Elementary Surveying: An Introduction To Geomatics (15th Edition)
15th Edition
ISBN: 9780134604657
Author: Charles D. Ghilani
Publisher: PEARSON
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Chapter 18, Problem 18.14P
To determine
How maximum effectiveness is achieved in maps design
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A3.1- A simply supported beam is subjected to factored concentrated load of 400 kN at
mid-span. The beam has a 10m span and a rectangular cross-section with bw = 350mm,
effective depth d = 520mm, and total height h = 620mm.
a) Ignor the self-weight of the beam and design the required shear reinforcement for the
beam. Use 10M U-stirrups.
b) Sketch the beam elevation and show the stirrups.
Given: The beam is reinforced with 5-25M longitudinal bars
f'c = 30 MPa
fy = 400 MPa
Maximum aggregate size: 20mm
Figure 1
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A=Sin² E=290ooks for
diagonal members
A= 30.25in² E = 1800 ksi
for hoizontal &
Vertical members
roller@G, make Da roller
An urban freeway is to be designed using the following information.
AADT = 52,600 veh/day
K (proportion of AADT occurring during the peak hour):
D (proportion of peak hour traffic traveling in the peak direction):
Trucks:
0.11
0.65
8% of peak hour volume
PHF = 0.94
Lane width:
Shoulder width:
Total ramp density:
Terrain:
12 ft
10 ft
0.5 interchange/mile; all interchanges are to be cloverleaf interchanges
rolling
Determine the number of lanes in the peak direction required to provide LOS C. (Assume commuter traffic and assume no RVs.)
lanes
Show all calculations required. (Calculate your answers for the peak direction only. Enter fy the peak hour volume in veh/h, the free flow speed in mi/h, the demand flow rate in pc/h/In, the mean speed in mi/h, and the density in pc/mi/In.)
fHV
peak hour volume
free flow speed
demand flow rate
mean speed
veh/h
mi/h
pc/h/In
mi/h
density
pc/mi/In
Chapter 18 Solutions
Elementary Surveying: An Introduction To Geomatics (15th Edition)
Ch. 18 - Prob. 18.1PCh. 18 - Prob. 18.2PCh. 18 - Prob. 18.3PCh. 18 - Prob. 18.4PCh. 18 - Prob. 18.5PCh. 18 - Prob. 18.6PCh. 18 - Prob. 18.7PCh. 18 - Prob. 18.8PCh. 18 - Prob. 18.9PCh. 18 - Prob. 18.10P
Ch. 18 - Prob. 18.11PCh. 18 - Prob. 18.12PCh. 18 - Prob. 18.13PCh. 18 - Prob. 18.14PCh. 18 - Prob. 18.15PCh. 18 - Prob. 18.16PCh. 18 - Prob. 18.17PCh. 18 - Prob. 18.18PCh. 18 - Prob. 18.19PCh. 18 - Prob. 18.20PCh. 18 - Prob. 18.21PCh. 18 - Prob. 18.22PCh. 18 - Prob. 18.23PCh. 18 - Prob. 18.24PCh. 18 - Prob. 18.25PCh. 18 - Prob. 18.26PCh. 18 - Prob. 18.27PCh. 18 - Prob. 18.28PCh. 18 - Prob. 18.29PCh. 18 - Prob. 18.30PCh. 18 - Prob. 18.31P
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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
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