The torque you will see at any point ##x## on the rotating components when the blade hits a rock will be:
T_x = T_{in} - I_x\alpha
Where ##I_x## is the moment of inertia of the components from the shear bolt up to point ##x## and ##T_{in}## is the maximum torque handled by the shear bolt.
When hitting a rock, the blade will decelerate, so ##\alpha## will be negative thus - you are right - the torque at ##x## is greater than ##T_{in}##.
How big is ##\alpha##? That is the million dollar question which can only be answered by proper testing. And the results are valid for only the conditions tested. The value of ##\alpha## depends on the amount of deformation of both your machine and the object that was hit, which depends the elasticity of the materials and the shapes of the machine components and the object.
Are all this trouble worth it? Let's take the bolts holding the blade to the shaft. ##I_x## will be basically the shaft, so probably relatively small compared to the blade itself. Even then, estimating the time (##\alpha = \frac{\Delta\omega}{\Delta t}##) would be a wild guess. Certainly not more precise than estimating a safety factor.
Estimating deceleration
Another way to look at it (that doesn't involve knowing the time taken to stop) is to estimate the work done to absorb the energy of the hit. The energy equation when completely stopping the blade from angular velocity ##\omega## will be:
\frac{1}{2}I\omega^2 = T\theta
Where ##T## is the
average torque (may reach some higher peaks) and ##\theta## is the combined angular deformation of the rotating components. ##\theta## can be found based on the
torsion definition (##\theta = \frac{Tl}{JG}##; which do not include the deformation of the hit object, so it gives a safety margin), such that:
T = \sqrt{\frac{I\omega^2JG}{2l}}
That would be the torque needed to decelerate the rotating components. So it should be equal to ##I\alpha##, therefore:
\alpha = \omega\sqrt{\frac{JG}{2Il}}
Which could be used in the previous equation for any component ##x##.
I did not test that, but the math looks good.
