Now we will seek simplifications to Equation (6.36) as follows. The strength tensors can be written in the form

(6.41)

and

(6.42)

where, both strength tensors are assumed to be symmetric with 6 and 21 independent constants for  and , respectively.
           
The number of independent strength parameters can be further reduced if we have some form of material symmetry. We consider a special case of specially orthotropic material. Thus, for specially orthotropic material the terms  and  will vanish. Further, the off-diagonal terms which give normal shear coupling like , etc. will also become zero if we assume that the sign of shear stress does not change the failure stress. Further, with same reasoning we assume that the shear strengths are also uncoupled leading to . Thus, with this material symmetry, we have

and

The number of independent strength parameters are now 3 and 9 for  and , respectively. Thus, for orthotropic material the criterion becomes

(6.43)

Now we will determine the strength parameters by thought experiments. First, we apply  and other stress components being zero. For this state of stress, the failure in tension requires . Thus, Equation (6.43) becomes

(6.44)

Similarly, for this state of stress, the failure in compression requires . This results in Equation (6.43) to become

(6.45)

Equation (6.44) and Equation (6.45) are two simultaneous equations with  and  as two unknowns.

Solution of these two equations gives

(6.46)