Sheet Metal Working: Series of operaons that involve cung, bending and drawing sheet metal. Sheet metal (from 0.4mm or 1/64in to 6mm or 1/4in thickness); Plate (from 6mm upwards). Operaons are usually performed as cold working. Advantages of Sheet Metal: Thickness= 0.4 to 6 mm (1/64 to 1/4 in) • Relavely low cost. • Good dimensional accuracy and good surface finish. • High strength Basic Types of Sheet Metal Operaons Cung: It involves processes such as punching, shearing and blanking. Bending: Deform the sheet around a straight axis. Drawing: Deform the sheet into convex or concave shapes. Mechanical Properes of Sheet Material

www.nzsteel.co.nz

When subjected to a tensile force there are three deformaons to be measured: the longitudinal strain, the strain in the width direcon and the strain in the thickness direcon. The material is anisotropic.

finance.pipex.com Shearing: Shearing of sheet metal between two cung edges: (1) just before the punch contacts work (2) punch begins to push into work, causing plasc deformaon; (3) punch compresses and penetrates into work causing a smooth cut surface; (4) fracture is iniated at the opposing cung edges which separates the sheet.

Forces in a Shearing Operaon L = Total length Sheared t = thickness Force = 0.7×UTS × t × L

Shearing: A cung metal operaon usually along a straight line, between two cung edges. Punching and Blanking

Punching and blanking are very similar. In punching the cut piece is scrap, while in blanking the cut piece is the desired product. Convenonally sheared surface showing the disnct regions of deformaon and fracture and (boom) magnified view of the sheared edge. (Courtesy of Feintool Equipment Corp., Cincinna, OH.)

Percent Penetraons Penetraon=Roll over + Burnish Material % Penetraon Characteriscs of a Die Cut edge Silicon Steel 30 Roll Over – Flow of material around the punch and die .The larger Aluminum 60 the clearance the greater the roll over .10 C Steel Annealed 50 Burnish – The rubbed or “cut” poron of the edge. The sharper .10 C Steel Cold Rolled 38 the punch the wider the burnish .20 C Steel Annealed 40 Fracture – The angled surface where the material separates from .20 C Steel Cold Rolled 28 the parent material .30 C Steel Annealed 33 Burr – The very sharp projecon caused by a dull cung on the .30 C Cold Rolled 22 punch or die.

E.V. crane, Plasc Working in Presses, John Wiley and Sons, Inc., New York, 1948, p. 36 Fineblanked surface of the same component as shown. (Courtesy of Feintool Equipment Corp., Cincinnati, OH.) Clearance: It is defined as the distance between the punch cung edge and the die cung edge. It depends on the hardness and thickness of the material. As the thickness increases, the clearance must increase. The clearance typical values ranges from 4% to 8% of the thickness of the material. The recommended clearance is calculated by: c=a.t where c=clearance, t=thickness and a=allowance Metal group a _ aluminum alloys (1100, 5052) 0.045 Die size determines blank size Db; punch aluminum alloys (2024 and 6061); brass, 0.060 so cold rolled steel, so stainless steel size determines hole size Dh.; c = clearance cold rolled steel, stainless steel, (hard & half-hard) 0.075

For a round blank of diameter Db is determined as: Blank punch diameter = Db - 2c Blank die diameter = Db For a round hole (piercing) of diameter Dh is determined as: Hole punch diameter = Dh Hole die diameter = Db + 2c Forces in a Punching Operaon Angular Clearance: Allows the blank or the slug to drop L = Total length Sheared off easily. Typical values t = thickness ranges from 0.25degrees to Force = 0.7×UTS × t × L 1.5 degrees Forming Properes of Sheet Metal Sheet metal, due to its manufacturing process is Typical Range of Average Normal Anisotropy (R ) for various Sheet Metals not an isotropic material. Anisotropy is caused by avg the thermal processing of the sheet. Two types, Zinc 0.2 namely crystallographic anisotropy and mechanical Hot rolled steel 0.8-1.0 fibering. Example: Low carbon steel exhibits an upper Cold rolled rimmed steel 1.0-1.35 and lower yield strength. As a result during Cold rimmed aluminum–killed steel 1.35-1.8 deformaon, it shows stretch strain bands

(Lueder’s bands). These can be eliminated by a Aluminum 0.6-0.8

Copper and Brass 0.8-1.0 reducon of thickness of 0.5 to 1.5% by cold rolling. Titanium 4-6

Example Esmate the force required for punching a 25mm diameter hole through a 3.2mm thick annealed tanium alloy Ti-6Al-4V sheet at room temperature. Data: UTS = 1000MPa F = 0.7×UTS × t × L = 0.7×1000MPa × 0.0032m × π × 0.025 =176kN Cutlery manufacturing: Cutlery manufacture involves blanking the stainless steel or sterling silver to the proper shape. A series of rolling operaons then gives the piece the correct thickness. Aer heat treatment and trimming, the piece has a paern embossed on it in a stamping operaon. Finally, the piece is buffed and polished. Blanking: The metal inside a closed contour is the desired part

Punching: The metal inside the contour is discarded.

