review notes

Installation

1.Run PhotoToCartoon.exe2.After setup complete, exit program3. Open bean.reg 4. When prompted after opening bean.reg Click YES to Register Full Version

5. Done!

Problem 814Find the moment at R2 of the continuous beam shown in Fig. P-814.

Solution 814

HideShow or hide the solution

Where

Thus,

answer

Problem 813Determine the moment over the support R2 of the beam shown in Fig. P-813.

Solution 813


Where

Thus,

answer

Beam DeflectionSubmitted by Romel Verterra on Sat, 09/10/2011 - 23:39

The deflection of beam is the deviation of the elastic curve from the undeformed section called neutral axis. Understanding how the beam deflects from the neutral axis, allow us to solve not only the deflection of beam but also the support reactions of indeterminate beams. Equations of static equilibrium allow us to solve determinate beams with up to two unknowns. With the addition of geometry of elastic curve to the static equations, there is no such thing as indeterminate beam.

There are several ways of finding the deflection of beams, area-moment method (also called moment-area method) is so far the most convenient, and maybe the most powerful method of solving for the beam deflection.

Methods of solving beam deflection

1. Double-integration method2. Moment-area method3. Strain-energy method (Castigliano’s Theorem)4. Method of superposition5. Conjugate-beam method

6. Area of water jet at the highest point7. Submitted by Romel Verterra on Thu, 01/12/2012 - 05:59

8. ProblemA nozzle inclined at an angle of 60° with the horizontal issues a 50-mm diameter water jet at the

rate of 10 m/s. Neglecting air resistance, what is the area of the jet at the highest point of the projectile?

9. A. 0.00311 m2

B. 0.00322 m2

C. 0.00231 m2

D. 0.00393 m2

SolutionVelocity of the jet at the highest point

The flow is continuous, thus,

Answer: [ D ]

Total thrust on the 90 degree bendSubmitted by Romel Verterra on Tue, 01/17/2012 - 22:36

ProblemA horizontal 90 degree bend connects two 400-millimeter diameter pipe carrying water at 150 Liters per

second. The pressure at the inlet and outlet of the bend is 120 kiloPascal. Determine the total thrust acting on the bend.

A. 32.11 kiloNewtonB. 25.55 kiloNewtonC. 27.89 kiloNewtonD. 21.58 kiloNewton

Solution

Pressure force

Velocity of flow

Dynamic force

Answer: [ D ]

Releasing flood water from a reservoirSubmitted by Romel Verterra on Sat, 01/21/2012 - 08:16

ProblemA spillway with permanent crest elevation of 42 m and 10 m long, releases flood water from a reservoir

whose surface area is 6.5 hectares. The initial water surface elevation is 44.3 m. What is the water surface elevation after 0.75 hour? Use Francis formula neglecting velocity of approach.

A. 43.21 mB. 42.92 mC. 42.33 mD. 42.64 m

Solution

HideShow or hide the solutionFrancis formula for variable head

Where

Answer: [ B ]

Determinate and unstable structuresSubmitted by Romel Verterra on Mon, 04/16/2012 - 17:22

Situation

For the structures shown in Figures ME-01A, ME-01B, and ME-01C, identify whether the structure is unstable, statically determinate, statically indeterminate to the first degree, and statically indeterminate to the second degree.

Part 1: The structure shown in Figure ME-01A is:A. UnstableB. Statically determinateC. Statically indeterminate to the first degreeD. Statically indeterminate to the second degree

Part 2: The structure shown in Figure ME-01B is:A. UnstableB. Statically determinateC. Statically indeterminate to the first degreeD. Statically indeterminate to the second degree

Part 3: The structure shown in Figure ME-01C is:A. UnstableB. Statically determinate

C. Statically indeterminate to the first degreeD. Statically indeterminate to the second degree

Vertical deflection at a joint of cantilver trussSubmitted by Romel Verterra on Sat, 02/18/2012 - 05:43

SituationFor the truss shown in Figure EM(01)-5, the product of the cross-sectional area and modulus of elasticity is 1 000 000

Newton.

Part 1: Which of the following gives the axial force on member BC due to the given loads.A. 500 NewtonB. 200 NewtonC. 400 NewtonD. 600 Newton

Part 2: Which of the following gives the axial force on member BC due to a vertical unit load at C.A. 1B. 0.75C. 1.25D. 1.5

Part 3: Which of the following gives the vertical deflection at C.A. 5.35 mmB. 6.75 mmC. 8.45 mmD. 3.25 mm

Solution

HideShow or hide the solutionStress due to actual loads:At joint C:

Answer for Part 1: [ A ]

At joint A:

Stress due to unit load:At joint C:

Answer for Part 2: [ C ]

At joint A:

Vertical deflection at C:

Answer for Part 3: [ B ]

Discharge and slope of open channelsSubmitted by Romel Verterra on Wed, 11/30/2011 - 22:17

CE Board Exam Problems (May 1994) - Manning's formula for open channels

Problem 1A circular concrete sewer pipe with coefficient of roughness n = 0.013 is 1.60 m in diameter and flowing half-full has a slope of 4 m per 5 km. Compute the discharge on the sewer pipe.

A. 3.456 m3/sB. 2.674 m3/sC. 1.190 m3/sD. 4.324 m3/s

Solution 1


Answer: [ C ]

Problem 2A trapezoidal canal is to carry 50 cu. m./sec with a mean velocity of 0.40 m/sec. One side is vertical, the other has a slope of 2H to 1V. Compute the minimum hydraulic slope if the canal is lined with rubble masonry assuming n = 0.016. Use Mannig’s C and check by Kutter’s C.

Solution 2

HideShow or hide the solutionArea of flow

For minimum hydraulic slope, the section must be most efficient. From the figure,

For the most efficient trapezoidal section, the top width “x” equals the sum of the sides.

Depth of flow

Hydraulic radius

for most efficient trapezoidal section

Slope of energy grade line (hydraulic slope)

answer

Couple resultant of parallel forcesSubmitted by Romel Verterra on Sun, 11/27/2011 - 16:09

Board Exam Problem November 1994 - Resultant of parallel force system

Find the resultant of four parallel forces acting on a horizontal bar AB with length of 11 m; 400 kN downward at the left most end, 900 kN upward 3 m from the left end, 800 kN downward at 9 m from the left end and 300 kN upward at 11 m from the left end.

A. 1200 kN force acting upwardB. 1200 kN force acting downwardC. 1200 kN couple acting counter clockwiseD. 1200 kN couple acting clockwise

Solution


Since R = 0, the forces creates a couple. The resultant couple can be solved by taking moments about any point on the bar.

Resultant couple

Answer: [ D ]

Check:

Finding the 3rd vertex of the triangle with centroid and other vertices givenSubmitted by Romel Verterra on Tue, 04/24/2012 - 22:21

ProblemTwo vertices of a triangle are (6, -1) and (7, -3). Find the ordinate of the third vertex such that the centroid

of the triangle will lie at the origin.

