Determine the maximum and minimum stress intensities at the base:

A masonry chimney 20 m high of uniform circular section, 5 m external diameter & 3 m internal diameter ought to withstand a horizontal wind pressure of intensity 2 kN/m² of the projected area. Determine the maximum and minimum stress intensities at the base. Take unit weight of masonry as 21 kN/m³.

Solution

Height of the chimney, H = 20 m External diameter, D = 5 m Internal diameter, d = 3 m

Unit weight of masonry, γ = 21 kN/m³

Direct compressive stress because of self weight on the base of the chimney,

f₀  = γ H = (21 × 20) = 420 kN/m²

Wind pressure, p = 2 kN/m²

Projected area, A = DH = 5 × 20 = 100 m²

Wind force, P = pA = 2 × 100 = 200 kN

Distance of centre of gravity of the wind force through base, = H/ 2 = 10 m

Bending moment, M = PH/ 2= 200 × 10 = 2000 kNm

Section modulus, Z =  (π/32) × (D4  - d 4 )/ D =  (π/32) ((5 ⁴  - 3⁴ )/5)

= 10.68 m

Bending stress,

f _b  =± M/ Z

=± 2000/10.68

 = ± 187.266 kN/m²

Maximum stress induced = 420 + 187.266 = 607.266 kN/m²

Minimum stress induced = 420 - 187.266 = 232.734 kN/m² .