I could have given you the sizes of what you need but on second thought, I decided to show you how this is calculated so that if you need to do some adjustments, then you can later do it on your own. Here we go.

1. First we calculate the loading of the beam

a. Live load- for a residential building the normal design live load is 40 lbs per square foot. This translates to a load on the beam supporting joist with 12 ft span of:

LL = 40 x 12 = 480 lbs per foot

b. For the weight of the flooring, we can assume this to be a 1" thick plywood which weights about 5 lbs per square foot . This translates to a dead load on the bem of:

DL1 = 5 x 12 = 60 lbs per foot

c. For the joists, each 12 feet of 2" x 8" x 12 ft joist weighs about 64 lbs. But since the spacing of the joists is 16" or 1.333 ft, this translates to a load on the beam of:

DL2 = 64/1.333 = 48 lbs per foot

d.. The other dead load on the beam is the weight of the beam itself. Let's take a wid guess on this and assume it to be;

DL3 = 20 lbs per foot.

2. The total load on the beam therefore is:

W = LL + DL1 + DL2 + DL3 = 480 + 60 + 48 + 20 = 608 lbs/ft

4. The maximum bending moment on the beam is:

M = 12(WL^2)/8 = (12 x 608 x 17^2)/8 = 263568 in- lbs

5. The required moment of inertia of the beam section is:

I = Mc/S

Where;

M is the maximum bending moment

c = (1/2) the depth of the beam = 8/2 = 4 inches

S = the allowable stress for the beam material which is 20,000 psi for steel beams.

Therefore the required beam should have a moment of inertia of not less than:

I = 263568 x 4/20,000 = 52.714 in^4

The lightest W8 beam that meets this requirement is a W8 x 18 lbs per foot which has a moment of inertia of 61.9 in^4.

6. For the columns, the load on each column is:

P =WL/2 = (608 x 17)/2 = 5168 lbs

Assuming that the length of the column is 8 ft or 96 inches, the required slenderness ratio is between 60 to 120. Using a slenderness ratio of 80, the radius of gyration of the column section should be about:

r = 96/80 = 1.2

We now try a 3" schedule 40 pipe. The properties of this pipe section are:

I = 3.006 in^4

A = 2.219 in ^2

r = (3.006/2.219)^.5 = 1.164 inches

l/r = 96/1.164 = 82.47

Using the old AISC formula for allowable compressive stress:

Sc = 17000 - .485(l/r)^2 = 17000 - .485(82.47)^2 = 13700psi

P/A = 5168/2.219 = 2329 psi which is less than 13700 psi.

Use a 3" schedule 40 pipe for the columns.