Residential Construction Academy: House Wiring (MindTap Course List)
5th Edition
ISBN: 9781337402415
Author: Gregory W Fletcher
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Question
Chapter 6, Problem 10RQ
To determine
The reason for both the electric heating load and the air conditioning load are not used in a service entrance calculation and the section of
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Need help!! Add martin luther king jr day as a holiday so it won't be a work day
ضهقعفكضكشتبتلتيزذظظؤوروىووؤءظكصحبت٢٨٩٤٨٤ع٣خ٩@@@#&#)@)
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
lb/ft3
.
Chapter 6 Solutions
Residential Construction Academy: House Wiring (MindTap Course List)
Knowledge Booster
Similar questions
- ضهقعفكضكشتبتلتيزذظظؤوروىووؤءظكصحبت٢٨٩٤٨٤ع٣خ٩@@@#&#)@)arrow_forwardضهقعفكضكشتبتلتيزذظظؤوروىووؤءظكصحبت٢٨٩٤٨٤ع٣خ٩@@@#&#)@)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_forward
- Need help!!!arrow_forward2 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_forward
- Refer 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_forwardPoint 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_forward
- A 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_forwardIn 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_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Residential Construction Academy: House Wiring (M...Civil EngineeringISBN:9781337402415Author:Gregory W FletcherPublisher:Cengage Learning
Residential Construction Academy: House Wiring (M...
Civil Engineering
ISBN:9781337402415
Author:Gregory W Fletcher
Publisher:Cengage Learning