Using the grade stake elevations in the chart below, prepare a grade sheet showing cuts/fills (ft and in.) for the proposed curb from 0 + 52 to 8 + 38. Grade Stake ElevationsРС0 + 52504.714 + 50508.461 + 00504.785 + 00508.771 + 50504.935 + 55510.612 + 00504.986 + 00511.732 + 50505.826 + 50512.003 + 00506.997 + 00512.023 + 50507.617 + 50512.113 + 75507.878 + 00512.244 + 00508.26 PC 8 + 38512.73

As we know that for economy the cuts and fills must be equal So Average Elevation is taken for the…

Grade Stake Elevations PC 0 + 52 504.71 4 + 50 508.46 1 + 00 504.78 5 + 00 508.77 1 + 50 504.93 5 + 55 510.61 2 + 00 504.98 6 + 00 511.73 2 + 50 505.82 6 + 50 512.00 3 + 00 506.99 7 + 00 512.02 3 + 50 507.61 7 + 50 512.11 3 + 75 507.87 8 + 00 512.24 4 + 00 508.26 PC 8 + 38 512.73

Grade Stake Elevations PC 0 + 52 504.71 4 + 50 508.46 1 + 00 504.78 5 + 00 508.77 1 + 50 504.93 5 + 55 510.61 2 + 00 504.98 6 + 00 511.73 2 + 50 505.82 6 + 50 512.00 3 + 00 506.99 7 + 00 512.02 3 + 50 507.61 7 + 50 512.11 3 + 75 507.87 8 + 00 512.24 4 + 00 508.26 PC 8 + 38 512.73

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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400 Hole of radius 200 300 --> x' 600 - 750
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Table: Profile Elevation STATION BS BM1 0+000 +005 0+010 0+015 0+020 0+025 0+030 0+035 0+040 0+045 0+050 0+055 0+060 0+065 0+070 0+075 0+080 0+085 0+090 0+095 0+100 0+105 TP1 0+110 0+115 0+120 0+125 0+130 0+135 0+140 0+145 0+150 0+155 0+160 0+165 0+170 0+175 1.515 1.700 HI FS 1.109 IFS 1.520 1.508 1.490 1.470 1.455 1.408 1.418 1.423 1.428 1.403 1.396 1.420 1.405 1.340 1.323 1.306 1.260 1.234 1.229 1.201 1.130 1.109 1.673 1.636 1.647 1.455 1.472 1.430 1.392 1.352 1.320 1.292 1.255 1.210 1.770 1.120 ELEVATION 100 m 0+180 0+185 0+190 0+195 TP2 0+200 0+205 0+210 0+215 0+220 0+225 0+230 0+235 0+240 0+245 0+250 0+255 0+260 0+265 TP3 0+270 0+275 0+280 0+285 0+290 0+295 0+300 0+305 0+310 0+315 0+320 0+325 0+330 0+335 0+340 0+345 0+350 0+355 0+360 0+365 1.660 1.977 O 1.030 0.913 1.085 1.050 1.017 1.030 1.630 1.648 1.638 1.554 1.389 1.285 1.216 1.161 1.478 1.086 1.039 0.968 0.922 0.913 1.913 1.871 1.814 1.758 1.668 1.545 1.444 1.412 1.353 1.354 1.335 1.328 1.297 1.273 1.241 1.156 1.155 1.101…
3. An A36 Steel Cantilever Beam is shown below, also showing its cross section. Determine the following: a. The Support Reactions b. Centroid of the Cross Section c. Centroidal Moment of Inertia of the Section parallel with the x-axis d. Centroidal Moment of Inertia of the Section parallel with the y-axis Given the formula of the Maximum allowable bending stress of steel, fb = 0.66Fy. Where Fy is the Yield Stress of Steel. Where the bending stress for a beam is (fb) = Mc/I, where c is the distance of the centroid to the extreme most fiber of the beam (either top-most or bottom-most distance). Does the Steel fail under the loading given? Use Fy = 450 Mpa. Use the Moment in (a) e. and the moment of inertia in (c).
(b)- A page of a level with some missing readings is reporduced below. Fill in the missing entries along with the necessary arithmetical checks. Station I.S F.S Rise 1 2 3 4 5 6 7 8 9 10 11 Total = B.S 3.650 X -2.170 X 6.490 X 2.830 3.640 2.410 2.320 3.000 X X 2.420 2.750 Fall 1.320 2.750 1.320 R.L 108.260 Remark
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