Peter Ulintz
Technical Director

# Managing Horizontal Forces in Stamping Dies—Part 2

June 1, 2020

Last month’s column addressed the management of horizontal forces in the dieset. This month: The prevention of die-component shifting due to horizontal die forces.

Dies experiencing significant side loads often have multiple causes, including: poor alignment of die components during construction; misalignment resulting from a bad hit or die crash; shifting of components due to angular contact between angular form steels; nonsymmetrical forms or draws where the punch and die are loaded off-center at initial contact; and cutoff, trim, bending and flanging operations where forces act on only one side of the die steel.

Shifting of Die Components

Maintaining alignment and designed clearances between die components is paramount for long die life and consistent part quality. When a die component shifts, alignment and clearances can alter significantly enough to cause galling or shearing damage to the tooling.

Shifting of die steels is prevented through a combination of forces applied normal (perpendicular) to the mounting surface and the friction generated between the two surfaces. Friction occurs because the surfaces in contact are not smooth. Small ridges on the mating surfaces lock together, restricting movement. For movement to occur, these ridges must be broken off (sheared) by a horizontal force of sufficient magnitude. The force required to move two sliding surfaces over each other, divided by the force holding them together, is known as the friction coefficient (µ).

Intuitively, we know that higher frictional forces better prevent the shifting of die components. Now let’s covert this intuitive understanding into something more useful and easy to remember:

Friction is fun, or F = µ N

Where F is the static frictional force, µ the friction coefficient and N the normal force.

The friction coefficient for two pieces of steel in contact, clean and dry, is approximately 0.75. With each surface lightly oiled, the friction coefficient could approximate 0.15.

In general, cutting and forming forces act together with the mounting screws to produce a normal force great enough to prevent shifting of the die components. This is achieved through the selection of screw size and quantity, and the torque applied to each screw.

Fastener Selection

In the accompanying illustration, assume a cutting thrust of 8000 lb.-force (lbf) exerted horizontally on the upper trim section. Considering that most die sections feature a light oil application to prevent rust and the fact that we also want a safety margin, assume a friction coefficient of 0.15. How many socket head cap screws are required to prevent it from shifting?

Let X equal the total clamping force required

X = 7500 lbf divided by 0.15

X = 50,000 lbf total

The accompanying table depicts the calculated pretension stress and the tightening torque values for several 3-in.-long socket head cap screws. Several viable options exist to meet the 50,000 lbf requirement: Five 3⁄8-24 alloy-steel cap screws, torqued to 67 ft.-lbf; four 7⁄16-14 alloy-steel cap screws, torqued to 94 ft.-lbf; or three 1⁄2-13 alloy-steel cap screws torqued to 144 ft.-lbf.
Be careful when selecting stainless steel cap screws, which feature tensile-strength, pretension and torque values considerably less than those of alloy steel. In the example, generating 50,000 lbf requires 12 of the 3⁄8-24 stainless steel cap screws, torqued to 27 ft.-lbf.

Should stainless steel cap screws be required to help prevent in-die magnetism, the quantity of screws could be reduced by increasing the friction coefficient. Suppose that the mounting surface of the die component is ground to a very high finish, lightly oiled and mounted on a smooth, cold-rolled sub-plate. The friction coefficient can be increased by leaving a rough ground surface on the bottom of the die section and ensuring completely dry and oil-free mounting surfaces. Under these conditions, let’s assume a friction coefficient of 0.75.

Let X equal the total clamping force required

X = 7500 lbf divided by 0.75

X = 10,000 lbf total

Now, four 3⁄8-16 stainless steel cap screws, torqued to 24-ft.-lbf would suffice. However, it is imperative to assure that the friction conditions do not change. Any decrease in friction requires an increase in the number of screws or a change in the screw material in order to apply a higher torque.

Also, remember that torque requirements increase for dry, rough (or damaged) tapped holes or when dirt or chips are present in the threads. Be diligent in your maintenance practice.

Other difficulties may be encountered even when using the correct size and quantity of screws. For example, the diemaker does not torque the screws properly, if at all. My first-hand experience reveals that a torque wrench seldom is found in a diemaker’s toolbox. Instead, very long “cheater” pipes are slipped over the Allen wrench and pulled with as much force as the die-maker can exert or thinks is needed.

Only when using a torque wrench will the full value for each screw be obtained. When randomly tightened, only the tighter screws carry the load. Screws tightened without a torque wrench may overstretch and permanently yield. Many screws have broken from fatigue in die operations due to overstressing.

Next month, this series on managing horizontal forces concludes by addressing dowel pins, backup keys and methods to prevent component tipping. MF

Industry-Related Terms: , , , , , , ,
View Glossary of Metalforming Terms

Technologies: Materials, Tooling