Elementary Surveying (14th Edition)
14th Edition
ISBN: 9780133758887
Author: Charles D. Ghilani, Paul R. Wolf
Publisher: PEARSON
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Chapter 21, Problem 21.4P
To determine
The purpose that may be served by the phrase “more or less” in a description by metes and bounds.
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Determine the minimum of the appropriate yellow interval (Y
.
approach speed limit: 45 mi/h
approach grade: 3.2% downgrade
assumed perception-reaction time: 1.0 sec
assumed deceleration rate: 11.2 ft/sec²
assumed average vehicle length: 20 ft
width of intersection to be crossed: 56 ft
min'
in s) for a signal phase under the following conditions.
Compute the nominal shear strength of an M10 × 7.5 of A572 Grade 60 steel (Fy = 60 ksi).
For M10 × 7.5: d = 9.99 in., tw = 0.13 in., h/tw
Vn
=
x
kips
= 71.
A flexural member is fabricated from two flange plates 1/2 × 71/2 and a web plate 3/8 × 20. The yield stress of the steel is 50 ksi.
a. Compute the plastic section modulus Z and the plastic moment Mp with respect to the major principal axis.
(Express your answers to three significant figures.)
Z =
Mp
=
in. 3
ft-kips
b. Compute the elastic section modulus S and the yield moment My with respect to the major principal axis.
(Express your answers to three significant figures.)
S =
My
=
in.3
ft-kips
Chapter 21 Solutions
Elementary Surveying (14th Edition)
Ch. 21 - Prob. 21.1PCh. 21 - Prob. 21.2PCh. 21 - Prob. 21.4PCh. 21 - Prob. 21.5PCh. 21 - Prob. 21.6PCh. 21 - Prob. 21.7PCh. 21 - Prob. 21.8PCh. 21 - Prob. 21.9PCh. 21 - Prob. 21.10PCh. 21 - Prob. 21.11P
Ch. 21 - Prob. 21.12PCh. 21 - Prob. 21.13PCh. 21 - Prob. 21.14PCh. 21 - Prob. 21.15PCh. 21 - Prob. 21.16PCh. 21 - Prob. 21.17PCh. 21 - Prob. 21.18PCh. 21 - Prob. 21.19PCh. 21 - Prob. 21.20PCh. 21 - Prob. 21.21PCh. 21 - Prob. 21.22PCh. 21 - Prob. 21.23PCh. 21 - Prob. 21.24PCh. 21 - Prob. 21.26PCh. 21 - Prob. 21.27PCh. 21 - Prob. 21.28PCh. 21 - Prob. 21.29PCh. 21 - Prob. 21.30P
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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-kipsarrow_forwardGiven 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.arrow_forwardFor 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)arrow_forward
- 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
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