Since you have numerical parameters I am guessing that you are not necessarilly interested in an analytical solution. If that is true then you can use NDSolve with a range for parameter r that avoids the potential singularity at r=0. (I am not saying that there is indeed a singularity there, but if your range of values that you are interested in does not include r=0, then that will be a moot question. Also you can use NDSolve to explore whether there is indeed a singularity at r=0.)
On looking at this further I notice that there still are w/0 errors generated when using NDSolve... stay tuned....
Ok, change your value for v to
v = 3/10;
instead of a floating point value. Then DSolve works for you. I am not completelysure why the use of the floating point value causes DSolve to fail, but this does create a result.
p = 10;
E1 = 210000;
R = 100;
a = 30;
s = 10;
v = 0.3;
t1 = p*r/2;
t2 = p*(R^2 - r^2)/(2*r);
B = E1*s^3/(12*(1 - v^2));
sol = DSolve[{D[1/r*D[(r*w1'[r]), r], r] == t1/B,
D[1/r*D[(r*w2'[r]), r], r] == t2/B, w1'[0] == 0, w1[a] == 0,
w2[a] == 0, w1'[a] == w2'[a],
w1''[a] + v*(w1'[a]/a) == w2''[a] + v*(w2'[a]/a),
w2''[R] + v*w2'[R]/R == 0}, {w1[r], w2[r]}, r];
And this yields the following for the value of sol:
{{w1[r] -> (1/
1600000000)(-288928800 + 309332 r^2 + 13 r^4 - 936000000 Log[30] +
1040000 r^2 Log[30] + 936000000 Log[100] - 1040000 r^2 Log[100]),
w2[r] -> (1/
1600000000)(583891200 - 637068 r^2 - 13 r^4 - 1787760000 Log[30] +
936000000 Log[100] - 1040000 r^2 Log[100] + 851760000 Log[r] +
1040000 r^2 Log[r])}}
If you now explore your functions near r=0 you see that W2 has a singularity there:
Plot[Evaluate[{w1[r], w2[r]} /. sol[[1]]], {r, 0, 10},
PlotRange -> All]
