appendix b_ calculations_ structures portion (1-17-2015)

Rice Univeristy Engineers Without borders

Appendix D : Calculations

Hydraulic, Structural, and Electrical

Project Leaders: Edgar Silva and Rachel Sterling

9/14/2014

Given Values Value Origins

GeneralTotal Population (people) 1,200 Census from communityGrowth Rate 7% UN Nicaraguan growth rate with safety factor of 5Location Of Tank(Elevation, M) 825 Given by ENACAL, determined using GPSLocation Of Pump (Elevation, M) 724 Given by ENACAL, determined using GPSHydraulicsGallons Per Person (Gallons/Day) 24 Provided by handbook from engineer at ENACAL

Needed Tank Capacity (Days) 1 Community request

Pump Functioning Hours 5.5 - 9.5From pressure test completed on assessment trip, only accounting

for time that PSI at this elevation is >10(min pressure for pump is < 3 psi)

Number of tanks (22,000L each) Max 8 This number is capped at 8 because it was determined to be

StructuralSoil Type Clay Determined by assessment tripPsi Strength of Concrete 3000 Determined by mix ratiosPsi Strength of Steel 60000 Assuming grade 60 steel is usedMass of pump (lbs) 201 From pump spec sheetMass of filled tank (kg) 22000 Determined by volume of tank* More values given in structural sectionElectrical

Power Supply Single Phase This is determined from the phase of the Nicaraguan electrical grid

Structural Calculations

Soil Bearing Capacity Calculations

To calculate the ultimate bearing capacity of the soil, we used the following equation, the Terzaghi Bearing Capacity equation. The highest pressures will be experienced by the slabs holding the 22000 liter water tanks.

qult=(c)(Nc)(sc)+(ɤ )(D)(Nq)(sq)+(0.5)(B)(ɤ )(Ng)(sɤ )

The following chart contains important values:

Cohesive Strength of Soil (c)239.

4 kPa

Unit Weight of Soil (ɤ)8.56

8 kN/m3

Depth of Foundation (D) 0 mWidth of Foundation (B) 4 mSafety Factor 3Angle Of Internal Friction 17 degreesBearing Capacity Factor (cohesion) Nc 12.3Bearing Capacity Factor (surcharge and friction) Nq 4.8Bearing Capacity Factor (self weight and friction) Nɤ 1.7Shape Factor (Sc) 1.3Shape Factor (Sq) 1Shape Factor (Sɤ) 0.8

The cohesive strength of soil was determined by using a pocket penetrometer at the surface and 6 inches below the surface. A value of 239.4 kPa (2.5 tons per square foot) was measured at the surface. A value of 430.9 kPa (4.5 tons per square foot) was measured 6 inches below the surface. From these measurements we know the soil strengthens as depth increases. For our calculations we will be on the safe side and use the weakest measurement of 239.4 kPa.

The unit weight of the soil was determined by taking a 600 ml sample of soil in a bottle, and measuring its mass.

The depth and width of foundation were determined by our design criteria. The slabs need to be at least 4 meters wide to accommodate the water tanks. For simplicity of construction the slabs will just lay on the surface of the ground.

The safety factor was a value recommended to us by our mentor.The internal angle of friction and bearing capacities were determined by identifying the

type of soil at our construction site. By performing the ribbon test, the soil was identified as high plasticity clay soil. According to Swiss Standard SN 670 010b, this soil has an angle of friction between 17 and 31 degrees. We will use an angle of 17 degrees in our calculations to assume worst case soil conditions.

From Terzaghi’s tables, an angle of 17 degrees correspond to Nc = 12.3, Nq = 4.8, and Nɤ = 1.7.

Shape factors were derived from Terzaghi’s tables by assuming a square foundation. A square foundation was used for simplicity of construction.

Qult=(239.4)(12.3)(1.3)+(8.568)(0)(4.8)(1)+0.5(4)(8.568)(1.7)(0.8)=3851.3 kPa

Safe BearingCapacity=Qs= QultSafety Factor

=3851.33

=1283.8 kPa

Theoretical Load supported by Slab=1000(Qs)(B2)g

=1000(1283.8)(42)

9.81=2093814kg

The mass of the 22000 liter tank is 22000 kg. Therefore, the soil is more than strong enough to support the tank.

