1.1) Pressure Release
1.1.1) Internal pressure at the beginning
b = 12.67 bar = 186.2 psi
1.1.2) Internal pressure at the end
b = 1.0 bar = 14.7 psi
1.1.3) Duration of the pressure release from start
to end
T = 8.04 second
1.2) Property of the 19.0 thick steel plate.
Consider 12inches wide strip of the calculation
S = bd2 / 6 = 12 * 0.752 / 6 = 1.125 in3
I = bd3 / 12 = 12 * 0.753 / 12 = 0.422 in4
1.3) Wide of Doors L= 36.0 inch
2. Calculation of the bending stress and Deflection
2.1) Bending moment and stress of the steel plate
Assume pressure shall be linearly released at the speed
of 0.5 seconds interval over the entire pressure release
2.1.1) Uniform pressure release
W = 12 (186.2-14.7)/(8.04/0.5)= 128 lbs/in/0.5 second
2.1.2) Bending moment V
M = wL2/8 = 128 * 362/8 = 20,736 in-ibs
2.1.3) Bending stress
f= M/S = 20,736/1.125 = 18,432 lbs/in2
= 18.432ksi < 0.6*36(=21.6ksi) : O.K
2.1.4) Deflection of the plate = 5wL4/384EI
= 5*128*364/(384*29000000*0.422)
= 0.228 in = L/157 < L/120 : O.K
3. CONCLUSION
Based on the result of the calculation, the thickness of the plate, 19 mm for the Blast resistant doors are adequate.
