Forging Analysis - 1

1 Forging Analysis - 1. ver. 1. Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Overview Slab Analysis frictionless with friction Rectangular Cylindrical Strain hardening and rate effects Flash Redundant work Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Slab Analysis assumptions Entire Forging is plastic no elasticity Material is perfectly plastic strain hardening and strain rate effects later Friction coefficient ( ) is constant all sliding, to start Plane strain no z-direction deformation In any thin slab, stresses are uniform Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Open die Forging Analysis . rectangular part F. unit depth into figure y h x dx w F. Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Expanding the dx slice on LHS. p f p = die pressure x, d x from material on side x x + d x friction = friction force = p f p Prof.

2 Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Force balance in x-direction hd x + 2 friction dx = 0. 2 friction d x = dx h k Mohr's circle 2 p x x + p = 2k = flow = flow 3. (distortion energy (von Mises) criterion, all done on a per plane strain) unit depth basis Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Force balance Differentiating, and substituting into Mohr's circle equation d (2k ) = d ( x + p ) dp = d x 2 friction 2 friction . d x = dx dp = dx h h . noting: friction = p 2 dp 2 . dp = pdx = dx h p h Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Sliding region px 2 . x dp 2k p = 0 h dx Noting: @ x = 0, x = 2k = flow Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging pressure sliding region x ln px ln (2k ) = 2 . h Sliding region result (0 < x < xk). px 2 x . = exp . 2k h . 2 x . p x = flow exp done on a h per unit depth basis Prof.

3 Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging pressure . approximation Taking the first two terms of a Taylor's series expansion for the exponential about 0, for x 1. 0 0. 2 3 n n xk . x x x exp(x ) = 1 + x + + +L+ =. 2! 3! n! k! k =0. yields p x 2 x 2 x . = 1 + p x = flow 1 + . 2k h h . Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Average Forging pressure . all sliding approximation using the Taylor's series approximation w w w 2. px 2 x . 2 2 x 2. 2.. pave 0 2k dx 1 +. 0 . dx x +. h 2h 0.. = = =. 2k w w w 2 2 2. pave w . = 1 + . 2k 2h . w done on a pave = flow 1 + per unit depth basis 2h . Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging force . all sliding approximation F Forging = pave width depth w . F Forging = flow 1 + w depth 2h . Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Slab - die interface Sliding if f < flow Sticking if f flow can't have a force on a material greater than its flow (yield) stress deformation occurs in a sub-layer just within the material with stress flow sliding region sticking region Prof.

4 Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Sliding / sticking transition Transition will occur at xk using k = p, in: px 2 x k 2 xk . = exp = exp . 2k h 2 k h . hence: xk 1 1. = ln h 2 2 . Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Sticking region 2 . dp = pdx h Using p = k/ . 2 k dp = dx h . px x 2k 2k p dp = x h dx p x p xk = ( x xk ). kx k h Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Sticking region We know that at x = xk, px = k/ . k and xk 1 1. = ln h 2 2 . Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging pressure - sticking region Combining (for xk < x < w/2). px 1 1 x = 1 ln . +. 2k 2 2 . h 1 1 x . p x = flow 1 ln + . 2 2 h . Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging pressure . all sticking approximation If xk << w, we can assume all sticking, and approximate the total Forging force per unit depth (into the figure) by: p pedge 0.

5 X=0 x=w/2. Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging pressure . all sticking approximation pedge = 2k px x 2k 2k 2k dp = 0 h dx p x 2k = (x ). h px x . = 1 + . 2k h . x . p x = flow 1 + . h . Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Average Forging pressure . all sticking approximation w w w 2. px x 2 x . 2 2.. pave 0 2k dx 1 + dx 0 . h . x + . 2h 0. = = =. 2k w2 w2 w2. pave w . = 1 + . 2k 4h . w . pave = flow 1 + . 4h . Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging force . all sticking approximation F Forging = pave width depth w . F Forging = flow 1 + w depth 4h . Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Sticking and sliding If you have both sticking and sliding, and you can't approximate by one or the other, Then you need to include both in your pressure and average pressure calculations.