Notching: Edges or corners of a material is punched.

Trimming: Cung scrap or excess material for a fully or parally shaped part Shaving: Finishing operaon of a previously cut edge by removing a minimum amount of material. Progressive Stamping Dies Common method to handle complex parts Deep Drawing Complex 3D shapes can be made out of sheet metal. Blanking

Deep Drawing

Re-drawing

Ironing

Doming

Necking

Seaming Usually a cold working process. A punch forces a flat sheet metal into a deep die cavity. The die cavity is usually circular or rectangular. When the depth of the product is greater than its diameter, it is known as Deep Drawing and when the depth of the product is less than is diameter, it is known as Shallow Drawing. The sheet metal is supported on both sides by the blankholder, to avoid wrinkling If the Hold-down pressure (blankholder force) is too high the www.endo-mfg.co.jp sheet will tear and if it is too low it will wrink.

Draw beads may be used to control metal flow.

Esmaon of the Blank Diameter

π 2 π 2 2 D = d + πd h ⇒ D = d + 4d h 4 o 4 1 1 o 1 1 Formability Test: Deformaon in sheet materials are carried out by either stretching and/or drawing. The ability of the sheet to withstand large degrees of streching or drawing deformaon (shape change) without failure is known as formability.

Erichsen Test – Cupping Test A round punch is forced into a clamped sheet unl a crack (sudden drop in force) appears. Forming Limit Diagram (FLD) A FLD shows what combinaons of the major and minor strains produce failure in a sheet metal. To develop the FLD, the major and minor engineering strains are obtained. The curves represent the limits of drawing between failure and safe regions, By measuring the ellipses on the deformed paern, the largest and shortest direcons of the ellipses are the major strains and minor strain respecvely. Please note that the axes for these strains are 900 apart. This can be carried out at many different locaons in the work-piece. If both major and minor strains are posive, the deformaon are stretching, and the sheet metal will decrease in thickness. If the minor strain is negave, this contracon may parally or whole compensate any posive stretching in the major direcon. The combinaon of tension and compression is known as drawing, and the thickness may decrease, increase, or stay the same, depending on relave magnitude of the two strains. Example: A grid of 2.5mm circles is electroetched on a blank of AK sheet steel. Aer forming into a complex shape the circle in the region of crical strain is distorted into an ellipse with major diameter of 4.5mm and minor diameter of 2.0mm. Is the component close to failure??

4.5− 2.5 Major strain ε = ⋅100 = 80% 1 2.5 2.0 − 2.5 Minor strain ε2 = ⋅100 = −20% 2.5

The coordinates indicate that the part is in imminent danger.

Maximum − D Limit Drawing Rao LDR = blank It is defined as the rao between the largest diameter of Dpunch the blank that can be drawn into a specific punch diameter D without failure: DR = blank The recommended drawing raos are the following: D • for the first drawing: ~2 punch • for the second drawing: 1.2 – 1.25 • for the third drawing: 1.15 -1.18 • for further drawings: 1.1 -1.12

Reducon: Dblank − Dpunch The value of the reducon (r) should be less than 0.5 for reduction = r = a cylinder. Dblank

Thickness to Diameter rao t thickness − diameter − ratio = blank The thickness of the starng blank divided by the D blank blank diameter. The recommended values for this rao are greater than 1%. As the rao decreases, the tendency for wrinkling increases.

The starng diameter of the blank must be of the right size for the final dimensions of the cup to be correct. Assume constant volume and neglect any thinning during the process. Example: Determine if the following is feasible for manufacturing: A cylindrical cup with an inside diameter of 3.0in and height of 2.0in. Its starng blank size id 5.5in and its thickness 3/32in.

