Exam 2.pdf

7/21/2019 Exam 2.pdf

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Practice Exam 2

AM (B)

Solutions

Practice Exam 2AM (B)

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 1

Operating at a critical velocity of 15 ft /sec, the critical flow, in cubic feet per second, for a

rectangular 10-foot wide channel, is most nearly:

(A) 800(B) 1000

(C) 1200(D) 1400

Solution

Since critical velocity is provided, the critical depth can be solved and then the flow at the

critical depth can be calculated since the width of the rectangular channel is known.

vc = √ gd c = 15 ft /sec

Solving for d c = vc2

/ g = 152

/ 32.2 = 7.0 feet

Then using the critical depth relationship for a rectangular channel, the flow is determined.

23

2c

Qd

gw

Q2

= dc3

* g * w2

= 73

* 32.2 * 102

= 1104460 cfs2

Q = 1051 cfs

THE CORRECT ANSWER IS ( B )

Q. 2

A traffic stream has mean time headways of 2.4 seconds. If the jam density is 51.06 vehicles per mile, the optimum speed is 60 miles per hour, and the flow is 1500 vehicles per hour, the

capacity of the highway, in vehicles per hour, is most nearly:

(A) 1500

(B) 1530(C) 1575

(D) 1675

Solution

Optimum speed, uo = 60 mph

Free speed, u f = 2 x uo = 120 mph

max

* 120*51.061,531.8 /

4 4

f ju k q vehicles hour

THE CORRECT ANSWER IS: (B)

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 3

A mass concrete gravity retaining wall weighs 10 tons and supports soil as shown below.

The Factor of Safety for overturning is most nearly:

(A) 1.5(B) 2.0

(C) 3.0

(D) 3.5

Solution

Step 1: compute K A and K P

The coefficient of active pressure, K A is provided by:

1 sin

1 sin A K

The coefficient of passive pressure, K P is provided by:

1 sin

1 sin p K

where,

= angle of friction

1 sin(30)0.333

1 sin(30) A K

1 sin(30)3.000

1 sin(30) p K

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Step 2: compute horizontal forces

Sliding resistance,

T = W tan ( = 20000 x tan (30) = 11,547 lb/ft

P A = ½ K A z 2 = ½ x 0.333 x 120 x 212= 8,820 lb/ft

P P = ½ K P z 2 = ½ x 3.000 x 120 x 62= 6,480 lb/ft

Step 3: compute stabilizing and destabilizing moments

Assume that overturning would be about the toe, point P.

F = (20,000 x 6.5) + (6,480 x 2) / (8,820 x 7) = 2.32

Alternative question: determine the factor of safety against forward movement

Compute ratio of stabilizing and destabilizing forces

F = (T + P P )/ P A = (11,547+6,480)/ 8,820 = 2.04

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 4

A BOD sample was incubated at 27C for 7 days and was found to measure 100 mg/L. Given k 20

= 0.23 d -1

, base e. The concentration of BOD (mg/L) for this sample that would be measured in

an incubation time of 5 days and a temperature of 20C would be most nearly:

(A) 125(B) 76

(C) 35

(D) 20

Solution

20

27 0 1.047T k k

27-20

27 0.23 1.047 0.32k

0.32(7)100 112 /(1 ) 1

T L kT BOD BOD mg L

e e

0.23(5)(1 ) 112 1 76.5 /kt

T L BOD BOD e e mg L


7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 5

You are to evaluate the impact of moisture in a municipality’s solid waste stream on their processing and disposal systems. The table below is an analysis of the municipality’s wastestream based on records for the past 3 years. All values are expressed as percentages by weight.

The moisture content of each component is relatively constant.

The moisture content (%) in the average waste stream is most nearly:

(A) 36(B) 30

(C) 25

(D) 21

Solid Waste Characteristics

Component

Stream Composition,

Percent by Weight

Average

Average Moisture,

Percent by Weight

Organic

FoodPaper

Cardboard

Plastic

ClothRubber

Leather

Grass, Plants, etcWood

Inorganic

GlassMetals

Dirt, Ash, etc

8.039.0

6.0

5.0

2.51.0

0.3

20.53.0

6.05.7

3.0

65.07.0

6.0

2.5

10.52.5

10.0

55.027.5

2.53.0

9.0

100%

Solution

From the table in Question 515, Solid Waste Characteristic, stream composition and average

moisture % by weight are given. Consider 100 lb of the average solid waste stream. Determine

the pounds of moisture in each component (% by weight of component = lb of component timesthe moisture % by weight). Sum the total weight of moisture and divide by the 100 lb of the

average solid waste. Calculations are shown in the table below.