A. -13B. 4C. 13D. -4

Solution

HideShow or hide the solutionCentroid by coordinates formula

Answer: [ B ]

Although the abscissa is not required, we present the solution below for the sake of discussion. From the formula for abscissa of the centroid:

Solid geometry problems involving lateral area (cone and cylinder)Submitted by Romel Verterra on Mon, 04/23/2012 - 19:22

Problem 1: Slant height of a coneThe lateral area of a right circular cone of radius 4 cm is 100.53 sq. cm. Determine the slant height.

A. 8 cmB. 9 cmC. 6 cmD. 10 cm

Problem 2: Volume of cylinderFind the volume of a right circular cylinder whose lateral area is 25.918 m2 and base area of 7.068 m2.

A. 19.44 m3

B. 15.69 m3

C. 20.53 m3

D. 18.12 m3

Solution for Problem 1

HideShow or hide the solutionThe formula for the lateral area of right circular cone is

Answer: [ A ]

Solid geometry problems involving lateral area (cone and cylinder)Submitted by Romel Verterra on Mon, 04/23/2012 - 19:22

Problem 1: Slant height of a coneThe lateral area of a right circular cone of radius 4 cm is 100.53 sq. cm. Determine the slant height.

A. 8 cmB. 9 cmC. 6 cmD. 10 cm

Solution for Problem 1

Base radius of the cylinder

Altitude of the cylinder

Volume of the cylinder

Answer: [ A ]

Problem 2: Volume of cylinderFind the volume of a right circular cylinder whose lateral area is 25.918 m2 and base area of 7.068 m2.

A. 19.44 m3

B. 15.69 m3

C. 20.53 m3

D. 18.12 m3

solution for Problem 2

HideShow or hide the solutionBase radius of the cylinder

Altitude of the cylinder

Volume of the cylinder

Answer: [ A ]

Answer: [ A ]

Allowable compressive stress of a steel columnSubmitted by Romel Verterra on Sun, 11/27/2011 - 11:59

ProblemGiven the following properties of a steel column:

Modulus of elasticity, E = 200 GPaYield strength, Fy = 200 MPaLength, L = 12 mMoment of inertia, I = 37.7 × 106 mm4 andCross-sectional area, A = 8000 mm2

Find the allowable compressive stress.

A. 120 MPaB. 67.4 MPaC. 33.7 MPaD. 91.1 MPa

Solution


Radius of gyration

Slenderness ratio

Critical slenderness ratio

Allowable compressive stressSince KL/r > Cc, the column is a long column. For long columns,

Answer: [ C ]

Fixed End Moment - CE Board November 1994Submitted by Romel Verterra on Sun, 11/27/2011 - 10:04

Board Exam Problem November 1994 - Fixed end moment of fully restrained beam

A beam with both ends fixed and with a span of 7 m is carrying a load of 12 kN/m. Calculate the maximum moment of the beam.A. 49 kN·mB. 62 kN·mC. 32 kN·mD. 86 kN·m

Solution


The maximum moment for fixed ended beam with a uniform loading is at the fixed supports. It is given by the formula

Answer: [ A ]

Length of cut in the ground for a proposed road and angular unit conversionSubmitted by Romel Verterra on Sat, 04/21/2012 - 20:02

Problem 1: Length of cut for a proposed roadThe ground makes a uniform slope of 4.8% from STA 12 + 180 to STA 12 + 240. At STA 12 + 180, the center height of the roadway is 1.2 m fill. At the other station, the center height is 2.5 m cut. Find the

length of cut.

A. 30.85 metersB. 40.54 metersC. 46.32 metersD. 50.28 meters

Solution


From the figure shown above:

Answer: [ B ]

Problem 2: Angular unit conversionConvert 405° to mils.

A. 2 800 milsB. 10 200 milsC. 7 200 milsD. 6,200 mils

Solution


Answer: [ C ]

Area of a spherical triangle with given interior anglesSubmitted by Romel Verterra on Sat, 04/21/2012 - 17:17

ProblemFind the area of a spherical triangle of whose angles are 123°, 84°, and 73°. The radius of the sphere is

30 m.

A. 1863.3 square meterB. 1570.8 square meterC. 1958.6 square meterD. 1480.2 square meter

Solution

HideShow or hide the solutionArea of spherical triangle

Where → E is called the spherical excess

Answer: [ B ]

Book value of copier machine after making 300 thousand copiesSubmitted by Romel Verterra on Sat, 04/21/2012 - 11:35

ProblemA certain copier machine cost P150,000 with a trade-in value of P15,000 after making 800,000 copies.

Using the declining balance method, what is the book value when the machine had made 300,000 copies?

A. P68,111B. P64,896C. P62,531D. P63,254

Solution

HideShow or hide the solutionConstant percentage

Book value by declining balance method

Answer: [ D ]

Book value of the machine after 5 years by declining balance formulaSubmitted by Romel Verterra on Sat, 04/21/2012 - 09:02

ProblemA machine costing P480,000 has a life expectancy of 12 years with a salvage value of 10% of the first

cost. What is the book value after five years using the declining balance method.

A. P183,897B. P152,758C. P196,432D. P214,785

Solution

HideShow or hide the solutionBy declining balance method

First cost

Salvage value

Economic life

The constant percentage, k

mth year

Thus,

Answer: [ A ]

Another solution

HideShow or hide the solutionA direct formula for Book Value using constant percentage method can also be used

okay!

Finding the amount of annual deposits after 15 yearsSubmitted by Romel Verterra on Fri, 04/20/2012 - 21:11

ProblemFive thousand dollars is deposited at the end of each year for 15 years into an account earning 7.5%

compounded continuously. Find the amount after 15 years.

A. $133 541.3B. $152 754.2C. $148 365.9D. $112 854.1

Solution


Convert 7.5% compounded continuously to annually:

Answer: [ A ]

Compute the time for money to 4 times itselfSubmitted by Romel Verterra on Fri, 04/20/2012 - 18:35

ProblemHow long will it take for money to quadruple itself if invested at 20% compounded quarterly?

A. 10.7 yearsB. 6.3 yearsC. 9.5 yearsD. 7.1 years

Solution

HideShow or hide the solutionFrom the compounded interest formula, let t be the number of years:

Answer: [ D ]

Hydraulic jump, oxidizing agent, and void ratioSubmitted by Romel Verterra on Thu, 04/19/2012 - 15:36

Problem 1Water flows over a spillway and into a horizontal canal. If the water undergoes a jump, what is the depth of flow after the jump?A. criticalB. supercriticalC. subcriticalD. none of these

Problem 3A strong oxidizing agent. It improves the effectiveness of subsequent coagulation, apparently by polymerization of metastable organics; aids in removal of musty, earthy, fishy, and muddy tastes and odors. It breaks down the unstable gas into two atoms of oxygen.

A. ozoneC. chlorideB. ironD. nitrate

Problem 2The ratio of the volume of water to the volume of voids.

