Generally not. If they're the same thickness and size, you could figure each carried half the load. But that's not the same as figuring one plank twice as thick.
Consider why a cable if flexible, but a bar of the same size is not.
For two beams combined, the bending stiffness will be determined by the addition of the 2nd Moment of Inertias, ie. bd^3/6. This assumes the worst case that there is no friction between them and they slide. If you bolt the two together, however, then the MOI will be based upon the combined thickness, ie. (2/3)bd^3
we're assuming this is a beam ... what if this is a column ?
if it is a beam, and they're just stacked on top of each other, and there's a point load on the upper one, then doesn't the lower beam provide a distributed reaction to the upper ? ... IMHO not the same as saying each reacts 1/2
It seems that technicalgirlca is not actively following this thread. A clarification of the specific application is in order - beam or column? orientation of the (2) 8x10s as rb1957 stated?
You can make them a composite wood beam using the shear flow equation VQ/Ib to determine the ingerstitial shear generated at the intersection from your particular loads. Based on this shear value, you can determine the size and spacing of the thru-bolts to accomplish the composite action. This is assuming that the members are placed one on top of the other.
However, if they are placed side by side, they are just the two original beams. They should still be stitch bolted together, but this is mainly to get them to deflect uniformly, sharing the same load.
if they're on top of each other, no. unless you connect them and check horizontal shear.
if they're side by side, technically no.....but you'd be splitting the load in half anyway between the two members, so numerically, it'll come out that way.