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Sum of the moisture -21.42 lb; 21.42 lb moisture/100lb waste = 21.42% moisture by weight.

Column 1 Column 2 Column 3

Component % Moisture Content Column 1x Column 2

839

65

2.51

0.3

20.53

6

5.73

0.650.07

0.060.025

0.1050.025

0.1

0.550.275

0.025

0.030.09

5.22.73

0.361.125

0.2650.025

0.03

11.2751.825

0.15

0.1710.27

THE CORRECT ANSWER IS: (D)

Q. 6

A retaining wall is constructed by pouring concrete between temporary wood forms. The

concrete is at 70F, has a setting time of 1½ hr, and is placed at the rate of 2 ft (vertical) per

hour. The wall sheathing is constructed of 1-in x 6-in (nominal) boards, wales, and studs. ½ indiameter threaded-steel (yield strength = 36 kips/in

2) tie rods (tie backs) are used to connect the

wales between the two walls. The wales are spaced 2 ft 6 in apart vertically. The followingcharacteristics can be used for all lumber:

6 2

2

2

2

1.6x10 lbf/in

1700 lbf/in

100 lbf/in

1200 lbf/in

b

v

c

E

F

F

F

If the studs are spaced every 2 ft, the tie capacity, in lbf is most nearly:

(A) 2000

(B) 3000(C) 4000

(D) 5000

Solution

Assume that the allowable tensile stress in the ties is 0.6 F y

Tie capacity = 0.6 F y Atie

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions2

2

lbf 0.5in(0.6) 36, 000

in 2

= 4241.1 lbf

THE CORRECT ANSWER (C)

Q. 7A police officer used his radar detector to monitor a stream of vehicles. He recorded the

following point speeds:

47, 49, 51, 48, 50, 52, 46, 49, 53, 52, 53, 48, 52, 51, 48

Given the distance headway is 131.58 ft and the optimal mean speed is 55 miles per hour, the

capacity of the road, in vehicles per hour, is most nearly:(A) 1990

(B) 2000

(C) 2010(D) 2020

Solution

The number of observations, N = 15

( ) 74949.93

15

it

uu mph

1549.84

(1/ ) 0.300976 s

i

N u mph

u

1 1 140 /

131.58/ 5280 0.02492d

k vehicles mileh

* 49.84 * 40.128 2000 / sq u k vehicles hour

02 * 2*55 110 f u u mph

* 110*40.12849.84 110

f

s f

j j

u k

u u k k

73.37 / jk vehicles mile

max

110*73.372017.675 /

4 4

f ju k q vehicles hour

THE CORRECT ANSWER IS (D)

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 8

A concrete pipe has to be designed to drain an unfinished concrete parking lot of 5 acres. The peak flow rate for a 5 year return period is 28 cubic feet per second. If the Manning’s coefficientfor the pipe is 0.015, the pipe length is 1,000 feet, and the slope is 0.02 feet per foot run, the

diameter of pipe, in inches, is most nearly:

(A) 18

(B) 24(C) 30

(D) 36

Solution

Note: Standard diameters in United States starts at 12 inch and increase in 6-inch increments to60 inch, then continues to increase in 12-inch increments.

Q = Peak flow rate

n = Manning coefficient

A = Cross sectional area ( D2/4)

r H = Hydraulic radius = A/P

P = wetted perimeter

S = Bottom slope

Assuming the pipe runs full,

A = D2/4

P = D

r H = A / P = ( D2/4) / D = D/4

2 / 3 1/ 21.49( ( ) H Q A r S

n

28= (1.49/0.015)( D2/4)( D/4)

2/3(0.02)

1/2

D = 2.0 ft

So, a 24 in. pipe is required.

THE CORRECT ANSWER IS (B)

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 9

A 2,000 foot long rectangular channel carries 500 cfs and has a 10-foot wide bottom. The water

cross-section is 50 ft2 and the Manning’s roughness coefficient is 0.012. The energy loss due to

friction, in feet, for the section is most nearly:

(A) 1(B) 2

(C) 3

(D) 4

Solution

Using Manning’s equation, solve for slope and then use the slope-length relationship to

determine the energy loss due to friction.

