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

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. 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 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

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

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. a) Tensile Failure b) Joint separation From table 8-1 & 8-11 𝐹 𝑖 =0.75 𝑆 𝑝 𝐴 𝑡 𝐹 𝑖 = 32.775 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

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

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, 𝐷 𝑏 .

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 = 8836 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 =245.4 𝑚𝑚 2

Solution (cont.) 𝜋 𝐷 𝑏 𝑁𝑑 𝜋1100 36∗20 =4.8 for 𝑁= 36 and 𝐴 𝑡 =245.4 𝑚𝑚 2 , we can select M20x1.5 from Table 8-1 𝐷 𝑏 can be estimated as 𝐷 𝑏 =1000 + 2(3)+2(47) = 1100 𝜋 𝐷 𝑏 𝑁𝑑 𝜋1100 36∗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)

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

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

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

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)

Example of bolt under dynamic load 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.

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

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 𝜋1160 26∗24 =5.8 Check the bolt spacing with N is 30: 𝜋1160 30∗24 =5.06