A. moisture contentB. void ratioC. degree of saturationD. porosity

Equivalent pipe for pipes in seriesSubmitted by Romel Verterra on Sun, 01/29/2012 - 21:40

ProblemTwo pipes 1 and 2 having the following properties are connected in series:

Pipe 1: Length = 1500 m, Diameter = 850 mm, f = 0.025Pipe 2: Length = 1200 m, Diameter = 650 mm, f = 0.020

It is required to replace these two pipes with a single pipe whose length is 2700 m. Assuming that the friction factor for the new pipe is 0.015, what is the required pipe diameter?

A. 674 mmB. 578 mmC. 732 mmD. 512 mm

Solution

HideShow or hide the solutionFor the original pipes in series:

Head lost:

For the equivalent single pipe:

Answer: [ A ]

Effective stress at the mid-height of clay layer for sand and clay strataSubmitted by Romel Verterra on Sun, 01/29/2012 - 19:58

ProblemA clay layer 25 feet thick is overlain with 50 feet thick of sand (G = 2.71). The water table is 20 feet below the sand (ground) surface. The saturated unit weight of clay is 141 pcf. The sand below the water table

has a unit weight of 128 pcf. The sand above the water table has average moisture content of 20%. After drying, the sand was found to have a dry unit weight of 92 pcf. Determine the effective stress at the mid-

height of the clay layer.

A. 5010 lb/ft2

B. 4788 lb/ft2

C. 5480 lb/ft2

D. 5160 lb/ft2

Solution

HideShow or hide the solutionFor the sand above the water table

Effective stress

Answer: [ D ]

Porosity of soil with moist and saturated unit weights knownSubmitted by Romel Verterra on Thu, 01/26/2012 - 15:40

ProblemA clay sample has unit weight of 20.06 kN/m3 with moisture content of 8.2%. The saturated unit weight of

the sample is 21.58 kN/m3. Determine the porosity of the soil.

A. 45%B. 31%C. 47%D. 28%

Solution

HideShow or hide the solutionMoist unit weight

Saturated unit weight

Porosity

Answer: [ B ]

Finding the hydraulic radius of a trapezoidal canalSubmitted by Romel Verterra on Sat, 01/21/2012 - 20:22

ProblemWhat is the hydraulic radius of a trapezoidal canal of side slope 2 horizontal to 1 vertical and base width 4

m if the depth of flow is 1.5 m?

A. 0.98 meterB. 0.45 meterC. 0.76 meterD. 0.83 meter

Solution


Area of flow

Wetted perimeter

Hydraulic radius

Answer: [ A ]

Releasing flood water from a reservoirSubmitted by Romel Verterra on Sat, 01/21/2012 - 08:16

ProblemA spillway with permanent crest elevation of 42 m and 10 m long, releases flood water from a reservoir

whose surface area is 6.5 hectares. The initial water surface elevation is 44.3 m. What is the water surface elevation after 0.75 hour? Use Francis formula neglecting velocity of approach.

A. 43.21 mB. 42.92 m

C. 42.33 mD. 42.64 m

Solution

HideShow or hide the solutionFrancis formula for variable head

Where

Answer: [ B ]





Solution


Pressure force

Velocity of flow

Dynamic force

Answer: [ D ]





Solution

HideShow or hide the solutionPressure force

Velocity of flow

Dynamic force

Answer: [ D ]

Rectangular block floating in oil and waterSubmitted by Romel Verterra on Sun, 01/15/2012 - 21:16

ProblemThe block shown in Figure-HD4(11-00) weighs 35,000 lbs. Find the value of h.

A. 1.678 feetB. 1.495 feetC. 1.123 feetD. 1.988 feet

Solution

HideShow or hide the solutionBuoyant forces

Sum up vertical forces

Answer: [ B ]

Time for the water from orifice to hit the groundSubmitted by Romel Verterra on Sun, 01/15/2012 - 20:00

ProblemAn orifice located at the vertical side of a large tank is 8 m from the level ground. How long will it take for

the water to reach the ground after opening the orifice?

A. 1.87 secondsC. 1.77 secondsB. 1.52 secondsD. 1.28 seconds

Solution

HideShow or hide the solutionFrom the formula:

→ vertical motion of projectile

Where,s = yvo = voy

a = -g

Thus, the formula will become

Note: voy = 0 at the highest point of the projectile, thus,

Answer: [ D ]

Theoretical velocity of flowProblemWhat is the theoretical velocity of flow through an orifice located 3.2 m below the water surface?

A. 10.67 m/sB. 12.12 m/sC. 7.92 m/sD. 8.45 m/s

Solution


Answer: [ C ]

Difference in head after 2 minutes of flowSubmitted by Romel Verterra on Sun, 01/15/2012 - 10:58

ProblemA vertical rectangular water tank is divided into two chambers whose horizontal sections are 3 m2and 5 m2, respectively. The dividing wall is provided with a 100 mm × 100 mm square hole located 0.5 m from the bottom and whose coefficient of discharge is 0.60. Initially there is 5 m deep of water in the smaller

chamber and 1 m deep of water in the larger chamber. What is the difference in the water level in the two chambers after 2 minutes?

A. 1.01 mB. 2.54 mC. 1.32 m

D. 1.87 m

Solution


From the formula:

WhereH1 = h1 = 4 mH2 = h2 = requiredt = 2 min = 120 sec

Answer: [ C ]

Flow of water in a pipe systemSubmitted by Romel Verterra on Thu, 01/12/2012 - 14:34

ProblemFor the pipe system shown in Figure 03, determine the flow in pipe 3 if the flow in pipe 4 if 5 MLD. The

properties of the pipes are as follows:

Pipe Length (m) Diameter (mm) Friction factor, f

1 600 600 0.020

2 500 350 0.025

3 400 400 0.035

4 550 600 0.030

A. 2.84 MLDB. 2.54 MLDC. 3.97 MLDD. 4.51 MLD

Solution

HideShow or hide the solutionNote: MLD = million liters per day

Pipes (2) and (3) are in parallel connection, thus,

From the figure above:

Substitute Q2 = 0.7579Q3

Answer: [ A ]

Area of water jet at the highest pointSubmitted by Romel Verterra on Thu, 01/12/2012 - 05:59

ProblemA nozzle inclined at an angle of 60° with the horizontal issues a 50-mm diameter water jet at the rate of 10

m/s. Neglecting air resistance, what is the area of the jet at the highest point of the projectile?

A. 0.00311 m2

B. 0.00322 m2

C. 0.00231 m2

D. 0.00393 m2

Solution

HideShow or hide the solutionVelocity of the jet at the highest point

The flow is continuous, thus,

Answer: [ D ]

Oil and water in a tank of two chambersSubmitted by Romel Verterra on Tue, 01/10/2012 - 21:24

ProblemFor the tank shown in Figure HD-22(00), h1= 3m and

h3 = 4 m. Determine the value of h2.

A. 1.19 mB. 1.11 mC. 1.87 mD. 1.93 m

Solution

HideShow or hide the solutionSum-up pressure head from A to B in meters of water

Answer: [ A ]

Dynamic force of water jet from the orificeSubmitted by Romel Verterra on Tue, 01/10/2012 - 14:43

ProblemA tank containing a liquid of unit weight 'w' has an orifice on its vertical side. The orifice has an area of 'Aj' with its center 'H' meters below the free liquid surface. Determine the dynamic force at the jet immediately

after the orifice.