Pump Station Cross-Section Calculations

a. Design slab thickness:

The thickness of the slab will be controlled by the shear forces it needs to resist. From our previous calculations, we know the soil is more than strong enough to support our structures. However, the soil is expansive, and will expand and contract based on precipitation in the area. The the slab will trap moisture in the soil below it, and that soil will remain at a relatively constant moisture content, while the soil at the edges of the slab will quickly lose moisture in the dry season and shrink, and will quickly absorb moisture in the wet season and expand. This leads to two critical conditions for the slab, called dishing and doming, or also called edge list condition and center lift condition. Dishing occurs in the wet season when the slab is only supported by the expanded soil on the edges. Doming occurs in the dry season when the slab is only supported by the still moist soil under the slab. For our calculations, we will treat the slab as a wide beam. Additionally, we will assume worse case soil expansion and contraction. We will assume the slab dimensions of 5.80 x 2.60 due to the amount of spacing needed for the pump and valves.

Overhead view of forces:

Doming force diagram:

Dishing force distribution:

Table of ConstraintsDimensions of Slab 5.80 x 2.60 metersMass of pump 91.17 kgDimensions of pump platform 1.00 x 0.40 x 0.50 metersDensity of concrete and CMU blocks 2400 kg/m3

Strength of concrete, f’c 13.8 MPaShear Strength Safety Factor 0.75Height of CMU wall 2.40 metersWidth of CMU wall 0.15 metersAcceleration due to gravity (g) 9.81 m/s2

i. Doming slab thickness calculation:

Concrete shear strength∈newtons per meter squared=2√ f ' c

6 (1000000)

The slab will be designed to have no vertical web reinforcement. By ACI code, the design shear strength of concrete without web reinforcement is (Design shear stress)=(0.5)(safety factor)(concrete shear strength)

The maximum shear will occur in the center of the slab, where the soil supports the slab.

(0.5) (safety factor ) (Concrete shear strength ) (widthof slab ) (height of slab )=outside wall weight force+distributed force of slabweight+distributed force of perimeter walloutside wall weight force=(widthof slab)(thickness of wall)(height of wall)(density of wall)(g)

outside wall weight force=(2.6 ) ( .15 ) (2.4 ) (2400 ) (9.81 )=22037newtons

distributed force of slabweight=(width of slab)(height of slab )(density of concrete )(g)(lengthof slab)/2

distributed force of slabweight= (2.6 ) (height of slab ) (2400 ) (9.81 ) (5.8 )2

=177522(height of slab)

distributed force of perimeter wall=2(thickness of wall)(height of wall)(density of wall)(g)(lengthof slab)/2

distributed force of perimeter wall=2 ( .15 ) (2.4 ) (2400 ) (9.81 ) (5.8 )2

=49160newtons

(0.5) (0.75 )( 2√13.86 )(1000000)(2.6 ) (height of slab )=22037+177522 (height of slab )+49160

Height of slab = .167 meters

ii. Dishing slab thickness calculation:

The maximum shear will occur at the left and right of the slab, where it is supported by the soil. Calculations will find shear at the left support.

(0.5 ) ( safety fac tor ) (Concrete shear strength ) (widthof slab ) (height of slab )=force ¿ pump platform+distributed force of slabweight+distributed force of perimeter wall+¿outside wall weight force−¿ reaction force

Force ¿ pump platform=(pumpweight )(g)+(volumeof pump platform)(density of concrete)(g)

Force ¿ pump platform=(91.17 ) (9.81 )+(1 ) (0.4 ) (0.5 ) (2400 ) (9.81 )=5603newtons

¿outside wall weight force=(widthof slab)(thickness of wall)(height of wall)(density of wall)(g)

