𝐴 𝑡 − 𝐹 𝑖 𝐶𝑃
Load factor of safety guarding against overloading
𝑛 0 = 𝐹 𝑖 𝑃(1−𝐶
Load factor for joint separation

8. BOLT FAILURE MODES a: tensile failure b: Joint Separation
𝜎 𝑏 = 𝐹 𝑏 𝐴 𝑡
𝑛 𝑝 = 𝑆 𝑝 𝐴 𝑡 𝐶𝑃+ 𝐹 𝑖
𝑛 𝑏 = 𝑆 𝑏 𝜎 𝑏
𝑛 𝐿 = 𝑆 𝑝 𝐴 𝑡 − 𝐹 𝑖 𝐶𝑃
b: Joint Separation
To separate the joint:
𝑛 0 = 𝐹 𝑖 𝑃(1−𝐶
𝑛 0 = 𝑃 𝑚𝑎𝑥 𝑃

9. Example of bolt under static load
An M14x2 grade 5.8 bolt is used in a bolted connection. The joint constant is C = 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.

10. Solution. a) Tensile Failure b) Joint separation From table 8-1 & 8-11
𝐹 𝑖 =0.75 𝑆 𝑝 𝐴 𝑡
𝐹 𝑖 = kN
a) Tensile Failure
b) Joint separation
𝑛 0 = 𝐹 𝑖 𝑃(1−𝐶
𝑛 𝐿 = 𝑆 𝑝 𝐴 𝑡 − 𝐹 𝑖 𝐶𝑃
𝑃 = 15.6 kN
𝑃 = 14.4 kN

11. Gasketed joints Where: N is the no. of bolts
𝐷 𝑏
𝑛 𝑝 = 𝑆 𝑝 𝐴 𝑡 𝐶( 𝑃 𝑡𝑜𝑡𝑎𝑙 /𝑁)+ 𝐹 𝑖
𝑛 𝐿 = 𝑆 𝑝 𝐴 𝑡 − 𝐹 𝑖 𝐶( 𝑃 𝑡𝑜𝑡𝑎𝑙 /𝑁)
𝑁 𝐴 𝑡 = 𝑛 𝐿 𝐶𝑃 (1−𝛼) 𝑆 𝑃 OR
𝑛 0 = 𝐹 𝑖 ( 𝑃 𝑡𝑜𝑡𝑎𝑙 /𝑁)(1−𝐶
𝑁 𝐴 𝑡 = 𝑛 0 (1−𝐶)𝑃 𝛼 𝑆 𝑃 OR
Where:
N is the no. of bolts
𝐷 𝑏 is the diameter of the bolt circle
𝛼 is 0.75 or 0.9
3≤ 𝜋 𝐷 𝑏 𝑁𝑑 ≤6

12. 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 Give a suitable bolt pitch circle diameter, 𝐷 𝑏 .

13. Solution 𝑁 𝐴 𝑡 = 𝑛 𝐿 𝐶𝑃 (1−𝛼) 𝑆 𝑃 = 8836 𝑚𝑚 2 𝐶=𝑝𝑖 𝐷 𝐴 𝑡 =245.4 𝑚𝑚 2
From table 8-1 & 8-11
𝑁 𝐴 𝑡 = 𝑛 𝐿 𝐶𝑃 (1−𝛼) 𝑆 𝑃 = 𝑚𝑚 2
𝐴 𝑡 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
𝐴 𝑡 = 𝑚𝑚 2

14. Solution (cont.) 𝜋 𝐷 𝑏 𝑁𝑑 𝜋1100 36∗20 =4.8
for 𝑁= 36 and 𝐴 𝑡 = 𝑚𝑚 2 , we can select M20x1.5 from Table 8-1
𝐷 𝑏 can be estimated as
𝐷 𝑏 = (3)+2(47) = 1100
𝜋 𝐷 𝑏 𝑁𝑑
𝜋 ∗20 =4.8
NUMBER OF BOLTS = 36
BOLT DESIGNATION : M20 X 1.5
BOLT PITCH CIRCLE DIAMETER = 1100 mm
(NOTE: This is only one possible solution)

15. Load variation Per Bolt
Dynamic Loading
a) Tensile Failure t
Load variation Per Bolt

16. Dynamic Loading (cont.)
a) Tensile Failure
The alternating and mean forces per bolt are, respectively,
where
where
Applying 𝑛 𝐿 to the external load at both equations and substitute in Goodman line and rearranging:
where 𝑛 𝑓 & N are the fatigue factor of safety and no. of bolts respectively

17. Load variation Per Bolt
Dynamic Loading (cont.)
b) Joint Separation t
Load variation Per Bolt

18. Dynamic Loading (cont.)
b) Joint Separation
where 𝑛 𝑓2 is the seperation factor of safety
Table 8-17 Material Properties (Endurance Strength) for Standard Bolts (SI)

19. Example of bolt under dynamic load
The fluctuating pressure in the cylinder shown is given by: Using Class 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.

20. Solution. 𝑆 𝑢𝑡 =1040 MPa From table 8-1 & 8-11 From table 8-17
𝛼 = 0.75
𝐶=0.25

21. Solution (cont.) 𝜋 𝐷 𝑏 𝑁𝑑 𝜋1160 26∗24 =5.8 𝜋1160 30∗24 =5.06
a) Tensile Failure
b) Joint separation
Check the bolt spacing with N is 26:
𝜋 𝐷 𝑏 𝑁𝑑
𝜋 ∗24 =5.8
Check the bolt spacing with N is 30:
𝜋 ∗24 =5.06