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Powerhouse Design- Natick Dam Web Page Design of Powerhouse Substructure W. Fay P.E. Natick Hydroelectric Project 3/28/90 West Warwick, Rhode Island FERC License Project No. 3013-RI
Stamp I) Methodology:
A) Floor slabs to be designed as one-way floor slabs using the "ACI-Ultimate Strength Design Methods"
B) Walls to be designed as one-way floor slabs using the "ACI-Ultimate Strength Design Methods" and assuming hydrostatic load distributions for saturated soils.
C) Beams to be designed using ASI Code.
II) Design Approach:
A) Design assumes waterbox floor slab is supported on three sides by reinforced concrete walls and by a steel beam on the downstream, tailrace side. Additionally, the discharge pit will be subdivided into two equal sized compartments by a wall running in the streamwise direction and this reinforced wall will support the midspan of the waterbox floor.
Sketch:
B) Waterbox floor slab loads will consist of uniformly distributed dead loads of water, rebar and concrete and the point loads of the turbine gate cases. The design will incorporate four (4) transverse beams, one on each side of the gate cases to support the gate case point load.
C) The discharge pit walls will be designed for external loads for when the pit may occasionally be dewatered by a coffer dam to maintain the draft tubes.
(1) D) Waterbox walls will be designed for the internal pressure due to the headwater and for external pressure due to dewatering and emergency repair work during highwater.
E) Generator floor slabs will be designed for a uniform floor loading of 300 lb/sq.ft and for the point loads of the generator, turbine shaft, turbine runner and hydraulic thrust. This point load will be supported by transverse beams similiar to B) above. These beams will be designed according to ASI Code to have proper strength, with minimal deflection.
F) The upstream side of the generating floor will be supported by a steel beam spanning the waterbox inlets.
G) The waterbox will be divided into two (2) seperate chambers by a structural wall.
H) The main steel beam supporting the waterbox floor slab will be designed conservatively to support the weight of the waterbox rear wall, powerhouse rear wall and one half (1/2) the weights of the waterbox floor slab, the generating room floor slab, the turbine/generator hydraulic thrust point load and the overhead crane point load (assumed to be in the most compromising position).
I) The main steel beam supporting the generating room floor slab and spanning the waterbox inlets will be designed conservatively to support the weight of the generating room upstream wall and one half (1/2) the weights of the generating room floor slab, the turbine/generator hydraulic thrust point load and the overhead crane point load (assumed to be in the most compromising position).
J) The following elevations will be assumed:
****(Datum is USGS-MSL)****
1) Top of discharge pit floor slab - 21.0 msl
2) Mean tailwater elevation - 31.5 msl
3) Waterbox floor - 40.0 msl
4) Mean headwater elevation - 49.1 msl
5) Top of generating room floor slab - 54.0 msl
Assume main beams forming the discharge pit/tailrace and water box inlet arches are WF-30"x 124 #/ft.
(2)
III) Design of the Waterbox Floor Slab:
A) Dimensions- from Fay Engineering Services layout dated 12/27/89 entitled "Powerhouse Longitudinal Section". Assume these dimensions are correct.
1) External dimensions:
a) length = 40.0'
b) width = 18'
2) Internal dimensions:
a) length = 36.0'
b) width = 16.0'
B) Loading:
1) Water, total load = 62.4 #/ft3 * 16' * (38'* (50.5-40.0))
= 398,361 lbs
2) Water, load per foot width of box = 10,483 lbs/ft-width of flume
3) Water, load per foot width of box per foot width of beam = 655 lb/ft
4) Concrete, total load = 150 #/ft3 * 16' * 38.0'*
(9"/(12"/ft)) = 68,400 lbf
5) Concrete, load per foot width of box = 68,400 lbf/38.0' = 1800 lbf/ft width of box
6) Concrete, load per foot width of box per foot width of beam =
1800 lbf/ft-width/16' = 113 lbf/ ft
7) Since the waterbox will be periodically dewatered, treat the water as a live distributed load and use the larger live load overload factor ACI ultimate strength multipliers.