¿outside wall weight force=(2.6 ) ( .15 ) (2.4 ) (2400 ) (9.81 )=22037newtons

distributed force of slabweight=(width of slab)(height of slab )(density of concrete )(g)(lengthof slab)

distributed force of slabweight= (2.6 ) (height of slab ) (2400 ) (9.81 ) (5.8 )=355044(height of slab)

distributed force of perimeter wall=2(thickness of wall)(height of wall)(density of wall)(g)(lengthof slab)

distributed force of perimeter wall=2 (.15 ) (2.4 ) (2400 ) (9.81 ) (5.8 )=98320newtons

¿reaction force= totaldownward force2

¿reaction force=5603+(2 )22037+98320+355044 (height of slab)

2¿reaction force=73999+177522(hei ght of slab)

(0.5 ) (0.75 )( 2√13.86 ) (1000000 ) (2.6 ) (height of slab )=5603+355044 (height of slab )+98320+22037−73999−177522(height of slab)


iii. Final slab thickness

Bowling requires a thicker slab, so that is our controlling case for shear resistance. Therefore, the slab must be at least .167 meters thick. For ease of construction, the slab will be designed as 0.18 meters thick

b. Design moment calculation:

The slab is relatively thin, so bending moments in the slab should be considered for the dishing and bowling case. These moments will be used to calculate the needed steel reinforcement. To calculate the design moment, we need to determine what part of the slab will experience the most moment. We will assume the slab dimensions of 5.80 x 2.60 x 0.18 meters as given, due to the amount of spacing needed for the pump and valves. The previous figures and table of constraints will be used.

i. doming moment calculation

In the doming case, the slab acts like a cantilever, so maximum moment will be at the center of the slab.Bendingmoment=(outsidewall weight )(distance)+(force¿ perimeter wall)(distance)+( force¿slabweight)(distance)

From doming shear calculationsoutside wall weight force=22037newtons

force¿ perimeterwall=49160newtons

force¿ slab weight=177522 (0.18 )=31954newtons

Bendingmoment=(22037 ) (2.9 )+( 49160 )( 2.92 )+31954( 2.9

2)

Bendingmoment=181523newtonmeters

Designmoment=Maximummoment∗safety factor

Designmoment=181523∗1.2=217827N∗m

ii. dishing moment calculation

In the dishing case, the slab acts like a simply supported beam, so maximum moment will be at the center of the slab.Bendingmoment=¿wall weight ¿¿ (distance )+(¿ perimeter wall force)(distance )+¿weight force¿¿ (distance )−(¿ reaction force)(distance) From dishing shear calculations¿outside wall weight force=22037new tons

¿ force¿ perimeterwall=983202

=49160newtons

¿ force¿ slabweight=355044 (0.18 ) /2=31954newtons

¿reaction force=73999+177522 ( .18 )=105953newtons

Bendingmoment=(22037 ) (2.9 )+( 49160 )( 2.92 )+31954( 2.9

2 )−105953(2.9)

Bendingmoment=−125741newtonmeters


Designmoment=125741∗1.2=150889N∗m

c. Rebar design calculations

For rebar design, Whitney Stress Block theory will be used. We need to solve for the area of steel to be used. Tensile stresses will be at the top of the slab during doming, and tensile stresses will be at the bottom of the slab during dishing. A concrete skirt will extend into the ground around the perimeter of the slab, to prevent the erosion of soil under the slab. It is submerged in the soil, so its weight has been neglected in moment calculations, but it will provide a compressive surface for the doming case. We will assume the bars are placed 5 cm from the faces of the slab Design InputsConcrete Design Compressive Stress (f'c) 13.8 MPaYield Stress of steel (fy) 420 MPaBase (b) 0.18 meterDoming Rebar Design Moment 217827 N-m

Dishing Rebar Design Moment 150889 N-mSafety Factor (phi) 0.9Depth (d) 0.13 meterWidth 2.60 meterModulus of elasticity of steel (Es) 200 GPaa to c ratio (β) 0.85

Dishing force diagram

Doming force diagram

Doming force cross section

βc=a

Dishing Force Equilibrium: concrete compressive force=steel tensile force

( f 'c )(βc1)(width)=(As)( fy)(13800000)(0.85∗c1)(2.6)=(As)(420000000)

(30498000)(c1)=(As)(420000000)

