# ME311-HW5 - Homework#5 Solution ME 311-Winter 2016 Due Date...

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Homework #5 Solution ME 311-Winter 2016 Due Date 02/19/2016
1 4.5. The C 200 × 20 channel section of Figure P4.5 has second moments of area I x = 15 × 10 6 mm 4 , I y = 0 . 637 × 10 6 mm 4 . It is loaded by the bending moments M x = 2200 Nm and M y = - 350 Nm. Find the magnitude and location of the maximum tensile stress in the beam. O x y -350 Nm 15.6 2200 Nm 203 59.5 9.9 Figure P4.5 We first calculate E R x = M x I x = 2200 × 10 3 15 × 10 6 = 0 . 147 N/mm 3 E R y = M y I y = - 350 × 10 3 0 . 637 × 10 6 = - 0 . 549 N/mm 3 , from equations (4.16, 4.17) and hence σ zz = Ey R x - Ex R y = 0 . 147 y + 0 . 549 x in MPa if x , y are in mm. The maximum tensile stress will occur when both x and y are at their most posi- tive — i.e. at the top right hand corner whose coordinates are x C = 59 . 5 - 15 . 6 = 43 . 9 ; y C = 203 2 = 101 . 5 . We therefore have σ max zz = 0 . 147 × 101 . 5 + 0 . 549 × 43 . 9 = 39 MPa .
1 4.8. The Z-section of Figure P4.8 has second moments of area I x = 560 , 000 mm 4 , I y = 290 , 000 mm 4 , I xy = 300 , 000 mm 4 . It is used for a beam of length 2 m which is simply supported at its ends and loaded by a uniformly distributed vertical load of 1000 N/m. Find the location and magnitude of the maximum tensile stress. x y 10 10 10 O 40 40 50 all dimensions in mm Figure P4.8 Figure P4.8.1(a) shows a free-body diagram of the beam, from which, using sym- metry, we deduce that the reactions at each end must be R 1 = R 2 = 1000 × 2 2 = 1000 N . 1000 N/m 2 m R 2 R 1 1000 N/m 1000 N y V 1 m M x (a) (b) Figure P4.8.1 The maximum bending moment will occur at the mid-point and can be deter- mined from the free-body diagram of the beam segment shown in Figure P4.8.1(b). We find M max x = - 1000 × 1 + 1000 × 0 . 5 = - 500 Nm .
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