Transportation Assignment-1 Submitted by: 1. Sudheendra CE13B0952. Karthikeyan CE13B087QuestionThe random numbers generated are:E1 = 1800 MPaE2 = 380 MPah1 = 18 cmh2 = 38 cmHence, the height of the first layer is 18 cm and height of second layer is 38 cm. The modulus of elasticity for first layer is 1800 MPa and 380 MPa for second layer. For third layer, it is 75 MPa.a) From Kenpave analysis,The vertical stress at various depth under the center is given in the following table:Depth (cm)Vertical Stress (MPa)

0916.000

9 (Mid-point of first layer)631.573

18 (On top of second layer)230.947

37 (Mid-point of second layer)70.650

56 (On top of third layer)24.206

b) The percentage stress decrease over 1st layer = (916.000 230.947)/ (916-0) *100 = 74.78 %The percentage stress decrease over 2nd layer = (230.947 - 24.206)/ (916-0) * 100 = 22.57 %The percentage stress decrease over 3rd layer = 24.206 / 916 * 100 = 2.64%Questiona) Radius of contact area = (( 40*103)/3.14*600*103)0.5= 14.56 cm Hence, r1 = r2 = 14.56 cm (r/a) factor for load 1 = (30/14.56) = 2.06 (r/a) factor for load 2 = 0 z/a factor for load 1 = z/a factor for load 2 = 30/14.56 = 2.06i) for z/a = 2 and r/a = 2, r/q = 0.06 implying r = 0.06*q = 0.06*600 = 36 kPa ii) For z/a = 2 and r/a = 0, r/q = 0.018 implying r = 0.018*q = 10.8 kPaTotal stress = 36+10.8 = 46.8 kPab) Radius of contact area = (40*103 / 3.14*600*103)0.5 = 14.56 cmHence, r1 = 14.56 cm and r2 = 14.56 cm r/a factor for load 1 = (30/14.56) = 2.06 r/a factor for load 2 = 0z/a factor for load 1 = z/a factor for load 2 = 30/14.56 = 2.06i) For z/a = 2, r/a = 2, = 0.06*q = 0.06 * 600 = 36 kPa ii) For z/a = 2, r/a =0, = 0.018*q = 16.2 kPa Total stress = 36 + 16.2 = 52.2 kPac) Radius of contact area, r1 = 14.56 cm and r2 = 12.61 cmr/a factor for load 1 = 2.06r/a factor for load 2 = 0z/a factor for load 1 = 2.06 and z/a factor for load 2 = 2.38i) For z/a = 2, r/a = 2, r = 0.06*q = 36 kPaii) For z/a = 2.38, r/a = 0, r = 0.01*q = 12 kPaTotal stress = 36 kPa +12 kPa = 48 kPa

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