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The Delft Sand, Clay & Rock Cutting Model, 2019a

The Delft Sand, Clay & Rock Cutting Model, 2019a

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<strong>The</strong> <strong>Delft</strong> <strong>Sand</strong>, <strong>Clay</strong> & <strong>Rock</strong> <strong>Cutting</strong> <strong>Model</strong>.<br />

<br />

N1<br />

N1<br />

sin<br />

<br />

h w<br />

i<br />

<br />

(7-65)<br />

Substituting equation (7-64) in equation (7-65) gives for the normal stress on the shear plane:<br />

s c hi w s a hb<br />

w<br />

cos( )<br />

<br />

sin sin sin<br />

N1<br />

<br />

h w sin( )<br />

i<br />

c<br />

<br />

s<br />

sin<br />

cos( ) r sin<br />

sin( )<br />

<br />

<br />

(7-66)<br />

Assuming a fixed strain rate factor λs for cohesion and tensile strength, the normal stress minus the shear strength<br />

(cohesion) has to be bigger than the tensile strength, where the tensile strength is negative (compressive stresses<br />

are positive).<br />

N1 s c<br />

s T<br />

(7-67)<br />

Substituting equation (7-66) into equation (7-67) gives the condition for ductile failure:<br />

<br />

<br />

sin<br />

cos( ) r sin<br />

c c <br />

sin( )<br />

s s s T<br />

(7-68)<br />

<strong>The</strong> transition from the Flow Type mechanism to the Tear Type mechanism is at the moment where the equal<br />

sign is used in the above equation, resulting in a critical ratio between the tensile strength and the shear strength,<br />

still also depending on the ac ratio r according to:<br />

<br />

<br />

sin<br />

<br />

r cos( ) sin( )<br />

<br />

sin<br />

<br />

c <br />

sin( )<br />

<br />

<br />

<br />

<br />

<br />

T<br />

<br />

(7-69)<br />

Figure 7-27 shows the critical ratio curves of the ratio of the tensile strength to the shear strength (cohesion) of the<br />

transition of the Flow Type mechanism to the Tear Type mechanism. Since the tensile strength is considered to<br />

be negative, the more negative this ratio, the higher the relative tensile strength. Below a curve the Flow Type<br />

may be expected, above a curve the Tear Type. Only negative ratios should be considered, since the tensile<br />

strength cannot be positive. <strong>The</strong> figure shows that for r=1 (high adhesive forces) the curve just touches a ratio of<br />

zero, but never becomes negative, meaning the Tear Type will never occur. For smaller r values the curves are<br />

more negative for a decreasing r value. <strong>The</strong> minimum for r is zero (no adhesion). <strong>The</strong> figure also shows that all<br />

curves (except the r=0 curve) start with a positive value, then decrease with an increasing blade angle to a minimum<br />

value and with a further increasing blade angle increase again to positive values. For blade angles larger than 90<br />

degrees tensile failure will never occur. Because of the choice of the parameter hb, the blade height, at constant<br />

blade height the length of the blade is increasing with a decreasing blade angle. This means that the adhesive force<br />

on the blade increases with a decreasing blade angle, resulting in increasing normal stresses on the shear plane.<br />

Higher normal stresses suppress tensile failure. On the other hand, an increasing blade angle will increase the<br />

normal stress on the shear plane because of the force equilibrium. So we have two effects, the normal stresses on<br />

the shear plane will decrease with an increasing blade angle because of the decrease of the adhesive force and the<br />

normal stresses will increase with an increases blade angle because of the force equilibrium. <strong>The</strong> result is a curve<br />

with a minimum.<br />

Page 216 of 454 TOC Copyright © Dr.ir. S.A. Miedema

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