C) Sketch of typical section through proposed waterbox floor:
(3) D) Design Calculations: - reference, "Design of Concrete Structures", 9th ed., Winter & Nilson.
1) Assume the yield strength of the steel reinforcing is 30,000 psi and that the compressive strength of the concrete is 3000 psi.
2) Select the trial thickness of the slab, use L/20 from Table 5.1, p.206 in Winter & Nilson.
T= (12 "/ft * 18')/20 = 10.8" approximately = 11"
3) The slab weight is 150 #/ft3 * (11/12) = 138 PSF
4) Apply the ACI load multipliers and obtain the factored load:
Dead Load = 113 PSF * 1.4 = 158.2 PSF
Live Load = 655 PSF * 1.7 = 1113.5 PSF
Total Factored Load = 1272 PSF
5) Use the ACI moment coefficients to determine the design moments at the critical sections:
a) Since the floor slab is being designed as a one-way slab in the short direction (ie: from the inlet end to the tailrace end), the slab will be resting on the main 30" beam which acts as the arch at the rear of the poerhouse over the tailrace and will be built into the top of the rear (upstream wall) of the discharge pit. At the tailrace end, the floor slab is simply supported and the beam is free to twist and cannot be assumed to be rigid, so use:
(1/11)*Wu*ln^2 >>>>>>>>from Table 8.1 W&N
Sketch:
b) At the upstream end, the slab is to be built into the discharge pit wall and can be assumed to be rigid, so use:
(1/14)*Wu*ln^2>>>>>>>>from Table 8.1 W&N
Sketch:
(4) c) At the interior span use:
(1/14)*Wu*ln^2>>>>>>>>from Table 8.1 W&N
d) At the tailrace: -M = 1/11*1.27 KSF *18'^2 = 37.4 ft-kips
e) At the upstream end: -M = 1/14*1.27 KSF *18'^2 = 29.4 ft-kips
d) At the midspan: -M = 1/16*1.27 KSF *18'^2 = 25.7 ft-kips
5) Determine the maximum steel ratio permitted by the ACI Code:
Pmax = 0.75*Pbalanced = 0.75*0.85*B1*(fc'/fy)*(87,000/(87,000+fc'))
this formula for fc'<4000 psi and B1=0.85
Pmax=0.75*0.85*0.85*(3000 psi/30,000 psi)*(87,000/(87,000 + 30,000))
= 0.04
6) Determine the minimum required effective depth: (This is controlled by the largest moment at the tailrace)
d^2 = Mu/(phi*p*fy*b*(1-(0.59*p*fy/fc))) note: phi=0.9 for bending
= (37.4 ft-kips*(12"\ft))/(0.9*.04*30*12*(1-.59*0.04*(30,000/3000)))
= 45.3 in^2
Therefore, d = 6.7 inchs
7) Determine the minimum effective depth using code restrictions:
dm= 11"- 1" = 10"
8) Since the calculated value of 6.7 inches is less then the coded effective depth, use d= 10 inches
9) At the tailrace end, assume the stress block depth a = 1.00 inch. Then the area of steel required per foot width in the top of the slab is:
As= Mu/(phi*fy*(d-a/2))= (37.4 ft-kips*12"/ft)/0.9*30*(10-1/2)=
1.75 in^2
10) Check the assumed depth:
a= As*fy/(0.85*fc'*b)= 1.75 in^2 *30,000/(0.85*3000*12"/ft)= 1.72 in
11) The assumed area of steel and the calculated area of steel are reasonably close so use 1.72 in^2 of rebar per foot width of floor slab.
