ADDITIONAL SHEAR RESISTANCE CAUSED BY NORMAL PRESSURE

I. R. Konyakhin

(Presented by the scientific seminar of the Department of Strength of Materials, TPI)

Microasperities of a rough surface operating in sliding undergo shear deformation during the process of friction between surfaces.

If, in addition to the shearing force, another load is applied to a protrusion on the surface in a direction perpendicular to the shear plane, then the shear strength of the protrusion will increase.

In particular, during friction, the microasperities of a rough surface always operate in the presence of such pressure.

B. V. Deryagin [3], in one of his works, while considering the law of friction advanced by him, expresses the idea that, in the case of plastic shear deformation, normal pressure must increase the resistance to this deformation.

P. Bridgman [2] subsequently discovered this phenomenon. He carried out experiments involving shear deformation of material under high specific pressures, and in doing so observed an increase in the resistance of the material to deformation.

I. V. Kragelsky [5], in proposing his theory for calculating the coefficient of friction, takes into account the increase in the shear resistance of asperities caused by normal pressure acting on the contact area.

The purpose of the present article is to show that the additional resistance to shear caused by the presence of a normal force is due to two principal factors: strengthening of the material on the shear plane, occurring as a result of its plastic compression by the normal load, and the special state of the substance caused by the action of normal pressure.

This state of the substance may be called a “constrained” state. It consists in the fact that a material which, in the absence of a normal force, works in shear within the elastic range, in the presence of a normal load undergoes shear according to the laws of plastic deformation; in this case its resistance is increased. When the force is removed, this state of “constraint” disappears, and the material again acquires the ability to work elastically in shear.

The proposition stated above was studied by us in experiments involving the phenomenon of preliminary displacement.

Each of the two factors indicated produces its own share of additional shear resistance.

In the present article we shall confirm this by the results of experiments conducted by us on wooden and steel specimens, not by the method of preliminary displacement, but by ordinary methods of deformation.

For the experiments, specimens were taken whose shape was borrowed from I. I. Bobarykov [1] (Fig. 1).

Shearing along two surfaces was performed on a tensile-testing machine by pulling one part of the specimen out of the other.

In order to reveal separately the influence of strengthening and then of “constraint,” the specimens were tested in three ways.

1. Shearing was performed without applying normal pressure to the shear surfaces, in order to obtain pure shear, i.e., without the participation of normal stresses. Testing of several specimens gave, on average, $\tau_0 = 29\ \text{kg}/\text{cm}^2$.

Fig. 1. Wooden specimen for testing in shear along two surfaces by tension.

1. In the second experiment the influence on shear resistance of the strengthening of the material caused by the action of normal pressure was revealed.

A compressive force $N$ was applied to the specimen perpendicular to its shear surfaces by clamping the test bar in a bench vise. The clamping force was measured by means of a conical crusher [4], inserted between one of the vise jaws and a hardened steel plate placed on the side surface of the specimen, exactly opposite the middle of the surface of the future shear. The bar was left in the vise for two hours, after which it was subjected to shear along the fibers in the same way as in the preceding experiment.

The stress now obtained was already $\tau_1 = 42\ \text{kg}/\text{cm}^2$, i.e., compared with the preceding values, greater by $13\ \text{kg}/\text{cm}^2$, or by 45%. This increase in resistance occurred as a result of strengthening of the material caused by preliminary compression in the vise by a normal pressure $N = 250\ \text{kg}$.

1. To reveal the effect of “constraint” of the material on the shear surface by normal pressure, shear was now performed in the presence of such pressure.

For this purpose the specimen was compressed in its transverse direction by means of a clamp having the form of a rectangular frame provided with a screw, by means of which the transverse compression was produced. The force of pressure by the clamp screw was measured in the same way as in the second experiment, with the aid of a conical crusher.

This force was chosen equal in magnitude to that in the preceding case, so that the strengthening effect from compression would be the same, i.e., $\tau_1 = 42\ \text{kg}/\text{cm}^2$.

But the shear stress, now already in the presence of a normal load, proved to be $\tau_2 = 52\ \text{kg}/\text{cm}^2$, i.e., greater by 35%.

The increase in the resistance of the material due to the influence of both factors, that is, strengthening and “confinement,” amounts in all to 23 kg/cm², or 80%.

We were able to obtain a similar increase in shear resistance—more than 100%; however, in this case the outer surfaces of the specimens proved to be crushed, which made it impossible to raise the normal load still further. In addition, during testing the specimen did not withstand rupture.

In addition to the study described above, we also carried out tests on grade-3 steel. In this case, specimens of the previous shape proved unsuitable, owing to the small difference between the tensile and shear strength of the steel. In order to ensure that the specimen would shear rather than rupture, it was necessary to assign a very small length for the shear plane. As a result, during testing there occurred strong bending of those parts of the specimen which, after withstanding the sheared portion situated between them, form a fork with ends bent outward, and with the parts opposite in length bent inward. Consequently, at one end of the shear plane there act compressive bending stresses, and at the other end—tensile stresses of the same bending. The indicated compressive stresses create on the shear plane a “confined” state, which in the best case greatly distorts the experimental results, and in the worst case leads to rupture of the specimen.

