Answer:
Beam of 25" depth and 12" width is sufficient.
I've attached a detailed section of the beam.
Explanation:
We are given;
Beam Span; L = 20 ft
Dead load; DL = 0.50 k/ft
Live load; LL = 0.65 k/ft.
Beam width; b = 12 inches
From ACI code, ultimate load is given as;
W_u = 1.2DL + 1.6LL
Thus;
W_u = 1.2(0.5) + 1.6(0.65)
W_u = 1.64 k/ft
Now, ultimate moment is given by the formula;
M_u = (W_u × L²)/8
M_u = (1.64 × 20²)/8
M_u = 82 k-ft
Since span is 20 ft, it's a bit larger than the average span beams, thus, let's try a depth of d = 25 inches.
Effective depth of a beam is given by the formula;
d_eff = d - clear cover - stirrup diameter - ½Main bar diameter
Now, let's adopt the following;
Clear cover = 1.5"
Stirrup diameter = 0.5"
Main bar diameter = 1"
Thus;
d_eff = 25" - 1.5" - 0.5" - ½(1")
d_eff = 22.5"
Now, let's find steel ratio(ρ) ;
ρ = Total A_s/(b × d_eff)
Now, A_s = ½ × area of main diameter bar
Thus, A_s = ½ × π × 1² = 0.785 in²
Let's use Nominal number of 3 bars as our main diameter bars.
Thus, total A_s = 3 × 0.785
Total A_s = 2.355 in²
Hence;
ρ = 2.355/(22.5 × 12)
ρ = 0.008722
Design moment Capacity is given;
M_n = Φ * ρ * Fy * b * d²[1 – (0.59ρfy/fc’)]/12
Φ is 0.9
f’c = 4,000 psi = 4 kpsi
fy = 60,000 psi = 60 kpsi
M_n = 0.9 × 0.008722 × 60 × 12 × 22.5²[1 - (0.59 × 0.008722 × 60/4)]/12
M_n = 220.03 k-ft
Thus: M_n > M_u
Thus, the beam of 25" depth and 12" width is sufficient.