Doming Force Equilibrium:concrete compressive force=steel tensile force

( f 'c )(βc2)(width)=(As)( fy)(13800000)(0.85∗c2)(2∗.15)=(As)(420000000)

(3519000 ) (c2 )=(As)(420000000)

Dishing Moment Equilibrium:

( f 'c ) (βc1 ) (width )(d−βc 12 )=dishmoment

safety factor

(13800000)(0.85∗c1)(2.6)(.13−0.85∗c 12 )=150889

0.9

(30498000)(c1) (.13−.425(c1))=167654

Doming Moment Equilibrium:

( f 'c ) (βc2 ) (width )(d−βc 22 )=domemoment

safety factor

(13800000)(0.85∗c2)(2∗.15)(.63−0.85∗c22 )=217827

0.9(3519000)(c 2) (.63−.425 (c2))=242030

Using the Dishing Moment Equation, c1=.0507 mUsing the Dishing Force Equation, As = .0037 m2

The moment capacity in the doming case must be checked for this area of steel.

Using the Doming Force Equation, c2 = .440 mThe compressive block remains within the concrete skirt, so the assumptions made in the force balance equation are correct.Using the Doming Moment Equation, 685382 n*m>242030 n*mThe moment capacity exceeds the expected moment during doming.

Therefore, the design is good and As = .0037 m2

Total Density of steel= AsCross sectional area of concrete

= .0037.18∗2.6

=0.79 %

Areaof steelneeded along short side=.0037m2

Areaof steelneeded a longlong side=.0037∗5.82.6

=.083m2

Areaof ¿4 rebar=0.000129032m2

number of ¿4 rebar short side= .0037.000129032

=29

number of ¿4 rebar longside= .0083.000129032

=65


number of ¿5 rebar short side= .0037.0002

=19

number of ¿5 rebar long side= .0083.0002

=42

Calculations such as this can be easily repeated for other rebar sizes. Selection of rebar will depend on local prices of steel. Drawings will be for #4 rebar, due to its known price in Nicaragua. Rebar will be equally spaced, with 10 cm or greater clearance from the edges. The density of steel of 2.05% is greater than the recommended value of 0.1% of steel needed to prevent cracking from ACI committee 360, so this design is good.

Pump Station Construction Notes

a. Concrete skirt

The purpose of the concrete skirt is to prevent the slab from being undermined by erosion. The subterranean concrete skirt will be poured before the slab is poured and the slab rebar is placed. Ensure that the rebar that is inside the CMU walls is vertically placed in the concrete skirt and sticking out of the ground and to the proper measurements. This rebar will help secure the skirt and the CMU walls.

b. Pump platform

Ensure that rebar is sticking up in the middle of the slab before pouring. Also ensure that this rebar sticks out at least 0.45 meters from the surface of the slab. The pump will be placed on a raised concrete platform, a 1.00x0.50x0.50 meter block of concrete. This platform will be poured separately from the underlying slab, so it needs to be secured to the slab through the rebar. This rebar will secure the platform, and therefore the pump, to the slab, and the rebar will minimize cracking so that the pump remains level.

c. Thrust block

The thrust block is a block of concrete placed around the pipes in front of the discharge end of the pump so that the pipes are protected from hydraulic forces. The thrust block adds mass to the

pipes, so for a given force they are deflected a shorter distance. This reduction in deflections reduces the stress on the pipes and reduces the risk of the pipes breaking. It is poured after the slab is poured and after the pipes are set.

d. Top of wall slope

After the wall is complete, ensure that the top of the wall has a slope. This can be achieved by mixing concrete and placing it on the top of the CMU wall. Make sure the 0.2 meter high section is on the side of the facility facing uphill and that the top of the wall slope is in the same direction as the surrounding terrain. This ensures that rain runs off of the roof and away from the building and not back towards the foundation.

e. CMU wall and roof interface

Ensure that the rebar that is placed in the CMU walls penetrates through the top of the wall and the sloped concrete layer. This allows the welders to weld metal beams directly onto the structure. These beams go across the roof, which provides an anchor point for the sheet metal roof.

