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Tài liệu cung cấp cho mọi người phương thức tính toán cũng như định hướng thiết kế trong công nghệ dập thủy tĩnh phôi ống. Trong đó có nêu một số tính toán và thiết kế của một số dạng biên dạng ống cơ bản.
Pipe products – Various shapes of pipes Connecting parts Cam shaft Crank shaft Classification HYDROFORMING OF TUBE HYDROFORMING OF TUBE with one axis load HYDROFORMING OF TUBE with more axis loads HYDROFORMING OF TUBE with one axis load Fluid pressure, vertical and horisontal forces HYDROFORMING OF TUBE with more axis loads Horisontal force, counter force, bending force, force to close the dies Stress state by Hydroforming of tube Stress state of tube hydroforming with one axis load Direction of long the axis Radius direction Tangential direction Stress state of tube hydroforming with vertical and horisontal loads Stress state of tube hydroforming with horisontal and bended loads Calculation the process parameters Calculation the process parameters p – fluid pressure for shaping of tube Fs – counter force Fc – closing force Fa, Fb, Fp, Fs – Forces for forming process (axial Force, bending Force…) Calculation the process parameters in case of one axis load The equation to describe the shaping of tube: σθ σ Z p + = Rθ R Z ti Where: σ θ , σ Z Tangential and along axial stress appeared by fluid pressure p Rθ,RZ Radius of horizontal and vertical sections of tube ti – currently thickness of tube 10 Due to Rz is very big, hence: σθ p = R θ ti Base on plasticity rule p = σ θ t i / R θ σ θ = σS Where σS - Yield stress Therefore, the minimal fluid pressure needed for shaping of tube is: p ti = σS Rθ The material has tensile strength σ B p max ty = 2.σ B d Where: ty – calculated thickness = 87.5 % t (according to experiment) t - initial thickniss d – outer diameter of initial tube 11 Calculation the process parameters in case of more axis loads Process parameters are: p – fluid pressure Fs – counter force (Counterholding force) Fc – closing force (Press force) Fa – axial Force 12 Calculation fluid pressure p: In the area of knot (vung dinh vau noi) the work piece will be strongly thined down That is serious area to create the crack Needed fluid pressure for shaping of tube: p ti = σS R 'θ As experiment: p = 0,13.σS + 1,15.σS t / d Fluid pressure by cracking of tube: p max ti = σB R 'θ 13 Calculation axial force Fa: Axial forces are very important during shaping process They assist the work pices flowed, displaced in to the area of knot Fa = Fσ + FP + Fµ Fσ - Forming force Fµ - Friction force Fσ Calculation d Fσ = FP – Axial punch force ∫σ Z 2π.p.dp d −t i d2 − ( d − ti ) 3 d '2 d' Fσ = πp + β.σS ( d '− t i ).t i + ln d'−2.t i 4 β =1,15 for flat stress state d’ – diameter of knot 14 Calculation the axial punch force FP FP ( d − 2.t i ) = p.π Calculation the friction force on the contact surface between work piece and die Fµ Fµ = ∫ µ.σ K dS µ - Friction coefficient S – Area of contact surface For T – Part, we have: li − d ' β.σS Fµ = µ.π.d. p + t i . .d d − 2.t i li, ti – currently length and thickness of tube 15 Finally, the axial force cam be calculate as follows: d d2 d t ( )( ) Fa = πp + βσ S 0,75( d − t ) − ln + , µ d l − d ' p + βσ S d − 2t d − 2t Where l, t – length and thickness of tube by finishing of process d’ – diameter of knot Calculation counter force (Counterholding force) Fs: πd '2y p − FS = σ Z '.π.d'cp t d 'cp and d 'y - middle and inner diameter of knot π( d'−2.t ) FS = p − ( 0,6 ÷ 0,7 ) σ B π.( d'− t ).t 16 Calculation closing force (press force) Fc: n Fc = ∑ σ K Si i =1 σK i i - contact stress on the contact surface Si - area of projection (dien tich hinh chieu) For the T-Part, we have: βσ S t i d( li − d ') + pd ' ( d + nh i ) Fc = p + d − 2t i Where li, hi, ti – currently length of tube, currently height of knot, currently thickness of tube n – number of knot 17 Example Material: C10 σ B = 340 N / mm , σS = 210 N / mm Thickness: mm Calculation : p – fluid pressure Fs – counter force Fc – closing force Fa – axial Force 18 Calculation fluid pressure p: Needed fluid pressure for shaping of tube: p = 0,13.σS + 1,15.σS t / d (d = 32 mm – mm = 30 mm) p = 0,13.210 + 1,15.210.2 / 30 = 43,4 N/mm Fluid pressure by cracking of tube: p max ti = σB R 'θ We have: R 'θ - minimal radius of knot R 'θ = d ' / = 26 / = 13 mm Pmax = 340.2 / 13 = 52,3 N/mm 19 Calculation axial force Fa: Fa = Fσ + Fp + Fµ d2 − ( d − ti ) 3 d '2 d' Fσ = π p + β.σS ( d'− t i ) t i + ln 4 d ' − t i β = 1,15 32 − ( 32 − 2) 30 30 3 ⇒ Fσ = π 43,4 + 1,15.210 ( 30 − ) + ln 30 − 2.2 4 ⇒ Fσ = 51336 N ≈ 5,1 Tons Fp ⇒ Fp ( d − 2.t i ) = p.π ( 32 − 2.2 ) = 43,4.π ⇒ Fp ≈ 2,7 Tons = 26710 N 20 l i − d' β.σ S Fµ = µ.π.d. p + t i . .d d − 2.t i li,– currently length = 50 mm ti - thickness of tube = mm µ - friction coefficient = 0,16 1,15.210 50 − 30' ⇒ Fµ = 0,16.π.30. 43,4 + t i . .30 30 − 2.2 ⇒ Fµ = 312018 N ≈ 31,2 Tons Fa = Fσ + Fp + Fµ = 2,7 + 5,1 + 31,2 ≈ 39 Tons 21 Calculation counter force (Counterholding force) Fs: π( d '−2.t ) FS = p − ( 0,6 ÷ 0,7 ) σ B π.( d '− t ).t π( 30 − 2.2 ) ⇒ FS = 43,4 − ( 0,6 ÷ 0,7 ) 340.π.( 30 − ).2 ⇒ FS = −12841 N = -1,2841 Tons Calculation closing force (press force) Fc: n=1 βσ S t i d( li − d ') + pd ' ( d + nh i ) Fc = p + d − 2t i 1,15.210.2 ⇒ Fc = 43,4 + .32( 50 − 30 ) + 43,4.30( 32 + 1.25) 32 − 2.2 ⇒ Fc = 113030 N ≈ 11,3 Tons 22