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MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months)
5th Edition
ISBN: 9781305581159
Author: Nicholas J. Garber; Lester A. Hoel
Publisher: Cengage Learning US
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Chapter 15, Problem 24P
To determine
The factors that must be taken into account in the design of bicycle paths.
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- You are an engineer designing an aeration tank for a wastewater treatment plant receiving municipal waste. The activated sludge system (aeration tank and secondary clarifier) you design needs to remove 85% of the incoming BODs from the primary effluent, giving a final concentration of 30.0 mg BOD5/L exiting the system. Your design will maintain a concentration of 2500 mg VSS/L and F/M ratio of 0.5 g BOD/g VSS d in the aeration tank. In an effort to keep the waste activated sludge low in water, only about 0.1% of the primary effluent flow (Q) exits the WAS line. The secondary effluent has a flow rate of 9990 m³/d. What volume (in m³) will the aeration tank need to be? Write out all equations and state any assumptions as needed. Primary effluent: Aeration tank V, X, S Secondary clarifier Secondary effluent: Q. Xo. So Qe, Xe, S 0 po RAS line: Q, X., S Activated sludge control volume WAS line: Qw, X, Sarrow_forwardAum A waste incinerator stack emits 27,027 kg/yr of cadmium and has an effective stack might of 100 m. The wind speed is 5 m/s at an anemometer located at 10 m. It is a clear sunny day with the sun nearly overhead (Class B conditions). a. Calculate the ground-level cadmium concentration (in ug/m³) at a distance of 2 km directly downwind? [Refer to tables in the textbook to help with this problem] What is the concentration of cadmium (in µg/m³) inside of a house at the location in part (a) after two hours? Note that the initial concentration of cadmium in the house was zero, there are 0.25 air changes per hour (ach), there is no source of cadmium inside the house, and cadmium is considered a conservative pollutant.arrow_forward6. A simply supported beam is subjected to uniformly distributed loads: a service dead load WD = 5 kips/ft and a service live load w₁ =7 kips/ft. Use the load combination w₁ = 1.2WD + 1.6WL. Note that the shear force is expressed as V₂ = w₁ (-/-- x) where x is the distance from a support. (65pts total) The beam also has the following properties. f = 4ksi, fy = fyt = 60 ksi, b = 18 in, h = 26 in, d = 24.5 in, 20ft No. 4 bars, U-stirrups used, Normalweight concrete (1) Calculate the required spacing of stirrups. (15pts) (2) Calculate the maximum spacing. Also, determine the location along the beam where the maximum spacing can be applied (i.e., the distance from support). (10pts) (3) Show the stirrup designs along with the shear force diagram. Indicate the numbers and spacings of stirrups. A transition between the required spacing and the maximum spacing is unnecessary. (15pts) (4) Assume the live load decreases, and, as a result, the factored shear force is 30 kips. Discuss whether…arrow_forward
- 5. A singly reinforced beam has a width b = 16 in., a height h = 20 in., and an effective depth d = 18.5 in. The beam is reinforced with six No. 8 bars. The beam is subjected to a positive bending moment, causing the bottom of the beam to experience tension. (65pts total) The material properties are as follows: f = 5 ksi, fy = 60 ksi d 888 Six no. 8 bars (1) Find the bending moment [kips-in] that can cause tension cracks. Use the following parameters regarding the equivalent transformed section. Do not consider load and resistance factors. (15pts) - Neutral axis: 10.7 in. from the top face of beam - Moment of inertia: 12,550 in² around the neutral axis - Modulus of rupture: fƒ„ = 7.5√ƒ! [psi] (i.e., the concrete tensile stress for crack initiation) (2) Find the service moment strength [kips-in]. The allowable stress of concrete and steel is 0.45f, and 0.5fy, respectively. Do not consider load and resistance factors. (15pts) (3) Calculate the nominal moment strength M, and the design…arrow_forwardQ2) Determine the bar forces and reactions of the truss. ABD= 4 in², A= 2 in² and E=30000 kips/in². A D 20 ft 60 kips 15 ft Barrow_forwardL h Water Fig. P4 Hinge Farrow_forward
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- Select a W-shape for the following conditions: Beam spacing = 12 ft Span length = 25 ft Slab and deck combination weight = 43 psf Partition load = 20 psf Ceiling weight = 5 psf Flooring weight = 2 psf Live load 160 psf Fy = 50 ksi The maximum live load deflection cannot exceed L/340. Shape Mp (ft-kips) M₁/ (ft-kips) Vn (kips) Vn/v (kips) Ix (in. Vn/n, (kips) I. (in.) W14 × 61 383 254 156 104 640 W21 × 44 358 238 217 145 843 W16 × 50 345 230 186 124 659 W18 × 46 340 226 195 130 712 a. Use LRFD. Calculate the required moment of inertia, the required moment strength, and the maximum shear. (Express your answers to three significant figures.) Ix = in. 4 Μι = Vu Select a shape: -Select- b. Use ASD. ft-kips kips Calculate the required moment of inertia, the required moment strength, and the maximum shear. (Express your answers to three significant figures.) Ix Ma = Va = Select a shape: -Select- 4 in. ft-kips kipsarrow_forwardThe beam shown in the figure below has continuous lateral support of both flanges. The uniform load is a service load consisting of 70% dead load and 30% live load. The dead load includes the weight of the beam. 6 k/ft +9 18' If A992 steel is used, is a W12 × 35 adequate? For A992 steel: Fy = 50 ksi. bf h For W12 x 35: = 6.31, = 36.2 in., Zx 2tf tw (Express your answers to three significant figures.) a. Use LRFD. Mu Фомп = ft-kips ft-kips A W12 × 35 is -Select- b. Use ASD. Ma = ft-kips Mn ft-kips 26 A W12 × 35 is -Select- = +6→ 51.2 in. 3arrow_forwardVerify the value of Zx for a W10 × 30 that is tabulated in the dimensions and properties tables in Part 1 of the Manual. For W10 × 30: Ag Use the table below. - 8.84 in.2, d 10.5 in. AISC Manual Label y (in.) WT 9 × 25 2.12 WT 8 × 25 1.89 WT 7 × 24 1.35 WT 6 × 25 1.17 WT 5 × 15 1.10 (Express your answer to three significant figures.) Zx = 3 in.arrow_forward
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