Select the bolt material to determine the strength distribution: first select the bolt material, the bolt performance level, the strength distribution is normal distribution, find the tensile strength Rb, the yield limit Rs. From the empirical formula: R--1=0.23 (R-b +Rs)S--1=0.577R--1RS-1=0.08S--1(3) Determine the shear stress distribution of the bolt: 1 The average lateral load of the single bolt is F-: F-=F-6n( 2) The total lateral working load of the F-6-bolt group in the formula, N; n - the number of bolts.
The average cross-sectional area of ​​the bolt is A-: A-=Pd-24 where d-0—the diameter of the bolt shear surface, mm.
The standard deviation of area A, RA=Pd-Rd2, is measured according to the manufacturing tolerance of bolt diameter. The relationship between d-0 and Rd is found as: Rd=0.002d-. The average value of predicted shear stress is: (S-, RS)=F -, RFA-, RA bolt shear stress mean: S-=F-iA-=4F-iPd-20 (4) where i - the number of joint faces.
The standard deviation of shear stress is: RS=4iP(F-R2d2+(d-20)2R2F(d-20)412 where Rd2=2d-0Rd=2d-0×0.002d-0=0.004d-20(4) Solve the bolt diameter d-0 according to the joint equation: the coupling equation Z=-S--1-S- When the bolt rod or hole wall is squeezed, it is also a main failure mode to calculate the crushing of the bolt rod or the hole wall due to the extrusion. Especially when the material strength of the joint is low, it is often crushed. The design criterion is: The probability that the crushing stress Rp of the bolt or hole wall is less than the compressive strength Sb of its corresponding material must be greater than or equal to the reliability required by the design. R(t) is expressed as: P(Rp Determining the stress distribution of the bolt: For the value of the size L-min, if the L-min is too small, the extrusion stress Rp is too large, and if the L-min is too large, the contact surface is unevenly biased due to manufacturing errors. In order to find Rp and RRp, the pressing area A-=d-0L-min must be obtained. For the convenience of calculation, first find out the relationship between d-0 and L-min, as shown by the shear bolt connection. It is known that the working shear force is F=12kN, the reliability of bolting target is R(t)=0.9998, the bolt is 45 steel, the bolt performance level is 6.8, Rb=600MPa, Rs=480MPa, when the joint is cast iron HT250, press Shear strength design screw diameter d-0 = 12.5mm, according to the extrusion strength design, the ratio of L-mind-0 and screw diameter d-0. As shown, it is clear that the reliability design method is more reasonable than the safety factor method. The advantage is that it can quantitatively represent the safety and reliability of the part, and also has a predictable life of the part. The safety factor method does not. Generally, L-min=1.25d-0 is obtained. Then: Rp=F-id-0L-min=F-1.25id-20Lmin/d0 is related to d0. Standard deviation RRp=11.25iF-2R2d2+(d-20)2R2F (d-20) 412 where RF=0.06FZ=-SbRp(R2sb+R2Rb)12(8) According to the reliability R(t) value required for bolting, check the standard normal distribution table to obtain the coupling coefficient Z value. Substituting the relevant parameters Z, Sb, Rp, Rsb, RRp into the above formula, the bolt diameter d-0 is obtained, the standard value of the screw diameter d-0 is checked, and the nominal diameter d-.
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