D 5.5 DR = blank = =1.833 < 2 Dpunch 3.0

Dblank − Dpunch 5.5− 3.0 r = = = 0.4545 < 0.5 Dblank 5.5

3 The drawing operaon is feasible. t blank = ( 32) = 0.017 > 0.01 D blank 5.5 Anisotropy Rao There are two different types of anisotropy rao, namely, normal and planar anisotropy rao. Normal Anisotropy Rao (R) : Measured in a tensile specimen, it is the rao between the true strain in the width direcon and the true strain in the thickness direcon. The tensile specimen must conform specific technical standards. The longitudinal direcon of the tensile specimen can be parallel or to a certain angle with respect to the rolling direcon of the sheet. εwidth R0 + 2R45 + R90 R = Raverage = εthickness 4 Planar Anisotropy Rao: It determines the variaon of the R + R true strain in the plane of the sheet (rolling plane). ΔR = R − 0 90 45 2

The value of the normal anisotropy rao determines the liming drawing rao and the value of the planar anisotropy rao correlates with the material propensity to earing. High values of normal anisotropy combined with low values of planar anisotropy provides opmal drawability. The maximum value of the normal anisotropy also depends on the grain size of the material.

Note: If ΔR=0, no ears form. The height of ears increases as Δ R increases. Maximum blank diameter D LDR = = b Punch diameter Dp

The relaonship between average normal anisotropy and the liming drawing rao for various sheet metals. Source: Aer M. Atkinson. Example:

A special deep-drawing steel showed a 30% longitudinal elongaon and 16% decrease in thickness when it is subjected to a tensile test. Esmate the liming drawing rao (LDR) for this steel. l − l l o = 0.3 ⇒ =1.3 ⇒ ln(1.3) = 0.26236 lo lo w − w w o = −0.16 ⇒ = 0.84 ⇒ ln(0.84) = −0.1743 wo wo w w ln o ln o 1 w w ln R = ( ) = ( ) = ( 0.84) =1.98 h ln o ln wl ln(0.84⋅1.3) ( h ) ( wolo )

From the graph the LDR~2.7 Drawing Force

(! D $ + * b - Fmax = π Dpt(UTS) # &− 0.7 )*" Dp % ,- Wrinkling can be reduced if a blankholder is loaded by maximum punch force The force increases with increasing blank diameter, thickness, strength and the rao Bending Some sheet are bend along certain lines to produce a desired shape. Bending introduces plasc deformaon to the part and it should remain in the desired shape (angle) aer the load is released. Spring-back is the part of deformaon (the elasc part) that recovers in the plascally deformed material once the load has been released.

Original shape Desired deformed shape Springback

When the load is released there is a decrease in the bending angle (Springback) due to the elasc recovery of the material. Bend allowance: The amount of deformaon of the neutral axis of the sheet depends on the bend radius and bend angle. The final dimension of the neutral axis (in the bending area) is used to calculate the blank length for the bend part. R is the bend radius, α is the bend angle , t is the thickness and k is a constant. In an ideal case, the neutral axis remains at the center of the secon k=0.5. In pracce, k ranges from 0.33 (for R<2t) to 0.5 (for R>2t).

Lb = α (R + kt) = αo (Ro + kt) = α f (Rf + kt) As the part is bended, the longitudinal dimension of the flat length is increased. The bend allowance is the amount of material that need to be added to the flange dimensions (leg parts) in other to develop a flat paern. Example: suppose that flanges lengths of 2” and 3” with an inside radius of 0.250” at 90degrees are required. Then the flat dimensions are (2”-(0.25+0.125)) and (3”-(0.25+0.125)), i.e 1.625 and 2.625 respecvely. The length of the flat sheet (bend allowance) is 1.625 + 2.625 + 0.457 = 4.707” Bend deducon: It is the amount of material that has to be removed from the sum of the flanges to obtain a flat paern.

! 90 $ Lb = α (R + kt) = #π &(0.250 + 0.33× 0.125) = 0.457 " 180 %

Springback Aer releasing the pressure of the forming tool, the deformed work- piece experience a dimensional change (strain) due to the elasc recovery of the material. Sprinback is found in all forming operaons, but is is more pronounced in bending. As the yield strength of the material increases or as the Modulus of elascity decreases the springback deformaon increases.

