sheet metal forming · rapid sheet metal prototyping through the cutting, punching and bending of...
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Sheet metal forming
Shearing
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Mid-States Wool Growers
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Shearing of small and medium-sized metallic components
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Shearing M/c 2500mm x 8mm Lama Make Delhi
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Arm Front Suspension(Maruti Zen Car YE2)
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ClampsClamps
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Effect of die clearance
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Effect of a great blade clearance - sheet is bent off
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Die clearance• Optimizing the die clearance is the most important • Too large clearance leads to large burrs and poor part
quality. • Too tight clearance results in parts with
– Poor edge quality– Reduces tool life, and – Leads to more frequent tool component replacement.
• Optimal die clearance improves – Edge quality– Tool life – Regrind interval and – Reduces cost per hit.
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Penetration
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Punching force
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Punch Force (Fmax)
• Product of shear strength of metal sheet and cross-sectional area being sheared
• Friction between the punch and sheet– Because
• Shear zone subject to plastic deformation• Friction• Cracks• Various shape of punch-force vs. stroke curves
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Punch Force (Fmax)
• Punch Force = Fmax = 0.7(UTS) t L– Where
• UTS = Ultimate tensile strength of the sheet material• t = Thickness of the sheet material• L = Total length of sheared edge
– For round hole L = πD; D = Diameter of the round hole
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Problem
• A square hole of 25 mm side is to be punchedon a strip of 2 mm thickness of O-6061aluminium. Calculate the press force required.(see Table 3.7, Manufacturing processes forengineering materials by Kalpakjian &Schmid)
Bending
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Motorcycle Bicycle Car 90 Degree Bend Valve Extension
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Sell Metal Braided Flexible Tire Valve Extension- 210mm 90 Bend
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Flexible Valve Extensions and Brackets
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Rapid Sheet Metal Prototyping through the cutting, punching and bending of the sheet metal.
53Automotive sheet metal
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Sheet metal forming stamping bending welding parts
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WELDON-sheet metal forming stamping bending welding parts
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Sheet metal pressed components
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Bend allowance, Lb
• Lb = α ( R + kt)Where
α = bend angle, in radiansR = bend radius, is derived experimentally t = sheet thickness
k = a constant= 0.5, when neutral axis remains at the center= 0.33 for R < 2t
to = 0.5 for R > 2t
Minimum bend radii at room temperature
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Material Material conditionSoft Hard
Aluminium alloys 0 6tBeryllium, copper 0 4tBrass, low leaded 0 2tMagnesium 5t 13tSteel
Austenitic, stainless 0.5t 6tLow carbon, low alloy, HSLA 0.5t 4t
Titanium 0.7t 3tTitanium alloys 2.6t 4t
Engineering strain in outer/inner fibre
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innerouter
inner
outer
ee,llyTheoretica
1TR21
2TR
2TRR
e
1TR21
2TR
R2TR
e
=
−=
−
−−
=
+=
+
−
+
=
Where,R = Bend radius T = sheet thickness
Effect of R/T ratio
• With decrease in (R/T) ratio tensile strain in the outer fibre increases and it may crack.
• R, at which crack appears is the minimum bend radius
• Can be determined experimentally.
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Minimum bend radius
• Theoretically when
0.1tR2
0.1ee innerouter
+
==
Whereeouter and einner = strains at outer and inner fibres R = Bend radiust = thickness of sheetThe radius R, at which a crack appears on the outer surface of the bend is called minimum bend radius. It is to be determined experimentally.
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Plastic anisotropy
• Also called – Planar anisotropy– Strain ratio– E.g. Cold rolling of sheets
Estimation of maximum bending force, Fmax
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WLt)UTS(kF
2
max =
WhereUTS = Ultimate tensile strength of workL = Length of bendt = Thickness of the workW = Width of the die opening (see fig 7.22 , Manufacturing processes
for engineering materials by Kalpakjian &Schmid)k = The factor includes various factors, including friction
= 1.2 to 1.33 for V – die= 0.3 to 0.34 for wiping die
Problem
A specimen of 3 – mm and 200 mm width oftitanium alloy Ti 99.5 is bent in a V-die of 200 mmwidth. Calculate (a) the bend allowance and (b) thebending force. (see Table 3.14 & 7.2 etc,Manufacturing processes for engineeringmaterials by Kalpakjian & Schmid).
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Spring back
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Spring back
• Plastic deformation is followed by elastic recovery
• This recovery in bending is called spring back• Bend allowance before and after bending is
same
radiibendfinal&initial...R&R,Where
1TR2
1TR2
KfactorSpringback
2TR
2TRallowanceBend
fi
f
i
i
fs
ffii
−−
+
+
=αα
==−
α
+=α
+=−
Cont.........
• Ks = 1 indicates no spring back• Ks = 0 shows complete elastic recovery e.g.
Spring
• Negative spring back– Spring back occurs in reverse direction e.g. V-die
bending
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Estimate of spring-back
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1ET
YR3ET
YR4RRspringback.of.Estimate i
3i
f
i +
−
==
WhereRi = Radius before spring backRf = Radius after spring backY = Uni-axial yield stressE = Modulus of elasticity of work materialT = Thickness of the work
Compensation for spring back
• Over bending• Coining• Stretch bending• Bending at high temperature reduces spring
back
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Problem
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A straight bead is being formed on a 1.5 mm thickaluminium sheet in a 20 mm diameter die as shown infigure (Problem 7.72, Manufacturing processes forengineering materials by Kalpakjian & Schmid). If theyield stress of the material is Y = 250 MPa, taking anaccount the spring back, calculate the outsidediameter of the bead after its formation and removalfrom the die.
Deep Drawing
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Deep Drawing
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Steps in deep drawing process
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The Deep-Draw ProcessCup Operation: Converts the flat blank to tubular form that resembles a cup.First Draw: Reduces the diameter of the cup.Second Draw: Further reduces the diameter of the cup and lengthens the part.Final Draw: Brings the part to the desired outside and inside dimensions.Pierce: The bottom of the tubular component is punched out and the slug is extracted.Clip: The uneven edge is removed and the finished part is ejected.
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Deep drawing is done in several steps with a number of punches and drawing dies.
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Deep-draw stainless steel cylinders for automotive parts and home appliance parts.
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precision sheet metal deep drawn power window regulator electric motor shell
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Deep drawing products by Geerts
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Automotive Sheet Metal Part
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Deep Drawn Components in Faridabad.S. I. Engineering Products Pvt. Ltd.Faridabad
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Analysis of deep drawing process
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Deep draw ability (limiting drawing ratio, LDR)
• LDR = blank diameter, Di / punch diameter, Dp
• Failure occurs due to thinning.• Thinning is due to movement of the material
in the cavity• Hence, the material must be capable of
– Undergoing reduction in width– Resist thinning under tensile stresses in cup wall
• LDR can be calculated theoretically (Solved Example 7.8)
Empirical equation to estimate maximum force in deep drawing
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−= 7.0
DD)UTS(tDF
p
bbpmax π
WhereFmax = Maximum deep drawing forceDb = Blank diameterDp = Punch diametertb = Blank thickness
Problems
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Problem
• Estimate the force required for deep drawing with a blank diameter of 200 mm and a punch diameter of 125 mm, if the material is 3 mm thick and is of titanium alloy Ti 99.5. (see Table 3.14 , Manufacturing processes for engineering materials by Kalpakjian &Schmid)
Thanks
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