This document contains calculations for the design of an OS beam and WE beam. For the OS beam, the required moment capacity is calculated as 110 kNm. The required shear capacity is calculated as 35.2 kN/m. For the WE beam, the required moment capacity is calculated as 123.4 kNm. The required shear capacity is calculated as 39.48 kN/m. A variety of equations are used to calculate stresses, moments of inertia, shear and moment capacities.
2. Solution: for OS Beam a/b = 5000 / 300 = 16.67 From the graph, assume Є = 0, kb = 24 2 Mahmoud Shaaban
3. σb = 24 π2x 205000 (3/300)2 12 (1 – 0.32) σb = 444.6 Mpa σ b > Fsy; Design should be on Fsy in equation 11 for the moment For OS Beam: Pb = Σo.x.Fb where Σo = 150+2x300+2x55 = 860 = 860x2500x0.1 fb = 0.1 = 215000 N x = 5000/2 = 2500 3 Mahmoud Shaaban
4. Since Y = 0, Frt & Frc = 0 Using eq. 2 for Nc bc = bs = 2ts – 2 tsc = 150 – 2x3 – 0 = 144 α = 0.85 – 0.007 (30-28) = 0.836 Nc = 215000 0.85x30x0.836x144 Nc = 70.03 mm 4 Mahmoud Shaaban
5. Using eq 8 (y=0) to find Ns Ns = 350 x 3 (2x300 + 150 – 2x55) – 215000 4 (3x350) Ns = 108.8 mm Using eq 11 to find Mu Mu = 3x350 (3002 + 300x150 – 2x108.82) – 0.425 x 0.8362 x 70.032x150x30 Mu = 110 KNm W = 8 Mu/L2 = 8x 110 / 52 = 35.2 KN/m 5 Mahmoud Shaaban
6. WE BEAM Y should equal d/2 or d/4 Y = d/2 = 300/2 = 150 mm tse = 3 mm Ast = Asc = 0 Using Є = 0 a/b = 16.66 Kb = 24 σ b > Fsy; Design should be on Fsy in equation 11 for the moment 6 Mahmoud Shaaban
7. Pb = Σo.x.Fb where Σo = 860 + 4Y = 1460 mm = 1460x2500x0.1 fb = 0.1 = 365000 N x = 5000/2 = 2500 Using eq2. to find Nc bc= 150-2x3-2x3 = 138 Nc = 365000 0.85x30x0.836x138 Nc = 124.07 mm 7 Mahmoud Shaaban
8. Using eq 7. to find Ns Ns = 350x3 (2x300+150-2x55)+2x150x3x350-365000 4 (3x350 + 3x350) Ns = 74.05 mm Mahmoud Shaaban 8
9. Since Y > Ns then will use eq10 to find Mu Mu = 3x350 (3002+300x150-2x74.052) + 3x350 (1502 – 2x74.052) - 0.425x0.8362x124.072x138x30 Mu = 123.4 kNm W = 8Mu/L2 = 8x 123.4/52 = 39.48 kN/m 9 Mahmoud Shaaban