Question

4) 20 Marks A builder strengthens a 150 mmx 250 mm timber beam by connecting three 10 mm thick steel plates to the top and the two sides of the beam as shown below (there is no steel plate on the bottom). The beam acts compositely. It should be assumed that Eimber 10000 MPa and Esteel 200000 MPa and the materials exhibit linear elastic behaviour. The beam has a total length of 5 m. Steel has a density of 7800 kg/m3 and timber a density of 1000 kg/m. The beam is subjected to a UDL of 25 kN/m (the self weight is included in this value). What is the self-weight per unit length? Draw the bending moment diagram, and clearly identify the locations and magnitudes of the maximum bending moment. a) b) c) Consider the cross section of the beam that experiences the largest absolute value of moment. Draw d) the strain and stress distributions (values required) due to bending at this section. Comment Approximately page of text plus diagrams is expected. on the effectiveness/efficiency of the strengthening technique used by the builder 150 Steel 25 kN/m 250 Timber 5 m All dimensions are in mm Typical Beam Cross-Section

Answer 1

Given: Cross section of timber beam = 150mm times 250 mm Thickness of steal plate = 10 mm length of beam = 5 m = 5000 mm E_timber = 10000 mpa E_steal = 200000 varsigma_ = 7800 kg/m^3: varsigma_tim = 1000 kg/m^2 Self of beam = wt. of timber st. of steal plate = (0.15 times 0.25) times 1000 + (0.15 times 0.01 + 0.25 times 0.01 times 2) times 7800 = 88.2 kg/m = 865.2 N/m = 0.865 KN/m M_max = wL^2/8 = 25 times 5^2/8 = 78.125 KN - M Note Due to symmetric loading xi simply beam, M_max will lie at the . stain in steal = in timber or E_S = E_L (sigma/E)_s = (sigma/E)_t sigma_c = (E_s/E_t)sigma_t or (P/A)_s = m(P/A)_t (Because Dlodulor ratio m = E_s/E_t = 200000/10000 = 20) A_T = MA_s A_T = 20A_s Equivalent area of timber calculating cautia of gravity y bar = A_1y_1 + A_2y_2/A_1 + A_2 =