BENDING OF SHEET METALS

Dr.R.Narayanasamy,B.E.,M.Tech.,M.Engg.,Ph.D.,(D.Sc) Professor, Department of Production Engineering ,

National Institute of Technology,Tiruchirappalli - 620015,

Tamil Nadu, India.

Basic

• A straight length is converted into a curved length.

• Neutral axis passes through zero strain.• Neutral axis is at the center of the

thickness.• After bending, t < to

Where, t -metal thickness after bend

to - metal thickness before bend

Bending…

Continued….

• As radius of bend (r) increases, thinning increases.

• Bend length (B) = Where, α – bend angle

• Above neutral axis, stress is tensile.• Below neutral axis, stress is compressive.

tR 45.0360

2

Principal directions

• εθ is always greater. Where, ‘θ’ is longitudinal direction

• When α > 900, εθ is uniform. (Where,α is bend angle)

• εθ is greater at the edges.

• When w/t > 8, εtransverse is zero.

ε1Circumferential

ε2Transverse

ε3Radial

Bend radius

• For many parts, sharp bend radius is required.

• Sharp bend leads to fracture.• Bend radius < t, should be avoided

otherwise fracture takes place.• For a given metal and ‘t’, the bend radius

cannot be below some minimum value.

Continued….• Rmin depends on

1. angle of bend (α)2. width of bend (w)

• As bend angle increases, Rmin decreases.• As bend width (w) increases, Rmin decreases.• Rmin depends on sheet thickness (t).• Rmin/t is independent of the metal.• Rmin/t increases as sheet thickness (t) increases.• Rmin varies from metal to metal.

Annealed metals

• Rmin is equal to sheet thickness (t).

• For good annealed metals, Rmin < 0.5t

• For hard metals, Rmin

≈4 to 5t

Temper Rmin

Annealed1/4 hard1/2 hard3/4 hard

Full

0.5t-1.0t1.0t-2.5t2.5t-4t3t-5t4t-6t

Rmin for Austenitic stainless steels :

1

2

3

4

Circumferential Strain

Percentage

Angle of Bend30 60 90 120 1500

10

20

30

40

50

Minimum bend angle for producing constant strain

For Increasing

TR

R T1. 0.090 0.1252. 0.375 0.2503. 0.500 0.2504. 0.625 0.250

Minimum R for Austenitic Stainless Steels

Continued….

• When bend angle, α < 900, εθ is proportional to the bend angle.

• εθ reaches a maximum value when

α ≈900

• Rmin/t increases when εθ decreases.• As bend width (w) decreases, failure at smaller

strain.• When w > 8t, Rmin is constant.

• Rmin depends upon cut-edge condition.

Effect of edge

• Smooth edge permits larger Rmin

• Bad edge does not permit larger Rmin

• These affect particularly in Aluminum alloys

Effect of bend width on R min

TW

7075 - T

2024 - TTRmin

Effect of edge condition on Rmin

TRmin

TW

New Saw

Dull Saw7075 - T

Pinching method of sharp bending

Effect of ductility on Rmin

• High ductility increases Rmin

• High reduction in area in tensile test increases Rmin

Rmin/T

% Elongation

Rmin/T

% Reduction in area (q)

A1 AB

R1

R

Spring Back

Spring back• After release of load, bent part assumes different shape• Elastic distortion is called spring back• After spring back, the radius of curvature increases• Assume bend length (B) is constant,

B = (R + 0.5t) A → under loadB = (R1 + 0.5t) A1 → after release of load

Bend length (B) is equal for bothK = (R1 + 0.5t) / (R + 0.5t) = A / A1

Where ‘K’ is the spring back factor

Aluminium alloy 2024.0, 7075-0

Annealed Aust stainless steel (ASS)2024 T

¼ hand A.S.S.

½ hand A.S.S.

0.5

0.6

0.7

0.8

0.9

1.0

1.0 20

Sprin

g B

ack

K

R/T ratio in log scale

Continued….

• ‘K’ is independent of Rmin/t• For small bend radius, K≈ 1.0• For larger bend radius, K is very less.• If bend radius larger, deformation or

bending will be elastic.

Stress State

0 2 4 6 8 10 12

10

20

30

40

50

0

0.1

0.2

0.3

0.4

0.5

Bending Strain %

Bending Strain

w/t

σ2/σ1

σ2/σ1

Stress state

• (σ1/σ2)ratio decreases with the decrease in ductility.

• (σ1/σ2) ratio increases with increasing ratio of (w/t) where, σ2 → Transverse Stress σ1 → Circumferential (or) Longitudinal Stress

• For low (w/t) ratio, (σ1/σ2) decreases(i.e.) Stress state is pure tension

Continued….

• As (w/t) = 8, (σ1/σ2) is maximum

• (σ1/σ2) ratio is maximum of 0.50• As (w/t) increases, cracking takes place at

the centre• As (w/t) decreases, cracking takes place

at the edge

Distance along circumference

(ε1)C

ircum

fere

ntia

l St

rain

%

0

10

20

30

40

30°

60°

90°

120°

150°

Bend Angle

R/T = 0.7

2024 – T Al alloy

1/8” thick

Distance along circumference

Bend Angle

R/T = 2.5

2024T Al alloy 1/4” thick

30° 70° 110° 150°(ε1)

-20

0

20

40

% T

rans

vers

e st

rain

2024-T Al. alloy sheet

Distance across specimen (along width)

ε1ε2

30

10

2024 Al. alloy

- 0.5

- 0.4

- 0.3

- 0.2

- 0.1

0

ε2/ε1

Distance across specimen

12

w/t = 4

w/t =8

Al alloy 2024 T

Experimentalresults

w/t = 1

2

4

ε2

(theoretical circumferential strain)

Curvature T/R

Nat

ural

str

ain

Tool design aspects

• Spring back allowance depends on young’s modulus

• Spring back is 5 to 6° for cold rolled metals

• Spring back is 3 to 4° for annealed metals

Thank You