Architectural Drafting and Design (MindTap Course List)
7th Edition
ISBN: 9781285165738
Author: Alan Jefferis, David A. Madsen, David P. Madsen
Publisher: Cengage Learning
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Chapter 10, Problem 10.9Q
To determine
The four design criteria that should be considered in the planning of a dining room.
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Prob. Design the dimensions (rectangular)
and longitudinal reinforcements for the
beans sham. Design the beams as SRBS.
Given: fi= 21 MPa
fy= 275 hPa
X= 23.5 kaf. λ= 1.0
Please answer the following show me how to solve in your paper dont type thank you
Prob. Design the dimensions (rectangular)
and longitudinal reinforcements for the
beans sham. Design the beams as SRBS.
Given: fi= 21 MPa
fy= 275 hPa
X= 23.5 kaf. λ= 1.0
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- A simply supported rectangular RC beam is to carry a uniform factored dead load of 1.2 kip/ftand a concentrated factored live load of 16 kip at mid-span. The beam self-weight is not includedin these loads. The concrete weighs 135 pcf. The span length is 25 ft. Please find the smallestsection allowed by ACI and design accordingly. Use f c’ = 5,000 psi, f y = 75,000 psi. Theexposure is interior with no weather exposure.a. Assume an arbitrary self-weight/ft of the beam.b. Find the maximum factored bending moment in the beam.c. Set up the moment equation and solve for the beam section.d. Revise the assumption if needed.Hint: The beam section (b and h) and steel reinforcement are inversely proportional. The smallestallowable beam section will be for the largest allowable steel ration (ρmax), and vice versa. Sincethe steel ratio is fixed, two remaining variables (b, d) need to be found from the moment equations.Then, bd2 term can be solved to get an acceptable b and d combination.arrow_forwardFind: ftop and fbottom of (initial stage, construction phase, final stage)arrow_forward+150+ Assignment SW+ SLAB SDL = 250 150 - 3.3 kPa укра LL = 3 kPa 3 ୪ 8c = 23.6 kN/m² P = 3000 KN loss, = 9% Coss = 20% LBEAM = 9m COMPUTE AND DRAW THE STESS DIAGRAM (TRIBUTARY WIDTH= 600m 350mm FIND: f TOP & BOTTOM fe = 35Mpa 100mm f'c = 42.5 MPa 218 5m) EC = 4700 √ fc (MPa) (Initial, Const. phase, final stage)arrow_forward
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- 3k a 5 I IKLF d 25 5' S' E=29000ksi I = 400 in 4 Ex = ? Q = ?arrow_forwardA simply supported rectangular RC beam is to carry a uniform factored dead load of 1.2 kip/ft and a concentrated factored live load of 16 kip at mid-span. The beam self-weight is not included in these loads. The concrete weighs 135 pcf. The span length is 25 ft. Please find the smallest section allowed by ACI and design accordingly. Use f c’ = 5,000 psi, f y = 75,000 psi. The exposure is interior with no weather exposure. a. Assume an arbitrary self-weight/ft of the beam. b. Find the maximum factored bending moment in the beam. c. Set up the moment equation and solve for the beam section. d. Revise the assumption if neededarrow_forwardIdeal gas Problems 3-1 The molecular weight of carbon dioxide, CO2, is 44. In an experiment the value y for CO2 was found to be 1.3. Assuming that CO2 is a perfect gas, calculate the gas constant, R, and the specific heats at constant pressure and constant volume, Cp, Cv (0.189 kJ/kg.K; 0.63kJ/kg.K; 0.819kJ/kg.K) 3-2 Oxygen, O2, at 200 bar is to be stored in a steel vessel at 20°C the capacity of the vessel is 0.04m³. Assuming that O₂ is a perfect gas, calculate the mass of oxygen that can be stored in the vessel. The vessel is protected against excessive pressure by a fusible plug which will melt if the temperature rises too high. At what temperature must the plug melt to limit the pressure in the vessel to 240bar? The molecular weight of oxygen is 32 (10.5 kg; 78.6°C) 3-3 A quantity of a certain perfect gas is compressed from an initial state of 0.085m³, 1 bar to a final state of 0.034m³, 3.9 bar. The specific heats at constant volume are 0.724 kJ/kg.K, and the specific heats at…arrow_forward
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