A. B.

C.

D.

Solution

HideShow or hide the solutionVelocity of the jet

Dynamic force

Answer: [ C ]

Normal depth of flow in triangular canalSubmitted by Romel Verterra on Tue, 01/10/2012 - 00:02

ProblemA triangular canal is as shown in Figure HD-00(11). The canal is

laid on a slope of 0.001 with n = 0.012 and discharges 2.4 m3/s. If θ = 60°, what is the value of the normal depth dn?

A. 1.53 mB. 1.68 mC. 1.79 mD. 1.21 m

Solution

HideShow or hide the solutionVertex angle

Inclined depth

Area of flow

Wetted perimeter

Hydraulic radius

Discharge by Manning’s formula

Answer: [ B ]

Combined axial and flexure stresses in simply supported steel beamSubmitted by Romel Verterra on Mon, 01/09/2012 - 06:25

SituationA simply supported steel beam 6 m long carries a uniform load of 32 kN/m and an axial compressive

force of 320 kN. The properties of the steel section are as follows:Area, A = 14 700 mm2

Section Modulus, Sx = 1921 × 103 mm3

Flange width, bf = 280 mmFlange thickness, tf = 16 mmOverall depth, d = 390 mmWeb thickness, tw = 19 mm

According to Section 4.6.1 of the NSCP, for members subject to axial compression and bending,

.

fa = computed axial stress, MPaFy = yield strength of steel = 248 MPafb = computed bending stress, MPaFb = allowable bending stress = 0.66Fy

Part 1: Which of the following most nearly gives the computed axial stress in the beam due to axial force alone acting on the beam.A. 22 MegaPascalsB. 27 MegaPascalsC. 16 MegaPascalsD. 32 MegaPascals

Part 2: Which of the following most nearly gives the computed bending stress in the beam due to the uniform load alone acting on the beam.A. 64 MegaPascalsB. 84 MegaPascalsC. 70 MegaPascalsD. 75 MegaPascals

Part 3: Which of the following most nearly gives the value the interaction equation.A. 0.5B. 0.4C. 0.7D. 0.6

Solution

HideShow or hide the solutionAxial stress:


Bending stress:

Answer for Part 2: [ D ]

Interaction equation:


Analysis of circular timber beamSubmitted by Romel Verterra on Sun, 01/08/2012 - 20:47

SituationA circular timber beam 250 millimeters in diameter has a simple span of 4 m. The beam carries a

uniformly distributed load of w (kN/m) including its own weight. The allowable stresses are 18 MPa for bending and 2 MPa for shear parallel to grain. Allowable deflection is 1/240 of the span length. E = 6000

MPa.

Part 1: Which of the following most nearly gives the value of w so that the allowable bending stress will not be exceeded. Hint: Convert the circular section to square section having the same area.A. 16B. 21C. 12D. 8

Part 2: Which of the following most nearly gives the value of w so that the allowable shearing stress will not be exceeded.A. 40B. 37C. 52D. 28

Part 3: Which of the following most nearly gives the value of w so that the allowable deflection will not be exceeded.A. 14B. 9C. 6D. 12

Solution

HideShow or hide the solutionBending:

Equivalent square section: (b = d = x)


Shear:

for circular section


Deflection:


Awarding of bid per P.D. 1594Submitted by Romel Verterra on Sun, 01/08/2012 - 18:46

SituationA project has been bid out by the Department of Public Works and Highways. The approved agency

estimate (AAE) is 500 million pesos. The results of responsive bids are as follows:Bidder A - P550,234,451.98Bidder B - P610,345,763.12Bidder C - P454,218,557.98Bidder D - P389,122,897.44Bidder E - P284,758,426.54

The implementing rules and regulations of P.D. 1594 states that no award of contract shall be made to a bidder whose bid price is higher than the allowable government estimate (AGE) or the Approved Agency Estimate (AAE), whichever is higher, or lower than 70% of the AGE. The allowable government estimate (AGE) is defined as one half the sum of the AAE and the average of all responsive bids. For the purposes of determining the average of responsive bids, bids higher than 120% of the AAE or lower than 60% of the AAE shall not be considered. No negotiation will be allowed to bring down the bid to the level of the AAE/AGE.

Part 1: Which of the following most nearly gives the maximum bid price for the project in pesos?A. P500,000,000B. P600,000,000C. P550,000,000D. P650,000,000

Part 2: Which of the following most nearly gives the minimum bid price for the project in pesos?A. P350,000,000B. P250,000,000C. P300,000,000D. P400,000,000

Part 3: Which of the following gives the bidder to which the award can be made?A. Bidder DB. Bidder CC. Bidder AD. Bidder E

Solution

HideShow or hide the solutionResponsive bidders



Therefore, only bidders A, C, and D shall be considered as responsive bidders.

Average of responsive bids:

Allowable Government Estimate

According to the rules and regulation, no award of contract shall be made to a bidder whose bid price is higher than either the AAE or AGE, (P500,000,000) or whose bid is lower than 70% of AGE (P337,583,855.86).

Among the responsive bidders, the award cant be made to bidder A. It may either be awarded to bidders C or D, and among the two, Bidder D is more desirable. Answer for Part 3: [ A ]

Propped beam with moment load at simple supportSubmitted by Romel Verterra on Sun, 01/08/2012 - 09:07

SituationA propped beam is as shown in Figure TS-11(00). The moment applied at the simple supported end

causes a unit rotation at that end.

Part 1: Which of the following most nearly gives the value of the moment M?A. 670 kiloNewton-meterB. 690 kiloNewton-meterC. 700 kiloNewton-meterD. 710 kiloNewton-meter

Part 2: Which of the following most nearly gives the reaction at the simple support?A. 120 kiloNewtonB. 140 kiloNewtonC. 170 kiloNewtonD. 155 kiloNewton

Part 3: Which of the following most nearly gives the moment at the fixed end?A. -300 kiloNewton-meterB. -330 kiloNewton-meterC. -350 kiloNewton-meterD. -380 kiloNewton-meter

Solution

HideShow or hide the solutionM = moment in kN·mR = reaction in kNθAB = 1 radian

→ Equation (1)

→ Equation (2)

Substitute M = 4R to Equation (1)


From Equation (2)


Moment at the fixed support


Steel area of singly reinforced beamSubmitted by Romel Verterra on Sat, 01/07/2012 - 11:35

SituationA rectangular concrete beam has a width of 300 mm and an effective depth of 550 mm. The beam is

simply supported over a span of 6 m and is used to carry a uniform dead load of 25 kN/m and a uniform live load of 40 kN/m. Assume fc' = 21 MPa and fy = 312 MPa. Compression reinforcement if necessary

shall be placed at a depth 80 mm from the outermost compression concrete.