Area = wd = 50 ft 2

d = A/w = 50 / 10 = 5 ft v = Q/A = 500 / 50 = 10 ft /s

Manning’s equation: 2/ 3 1/ 21.49 H Q r S A

n

1/ 2 1/ 21.49( ) H r sv

n

Solving for the wetted perimeter, H r

* 10*5 502.5

2 * 10 2 *5 20 H

w d r

w d

solve for S

1/ 2 2

2 / 3 2 / 3

* 10 *0.0124.37 10

1.49( ) 1.49 * (2.5) H

v nS x

r

S = 0.0019

0.0019*2000 3.8 f h SL ft ft

THE CORRECT ANSWER IS: (D)

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 10

The mechanical and plasticity tests of a soil sample are shown below:Mechanical Analysis Plasticity

Sieve % finer Liquid Limit Plasticity Index

10 84 19 4

40 51

200 21

According to the AASHTO classification system, the classification is most nearly:

(A) A-2-4

(B) A-7-5

(C) A-1-b(D) A-3

Solution

Refer to the Standard Specifications for Transportation Materials and methods of Sampling and

Testing, 16th

edition, Washington D.C.: The American Association of State Highway and

Transportation Officials copyright 1993.

Note, this classification table is reproduced in most geotechnical textbooks. You will find it

typically titled as “The AASHTO Soil Classification System”.

The AASHTO Soil Classification system has two components:

1. Group Classification

2. Group Index

The selection of the Group Classification is by a process of elimination. The identification of the

Group Index is through the application of the Group Index empirical formula:

GI = ( F 200 – 35)[0.2 + 0.005( LL – 40)] + 0.01( F 200 – 15)( PI – 10)

Where GI = Group Index

F 200 = % soil passing the # 200 (0.075mm) sieve

LL & PI are the Liquid Limit and Plasticity Indices expressed as integers.

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Group Classification

Less than 35% passes # 200 sieve:

Group Elimination

A-1-a

A-1-b

A-2-4

A-2-5

A-2-6

A-3

A-4 x

A-5 x

A-6 x

A-7-5 x

A-7-6 x

% passing sieve #200 > 15:

Group EliminationA-1-a x

A-1-b

A-2-4

A-2-5

A-2-6

A-3 x

% passing sieve #40 > 50:

Group Elimination

A-1-b x

A-2-4

A-2-5

A-2-6

Liquid Limit < 41:

Group Elimination

A-2-4

A-2-5 x

A-2-6

Liquid Limit > 11:

Group Elimination

A-2-4

A-2-6 x

Therefore, soil can be classified as A-2-4

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Group Index

GI = ( F 200 – 35)[0.2 + 0.005( LL – 40)] + 0.01( F 200 – 15)( PI – 10)

GI = (21-35)[0.2 + 0.005(19 – 40)] + 0.01(21 – 15)(4 – 10)

GI = -1.69

i.e. GI = 0

The soil can be classified as A-2-4(0)

THE CORRECT ANSWER IS (A)

Q. 11

A steel beam supports a 6 in lightweight concrete masonry (CMU) wall around a mechanical

room. The CMU has a unit weight, , of 85 lbf/ft

3

. If the wall is 8 ft high and 10 ft long, theload on the beam is most nearly

(A) 340 lbf/ft

(B) 680 lbf/ft

(C) 3400 lbf/ft(D) 4100 lbf/ft

Solution:

Hint: The load on the beam is uniformly distributed.

The load on t he beam, w, is the weight of the wall per foot.

w ht

3

lbf 1 ft85 8 ft 6 in

ft 12 in

340 lbf/ft


Why Other Options Are Wrong

(B) This incorrect solution results from failing to multiply by the wall width.

(C) This incorrect solution is a result of calculating the total load rather than the load per foot.(D) This incorrect solution fails to perform the conversion from inches to feet.

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 12

A pump in the figure below is to deliver water through the existing system indicated in thefigure.

Which of the following will reduce the tendency for pump cavitations?

I. Increasing the discharge pipe diameter.II. Lowering the pump elevation.

III. Increasing the suction diameter.

(A) II and III only

(B) II only(C) I and II only

(D) I, II and III

Solution

Determine which statement will reduce the tendency for pump cavitaton.II. Lowering the pump elevation.

III. Increasing the suction diameter.


7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 13

Lightweight concrete (100 lbf/ft3) is used for the one-way slab shown.

compressive strength of concrete 3000 lbf/in2

yield stress of reinforcement steel 60,000 lbf/in2

A B C D E

12 in, typ

15 ft 12 ft

What is the minimum thickness for slab AB?

(A) 7.5 in

(B) 8.0 in(C) 8.6 in

(D) 10 in

Solution

Hint: Minimum thicknesses of one-way slabs not supporting partitions or other elements likely o be damaged by

deflection can be determined using ACI 318 Table 9.5 (a).

From ACI 318 Table 9.5(a), determine the minimum thickness, h, for a solid one-way slab withone end continuous.