Overbending, i.e. bending to a smaller radius of curvature than the required can compensate for the springback of the Original shape Desired deformed shape Springback material. For aluminum alloys and austenic steels the 3 springback can be approximated by the equaon Ro ! Roσ yield $ Roσ yield where Ro is the radius of curvature before = 4# & − 3 +1 releasing the load and Rf is the radius of Rf " Et % Et curvature aer releasing the load; t is the thickness, E is the Modulus of Elascity and σyield is the yield stress. Example

A 0.0359in thickness sheet (20-gage) is bent to a radius of 0.5in. Calculate the radius of the part aer it is bent and the required bend angle to achieve a 90o bend aer springback has occurred. Data: Yield Strength=40000psi; E=29x106psi

3 R ! Roσ yield $ Roσ yield o = 4# & − 3 +1 Rf " Et % Et 3 0.5 ! 0.5⋅ 40000 $ 0.5⋅ 40000 = 4# 6 & − 3 6 +1= 0.942 Rf " 29 ⋅10 ⋅ 0.0359 % 29 ⋅10 ⋅ 0.0359

Rf = 0.531in

Lb = αo (Ro + kt) = α f (Rf + kt) ! 0.0359 $ ! 0.0359 $ αo #0.5+ & = 90 ⋅#0.531+ & " 2 % " 2 % o ⇒ αo = 95.4 Bending Force K ⋅ L ⋅UTS ⋅t 2 The bending load can be calculated from the following equaon: Fb = Where UTS is the ulmate tensile strength of the material (psi); W L is the length of the bent part (in), t is the thickness (in); W is the width between the contact points (in) or 8t for V-bends K is 1.3 for die opening of 8t, 1.20 for die opening of 16t, 0.67 for U-bending, 0.33 for a wiping die Example: Esmate the force required for a 90 degrees bending of a St 50 steel of thickness of 2mm in a V die. The die opening can be taken as eight mes the thickness. The length of the part is 1m.

Die Opening W=8*2=16mm UTS=500MPa

2 K ⋅ L ⋅UTS ⋅t 2 1.33⋅1⋅ 500 ⋅ 0.002 F = = ( ) =166.25kN b W 0.016 Minimum Bend Radius: On the inside of neutral plane, the metal is compressed, while on the outside of the neutral plane is stretched. The outer layers under tension should not be excessively stretched as there is the possibility of rupture. The amount of stretching depends on the sheet thickness and the bend radius. There is a minimum bend radius that depends on the material properes. Minimum bend radius 0.5t for so materials, 3t for spring steels and 1t for the others. Example: A sheet material is to be bended according to the dimensions given in the figure. Determine the following: (a) starng blank size and (b) the bending force necessary if a V-die will be used with a die opening dimension of W=1.0”. Data: E=30000ksi ; Yield Strength=40000psi and a tensile strength=65000psi.

Length L = 1.75’, and the length of the part is: 1.5 +1.00 + BA. R/t = 0.187/0.125 = 1.5 < 2.0, so Kba = 0.33 For an included angle A’ = 1200, then A = 600 ! 60 $ Lb = α (R + kt) = #π &(0.187+ 0.33× 0.125) = 0.239" " 180 % Length =1.5+1.0 + 0.239 = 2.739"

2 2 Force: F= (KbfTSLt )/W = 1.33 (65,000)(1.75)(0.125) /1.0 = 2,364 lb

2 K ⋅ L ⋅UTS ⋅t 2 1.33⋅1.75⋅ 65000 ⋅ 0.125 F = = ( ) = 2364lb b W 1 Bending and Forming Tubes Stretch-Forming The form die is pressed into the work with force Fdie, causing it to be stretched and bent over the form. F = stretching force. It is used extensively in the aircra industry to produce parts of large radius of curvature. The materials used are very ducle. Spinning Ideal for • Lower producon volumes • Large parts • Inexpensive tooling

www.dissco.co.nz

www.ashfordmetalspinning.co.uk www.tradional-building.com (a) Schemac illustraon of the shear-spinning process for making conical parts. The mandrel can be shaped so that curvilinear parts can be spun. (b) and (c) Schemac illustraons of the tube-spinning process Roll Bending Large metal sheets and plates are formed into curved secons using rolls

Roll Forming Connuous bending process in which opposing rolls produce long secons of formed shapes from coil or strip stock High Energy Rate Forming Short me – High Energy forming processes. It includes explosive forming, electrohydraulic forming Explosive Forming and electromagnec forming. Use of explosive charge to form sheet (or plate) metal into a die cavity. Explosive charge causes a shock wave whose energy is transmied to force part into cavity. Applicaons: large parts, typical of aerospace industry. Electromagnec Forming Sheet metal is deformed by mechanical force of an electromagnec field induced in the work-piece by an energized coil. Presently the most widely used HERF process Applicaons: tubular parts

A pinched aluminum can, produced from a pulsed magnec field created by rapidly discharging 2 kilojoules from a high voltage capacitor bank into a 3-turn coil of heavy gauge wire. Source: Bert Hickman, Stoneridge Engineering. Hydroforming

Hydroforming uses water at high pressure to force the piece into a specific shape.