Part 1: Which of the following most nearly gives the maximum tension steel area for singly reinforced condition.A. 3810B. 4120C. 3960D. 3780

Part 2: Which of the following most nearly gives the required tension steel area in square millimeter.A. 3900B. 3750C. 3610D. 3860

Part 3: Which of the following most nearly gives the required number of 25-mm tension bars.A. 8B. 7C. 9D. 6

Solution

HideShow or hide the solutionBeta one factor:Since fc' < 30 MPa, β1 = 0.85

Balanced steel ratio:

Maximum steel ratio:

Maximum tension steel area:


Assuming singly reinforced beamFactored load and ultimate moment capacity

Coefficient of resistance

Required steel ratio

< ρmax (Singly reinforced only)

Required steel area


Number of 25-mm bars:


Variable force acting on the blockSubmitted by Romel Verterra on Fri, 01/06/2012 - 06:41

SituationA 5-kg block resting on a smooth surface is pushed horizontally by a force P as shown in Figure EM-

11(00). The graph of force P versus time is also shown in the figure.

Part 1: Which of the following most nearly gives the acceleration of the block during the first two seconds?A. 39 m/sec2

B. 45 m/sec2

C. 28 m/sec2

D. 32 m/sec2

Part 2: Which of the following most nearly gives the velocity of the block after five seconds?A. 113 m/sec

B. 102 m/secC. 118 m/secD. 129 m/sec

Part 3: Which of the following most nearly gives the total distance traveled by the block in five seconds?A. 395 mB. 380 mC. 420 mD. 405 m

Solution

HideShow or hide the solutionThe force P of the given graph of force versus time is in kilogram. To convert P to Newton, simply multiply P by gravitational acceleration g = 9.81 m/sec2.

For acceleration-time diagram, acceleration a = force/mass. The graph can easily be constructed as shown below.

The area under the acceleration–time curve between any two points represents the change in velocity between the points. Since the block starts from rest, the initial velocity is zero as shown in the velocity-time diagram. The area under the velocity-time curve represents the distance traveled.

From the acceleration-time diagram, acceleration during the first two seconds = 39.24 m/s2. Answer for Part 1: [ A ]

For the velocity-time diagram


The distance traveled is the area from velocity-time diagram


Reactions of Inverted L-frameSubmitted by Romel Verterra on Mon, 01/02/2012 - 11:11

SituationFor the frame shown in Figure EM-00NCE, the force F acting

upward at C causes a horizontal reaction of 100 kN at B.

Part 1: Which of the following most nearly gives the value of the force F?A. 100 kiloNewtonB. 75 kiloNewtonC. 135 kiloNewtonD. 120 kiloNewton

Part 2: Which of the following most nearly gives the reaction at A?A. 148 kiloNewtonB. 134 kiloNewtonC. 156 kiloNewtonD. 175 kiloNewton

Part 3: Which of the following most nearly gives the angle in degrees that the reaction at A makes with the horizontal axis (positive counterclockwise)?A. 260°B. 230°C. 140°D. 50°

Solution

HideShow or hide the solutionSummation of moment at hinge support A


Summ of horizontal forces

Sum of vertical forces

Total reaction at A



Moment at the inner support of continuous beamSubmitted by Romel Verterra on Thu, 12/29/2011 - 19:07

SituationA continuous beam is as shown if Figure AN-20. Using the moment distribution method and assuming E

and I to be constant:

Part 1: Which of the following most nearly gives the fixed end moment at A due to the loads on member AB.A. 10B. 12C. 15D. 18

Part 2: Which of the following most nearly gives the distribution factor at B on member BC, in percent. Use the modified K.A. 59B. 48C. 78D. 67

Part 3: Which of the following most nearly gives the moment at B in kiloNewton-meter.A. -13B. -10C. -18

D. -23

Solution

HideShow or hide the solutionFixed-end moments, FEM


Beam StiffnessAssume,

Modified

Modified

Distribution factors, DF


A B C D

1 1/3 2/3 2/3 1/3 1

FEM 12.00 -12.00 5.00 -5.00 3.56 -7.11

-12.00 → -6.00 3.56 ← 7.11

4.33 8.67 → 4.36

-2.16 ← -4.32 -2.16

0.72 1.44 → 0.72

-0.24 ← -0.48 -0.24

0.08 0.16 → 0.08

-0.053 -0.027

0.00 -12.87 12.87 -4.693 4.693 0.00

Moment at B = -12.87 kN·m Answer for Part 3: [ A ]

Approved Government Estimate per PD 1594Submitted by Romel Verterra on Sun, 12/25/2011 - 07:04

SituationA project has been bid out by the Department of Public Works and Highways. The approved agency

estimate (AAE) is 500 million pesos. The results of responsive bids are as follows:Bidder A - P550,234,451.98Bidder B - P610,345,763.12Bidder C - P454,218,557.98Bidder D - P389,122,897.44Bidder E - P284,758,426.54

The implementing rules and regulations of P.D. 1594 states that no award of contract shall be made to a bidder whose bid price is higher than the allowable government estimate (AGE) or the Approved Agency Estimate (AAE), whichever is higher, or lower than 70% of the AGE. The allowable government estimate (AGE) is defined as one half the sum of the AAE and the average of all responsive bids. For the purposes of determining the average of responsive bids, bids higher than 120% of the AAE or lower than 60% of the AAE shall not be considered. No negotiation will be allowed to bring down the bid to the level of the

AAE/AGE.

Part 1: Which of the following gives the responsive bidders for the project.A. Bidders A, C, D, and E onlyB. Bidders A, B, C, and D onlyC. Bidders A, C, and D onlyD. Bidders C and D only

Part 2: Which of the following most nearly gives the average of all the responsive bids in pesos.A. P489,356,200.00B. P505,789,600.00C. P443,678,100.00D. P464,525,300.00

Part 3: Which of the following most nearly gives the value of the approved government estimate (AGE) in pesos.A. P482,262,700.00B. P501,455,900.00C. P476,456,200.00D. P456,332,100.00

Solution


Bidder B is greater than P600,000 and bidder E is less than P300,000, therefore, only bidders A, C, and D shall be considered as responsive bidders.Answer for Part 1: [ C ]

Average of responsive bids:


Approved Government Estimate , AGE


Problem 01 | Bernoulli's Energy TheoremSubmitted by Romel Verterra on Thu, 03/08/2012 - 21:38

Problem 1The water surface shown in Figure 4-01 is 6 m above the datum. The pipe is 150 mm in diameter and the total loss of head between point (1) in the water surface and point (2) in the jet is 3 m. Determine the velocity of flow in the pipe and the discharge Q.

Solution 1

Hide Show or hide the solution Solve for velocity head at point (5)

Velocity of flow

Since the diameter of the opening is equal to the diameter of the pipe

http://cereview.info/book/fluid-mechanics-and-hydraulics/problem-01-bernoullis-energy-theorem

http://cereview.info/users/romel-verterra

answer

Discharge

answer

Problem 01 | Bernoulli's Energy TheoremSubmitted by Romel Verterra on Thu, 03/08/2012 - 21:38

Problem 1The water surface shown in Figure 4-01 is 6 m above the datum. The pipe is 150 mm in diameter and the total loss of head between point (1) in the water surface and point (2) in the jet is 3 m. Determine the velocity of flow in the pipe and the discharge Q.

