PLATE DESIGN - UDL : ROARK'S CHECK
MODULE : REFERENCES
PLATE ID :
REF. DRG :
Support at the edges of plate: = FIXED
Load Case: Live Load + Max.Acceleration (operating)
1.0 Material Properties:
Yield strength, Fy = 245 N/mm²
Modulus of Elasticity, E = 200000 N/mm²
Density of steel r = 7850 kg/m3
= 7850*9.81/1000
= 77.01 kN/m3
2.0 Physical Properties:
Length of the plate, a = 3.50 m
Width of the plate, b = 2.00 m
Assumed thickness of plate, t = 10.00 mm
Contingency factor, C.F = 1.15
Acceleration coeff in vertical direction, accln = 1.00
3.0 Design Loads:
Live Load, LL = 2.50 KN/mm²
Self weight of plate with Contingency Factor, Sw = t∗ρ∗C . F
= (10/1000)*77.01*1.15
= 0.89 KN/m²
Udl with acceleration due to self weight+LL, q = ( LL∗acc ln )+( sw∗acc ln)
= (2.5*1)+(0.89*1)
= 3.39 KN/m²
4.0 Check for Stresses:
4.1 Check for Bending:
a/b 1 1.2 1.4 1.6 1.8 2 ∞ From [R1] Table 11.4
b1 0.3078 0.3834 0.4356 0.4680 0.4872 0.4974 0.5000 case 8(a)
b2 0.1386 0.1794 0.2094 0.2286 0.2406 0.2472 0.2500
a 0.0138 0.0188 0.0226 0.0251 0.0267 0.0277 0.0284
The value of a/b = 3.5/2
= 1.75
The value of a/b lies between a/b 2 ∞
b1 0.4974 0.5000
b2 0.2472 0.2500
a 0.0277 0.0284
Interpolated value of b1 = 0.5000 REFERENCES
Interpolated value of α = 0.0284
2
Maximum bending moment at centre of long edge (b1 =1m) = β 1∗q∗b2 b 1∗t
2
∗
t 6
= 0.5 * 3.39 * 2² 1*(10/1000)²
*
(10/1000)² 6
Total Maximum bending moment at centre of long edge, M = 1.130 kNm/m
2
b∗t
Elastic Section Modulus ,S =
6
= 1000*10²
6
= 16666.67 mm3
b∗t 2 ###
Plastic Section Modulus ,Z =
4
= 1000*10²
4
= 25000.00 mm3
Mn = Mp = Fy*Zx ≤ 1.6*My R2, Eqn F6-1
Ωb = 1.67
1.6*My = 1.6*Fy*S
= 1.6*245*16666.67
= 6533334.64 N-mm
= 6.53 kN-m
Mp = Fy*Zx = 245*25000
= 6125000.00 N-mm
= 6.13 kN-m
Allowable moment Mal = (1/Ωb)*Min(Mp,1.6My)
= (1/1.67)*MIN(6.125,6.53)
= 3.67 kN-m
Unity check ratio = ( M / Mal )
= 1.13/ 3.67
= 0.308 <1, OK
4.2 Check for Shear:
Maximum shear force,R = 0 . 5∗q∗a
= 0.5 * 3.39 * 3.5
= 5.93 kN/m
Aw = b1*t
= 1000*10
= 10000.00
Ωv = 1.50 R2, Eqn G2-1
Cv = 1.00
Allowable shear strength, Rall = (1/Ωv)*(0.6*Fy*Aw*Cv) R2, Eqn G2-2
= (1/1.5)*(0.6*245*10000*1)/(1000)
= 980.00 kN/m
Unity check ratio = ( R / Rall)
= (5.93/ 980)
= 0.006
= 0.006 <1, OK
5.0 Check for Deflection:
Maximum deflection due to self weight,∂ = α∗q∗b 4
= E∗t 3
= ((0.0284* 3.39* 2^4)/(200000* (10/1000)^3))
= 7.70 mm
REFERENCES
Allowable deflection,∆ = b
240
= (2*1000)/240
= 8.33 mm
= Actual < Allowable,O.K
Unity check ratio = ∂
Δ
= 7.7/8.33
= 0.924 <1, OK
6.0 Summary:
Section Description UC Ratio Status
4.1 Bending Strenth 0.308 OK
4.2 Shear Strength 0.006 OK
5.0 Deflection 0.924 OK
7.0 Conclusion :
10 mm PLATE IS ADEQUATE
8.0 References :
[R1] - "Roark's Formulas for Stress & Strain" by W.C.Young (7th edition)
[R2] - Structural steel specification (ASD) - AISC 13th Edition.