6 Fforging = Fsliding + Fsticking F Forging = ( pave A)sliding + ( pave A)sticking Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Material Models Strain hardening (cold below recrystallization point). flow = Y = K n Strain rate effect (hot above recrystallization point). flow = Y = C ( & ) m 1 dh v platen velocity & = = =. h dt h instantaneous height Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-1. Lead 1 x 1 x 36 36 . y = 1,000 psi 1 . hf = , = 1 . Show effect of friction on total Forging force. Use the slab method. Assume it doesn't get wider in 36 direction. Assume cold Forging . Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-2. At the end of Forging : hf = , wf = 4 (conservation of mass). Sliding / sticking transition xk 1 1. = ln h 2 2 . 1. xk = ln = ". 2 2 Prof. Ramesh Singh, Notes by Dr.

7 Singh/ Dr. Colton Forging - Ex. 1-3. Sliding region: 2 x . px = flow exp . h . f . = 1150 exp(2 x ). Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-4. Sticking region 1 1 x . p x = flow 1 ln + . 2 2 h f . = 1150 ( + 4 x ). Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-5. sliding sliding sticking friction 12000 hill Forging pressure (psi). 10000. 8000 xk xk 6000. 4000 P(sticking). 2000 P(sliding). 0. 0 1 2 3 4. Distance from Forging edge (in). Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-6. Friction hill Forging pressure must be large ( ) near the center of the Forging to push the outer material away against friction Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-7. Determine the Forging force from: Force = p dA. since we have plane strain x F. = p x dx unit depth 0.

8 Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-8. We must solve separately for the sliding and sticking regions xk w/ 2 . F Forging = 2 p x dx .depth + 2 p x dx .depth . 0 x . sliding k sticking Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-9. Sliding first xk pave 2 x . flow 0 exp h dx h 2 x xk = = exp . unit depth xk 0 2 h 0. h 2 xk . exp 1 . 2 h . =. (xk 0). Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-12. Substituting values sliding pave 2 . exp 1 . flow 2 . = = unit depth ( 0). sticking pave 1 1 4 1 42 2 . 1 ln + . flow 2 2 2 2 4 . =. unit depth 4. 2. = Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-10. Sticking next w pave 1 . 2. 1 x . flow x 2 1 ln 2 + h dx = k unit depth w x 2 k w 1 2. 1 x2 . 1 ln x + . 2 2 2h . =. w x . 2 k .. xk Prof.

9 Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-11. pave flow 1 . 1 ln 2 . 1 w (. 2 xk +. 2 . 1 w2.. 2h 4. )2 . xk . = . unit depth w x 2 k Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-13. Now calculate the area/unit depth Asliding = 2 = Asticking = 4 2 = Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-14. Now calculate the forces F. unit depth (. = flow ( pave A)sliding + ( pave A)sticking ). F. = 1150 (( ) + ( )). unit depth = 21,110 lb / inch Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-15. Now, assume all sticking, so: F wf . = flow w f 1 + . unit length 4h . f . 4 . = 1150 4 1 + . 4 . = 23,000 lb / inch depth Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-16. or since the part is 36 deep: F(both) = 759,960 lbs = 380 tons w . F Forging = flow 1 + w depth 4h.

10 F(all sticking) = 828,000 lbs = 414 tons w . F Forging = flow 1 + w depth 2h . F(all sliding) = 496,800 lbs = 225 tons All sticking over-estimates actual value. Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging Effect of friction Effect of friction coefficient ( ) all sticking Friction coefficient Fmax (lbf/in depth) xk Stick/slide 0 4600 2 slide 11365 2 slide 19735 both 21331 both 22182 both 22868 both 23000 0 stick Friction is very important Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-17. Forging force vs. stroke all sticking 25000. Forging force (lbf/in mu=0. 20000 mu= mu= 15000 mu= depth). mu= 10000 mu= mu= 5000. 0. 0 Stroke (in). Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton Forging - Ex. 1-19. Maximum Forging force vs. friction coefficient ( ). all sticking 25000. 20000. Max Forging force (lbf/in depth).