Solution 1

Hide Show or hide the solution Solve for velocity head at point (5)

Velocity of flow



Since the diameter of the opening is equal to the diameter of the pipe

answer

Discharge

answer

Problem 02 | Bernoulli's Energy TheoremSubmitted by Romel Verterra on Fri, 03/09/2012 - 17:14

Problem 2From Figure 4-01, the following head losses are known: From (1) to (2), 0 m; from (2) to (3), 0.60 m; from (3) to (4), 2.1 m; from (4) to (5), 0.3 m. Make a table showing elevation head, velocity head, pressure head, and total head at each of the five points. How high above the center of the pipe will water stands in the piezometer tubes (3) and (4)?

Solution 2

Hide Show or hide the solution Given:

Total head loss from (1) to (5)



Note:

Sum up head from (1) to (5)

Since the diameter of the pipe is uniform and the opening for the jet is equal to the diameter of the pipe, the velocity heads at any point on the pipe are equal. Thus,





Tabulated result

Point Elevation head (m) Velocity head (m) Pressure head (m) Total head (m)1 6 0 0 6.02 0 0 6 6.03 0 3 2.4 5.44 0 3 0.3 3.35 0 3 0 3.0

Piezometric heights

answer

answer

Problem 03 | Bernoulli's Energy TheoremSubmitted by Romel Verterra on Sat, 03/10/2012 - 07:38

Problem 3A 300-mm pipe is connected by a reducer to a 100-mm pipe. See Figure 4-02. Points 1 and 2 are at the same elevation, the pressure at 1 is 200 kPa. The discharge Q is 30 liters per second flowing from 1 to 2 and the energy lost from 1 to 2 is equivalent to 20 kPa.

a. Compute the pressure at 2 if the liquid is water.b. Compute the pressure at 2 if the liquid is oil (sp gr = 0.80).c. Compute the pressure at 2 if the liquid is molasses (sp gr = 1.5).


Solution 3

Hide Show or hide the solution Discharge

Head loss

Velocity heads

Energy equation between 1 and 2


Part a: The liquid is water:

answer

Part b: The liquid is oil (sp gr = 0.80):

answer

Part 3: The liquid is molasses (s = 1.5):

answer


Problem 4In Figure 4-02, with 15 L/s of water flowing from 1 to 2 the pressure at 1 is 100 kPa and at 2 is 70 kPa. Compute the loss of head between 1 and 2.

Solution 4


Velocity head




answer


Problem 5With 30 L/s of water flowing in Figure 4-02, what pressure must be maintained at 1 if the pressure at 2 is to be 70 kPa and the loss of head between 1 and 2 is 5 percent of the difference in pressure head at 1 and 2.

Solution 5




Velocity heads

Head loss


answer



Problem 5With 30 L/s of water flowing in Figure 4-02, what pressure must be maintained at 1 if the pressure at 2 is to be 70 kPa and the loss of head between 1 and 2 is 5 percent of the difference in pressure head at 1 and 2.

Solution 5


Velocity heads

Head loss



answer

Problem 07 | Bernoulli's Energy TheoremSubmitted by Romel Verterra on Sun, 03/11/2012 - 19:27

Problem 7Compute the velocity head of the jet in Figure 4-03 if D1 = 75 mm, D2 = 25 mm, the pressure head at 1 is 30 m of the liquid flowing, and the lost head between points 1 and 2 is 5 percent of the velocity head at point 2.

Solution 7

Hide Show or hide the solution Velocity heads in terms of discharge Q



Head lost


Velocity head at point 2

answer


Problem 8In Figure 4-04, with 35 L/s of sea water (sp gr 1.03) flowing from 1 to 2, the pressure at 1 is 100 kPa and at 2 is -15 kPa. Point 2 is 6 m higher than point 1. Compute the lost energy in kPa between 1 and 2.


Solution 8


Velocity heads


answer




Problem 9The diameter of a pipe carrying water changes gradually from 150 mm at A to 450 mm at B. A is 4.5 m lower than B. What will be the difference in pressure, in kPa, between A and B, when 0.176 m3/s is flowing, loss of energy is being neglected.

Solution 9

Hide Show or hide the solution

Velocity heads

Neglecting head loss


answer


Problem 10The diameter of a pipe carrying water changes gradually from 150 mm at A to 450 mm at B. A is 4.5 m lower than B. If the pressure at A is 70 kPa and that B is 50 kPa, when 140 L/s is flowing.(a) Determine the direction of flow.(b) Find the frictional loss between the two points.

Solution 10


Velocity heads



Pressure heads

Total head

The flow is always from higher energy to lower energy. EA > EB, thus, the flow will be from A to B. answer

Energy equation between A and B

answer


Problem 11A horizontal pipe carries 30 cfs of water. At A the diameter is 18 in. and the pressure is 10 psi. At B the diameter is 36 in. and the pressure is 10.9 psi. Determine the head lost between the two points.

Solution 11


Velocity head



Pressure heads


answer


Problem 11A horizontal pipe carries 30 cfs of water. At A the diameter is 18 in. and the pressure is 10 psi. At B the diameter is 36 in. and the pressure is 10.9 psi. Determine the head lost between the two points.

Solution 11


Velocity head

Pressure heads




answer

Problem 12 | Bernoulli's Energy TheoremSubmitted by Romel Verterra on Mon, 03/12/2012 - 14:45

Problem 12In Figure 4-05, a 50 mm pipeline leads downhill from a reservoir and discharges into air. If the loss of head between A and B is 44.2 m, compute the discharge.

Solution 12


Note for circular pipes

Sum up energy head from A to B

answer

Problem 13 | Bernoulli's Energy TheoremSubmitted by Romel Verterra on Mon, 03/12/2012 - 21:36

Problem 13The 150-mm pipe line shown in Figure 4-05 conducts water from the reservoir and discharge at a lower elevation through a nozzle which has a discharge diameter of 50 mm. The water surface in the




reservoir 1 is at elevation 30 m, the pipe intake 2 and 3 at elevation 25 m and the nozzle 4 and 5 at elevation 0. The head losses are: from 1 to 2, 0; from 2 to 3, 0.6 m; from 3 to 4, 9 m; from 4 to 5, 3 m. Compute the discharge and make a table showing elevation head, pressure head, and total head at each of the five points.

Solution 13

Hide Show or hide the solution Head lost



answer

Velocity heads

Pressure heads



Tabulated result

Point Elevation head (m) Velocity head (m) Pressure head (m) Total head (m)1 30 0 0 30.02 25 0 5 30.03 25 0.2151 4.1849 29.44 0 0.2151 20.1849 20.45 0 17.4202 0 17.4


Problem 14Water discharges through an orifice in the side of a large tank shown in Figure 4-06. The orifice is circular in cross section and 50 mm in diameter. The jet is the same diameter as the orifice. The liquid is water, and the surface elevation is maintained at a height h of 3.8 m above the center of the jet. Compute the discharge: (a) neglecting loss of head; (b) considering the loss of head to be 10 percent of h.

