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 6, Problem 6.17Q
To determine
The shape of the window in the side view for the given condition.
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Given a portion of a pipe network below. Determine the true discharge in each pipe using the Hardy-Cross method. Use the Darcy-Weisbach formula with f = 0.02 for all pipes.
For the cantilever retaining wall shown in the figure below, let the following data be given:
Wall dimensions: H = 6.5 m, x1 = 0.3 m, x2 = 0.6 m, x3 = 0.8 m,
x4 =2m, x5 = 0.8 m, D= 1.5 m, a = 0°
Soil properties: 1 = 17.58 kN/m³, 1 = 36°, Y2 = 19.65 kN/m³,
215°, c230 kN/m²
For 2=15°: Ne 10.98; N₁ = 3.94; N₁ = = 2.65.
H
Calculate the factor of safety with respect to overturning, sliding, and bearing capacity. Use Y concrete = 24.58 kN/m³. Also, use k₁ = k₂ = 2/3 and Pp
FS (sliding)
(EV) tan(k102) + Bk2c₂ + Pp
Pa cos a
(Enter your answers to three significant figures.)
= 0 in equation
FS (overturning)=
FS(sliding)=
FS(bearing)
A W16×67 of A992 steel has two holes in each flange for 7/8-inch-diameter bolts.
For A992 steel: Fy
= 50 ksi, F₁ = 65 ksi.
For a W16×67: bƒ
=
10.2 in., tf
= 0.665 in., Zx =130 in.3 and S
= 117 in.3
a. Assuming continuous lateral support, verify that the holes must be accounted for and determine the nominal flexural strength.
(Express your answer to three significant figures.)
Mn
=
ft-kips
b. What is the percent reduction in strength?
(Express your answer to three significant figures.)
%
Reduction =
Chapter 6 Solutions
Architectural Drafting and Design (MindTap Course List)
Ch. 6 - Prob. 6.1QCh. 6 - Prob. 6.2QCh. 6 - Prob. 6.3QCh. 6 - Prob. 6.4QCh. 6 - Prob. 6.5QCh. 6 - Prob. 6.6QCh. 6 - Prob. 6.7QCh. 6 - Prob. 6.8QCh. 6 - Prob. 6.9QCh. 6 - Prob. 6.10Q
Ch. 6 - Prob. 6.11QCh. 6 - Prob. 6.12QCh. 6 - Prob. 6.13QCh. 6 - Prob. 6.14QCh. 6 - Prob. 6.15QCh. 6 - Prob. 6.16QCh. 6 - Prob. 6.17QCh. 6 - Prob. 6.18QCh. 6 - Prob. 6.19QCh. 6 - Prob. 6.20QCh. 6 - Prob. 6.21QCh. 6 - Prob. 6.22QCh. 6 - Prob. 6.23QCh. 6 - Prob. 6.24QCh. 6 - Prob. 6.25QCh. 6 - Prob. 6.26QCh. 6 - Prob. 6.27QCh. 6 - Prob. 6.28QCh. 6 - Prob. 6.29QCh. 6 - Prob. 6.30Q
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- A gravity retaining wall is shown in the figure below. Calculate the factor of safety with respect to overturning and sliding, given the following data: Wall dimensions: H = 6 m, x1 = 0.6 m, x2 = 2 m, x3 = 2m, x4 = 0.5 m, x5 = 0.75 m, x6 = 0.8 m, D= 1.5 m Soil properties: 1 = 17.5 kN/m³, ø₁ = 32°, 12 = 18 kN/m³, =22°, 40 kN/m² Y₁ H D x2 x3 x5 X6 Use the Rankine active earth pressure in your calculation. Use Yconcrete = 23.58 kN/m³. Also, use k₁ = k₂ = 2/3 and P₁ = 0 in the equation FS S(sliding) | tan(k102) + Bk₂c½ + Pp (Σν). Pa cos a (Enter your answers to three significant figures.) FS (overturning) FS (sliding)arrow_forwardQ4 Use b member Castigliano's second theorem to determine the structure shown Forces In figure below longer than + In IF required. For all members member EC IS 10mm E = 200 KN/mm² 200KN YE FV 100 KN A = 1800 mm² and 2 2m 3m B D 3m 8M *arrow_forwardFor the cantilever retaining wall shown in the figure below, let the following data be given: Wall dimensions: H = 8 m, x1 = 0.4 m, x2 = 0.6 m, x3 = 1.5 m, x4 3.5 m, x5 = 0.96 m, D= 1.75 m, a = 10° Soil properties: ₁ = 17.3 kN/m³, 1₁ = 32°, Y2 = 17.6 kN/m³, 2=28°, c₂ = 30 kN/m² The value of Ka is 0.3210. For 2=28°: N = 25.80; N₁ = 14.72; N₁ = 16.72. 3. Also, use k₁ = k₂ = 2/3 and Pp = 0 in the equation Calculate the factor of safety with respect to overturning, sliding, and bearing capacity. Use concrete = 24.58 kN/m³. A FS (sliding) (V) tan(k₁₂) + Bk₂c½₂ + Pp Pa cos a (Enter your answers to three significant figures.) FS (overturning) FS(sliding) FS (bearing)arrow_forward
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