Question

A steel channel section of length I has a built-in support at its lower end, and the upper end of the section has a rigid plate welded onto it, as illustrated in the figure below. The steel has a modulus of elasticity of E 200 GPa and a yield strength = 300 MPa. The section has a cross sectional area .d-1070 mm2 and second moments of area about its principal centroid axes of 1.-3.97 x 106mm4 and ly 0.658 × 106mm (a) Consider the case where the load Vacts on the rigid end plate and P-0 (1 If the line of action of the load V coincides with the centroid of the section (c as illustrated, then twisting of the section will occur because the load is not acting through the shear centre ofthe section. Produce a sketch showing the expected twisting of the section that will occur for this loading. 20 marks] (") To eliminate the twisting of the section, the line of action of V should be moved so that it passes through the shear centre of the section. lustrate the approximate location of the shear centre on a [20 marks] (iii) If L = 3500 mm and 1-21 kN (acting through the shear centre ofthe section), calculate the 20 marks] sketch of this section. deflection at the free end of the beam (b) Consider the case where the load P acts on the rigid end plate through the centroid of the section (c and V=0. (1) Consider the possibility of buckling of the column and identify the direction in which it is most likely 20 marks] (i) If the length L = 3500 mm determine the value of the force P corresponding to the critical buckling [20 marks] (iii) What is the maximum allowable length L that will ensure the column yields before it buckles as the 20 marks] to occur by sketching the column in its buckled orientation. load for this arangement. load P is increased from zero

Answer 1

e = Shear centre gf load is applied at G There will be no external element to resist twist due to shear flow through channel. Let us take a section z from free end of fiance horizonal shear is given as tau = v Q/I t_f = vt_f h/2/I t_f = vh/2I z This shows that shear stress variation is linear with respecto z At z = 0 z = 0 z = b z = vbh/2s m the web at y from centre of web vertical shear \r\n tau = vQ/I t_w = v/I t_w [b t_c h/2 + (h/2 - y) t_w (h/2 + y/2)] tau = v/2I t_w [b t_f h + t_w (h^2/4 - y^2)] t_w = t_f = t z = v/2I [bh + h^2/4 - y^2]