Does specifying constexpr on a constructor automatically make all objects created from it to be constexpr?

Having a constexpr constructor does not make declarations of that variable automatically constexpr, so t is not a constexpr. What is going on in this case is that you are calling a constexpr function, this line:

constexpr int b = t+test(); 

can be viewed as follows:

constexpr int b = t.operator+( test() ); 

So then the question is whether test() is a constant expression, which it is since the constructor is constexpr and does not fall under any of the exceptions under the draft C++11 standard section 5.19 [expr.const] paragraph 2 which says:

A conditional-expression is a core constant expression unless it
involves one of the following as a potentially evaluated subexpression
[…]

and includes the following bullet:

• an invocation of a function other than a constexpr constructor for a literal class or a constexpr function [ Note: Overload resolution
  (13.3) is applied as usual —end note ];

[…]

• an invocation of a constexpr constructor with arguments that, when substituted by function invocation
  substitution (7.1.5), do not produce all constant expressions for the constructor calls and
  full-expressions in the mem-initializers

• an invocation of a constexpr function or a constexpr constructor that would exceed the implementationdefined
  recursion limits (see Annex B);

We can see this more readily by making some small changes to test by introducing a member variable x:

class test{
    public:
    constexpr test(){

    }

    constexpr int operator+(const test& rhs) const {
        return x + 1  ;
    }

    int x = 10 ;
};

Attempting to access it in operator + and we can see that the following line now fails:

constexpr int b = t+test();

with the following error from clang (see it live):

error: constexpr variable 'b' must be initialized by a constant expression
constexpr int b = t+test();     // works at compile time!
              ^   ~~~~~~~~

note: read of non-constexpr variable 't' is not allowed in a constant expression
    return x + 1  ;
           ^

It fails because t is not a constexpr variable and therefore its subobjects are also not constexpr variables.

Your second example:

 constexpr int c = w + 2;  

does not work because it falls under one of the exceptions in the draft C++11 standard section 5.19 [expr.const] :

• an lvalue-to-rvalue conversion (4.1) unless it is applied to

  […]

  • a glvalue of integral or enumeration type that refers to a non-volatile const object with a preceding initialization, initialized
    with a constant expression, or