How to design a T-beam given f'c, fy, bw, h, Mu, Vu and verify?

Are you sure that's a T-beam? You haven't listed a depth of the slab?

http://composites.wsu.edu/qiao/teaching/CE433/ce43...

Otherwise, straight beam design

http://composites.wsu.edu/qiao/teaching/ce433.htm

look at the bottom of the page, that will at least get you started.

for a normal beam (not a T-beam) things are more straightforward, take a look at the PDF or the powerpoint for beam design, then...

estimate depth of reinforcement = 34" - 3/4" beam cover - 1/2" (#4 stirrup), -1/2" (to centroid of #8 bar) = 32.25"

Approximate area of steel = Mu / (4 * d) = 2.5 square inches (rounded up).

minimum steel reinforcement ratio = 200 / fy = 0.0033*width*depth = 1.5 square inches, so okay.

Given that steel yields - Tension in steel = fy * As = 150 kips.

Compression in concrete equal to Tension in steel = 0.85*fc' * width * depth (a) of stress block, find a.

a = 150*1000 (kips to psi conversion) / (0.85*4000concrete strength *14 width) = 3.15 inches.

Knowing the stress is uniform in the concrete, the distance between the couple (tension / compression) is d - a/2 = 30.67 inches.

To find the moment capacity of the section, we take the safety factor (0.9) times the Tension or compression force times the distance between the two forces, or

Mu = phi * T * d

Mu = 0.9*150 kips * 30.67 inches = 4140 kip inches, or 345 k-ft.

Check shear - normal shear = safety factor * sqrt(concrete strength) * width * depth to reinforcement = 0.75* sqrt(4000)*14*32.25 = 18547 lbs = 18.54 kips.

Since this is less than the ultimate shear required, we need shear stirrup reinforcement.

When we have shear stirrup reinforcement, we can double the shear stress in the concrete, so we only need to provide steel stirrups strong enough to resist 70k - 2*18.54k = 32.9 kips.

Steel stirrups, Vs = area of stirrups * 2 legs * fy * depth / spacing. Minimum stirrups spacing is d/2 or 24 inches, in our case, d/2 controls = spacing will be 16.125 inches, let's use 16 inches. We'll also use #4 stirrups, with an area per bar of 0.20 square inches.

Vs = 0.75 * Av * 2 * Fy * d / s = 0.75*0.20*2*60000*32.25/16 = 36.2 kips, so the minimum stirrup spacing is adequate.

Looks right to me. You can pick reinforcement for the flexure, based on 2.5 square inches or try to calculate the required steel more precisely, then you should have a concrete design book you can use to check the reinforcement will fit in the 14" width, but 3#8 bars should work pretty well and give you a final area of steel of 2.37, more than the 2.32 we roughly calculated. The presentation should have better steps on how to more accurately calculate moment capacity and required steel. Remember also to check strain on the steel to prove the 0.9 safety factor is valid.

yeah, your missing hf, the thickness of the slab...if "a" is greater than hf, you got t-beam action..otherwise...run as a rectangular beam