Storage Facility Concrete Cross Sections

a. Design slab thickness:

The thickness of the slab will be controlled by the shear forces it needs to resist. From our previous calculations, we know the soil is more than strong enough to support our structures. However, the soil is expansive, and will expand and contract based on precipitation in the area. The slab will trap moisture in the soil below it, and that soil will remain at a relatively constant moisture content, while the soil at the edges of the slab will quickly lose moisture in the dry season and shrink, and will quickly absorb moisture in the wet season and expand. This leads to two critical conditions for the slab, called dishing and doming, or also called edge list condition and center lift condition. Dishing occurs in the wet season when the slab is only supported by the expanded soil on the edges. Doming occurs in the dry season when the slab is only supported by the still moist soil under the slab. For our calculations, we will treat the slab as a wide beam. Additionally, we will assume worse case soil expansion and contraction. We will assume the slab dimensions of 3.15 x 3.15 due to the amount of spacing needed for the water tank.

Overhead view of forces:

Doming force diagram:

Dishing force diagram:

Table of ConstraintsDimensions of Slab 3.15 x 3.15 metersMass of water tank 22000 kgStrength of concrete, f’c 13.8 MPaShear Strength Safety Factor 0.75Acceleration due to gravity (g) 9.81 m/s2

i. Doming slab thickness calculation:

Concrete shear strength∈newtons per meter squared=2√ f ' c

6 (1000000)

The slab will be designed to have no vertical web reinforcement. By ACI code, the design shear strength of concrete without web reinforcement is (Design shear stress)=(0.5)(safety factor)(concrete shear strength)

The maximum shear will occur in the center of the slab, where the soil supports the slab.

(0.5 ) ( safety factor ) (Concrete shear strength ) (width of slab ) (height of slab )=distributed force of slabweight+distributed force of water tank

distributed force of slabweight=(width of slab)(height of slab )(density of concrete )(g)(lengthof slab)/2

distributed force of slabweight= (3.15 ) (height of slab ) (2400 ) (9.81 ) (3.15 )2

=116808 (height of slab)

distributed force of water tank=(massof tank)(g)/2

distributed f orce of water tank= (22000 ) (9.81 )2

=107910newtons

(0.5 ) (0.75 )( 2√13.86 ) (1000000 ) (3.15 ) (height of slab )=116808 (height of slab )+107910


ii. Dishing slab thickness calculation:

The maximum shear will occur at the left and right of the slab, where it is supported by the soil. Calculations will find shear at the left support.

(0.5 ) ( safety factor ) (Concrete shear strength ) (width of slab ) (height of slab )=distributed force of slabweight+distributed force of water tank−¿ reaction force

distributed force of slabweight=(width of s lab)(height of slab)(density of concrete )(g)(length of slab)

distributed force of slabweight= (3.15 ) (height of slab ) (2400 ) (9.81 ) (3.15 )=233615 (height of slab)

distributed force of water tank=(massof tank)(g)

distributed force of water tank=(22000 ) (9.81 )=215820newtons

¿reaction force= totaldownward force2

¿ reaction force=233615 (height of slab )+2158202

¿reaction force=107910+116808 (height of slab)

(0.5 ) (0.75 )( 2√13.86 ) (1000000 ) (3.15 ) (height of slab )=233615 (height of s lab )+215820−107910−116808 (height of slab)


iii. Final slab thickness

Both cases result in the same thickness slab. Therefore, the slab must be at least .176 meters thick. For ease of construction, the slab will be designed as 0.18 meters thick

b. Design moment calculation:

The slab is relatively thin, so bending moments in the slab should be considered for the dishing and bowling case. These moments will be used to calculate the needed steel reinforcement. To calculate the design moment, we need to determine what part of the slab will experience the most moment. We will assume the slab dimensions of 3.15 x 3.15 x 0.18 meters as given, due to the

amount of spacing needed for the pump and valves. The previous figures and table of constraints will be used.

i. doming moment calculation

In the doming case, the slab acts like a cantilever, so maximum moment will be at the center of the slab.Bendingmoment=( force¿water tank)(centroid distance)+(force¿ slabweight)(distance )