Solution 14


Neglecting head lost

answer

Considering head lost



answer


Problem 15A pump (Figure 4-07) takes water from a 200-mm suction pipe and delivers it to a 150-mm discharge pipe in which the velocity is 2.5 m/s. At A in the suction pipe, the pressure is -40 kPa. At B in the discharge pipe, which is 2.5 m above A, the pressure is 410 kPa. What horsepower would have to be applied by the pump if there were no frictional losses?

Solution 15Discharge

Velocity heads

Hide Click here to continue reading the solution



Neglecting head lost between A and B

Power delivered by the pump

answer


Problem 16A pump (Figure 4-07) takes water from a 200-mm suction pipe and delivers it to a 150-mm discharge pipe in which the velocity is 3.6 m/s. The pressure is -35 kPa at A in the suction pipe. The 150-mm pipe discharges horizontally into air at C. To what height h above B can the water be raised if B is 1.8 m above A and 20 hp is delivered to the pump? Assume that the pump operates at 70 percent efficiency and that the frictional loss in the pipe between A and C is 3 m.


Solution 16


Output power of the pump

Head Added


Velocity heads

Energy equation from A to C

answer

Problem 17 | Bernoulli's Energy TheoremSubmitted by Romel Verterra on Wed, 03/21/2012 - 15:25

Problem 17In Figure 4-08 is shown a siphon discharging water from reservoir A into the air at B. Distance 'a' is 1.8 m, 'b' is 6 m, and the diameter is 150 mm throughout. If there is a frictional loss of 1.5 m between A and the summit, and 1.5 m between the summit and B, what is the absolute pressure at the summit in kiloPascal? Also determine the rate of discharge in cubic meter per second and in gallons per minute.

Solution 17




Velocity head at B and C in terms of Q

Energy Equation between A and B

answer

answer

Thus, the velocity head at B and C is

Energy equation between A and C

answer

Problem 18 | Bernoulli's Energy TheoremSubmitted by Romel Verterra on Fri, 03/23/2012 - 21:43

Problem 18Figure 4-09 shows a siphon discharging oil (sp gr 0.90). The siphon is composed of 3-in. pipe from A to B followed by 4-in. pipe from B to the open discharge at C. The head losses are from 1 to 2, 1.1 ft; from 2 to 3, 0.7 ft; from 3 to 4, 2.5 ft. Compute the discharge, and make table of heads at point 1, 2, 3, and 4.


Solution 18


Head lost

Velocity heads in terms of Q


answer


Velocity heads at 2, 3, and 4


answer


answer

CheckingEnergy equation between 2 and 3

(check!)

Tabulated result

Point Elevation head (ft) Velocity head (ft) Pressure head (ft) Total head (ft)1 10 0 0 102 15 18.02 -24.12 8.93 15 5.7 -12.5 8.24 0 5.7 0 5.7

Physical Properties of SoilSubmitted by Romel Verterra on December 18, 2012 - 5:31pm

Soil is composed of solids, liquids, and gases. Liquids and gases are mostly water and air, respectively. These two (water and air) are called voids which occupy between soil particles. The figure shown below is an idealized soil drawn into phases of solids, water, and air.

Weight-Volume Relationship from the Phase Diagram of Soiltotal volume = volume of soilds + volume of voids

volume of voids = volume of water + volume of air

http://www.mathalino.com/users/romel-verterra

total weight = weight of solids + weight of water

Soil PropertiesVoid Ratio, eVoid ratio is the ratio of volume of voids to the volume of solids.

Porosity, nPorosity is the ratio of volume of voids to the total volume of soil.

Degree of Saturation, SDegree of saturation is the ratio of volume of water to the volume of voids.

Water Content or Moisture Content, wMoisture content, usually expressed in terms of percentage, is the ratio of the weight of water to the weight of solids.

Unit Weight, γUnit weight is the weight of soil per unit volume. Also called bulk unit weight (γ), and moist unit weight (γm).

Dry Unit Weight, γd

Dry unit weight is the weight of dry soil per unit volume.

Saturated Unit Weight, γsat

Saturated unit weight is the weight of saturated soil per unit volume.

Effective Unit Weight, γ'Effective unit weight is the weight of solids in a submerged soil per unit volume. Also called buoyant density or buoyant unit weight (γb).

Specific Gravity of Solid Particles, GSpecific gravity of solid particles of soil is the ratio of the unit weight of solids (γs) to the unit weight of water (γw).

Formulas for Properties of SoilSymbols and Notations = void ratio = porosity = moisture content, water content

= specific gravity of any substance = specific gravity of solids = degree of saturation = volume of soil mass

= volume of air = volume of water = volume of solids = volume of voids = total weight of soil

= weight of water = weight of solids = relative density = unit weight of soil mass, moist unit weight, bulk unit weight

= unit weight of soil solids = unit weight of water

= = buoyant unit weight, submerged unit weight = = dry unit weight

= saturated unit weight = liquid limit = plastic limit

= liquidity index

= plasticity index = group index

Basic FormulasUnit weight,

Weight,

Specific gravity,

Physical Properties of SoilTotal weight,

Total volume,

Volume of voids,

Void ratio, , Note:

Porosity, , Note:

Relationship between e and n, and

Water content or moisture content, , Note:

Degree of saturation, , Note:

Relationship between G, w, S, and e,

Moist unit weight or bulk unit weight,

, also or

Dry unit weight, and

Saturated unit weight,

Submerged or buoyant unit weight, or

Critical hydraulic gradient, or

Relative Density, or

Atterberg LimitsPlasticity index,

Liquidity index,

Shrinkage index,

Activity of clay, , where = soil finer than 0.002 mm in percent

Other Formulas

Volume of voids,

Volume of solids,

Volume of water,

Weight of water,

Weight of soil,

Dry unit weight,

Relationship between specific gravity of solids, moisture content, degree of saturation, and void ratioThe relationship between , , , and is given by the following

G = specific gravity of solid particlesw = moisture content or water contentS = degree of saturatione = void ratio

The formula above can be derived as follows:

Thus, as stated above.

Relationship between void ratio and porosityThe relationship between and is given by

and

Derivation is as follows

→ void ratio

→ n = Vv / V

(okay!)

→ porosity

→ e = Vv / Vs

(okay!)

Consistency of Soil – Atterberg LimitsSubmitted by Romel Verterra on December 18, 2012 - 6:52pm

Consistency is the term used to describe the ability of the soil to resist rupture and deformation. It is commonly describe as soft, stiff or firm, and hard.

Water content greatly affects the engineering behavior of fine-grained soils. In the order of increasing moisture content (see Figure 2 below), a dry soil will exist into four distinct states: from solid state, to semisolid state, to plastic state, and to liquid state. The water contents at the boundary of these states are known as Atterberg limits. Between the solid and semisolid states is shrinkage limit, between semisolid and plastic states is plastic limit, and between plastic and liquid states is liquid limit.


