A 20ft. Span, simple supported beam is uniformly loaded with 8kips/ft.

Question

1. Draw the shear and bending diagrams.
2. Find shear and bending stresses for a rectangular 10in x 20in cross section
3. Design it as a balanced reinforced concrete cross section. Use concrete fc’= 4 ksi, steel fy = 40 ksi.
4. Design it as a steel beam. Use plastic condition, steel Fy = 50 ksi.
5. Design as an unbalanced RFC beam. Use concrete fc’= 4 ksi, steel fy = 40 ksi. Use 14 x 20in cross-section.

Answer 1

w = 8kip/ft, L = 20ft. Va = Vb = wL/2 = 80 kips

1) SFD- Shear force is a linear diagram with shear varying as Va= +80 kips at one end to Vb=-80 kips at other end.

BMD - BMD is a parabola with max moment at center (10ft) of span, Mmax = wL²/8 = 400 ft-kips

2)Shear stress ζ= Va/bd = 400 psi or 0.4ksi

Bending stress (fb)= M*c/I ; I =bd³/12 = 6666.66 in³,

σ= 400*10³*10/ 6666.66 = 600 psi or 0.6ksi

3) Balanced RC beam design (assuming udl applied is factored load)

Mu = 400 ft-kips = 4800 in-kips,

For balanced beam, tensile steel reaches a value of εy = Fy/Es at the same time as concrete reaches εu =0.003. Hence, εy = 40/29000 = 0.00138.

Reinforcement ratio : ρb = 0.85β₁*(fc'/fy)* εu/(εu+εt). Substituting, ρb = 0.049.

Finding the required cross section:

Mu = ΦM_n ;Mn = ρfybd²(1-0.59fy/fc')

4800 = 0.9*0.049*40bd²*[1-(0.59*40/4)]

Solving, bd² = 3827.66 in³. Assuming b = 12", d = 17.85".

Hence provide cross section - 12"x20". Reinforcement = ρb*b*d =10.49 in². Provide 5 No. (14) bars as tension reinforcement.

4) Steel beam, plastic section design. For a plastic section,

Mn or Mp = Fy*Zx

ΦMn >Mu.

Mn req. = Mu/Φ = 4800/0.9 =5333.33 in-kips

Equate with capacity, 5333.33 = 50* Zx.

Zx _min = 106.66 in³.

Selecting a plastic section from AISC table 3-2. Selecting W 12x72 (Z =108 in³) or W10x88 (Z =113 in³).

5) Unbalanced RC beam - for cross section 14"x20".

Assume effective depth d = 17.5"

For tension-controlled section, ρb _max= 0.85β₁*(fc'/fy)* εu/(εu+εt_min)

ε_tmin = 0.00138 for Fy = 40 ksi, as seen.

ρb _max = 0.049,

As = 0.049* 14 * 17.5 =12 in².

Hence provide 3 No.(18) bars or 6 No.14 bars in two rows.