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) πΉπ = πΉπ + ππΆ πΉπ = −πΉπ + π(1 − πΆ 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 ππ π΄π‘ ππ = πΆπ + πΉπ Load factor of safety guarding against overloading ππΏ = ππ π΄π‘ − πΉπ πΆπ π0 = πΉπ π(1 − πΆ Load factor for joint separation BOLT FAILURE MODES a: tensile failure πΉπ ππ = π΄π‘ ππ = ππ ππ ππ π΄π‘ ππ = πΆπ + πΉπ ππΏ = ππ π΄π‘ − πΉπ πΆπ π0 = πΉπ π(1 − πΆ b: Joint Separation To separate the joint: Fm ο½ 0 ππππ₯ π0 = π 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 ο½ 115 mm 2 S p ο½ 380 MPa From table 8-1 & 8-11 πΉπ = 0.75 ππ π΄π‘ πΉπ = 32.775 kN a) Tensile Failure b) Joint separation πΉπ π(1 − πΆ ππ π΄π‘ − πΉπ ππΏ = πΆπ π0 = π = 15.6 kN π = 14.4 kN π·π Gasketed joints ππ = ππ π΄π‘ πΆ(ππ‘ππ‘ππ /π + πΉπ ππ π΄π‘ − πΉπ ππΏ = πΆ(ππ‘ππ‘ππ /π π0 = πΉπ (ππ‘ππ‘ππ /π (1 − πΆ ππ·π 3≤ ≤6 ππ OR ππ΄π‘ = ππΏ πΆπ (1 − πΌ ππ OR ππ΄π‘ = π0 (1 − πΆ π πΌππ Where: • N is the no. of bolts • π·π is the diameter of the bolt circle • πΌ 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, π·π . Solution S p ο½ 600 MPa ππ΄π‘ = ππΏ πΆπ (1−πΌ ππ From table 8-1 & 8-11 = 8836 ππ2 π΄π‘ can be estimated by calculating the cerconferance of the circle πΆ = ππ π· for a siutable spacing between bolts, π can be estimated as 36 bolts π΄π‘ = 245.4 ππ2 Solution (cont.) for π = 36 andπ΄π‘ = 245.4 ππ2 , we can select M20x1.5 from Table 8-1 π·π can be estimated as π·π =1000 + 2(3)+2(47) = 1100 ππ·π ππ 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 ο« N CPmin Fi ο« N Fi t Load variation Per Bolt Dynamic Loading (cont.) a) Tensile Failure The alternating and mean forces per bolt are, respectively, CP οΆ ο¦ CP ο¦ ο§ Fi ο« max ο· ο ο§ Fi ο« min N οΈ ο¨ N Pa ο½ ο¨ 2 οΆ ο· οΈ ο½ CPa N where Pa ο½ Pmax ο Pmin 2 CP οΆ ο¦ CP ο¦ ο§ Fi ο« max ο· ο« ο§ Fi ο« min N οΈ ο¨ N Pm ο½ ο¨ 2 οΆ ο· οΈ ο½ F ο« CPm i N where Pm ο½ Pmax ο« Pmin 2 Applying ππΏ to the external load at both equations and substitute in Goodman line and rearranging: nf ο½ S e NAt C ο¦ Sut ο ο‘S p οΆ ο§ο§ ο·ο· ο¨ Pa Sut ο« Pm S e οΈ At ο½ n f C ο¦ Pa Sut ο« Pm S e οΆ ο§ ο· ο§ S e N ο¨ Sut ο ο‘S p ο·οΈ where ππ & N are the fatigue factor of safety and no. of bolts respectively Dynamic Loading (cont.) b) Joint Separation (1 ο C ) Pmax ο Fi ο« N (1 ο C ) Pmin ο Fi ο« N Fm t Load variation Per Bolt Dynamic Loading (cont.) nf 2 ο½ b) Joint Separation NFi οΎ1 (1 ο C ) Pmax where ππ2 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 ο½ 10 (2 ο« cos ο·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 ο½ 384 mm 2 S p ο½ 830 MPa ππ’π‘ = 1040 MPa ππ = 162 MPa 2 2 6 ο©6ο°ο©* ο°1*1οΉ οΉ Pmaxο½ 3ο½*310 *10 πΆ οͺ=οͺ0.25 οΊ οΊN N max πΌ =P 0.75 οͺο« οͺο« 4 4 οΊο» οΊο» 2οΉ 2 οΉ2 ο© οΉ ο° * 1 6 1 6 ο©6ο°ο©* ο° * 1 Pmax Pο½min 3P*10 ο½ 1ο½*οͺ110 *410οͺ οΊοͺ N οΊ οΊN N min οͺο« οͺο« οΊο»οͺο« 4 4 οΊο» οΊο» From table 8-1 & 8-11 From table 8-17 2 6 ο© ο° *1 οΉ Pmax ο½ 3 *10 οͺ οΊ N οͺο« 4 οΊο» 2 6 ο© ο° *1 οΉ Pmin ο½ 1 *10 οͺ οΊ N οͺο« 4 οΊο» 2οΉ 2οΉ ο© ο© 2 2 P ο P ο° * 1 6 6 ο° *1 Pmax P 6 ο©6ο°ο©* Pο½max οmin Pοͺmin οΊ N ο°1*1οΉ οΉ Pa ο½ max min ο½ 1 *10 οͺ οΊ N Pmin 1ο *10 Pa Pο½a ο½ ο½ 1 * 10 N οͺ οΊ ο½ 1 * 10 N οͺ οΊ 2 4 οΊο» οͺο« 4 οΊο» 2 2 οͺο« 4 οͺο« οͺο« 4 οΊο» οΊο» 2 ο© ο° *12 οΉ P ο« P 6 Pmax ο Pmin 6 ο© ο° *1 οΉ min 2 οΉ2 P ο½ max ο½ 2 *10 οͺ ο© οΊ N ο© οΉ Pa ο½ ο½ 1 * 10 N οͺ οΊ m Pmax ο«P 1*1 6 6ο° * P ο« P ο° min max min 2 4*10οͺοΊο» οͺ οͺο« 4 οΊο» PmPmο½ ο½2 ο½ 2ο½οͺο«*210 οΊ οΊN N 2 4 2 P ο«P οͺο« οͺο« 4 οΊο» οΊο» 2οΉ 6 ο© ο° *1 Solution (cont.) a) Tensile Failure n f C ο¦ Pa Sut ο« Pm S e οΆ ο§ ο· Nο½ S e At ο§ο¨ Sut ο ο‘S p ο·οΈ ο½ 26 Check the bolt spacing with N is 26: ππ·π ππ π1160 = 5.8 26 ∗ 24 Check the bolt spacing with N is 30: π1160 = 5.06 30 ∗ 24 b) Joint separation (1 ο C ) Pmax N ο½ nf ο‘S p At ο½ 22.1