Calculation for tube HForming Tính toán thiết kế công nghệ dập thủy tĩnh phôi ống (IHU)

Pipe products – Various shapes of pipes
Connecting parts

Cam shaft

Crank shaft

1


Classification

HYDROFORMING OF TUBE

HYDROFORMING OF TUBE
with one axis load

HYDROFORMING OF TUBE
with more axis loads

2


HYDROFORMING OF TUBE
with one axis load

Fluid pressure, vertical and horisontal forces

3

HYDROFORMING OF TUBE
with more axis loads

Horisontal force, counter force, bending force, force to close the dies

4


Stress state by Hydroforming of tube
Stress state of tube hydroforming with one axis load

Direction of long the axis
Radius direction
Tangential direction
5


Stress state of tube hydroforming with vertical and horisontal loads

6


Stress state of tube hydroforming with horisontal and bended loads

7


Calculation the process parameters

8

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…)
9


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 min

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


1. 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 min

ti
= σS .
R 'θ
As experiment:

p min = 0,13.σS + 1,15.σS .t / d

Fluid pressure by cracking of tube:

p max

ti
= σB .
R 'θ
13


2. 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

2

∫σ

Z

.2π.p.dp

d −t
2 i

 d2 − ( d − ti ) 2
3

d '2
d'  
Fσ = πp.
+ β.σS . ( d '− t i ).t i + . ln


4
8
d'−2.t i  
4

β

=1,15 for flat stress state
d’ – diameter of knot

14


Calculation the axial punch force FP

FP

2
(
d − 2.t i )
= p.π

4

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   2 

li, ti – currently length and thickness of tube

15


Finally, the axial force cam be calculate as follows:

 d
d2
d 
t 


(
)(
)
Fa = πp. + βσ S 0,75( d − t ) − ln
+
0
,
5
µ
d
l
−
d
'
p
+
βσ

S

8 d − 2t 
d − 2t 

 4
Where l, t – length and thickness of tube by finishing of process
d’ – diameter of knot
3. Calculation counter force (Counterholding force) Fs:

πd '2y
p.

− FS = σ Z '.π.d'cp .t
4
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
4
2

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4. 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 2 , σS = 210 N / mm 2 .
Thickness: 2 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 min = 0,13.σS + 1,15.σS .t / d

(d = 32 mm – 2 mm = 30 mm)

p min = 0,13.210 + 1,15.210.2 / 30 = 43,4 N/mm 2
Fluid pressure by cracking of tube:

p max

ti
= σB .
R 'θ

We have:

R 'θ

- minimal radius of knot

R 'θ = d ' / 2 = 26 / 2 = 13 mm

Pmax = 340.2 / 13 = 52,3 N/mm 2

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Calculation axial force Fa:

Fa = Fσ + Fp + Fµ
 d2 − ( d − ti ) 2
3

d '2
d'  

Fσ = π p.
+ β.σS  ( d'− t i ) t i + . ln


4
4
8
d
'
−
2
t


i


β = 1,15
2

32 2 − ( 32 − 2)
30 2
30  
3

⇒ Fσ = π 43,4.
+ 1,15.210 ( 30 − 2 ) 2 +

. ln

4
8
30 − 2.2  
4

⇒ Fσ = 51336 N ≈ 5,1 Tons

Fp

⇒ Fp

2
(
d − 2.t i )
= p.π.

4

(
32 − 2.2 )
= 43,4.π.

⇒ Fp ≈ 2,7 Tons

4

2

= 26710 N
20



  l i − d' 
β.σ S
Fµ = µ.π.d. p +
t i .
.d
d − 2.t i   2 

li,– currently length = 50 mm
ti - thickness of tube = 2 mm
µ - friction coefficient = 0,16

1,15.210   50 − 30' 

⇒ Fµ = 0,16.π.30. 43,4 +
t i .
.30
30 − 2.2   2 

⇒ Fµ = 312018 N ≈ 31,2 Tons.

Fa = Fσ + Fp + Fµ
= 2,7 + 5,1 + 31,2 ≈ 39 Tons.
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Calculation counter force (Counterholding force) Fs:

π( d '−2.t )
FS = p.
− ( 0,6 ÷ 0,7 ) σ B .π.( d '− t ).t
4
2
π( 30 − 2.2 )
⇒ FS = 43,4.
− ( 0,6 ÷ 0,7 ) 340.π.( 30 − 2 ).2
4
⇒ FS = −12841 N = -1,2841 Tons
2

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