24

l h

l is the span length in inches

in(15 ft) 12 180 in

ftl

180 in

7.5 in24

h

For lightweight concrete the footnotes to the table indicate that the tabulated value must be

modified.(1.65 0.005 )ch w h

From ACI 318 Table 9.5(a), wc is 100. Therefore,

(1.65 (0.005)(100))(7.5 in)h

= 8.6 in

THE CORRECT ANSWER IS (C)

7/21/2019 Exam 2.pdf


7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 14

A conventional activated sludge plant treats 2.0 MGD as shown in the figure below. The rawwastewater contains 250 mg/L suspended solids. The mixed liquor suspended solids (MLSS) is

2,500 mg/L. The recycled sludge flow is 800,000 gpd containing 8,000 mg/L suspended solids.Suspended solids removal efficiency of the primary clarifier is 50%, and the primary sludgecontains 4% suspended solids. Effluent solids content = 25 mg/L.

The primary sludge and waste sludge flow rates (gpd) are, respectively, most nearly:

(A) 6,200 and 6,900

(B) 8,900 and 69,000

(C) 6,200 and 69,000(D) 6,200 and 620,000

AERATION TANK

SECONDARY

CLARIFIERPRIMARY

CLARIFIERRAW SEWAGE EFFLUENT

RECYCLED SLUDGE

WASTE

SLUDGE

PRIMARY

SLUDGE

ACTIVATED SLUDGE PLANT - SCHEMATIC

Solution

Solids in primary sludge (250x2x8.34)50% 2, 085 /lb day

Specific gravity of primary sludge 100

1.014 96

1.32 1.0

2,085Volume of primary sludge = 6, 200

0.04x1.01x8.34 gpd

Let secondary waste sludge flow = Qw

( ) ( ) /" ( )r r w r w e X Q Q Q Q X Q Q S

[2, 500x (2 0 .8)] [ (2 ) x25] [(0.8 )x8, 000]w wQ Q

69,000wQ gpd


7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 15

A dam of length 120 feet is constructed of an impermeable soil. The dam overlays a sand bed of thickness 6feet 6 inches as shown below. The permeability of the sand bed has been measured

and found to be 9.5 feet per day.

The mean discharge in gallons per minute is most nearly:

(A) 7.5(B) 9.62

(C) 10.5

(D) 11.0

Solution

Darcy’s Law states that the permeability of a soil is given by:

k = q / i A

where,

q = discharge

i = hydraulic gradient

A = cross sectional area

i = (945 – 905)/160 = 0.250

A = (6.5) x (120) = 780 square feet

q = kiA = (9.5) x (0.250) x (780) = 1852.5 cubic feet per dayq =(1852.5) x (7.48) gallons per day = 13,856.7 gallons per day

q =(13,856.7) / 1440 gallons per minute

q = 9.62 gallons per minute


7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 16

A field technician submits the results of a sand cone test for review. Prior to the test, thelaboratory determined the unit weight of the dry test sand to be 84.7 lbf/ft

3. The soil obtained

from the test hole was weighed at 3.65 lbf and had a moisture content at 17.3%. The initial

weight of the sand cone apparatus filled with suitable test sand was 13.36 lbf, and the final

weight after the sand filled the test hole was 11.13 lbf. Assuming the field moisture content iswithin the acceptable specification range for optimum moisture content, what is the in-place

percent compaction if the laboratory Proctor test results in a maximum dry density of 122.3

lbf/ft3?

(A) 97%

(B) 104%(C) 114%

(D) 133%

Hint: Determine the volume of the test hole by using the unit weight properties of the test sand.

SolutionThe unit weight of the test sand can be used to determine the volume of the test hole in the field.

The weight of dry soil obtained from the test hole can be divided by the volume of the test holeto determine the in-place dry unit weight.

Determine the quantity of test sand needed, by weight, to fill the hole.

est sand initial final _ W W W

= 13.36 lbf -11.13 lbf

= 2.23 lbf

Determine the volume of the test hole using phase relationships.

Vtest hole = test sand

test sand3

2.23 lbf

lbf 84.7

ft

W

= 0.0263 ft3

The weight of dry soil obtained from the test hole can be calculated using the relationship

fill soil, moist

fill soil, dry

3.65 lbf

1+w 1 0.173

W W

= 3.11 lbf

The in-place dry unit weight can be calculated by dividing the weight of dry soil obtained from

the test hole by the volume of the test hole.

fill soil, dry

fill soil, dry 3

test hole

3.11 lbf

0.0263 ft

W

V

= 118.3 lbf/ft3

The in-place percent compaction is

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

3fill soil, dry

dry, max3

lbf 118.3

ft100% 100%lbf

122.3ft

= 96.7% (97%)



(B) This incorrect solution is obtained by dividing the maximum Proctor dry unit weight by

the dry unit weight of the fill soil and multiplying by 100. The result is the reciprocal of the correct answer.