For this reason, for the shear tests, specimens were taken in the form of a cylinder provided at the middle of its height with a round bead of rectangular cross-section, forming a single whole with the specimen (Fig. 2).

Fig. 2. Specimen for shear testing of a steel bead:
a — shear in the absence of a confining ring,
b — shear in a confining ring.

The indicated bead was sheared from the cylinder by pushing the latter through an opening in a steel die in such a way that the cylindrical rod passed through the opening of the die, while the bead, being sheared off, remained on the upper face of the latter in the form of a removed ring.

To clarify the influence of strengthening of the material and its “confinement” on shear resistance, the tests were carried out in the same order as in the testing of wood in the preceding experiments.

1. Although the bead is sheared from the rod of the specimen, it also undergoes bending deformation; as a result the sheared ring is not flat, but somewhat turned out like the brim of a hat. This indicates that compressive and tensile bending stresses act in the shear plane. The former create confinement and strengthening of the material, and therefore the shear is not free.

To determine the magnitude of the pure-shear stress of the given material, specimens of rectangular cross section were tested by shearing on a Fremon machine. This gave \(\tau_0 = 2300\ \mathrm{kg/cm^2}\). Cutting off the collars with the aid of a die on a compression press gave \(\tau_{01} = 4700\ \mathrm{kg/cm^2}\); the difference \(\tau_{01} - \tau_0 = 2400\ \mathrm{kg/cm^2}\) was obtained owing to the strengthening and “constraint” of the material on the shear area by the normal bending stresses (Fig. 2b).

Moreover, the smaller the height of the specimen collar that is taken, the greater the bending and the higher the shear stress obtained. Thus, for example, for a collar 1 mm high and with an outside diameter of 16 mm, with the diameter of the rod part of the specimen equal to 10 mm, the shear stress reached 8000 \(\mathrm{kg/cm^2}\), instead of 2300 for free shear of the same metal.

Therefore, in the experiments specimens with a collar having a height of 3–4 mm were used. It was impossible to take a height greater than this, in order not to create excessive compressive stresses on the cylindrical part of the specimen while the latter was being pushed through the opening of the die.

1. The experiments to determine the influence of strengthening by compression were carried out as follows. The collar, together with the cylindrical part of the specimen, was first pressed into the slightly conical opening of a massive steel ring by pressing on its upper plane with a bushing. The pressing-in was done in order to create radial compression of the collar (Fig. 2c). The specimen was then pressed back out of the clamping ring and tested in shear in the same way as in the preceding experiment.

The shear stress obtained was \(\tau_1 = 5000\ \mathrm{kg/cm^2}\), that is, greater than before compression by 300 \(\mathrm{kg/cm^2}\). This additional stress appeared owing to strengthening, in addition to the strengthening that was caused by the normal bending stresses. They were added to the compressive stresses from the ring, and therefore the total compressive stress on the shear area was greater than the normal stresses from bending alone in the preceding experiment.

1. In the third experiment the collar was likewise pressed into the clamping ring with the same interference as before, but it was sheared directly in the ring, that is, in the presence of radial pressure normal to the cylindrical shear area.

In this case the shear stress was \(\tau_2 = 5700\ \mathrm{kg/cm^2}\), that is, 700 \(\mathrm{kg/cm^2}\) greater than in experiment 2. In comparison with free shear, this amounts to 29%.

Thus, the experiments show that the shear stress in general is composed of several stresses due to different factors:

\(\tau_0\)—the pure-shear stress, depending on the nature of the material;

\(\tau_y\)—the resistance due to strengthening of the material on the shear area caused by normal compressive bending stresses;

\(\tau_c\)—the resistance caused by the “constraint” of the material by normal bending stresses;

\(\tau_{yn}\)—the stress caused by strengthening by normal compressive stresses;

\(\tau_{cn}\)—the stresses caused by the “constraint” of the material by normal compressive stresses.

Consequently, the total shear stress must be determined from the equation

\[ \tau = \tau_0 + \tau_y + \tau_c + \tau_{yn} + \tau_{cn}. \]

Conclusion

1. Normal stresses acting on the shear area produce additional resistance to shear of two kinds: resistance due to strengthening of the material, and a second resistance due to the “constraint” of the substance. When the normal load is removed, the second resistance disappears.

2. In calculating the corresponding machine parts for shear, the additional resistances, which are of considerable magnitude, should be taken into account.

Literature

1. Bobarykov, I. I. On the question of the resistance of wood to shearing. Proceedings of the Tomsk Technological Institute, vol. 35, No. 3, 1914.

2. Bridgman, P. Investigation of Large Plastic Deformations and Fracture. Moscow, 1955.

3. Deryagin, B. V., and Lazarev, V. P. Application of the generalized law of friction to boundary lubrication. In: Friction and Wear in Machines, Academy of Sciences of the USSR, 1949.

4. Konyakhin, I. P., Sedokov, L. M., and Gorbenko, M. S. Determination of acting forces by means of conical crushers. Factory Laboratory, No. 5, 1958.

5. Kragelskii, I. V. Molecular-mechanical theory of friction. Second All-Union Conference on Friction and Wear in Machines, 1949.