Atterberg limits, then, are water contents at critical stages of soil behavior. They, together with natural water content, are essential descriptions of fine-grained soils.

Liquid Limit, LLLiquid limit is the water content of soil in which soil grains are separated by water just enough for the soil mass to loss shear strength. A little higher than this water content will tend the soil to flow like viscous fluid while a little lower will cause the soil to behave as plastic.

Plastic Limit, PLPlastic limit is the water content in which the soil will pass from plastic state to semi-solid state. Soil can no longer behave as plastic; any change in shape will cause the soil to show visible cracks.

Shrinkage Limit, SLShrinkage limit is the water content in which the soil no longer changes in volume regardless of further drying. It is the lowest water content possible for the soil to be completely saturated. Any lower than the shrinkage limit will cause the water to be partially saturated. This is the point in which soil will pass from semi-solid to solid state.

Determination of Liquid, Plastic, and Shrinkage LimitsCasagrande Cup Method for Liquid Limit Test

Casagrande CupCourtesy of MOHAN LAL AND SONS

The semispherical brass cup is repeatedly dropped into a hard rubber base from a height of 10 mm by a cam-operated crank.

The dry powder of the soil is mixed with distilled water turning it into a paste. The soil paste is then placed into the cup to a thickness of about 12.5 mm and a groove is then cut at the center of the paste using the standard grooving tool. The crank operating the cam is turned at the rate of 2 revolutions per second

lifting the cup and dropped it from a height of 10 mm. The liquid limit is the moisture content required to close a distance of 12.5 mm along the bottom of the groove after 25 blows.

The required closure in 25 blows is difficult to achieve in a single test. Four or more tests to the same soil at varying water contents are to be done for 12.5 mm closure of the groove. The results are then plotted on a semi-logarithmic graph with moisture content along the vertical axis (algebraic scale) and number of blows along the horizontal axis (logarithmic scale).

The graph is approximated by the best fit straight line, usually called the flow line and sometimes called liquid state line. The moisture content that corresponds to 25 blows is the liquid limit of the soil.

The slope of the flow line is called flow index and may be written as

Flow index,

where w1 and w2 are the water content corresponding to number of blows N1 and N2, respectively.

Plastic Limit Test

The plastic limit can easily be found by rolling a small soil sample into thin threads until it crumbles. The water content at which the threads break at approximately 3 mm in diameter is the plastic limit. Two or more tests are made and the average water content is taken as plastic limit. In this test, soil will break at smaller diameter when wet and breaks in larger diameter when dry.

Fall Cone Method for Liquid and Plastic Limit Tests

Cone PenetrometerCourtesy of SAIGON ISC

Fall cone method offers more accurate result of liquid limit and plastic limit tests. In this method, a cone with a mass of 80 grams and an apex angle of 30° is suspended above so that its pointed part will just in contact with the soil sample. The cone is permitted to fall freely under its own weight for a period of 5 seconds. The water content that allows the cone to penetrate for 20 mm during this period defines the liquid limit of the soil.

Like the cup method, four or more tests are required because it is difficult to find the liquid limit in a single test. The results are then plotted into a semi-logarithmic paper with water content along the vertical axis (arithmetic scale) and penetration along the horizontal axis (logarithmic scale). The best fit straight line is then drawn and the water content that corresponds to 20 mm penetration defines the liquid limit.

The plastic limit can be found by repeating the test with a cone of similar geometry but with a mass of M2 = 240 grams. The liquid state line of this cone will be below the liquid state line of the M1 = 80 grams cone and parallel to it.

The plastic limit is given as

Shrinkage Limit Test

The shrinkage limit is determined as follows. A mass of wet soil, m1, is placed in a porcelain dish 44.5 mm in diameter and 12.5 mm high and then oven dried. With oven-dried soil still in the dish, the volume of shrinkage can be determined by filling the dish with mercury. The volume of mercury that fills the dish is equal to the shrinkage volume. The shrinkage limit is calculated from

where m1 = mass of wet soil, m2 = mass of oven-dried soil, V1 = volume of wet soil, V2 = volume of oven-dried soil, and ρw = density of water.

Other FormulasShrinkage ratio

Specific gravity of solids

Unit Weights and Densities of SoilSubmitted by Romel Verterra on December 18, 2012 - 5:42pm

Unit Weights of SoilSymbols and Notationsγ = Unit weight, bulk unit weightγm = Moist unit weightγd = Dry unit weightγsat = Saturated unit weightγb, γ' = Buoyant unit weight or effective unit weightγs = Unit weight of solidsγw = Unit weight of water (equal to 9810 N/m3)W = Total weight of soilWs = Weight of solid particlesWw = Weight of waterV = Volume of soilVs = Volume of solid particlesVv = Volume of voidsVw = Volume of waterS = Degree of saturationw = Water content or moisture contentG = Specific gravity of solid particles

Bulk Unit Weight / Moist Unit Weight


Note: Se = Gw, thus,

Moist unit weight in terms of dry density and moisture content

Dry Unit Weight (S = w = 0)

From and , S = 0 and w = 0

Saturated Unit Weight (S = 1)

From , S = 100%

Buoyant Unit Weight or Effective Unit Weight

Unit weight of waterγ = 9.81 kN/m3

γ = 9810 N/m3

γ = 62.4 lb/ft3

Typical Values of Unit Weight for Soils

Type of soil γsat (kN/m3) γd (kN/m3)Gravel 20 - 22 15 - 17Sand 18 - 20 13 - 16Silt 18 - 20 14 - 18Clay 16 - 22 14 - 21

Densities of SoilThe terms density and unit weight are used interchangeably in soil mechanics. Though not critical, it is important that we know it. To find the formula for density, divide the formula of unit weight by gravitational constant g (acceleration due to gravity). But instead of having g in the formula, use the density of water replacing the unit weight of water.

Basic formula for density (note: m = W/g)

The following formulas are taken from unit weights of soil:

Wherem = mass of soilV = volume of soilW = weight of soilρ = density of soilρd = dry density of soilρsat = saturated density of soilρ' = buoyant density of soilρw = density of waterG = specific gravity of soil solidsS = degree of saturation of the soile = void ratiow = water content or moisture content

Density of water and gravitational constantρw = 1000 kg/m3

ρw = 1 g/ccρw = 62.4 lb/ft3

g = 9.81 m/s2

g = 32.2 ft/sec2

Relative DensityRelative density is an index that quantifies the state of compactness between the loosest and densest possible state of coarse-grained soils.

The relative density is written in the following formulas:

where:Dr = relative densitye = current void ratio of the soil in-situemax = void ratio of the soil at its loosest conditionemin = void ratio of the soil at its densest condition

γd = current dry unit weight of soil in-situ(γd)min = dry unit weight of the soil at its loosest condition(γd)max = dry unit weight of the soil at its densest condition

Designation of Granular Soil Based on Relative Density

Dr (%) Description0 - 20 Very loose

20 - 40 Loose40 - 70 Medium dense70 - 85 Dense

85 - 100 Very dense