Sustainable Energy
1st Edition
ISBN: 9781133108689
Author: Richard A. Dunlap
Publisher: Cengage Learning
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Chapter 5, Problem 2P
To determine
Find the total binding energy for a
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A steel rod 100 ft long holds a 200 lb weight as shown. If the diameter of the circular rod is ¼ inch, find the
maximum normal stress in the road, taking into account the weight of the rod itself. Use: density of steel = ϒ = 490
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Chapter 5 Solutions
Sustainable Energy
Ch. 5 - Prob. 1PCh. 5 - Prob. 2PCh. 5 - A sample of 137Cs (half-life = 30.1 years) decays...Ch. 5 - Prob. 4PCh. 5 - What are the decay products of the decay of the...Ch. 5 - Prob. 6PCh. 5 - Nuclide A has a half-life of 4 days, and nuclide B...Ch. 5 - 64Cu can decay by both decay and + decay. Draw an...
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- ضهقعفكضكشتبتلتيزذظظؤوروىووؤءظكصحبت٢٨٩٤٨٤ع٣خ٩@@@#&#)@)arrow_forwardA square flexible foundation of width B applies a uniform pressure go to the underlying ground. (a) Determine the vertical stress increase at a depth of 0.5B below the center using Aσ beneath the corner of a uniform rectangular load given by Aσ Variation of Influence Value I m n 0.5 0.6 0.8 1.0 0.2 0.4 0.2 0.01790 0.03280 0.03866 0.04348 0.05042 0.05471 0.4 0.03280 0.06024 0.07111 0.08009 0.09314 0.10129 0.5 0.03866 0.07111 0.08403 0.09473 0.11035 0.12018 0.6 0.04348 0.08009 0.09473 0.10688 0.12474 0.13605 0.8 0.05042 0.09314 0.11035 0.12474 0.14607 0.15978 1.0 0.05471 0.10129 0.12018 0.13605 0.15978 0.17522 (Enter your answer to three significant figures.) Ασ/90 = Activity Frame (b) Determine the vertical stress increase at a depth of 0.5B below the center using the 2 : 1 method equation below. 90 x B x L Aσ = (B+ z) (L+ z) (Enter your answer to three significant figures.) Δσ/90 = (c) Determine the vertical stress increase at a depth of 0.5B below the center using stress isobars in…arrow_forwardNeed help!!!arrow_forward
- 2 A flexible circular area is subjected to a uniformly distributed load of 450 kN/m² (the figure below). The diameter of the load area is 2 m. Estimate the average stress increase (Aσay) below the center of the loaded area between depths of 3 m and 6 m. H₂ 1.0 H₂ B 0.8 CHI HD DV 0.6 C 1.0 1.5 0.4 0.2 6.0 8.0. 10.0 2.0 2.5 3.0 4.0 5.0 H₁ (Enter your answer to two significant figures.) Δσαν τ kN/m² 6arrow_forwardRefer to the figure below. Using the procedure outlined in your textbook, determine the average stress increase in the clay layer below the center of the foundation due to the net foundation load of 45 tons. Use the equations: Aσ = and qo x B x L (B+ z)(L+ z) Aσ av (H2/H₁) Δσι +44 + Δσο net load 6 4:5 ft 10 ft 5ft x 5ft Sand Sand y=100 lb/ft³ Ysat 122 lb/ft³:" Ysat 120 lb/ft³: 0.7 C=0.25 Groundwater table C=0.06 Preconsolidation pressure = 2000 lb/ft² (Enter your answer to three significant figures.) Ασαν = lb/ft²arrow_forwardRefer to the figure below, which shows a flexible rectangular area. Given: B₁ = 4 ft, B₂ = 6 ft, L₁ = 8 ft, and L2 = 10 ft. If the area is subjected to a uniform load of 4100 lb/ft², determine the stress increase at a depth of 10 ft located immediately below point O. Use the table below. T B(1) 3 B(2) 2 L(1) * 4 L2) Table 1 Variation of Influence Value I n m 0.8 0.9 1.0 1.2 1.4 0.1 0.02576 0.02698 0.02794 0.02926 0.03007 0.2 0.05042 0.05283 0.05471 0.05733 0.05894 0.3 0.07308 0.07661 0.07938 0.08323 0.08561 0.4 0.09314 0.09770 0.10129 0.10631 0.10941 0.5 0.11035 0.11584 0.12018 0.12626 0.13003 0.6 0.12474 0.13105 0.13605 0.14309 0.14749 0.7 0.13653 0.14356 0.14914 0.15703 0.16199 0.8 0.14607 0.15371 0.15978 0.16843 0.17389 0.9 0.15371 0.16185 0.16835 0.1766 0.18357 1.0 0.15978 0.16835 0.17522 0.18508 0.19139 1.1 0.16843 0.17766 0.18508 0.19584 0.20278 (Enter your answer to three significant figures.) Aσ = lb/ft²arrow_forward
- Point loads of magnitude 100, 200, and 380 kN act at B, C, and D, respectively (in the figure below). Determine the increase in vertical stress at a depth of 6 m below point A. Use Boussinesq's equation. B 6 m A 6 m с 3 m D (Enter your answer to three significant figures.) Δαχτ kN/m²arrow_forwardTwo line loads q₁ = 30 kN/m and 92 = 44 kN/m of infinite lengths are acting on top of an elastic medium, as shown in the figure below. Find the vertical stress increase at A. 92 91 6 m 3 m 3 m Δσ A (Enter your answer to three significant figures.) Vertical stress increase at A = kN/m²arrow_forwardA flexible circular area is subjected to a uniformly distributed load of 144 kN/m² (see the figure below). The diameter of the load area is 2 m. Estimate the average stress increase (Aσay) below the center of the loaded area between depths of 3 m and 6 m. Use the equations: 1 Ασ = go 1 [1 + (2) ² ³/2 and Aσ av (H2/H1) Δσι + 41ση + Ασο 6 9 B/2 krark do Δε Aσ (Enter your answer to three significant figures.) Ασαν = kN/m²arrow_forward
- In construction what is the difference in general requirements specific project requirements?arrow_forwardRefer to the figure below. Determine the vertical stress increase Aσ at point A with the values q₁ = 90 kN/m, q2 = 410 kN/m, x₁ = 4m, x2 = 2.5 m, and z = 3 m. Line load = 91 Line load=92 Δε (Enter your answer to three significant figures.) Δατ kN/m²arrow_forwardRefer to the figure below. A strip load of q = 870 lb/ft² is applied over a width B = 36 ft. Determine the increase in vertical stress at point A located z = 15 ft below the surface. Given: x = 27 ft. Use the table below. B q = Load per unit area Aσ A Table 1 Variation of Ao/go with 2z/B and 2x/B 2x/B 2z/B 1.3 1.4 1.5 1.6 0.00 0.000 0.000 0.000 1.7 0.000 0.000 0.10 0.007 0.003 0.002 0.001 0.001 0.20 0.040 0.020 0.011 0.30 0.090 0.052 0.031 0.40 0.141 0.090 0.059 0.040 0.027 0.50 0.185 0.128 0.089 0.063 0.60 0.222 0.163 0.120 0.088 0.70 0.250 0.193 0.80 0.273 0.218 0.007 0.004 0.020 0.013 0.046 0.066 0.148 0.113 0.087 0.173 0.137 0.108 0.90 0.291 0.239 0.195 0.158 0.128 1.00 0.305 0.256 0.214 0.177 0.147 (Enter your answer to three significant figures.) lb/ft² Δοχ =arrow_forward
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