Question

Answers:

I_XX = 92.11×10⁶ mm⁴

w_max = 15.53 kN/m

1/R = 0.0085 m^-1

6. Calculate lxx for the steel I-section shown in Fig. Q6a. This section is to span a distance of 9 m and is loaded as shown in Fig. Q6b. Calculate the magnitude of the maximum UDL that can be sustained by the beam if the design stress of the steel is 239 N/mm2. Calculate the curvature of the beam at midspan under that maximum UDL, taking Estel 200 kN/mm2. ty 15mm w kN/m 250mm] x x 9m Fig. Q6b 15mm Fig. Q6a

Answer 1

I_xx = 150 times 18063/12 - 140 times (250)^3/12 Rightarrow = 274400000 - 182291666.7 Rightarrow I_xx = 92108333.33 mm64 Rightarrow I_xx = 92.11 times 10^6 mm^4 Let w_max be maximum value UDL Bending moment will be maximum at the centre of beam maximum bending moment at centre os span M_max = W_max L/2 times L/2 - W_max L/2 times 1/2 times L/2 Rightarrow M_max = W_max L^2/8 \r\n Bending stress will be maximum at top and bottom fibers and its magnitude is given by sigma_max = M_max/I_xx Y_max sigma_max = 239 N/mm^2 Y_max = 250/2 + 15 = 140 mm I_xx = 92.11 times 10^6 mm^4 Therefore 239 = M_max/92.11 times 10^6 times 140 Rightarrow M_max = 157244928.6 N_mm Rightarrow M_max = 157244928.6 times 1/10^3 times 1/10^3 KNm Rightarrow M_max = 157.2449286 kNM Now M_max = W_max L^2/theta Rightarrow 157.2449286 = W_max times 9^2/8 Rightarrow W_max = 15.53 kN/m Bending moment at centre of beam