# HW10 solution.pdf - 1 6.46 The unequal angle of Figure...

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1 6.46. The unequal angle of Figure P6.46 is loaded by a shear force of 6 kN, with the line of action shown. Find the twist per unit length, if the material of the beam is steel with G = 80 GPa. 6 kN 4 4 40 80 24 all dimensions in mm Figure P6.46 The shear centre must be at the intersection of the two legs, so the loading is equivalent to a load of 6 kN through the shear centre and a clockwise moment T = 6 × 10 3 × 26 × 10 - 3 = 156 Nm. The total perimenter of the section is b = 38 + 78 = 116 mm and it is all of uniform thickness 4 mm, so the maximum shear stress due to the torque is τ max = 3 × 156 × 10 3 116 × 4 2 = 252 MPa, from equation (6.62). There will be an additional stress due to the shear force through the shear centre which we estimate (conservatively) as τ = 3 2 V y A web = 3 × 6 × 10 3 2 × 80 × 4 = 28 MPa. This stress is additive to the torsional shear stress on the right side of the vertical leg near the corner, giving a maximum shear stress of τ max = 252 + 28 = 280 MPa. The torsional stiffness is
2 K = bt 3 3 = 116 × 4 3 3 = 2475 mm 4 and hence the twist per unit length is θ L = T GK = 156 × 10 3 80 × 10 3 × 2475 = 0 . 788 × 10 - 3 rad/mm = 0.788 rad/m (45.1 o /m).
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