Both, tensile loading and shear stress are represented in the Stress-Energy Tensor and hence are sources of gravity. As I understand it (please correct if wrong) tensile loading implies forces perpenticular to the surface, so that e.g. x-momentum flows in the x-direction across the unit surface. The same direction means that $T^{11}$ is pressure (negative or positiv depending on tension or compression). Whereas the non-diagonal elements are shear stress in case the force parallel to the surface.

In the FRW context negative pressure acts as repulsive gravity. Regarding a solid body, does negative pressure decrease the effective gravitational mass?

Can the non-diagonal shear stress components be negativ and positiv? What would a good example to show that, e.g. the deformation of a elastic ball if a gravitational wave passes by?

How does shear stress depending on it‘s sign act on gravity (in the sense of attractive vs. repulsive)?

Thanks

The statement that pressure exists is not frame-independent. For example, if you describe dust in its own rest frame, the only nonvanishing component of the stress-energy tensor is $T^{00}$, but if you do a Lorentz boost, now you can have a nonzero $T^{11}$ as well.
• @timm: The only reason that the word "pressure" needs to be there is in order to make contact with our Newtonian intuition and experience. Pressure is not a universal or fundamental category in relativity. The pressure of water in a swimming pool is isotropic, and that makes up part of our Newtonian experience: pressure acts like a scalar. In relativity, it is not true in general that $T^{11}=T^{22}=T^{33}$. This is only true for a perfect fluid, considered in the fluid's rest frame. Dark energy happens to behave as a perfect fluid. – user4552 Aug 27 '18 at 21:31