From doming shear calculationsforce¿water tank=107910newtons

force¿ slab weight=116808 (0.18 )=21025ne wtons

Bendingmoment=(107910 ) (0.6366 )+(21025 )( 1.5752 )

Bendingmoment=85253newtonmeters


Designmoment=85253∗1.2=102303 N∗m

ii. dishing moment calculation

In the dishing case, the slab acts like a simply supported beam, so maximum moment will be at the center of the slab.Bendingmoment=¿of water tank ¿¿ ( centroid distance )+¿weight ¿¿ (distance )−(¿ reaction force)(distance)

From dishing shear calculations

force¿¿half of water tank=2158202

=107910newtons

force¿¿ part of slab=233615 (0.18)2

=21025newtons

¿reaction force=107910+116808 (0.18 )=128935newtons

Bendingmoment=(107910 ) (0.6366 )+(21025 )( 1.5752 )−128935(1.575)

Bendingmoment=−117820newtonmeters


Designmoment=117820∗1.2=141384 N∗m

c. Rebar design calculations

For rebar design, Whitney Stress Block theory will be used. We need to solve for the area of steel to be used. Tensile stresses will be at the top of the slab during doming, and tensile stresses will be at the bottom of the slab during dishing. Due to the narrow slab and the existence of positive and negative moments, 2 rows of steel will be used. We will assume the bars are placed 5 cm from the faces of the slab Design InputsConcrete Design Compressive Stress (f'c) 13.8 MPaYield Stress of steel (fy) 420 MPaBase (b) 0.18 meterDishing Rebar Design Moment 141384 N-mDoming Rebar Design Moment 102303 N-mSafety Factor (phi) 0.9Depth (d) 0.13 meterWidth 3.15 meterModulus of elasticity of steel (Es) 200 GPaa to c ratio (β) 0.85

Dishing force diagram Doming force diagram

βc=a

Dishing force equilibrium:concrete compressive force+steel compressive force=steel tensile force( f 'c )(βc1)(width)+(AsTop)(Es)(.003)(c1−.05) /c1=(AsBottom)( fy)

(13800000)(0.85∗c1)(3.15)+(AsTop)(200000000000)(.003)(c1−.05) /c1=(AsBottom)(420000000)

(36949500)(c1)+(AsTop)(600000000)(c1−.05)/c 1=(AsBottom)(420000000)

Doming force equilibrium:concrete compressive force+steel compressive force=steel tensile force( f 'c )(βc2)(width)+(AsBottom)(Es)( .003)(c2−.05)/c 2=(AsTop )(fy)

(13800000)(0.85∗c2)(3.15)+(AsBottom)(200000000000)( .003)(c2−.05)/c 2=(AsTop )(420000000)

(36949500)(c 2)+(AsBottom)(600000000)(c 2−.05)/c2=(AsTop)(420000000)

Dishing Moment Equilibrium:

( f 'c ) (βc1 ) (width )(d−βc 12 )+(AsTop)(Es)(.003)(c 1−.05)(.05)/c1=dishmoment

safety factor

(13800000)(0.85∗c1)(3.15)(.13−0.85∗c12 )+(AsTop)(200000000000)(.003)(c1−.05)(.08)/c 1=141384

0.9(36949500)(c1) (.13−.425(c1))+(AsTop)(48000000)(c1−.05)/c 1=157093

Doming Moment Equilibrium:

( f 'c ) (βc2 ) (width )(d−βc 22 )+(AsBottom)(Es)(.003)(c 2−.05)(.08)/c2=domemoment

safety factor

(13800000)(0.85∗c2)(3.15)(.13−0.85∗c 22 )+(AsBottom)(200000000000)( .003)(c2−.05)(.05)/c 2=102303

0.9(36949500)(c 2) (.13−.425 (c2))+(AsBottom)(48000000)(c2−.05)/c 2=113670

With four equations and four unknowns, c1, c2, AsTop, and AsBottom, these values can be determined. Using Matlab:AsTop = .0020 m2

