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 42, Problem 42.12Q
To determine
The definition of a nested joists.
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The beam shown in the figure below has lateral support at the ends only. The concentrated loads are live loads. Use A992 steel and select a shape.
Do not check deflections. Use C = 1.0 (this is conservative). Suppose that PL
= 20 k.
PL
PL
Use the table below.
Shape
Mn Mn Vn Vn
Ω
Ων
W12 × 58
216
144 132 87.8
W12 × 65
269
179 142 94.4
W12 × 72
308
205
159 106
W14 × 61
235
157
156 104
W14 × 68
280
187
174 116
W14 × 74
318
212
192 128
W16 × 67
276 184
193 129
18'
a. Use LRFD.
Calculate the factored-load moment (not including the beam weight).
(Express your answer to three significant figures.)
Mu
=
ft-kips
Select a shape.
-Select-
b. Use ASD.
Calculate the required flexural strength (not including the beam weight).
(Express your answer to three significant figures.)
Ma
=
Select a shape.
-Select-
ft-kips
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)
Chapter 42 Solutions
Architectural Drafting and Design (MindTap Course List)
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- 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 =arrow_forwardA 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_forward
- For 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_forwardUse the Force Method to analyse the structure. After the analysis, draw Bending Moment and Shear Force diagrams. Redundants need to be put at 'j' and 'k' as moments (EI is constant all across the structure)I am so confused as to how to approach the problem. I would really appreciate an answer with a little explanation, or helpful working out!arrow_forwardI got 5.97 mm please show your work clearly. thank youarrow_forward
- A 7K SK-> VE 3 F T A=52 E=29000 ksi diagonal members 6' A=30.25.72 E=1800 ksi for horizontal & vertical member ↓ B Oc AD 8 Primary Structures remove roller @C make D a roller For Primary and Cut BF For redundant Ik ↑ ec Ik = @D Ik @BFarrow_forwardConsider the geometric and traffic characteristics shown below. Approach (Width) North South East West (56 ft) (56 ft) (68 ft) (68 ft) Peak hour Approach Volumes: Left Turn 165 105 200 166 Through Movement 447 400 590 543 Right Turn 162 157 191 200 Conflicting Pedestrian Volumes 900 1,200 1,200 900 PHF 0.95 0.95 0.95 0.95 For the following saturation flows: Through lanes: 1,600 veh/h/In Through-right lanes: 1,400 veh/h/In Left lanes: 1,000 veh/h/In Left-through lanes: 1,200 veh/h/In Left-through-right lanes: 1,100 veh/h/In The total cycle length was 283 s. Now assume the saturation flow rates are 10% higher, that is, assume the following saturation flow rates: Through lanes: 1,760 veh/h/In Through-right lanes: 1,540 veh/h/In Left lanes: 1,100 veh/h/In Left-through lanes: 1,320 veh/h/In 1,210 veh/h/In Left-through-right lanes: Determine a suitable signal phasing system and phase lengths (in s) for the intersection using the Webster method. (Enter the sum of green and yellow times for…arrow_forwardThe given beam has continuous lateral support. If the live load is twice the dead load, what is the maximum total service load, in kips / ft, that can be supported? A992 steel is used: Fy = 50 ksi, Fu=65 ksi. Take L = 30 ft. bf For W40 x 149: 2tf = 7.11, = = 54.3, Z 598 in.³ tw W W40 X 149 L (Express your answers to three significant figures.) a. Use LRFD. Wtotal = kips/ft b. Use ASD. Wtotal kips/ftarrow_forward
- The beam shown in the figure below is a W16 × 31 of A992 steel and has continuous lateral support. The two concentrated loads are service live loads. Neglect the weight of the beam and determine whether the beam is adequate. Suppose that P = 52 k. For W16 × 31: d = 15.9 in., tw = 0.275 in., h/tw = 51.6, and M = M₁ = 203 ft-kip, Mn/₁ = Mp/α = 135 ft-kip. P Р W16 x 31 a. Use LRFD. Calculate the required moment strength, the allowable shear strength, and the maximum shear. (Express your answers to three significant figures.) Mu = OvVn = ft-kip kips kips Vu = Beam is -Select- b. Use ASD. Calculate the required moment strength, the allowable shear strength, and the maximum shear. (Express your answers to three significant figures.) Ma = Vn/b - Va = Beam is -Select- ft-kip kips kipsarrow_forwardDetermine the smallest value of yield stress Fy, for which a W-, M-, or S-shape from the list below will become slender. bf/2tfh/tw Shape W12 × 72 8.99 22.6 W12 × 26 8.54 47.2 M4 × 6 11.9 22.0 M12 x 11.8 6.81 62.5 M6 × 4.4 5.39 47.0 S24 × 80 4.02 41.4 S10 × 35 5.03 13.4 (Express your answer to three significant figures.) Fy = ksi To which shape does this value apply? -Select- ✓arrow_forwardCompute the nominal shear strength of an M12 × 11.8 of A572 Grade 60 steel (Fy = 60 ksi). For M12 x 11.8: d = 12 in., tw = 0.177 in., h/tw = 62.5. Vn = kipsarrow_forward
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