(C) This solution is incorrect because the total weight of the fill soil is not converted to a dry

weight. The incorrect value is then divided by the test hole volume to obtain a greater

than correct unit weight and subsequent incorrect answer for percent compaction.

(D) This incorrect solution is obtained when the total weight of the fill soil is incorrectlyconverted to a dry weight (multiplying by 1 + w rather than dividing by it). The result is

much greater than the correct answer.

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 17

An aerial survey is to be flown over a region that varies in elevation from 750 feet to 1,250 feet.If the camera has a focal length 6.5 inches, and a scale of 1:25,000, the aircraft should be flown

at a height, in feet, that is most nearly:

(A) 5,000(B) 10,000

(C) 12,500(D) 15,000

Solution

The relationship governing aerial photogrammetry that is required is given by:

f S

H h

where S = photographic scale

f = camera focal length (feet)

H = aircraft height (feet)

h = average elevation of terrain (feet)

given S = 1/25,000

f = 6.5” = 0.542’

H = required

h = (750 + 1,250)/2 = 1,000’

substituting, H = 14,541.7 ft

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 18

An intersection with five entry points has witnessed a number of motor vehicle collisions. In2002, the number of crashes reported was 48, and the average 24-hour volumes entering the

intersection from its five approaches were 900, 1,230, 1,560, 1,435, and 980 vehicles,

respectively. The intersection’s crash rate is most nearly:

(A) 0.0079 crashes per vehicle

(B) 48 crashes per year

(C) 21.51 crashes per million entering vehicles(D) 7.86 crashes per million entering vehicles

(E) 2.87 crashes per million entering vehicles


7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 19

What is the volatile suspended solids concentration of the following wastewater sample?

sample volume filtered (VF) 200 mL

sample volume evaporated (VD) 100 mL

mass of dried crucible and filter paper (MS) 25.439 gmass of dry evaporation dish (MD) 275.41 g

mass of dried crucible, filter paper, and solids (MSS) 25.645 g

mass of dried evaporation dish and solids (MDS) 276.227 g

mass of ignited crucible, filter paper and solids (MSI) 25.501 g

mass of ignited evaporation dish and solids (MDI) 276.201 g

(A) 260 mg/L

(B) 310 mg/L

(C) 720 mg/L(D) 1000 mg/L

Hint: Find the solids that were ignited, not the ash that remains

Solution:

SS MSI VSS

VF

625.645 g 25.501 g 10 mL mg

200 mL L g

= 720 mg/L



(A) This incorrect solution calculates the volatile total dissolved solids (volatile TDS).6276.227 g-276.201 g 10 mL mg

260 mg/L100 mL L g

(B) This incorrect solution calculates the nonvolatile (fixed) solids.625.501 g-25.439 g 10 mL mg

310 mg/L200 mL L g

(D) This incorrect solution calculates the total suspended solids (TSS).625.645 g 25.439 g 10 mL mg

200 mL L g

1030 mg/L 1000 mg/L

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions

Q. 20

Given the phase diagram shown, what is most nearly the porosity?

(A) 0.42%

(B) 3.3%

(C) 9.1%(D) 30%

Hint: Start by determining the weight of water using the water content and weight of solids.

Solution

The porosity can be defined using the void ratio and is usually expressed as a percentage.

100%1

en

e

By assuming a unit value for the total volume, an equation for the void ratio can be simplified todetermine the porosity.

1 1 1

st w

wt w s

s s d

V V SG

eW W

The dry unit weight can be determined based on the total unit weight and the moisture content.

33

127.0

115.5 /1 1 0.1

t d

lbf

ft lbf ft

w

The void ratio can be determined.

3

3

165.4

1 0.43

115.5

lbf

ft e

lbf ft

The porosity can be determined.

0.43100% 30%

0.43 1n

THE CORRECT ANSWER IS (D)

7/21/2019 Exam 2.pdf


Practice Exam 2

AM (B)

Solutions


(A) This solution is incorrect because it is actually the void ratio expressed as a percentage.

(B) This incorrect solution is the reciprocal of the porosity without converting the number toa percentage. A common error is caused by a failure to understand the equation and the

reciprocal.(C) This solution makes the common mistake of dividing the weight of solids by the total

unit weight rather than by the solids unit weight to determine the volume of solids. Thisresults in a value that is greater than the correct value.

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