AsBottom = .0032 m2

Total Density of steel= AsTop+AsBottomCross sectional area of concrete

= .0020+.0032.18∗3.15

=0.917 %

Areaof steelneeded alongtop of both sides=.0020m2

Areaof steelneeded alongbottom of bothsides=.0032m2


number of ¿4 rebar topboth sides= .0020.000129032

=16

number of ¿4 rebar bottomboth sides= .0032.000129032

=25


number of ¿5 rebar topboth sides= .0020.0002

=10

number of ¿5 ℜ ¯bottom bothsides= .0032.0002

=16

Calculations such as this can be easily repeated for other rebar sizes. Selection of rebar will depend on local prices of steel. Drawings will be for #4 rebar, due to its known price in Nicaragua. Rebar will be equally spaced, with 10 cm or greater clearance from the edges. The density of steel of 0.917% is greater than the recommended value of 0.1% of steel needed to prevent cracking from ACI committee 360, so this design is good.

d. Landslide Risk Analysis

This storage facility will be placed on a sloped hill. The construction area will be flattened prior to construction, but a mass of soil will sit uphill of the facility. In heavy rains some of the hillside may flow downhill and impact the walls of the storage facility.

These calculations determine the resistance of the walls to a mass of fluid soil pushing against it. Specifically these calculations will determine the maximum height of flowing soil that the wall can resist. If soil is determined to be flowing near this height, then the wall must be reinforced and braced at the expected centroid of the fluid force.

Calculation InputsDensity of Soil 9270 N/m^3Yield Stress of Steel (fy) 420 MPaWidth of Wall 5.6 meterHeight of Wall 2.4 meterThickness of Wall 0.15 meterDepth of rebar in wall 0.075 meter

i. Moment Strength of the CMU wall

Cross section of wall:

Number of rebar = 12 #4 rebarAreaof steel=2.36¿2=.00152m2

Using Whitney Stress Block theory

Maximum force∈steel∗distance=moment

( fy)(Area of steel )(d−a2)=moment

a= (fy)(Area of steel)(0.85)( f ’ c)(width)

( fy)(As)(d−( fy)(Area of steel )(1.7)( f ’c )(width)

)=moment

(420000000)(.00152)(.075−(420000000)(.00152)(1.7)(21000000)(5.6)

)=Moment

Moment=45844N∗mii. Height of fluid soil needed to break the wall

Assuming the soil is perfectly fluid, then as it accumulates up against a wall segment it will exert a hydrostatic pressure against the wall. This pressure then exerts a moment on the fixed end of the wall in the ground.

Force diagram:

Force=(Average pressure)∗(ar ea)

Force=(unit weight )(height )2

∗(height )(Width)

Moment=Force∗distance

Moment=(unit weight)(height )2

∗(height )(Width)∗(height3

)

45844=(9270)(h)2

∗(h) (5.6)∗h3

h=1.74meters of fluid soilcentroid of force=1.74 /3=0.58meters

Storage Facility Construction Notes

a. CMU wall vertical rebar

Unlike the pump station rebar, there will be 2 pieces of rebar in each indicated cell. These walls are longer and taller than the pump station so the extra rebar is to provide greater strength.

b. CMU wall horizontal rebar

At specific heights horizontal rebar will be placed in the wall. This horizontal rebar prevents the wall from bending around a vertical axis. The CMU blocks must have notches cut into them so that the rebar can fit. Also, concrete must be poured around the rebar to secure it to the CMU wall.

c. CMU wall gaps

The wall is designed to have periodic vertical gaps where CMU blocks are not mortared together. A long continuous wall will randomly crack due to shrinkage, moisture, and thermal effects. Random cracks could compromise the structural strength of the wall and should be avoided. The designed gaps break the wall up into shorter segments that are less likely to randomly crack. Also, these gaps act like controlled cracks so we know where the weaknesses are located and we know where to place reinforcement.

d. Landslide risk

The walls as designed can withstand a landslide of perfectly fluid soil that is 1.74 meters tall. If soil accumulates past this height then additional reinforcement of the walls will be needed. This can be accomplished by adding steel reinforcement or steel bracing on the inside of the structure. It must be attached to a point on the wall that is 0.6 meters or higher off the ground to be effective because this is the height where the centroid of the landslide force will occur.

appendix b_ calculations_ structures portion (1-17-2015)

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