(5) 12) At the other critical sections use the same lever arm to determine the required cross sectional areas of steel rebar:
a) at the midspan: As= 25.7 ft-kips*12/(0.9*30*(11-1.72/2))= 1.13 in^2
b) at the upstream wall: As= 29.4 ft-kips*12/(0.9*30*(11-1.72/2))= 1.29 in^2
13) The minimum reinforcement required to control shrinkage is: see p. 207, W&N.
As= 0.002*12*11= 0.26 in^2/ 12" with strip
The required steel necessary for shrinkage is met by the steel required to meet the externally applied loads.
14) Determine the factored shear force:
Vu= 1.15 * (1272*18/2)-1272*(11/12)= 13,165-1166 = 11,999 lbs
15) The nominal shear strength of the slab is:
Vn= Vc= 2*b*d*fc'^0.5
= 2*12*11*(3000 psi^0.5)
= 14,460 lbs
16) The design shear strength is:
phi*Vc= 0.85*14,460 lbs= 12,291 lbs
17) Since the design shear strength is above the required shear strength, no additional steel is necessary to resist the internal shear forces.
IV) Design the Steel Main Support Beam at the Tailrace
A) Data for proposed Beam:
1) height= 30" weight= 124 lbs/ft-width length= 55 feet section modulus= 355 in^3 moment of inertia= 5360 in^4 depth= 30.16" width= 10.521" flange thickness= 0.93" web= 0.585"
2) Flat of flange= (10.521"-0.585")/2= 4.97" 90 % of the flat is 4.5"
(6) B) Determine the Beam Loading:
1) Point Loads:
a) generator= 225 rpm, 32 pole, 197 kva= 15,800 lbs
b) exciter= 1400 lbs
c) Shipping weight of turbine= 14,000 lbs
d) Hydraulic thrust= 21,636 #
2) Crane Load - assume a point load in the middle of the beam of 5 tons for a five ton bridge crane, this gives a 2.0 safety factor on the crane load.
3) Total point load is:
(15,800 # + 1400 # + 14,000 # + 21636 #)/2 + 5000 #= 31,418 lbs
4) Dead Loads:
a) Powerhouse backwall:
(40'*((69.0-54.0)*0.67) + 38*((53.0-40.0)*1.0))*150 lbs/ft^3= 134,400 #
b) Waterbox floor:
(18'*40'*1')*150 lbs/ft^3= 108,000 #
c) Generating room floor:
(18'*40'*1')*150 lbs/ft^3= 108,000 #
d) Water:
18'*34'*(50.5-31.5)*62.4 lbs/ft^3= 725,587 lbs
5) Total Distributed Load:
(134,400 # + 31,418 # + 216,000 #/2 + 725,587 #/2)/40'= 15,915 lbs/ft-width beam
6) Sketch the beam and the loading diagram:
(7) C) Determine the extreme fiber stress and the maximum deflection:
1) Even though the beam will be embedded into the walls on either side, the weight of the wall immediately above the beam will not counter act the loading and one should assume that the beam is simply supported. This is a conservative assumption and will result in the largest moment.
2) Maximun Moment:
15,915 lb/ft * (18.0'^2)/8 = 644,569 ft-lbs
3) The extreme fiber stress is:
Sigma = M/Z = (644,569 ft-lbs * 12"/ft)/355 in^3 = 21,788 psi
4) From the AISC "Manual of Steel Construction", 7th edition, P. 5-124, section 1.5.1.4.1, the allowable bending stress for W shapes is 0.66*Fy, where Fy is 36,000 psi.
Therefore, the recommended maximum load is:
Sigma max= 0.66*36,000 psi= 23,760 psi <<<<<<<<*****
15) Since the calculated stress is less then the maximum recommended by AISC Code, the 30" beam is alright for strength.
16) Determine the maximum deflection:
a) For uniformly distributed load with built in ends:
delta max= w*l^4/(384*E*I)
= 15,915 lb/ft * (18.0'*12")^4/(384*29,000,000 psi* 5360 in^4*12)= 0.58 inch
17) The calculated deflection is negligible and should not effect the concrete resting upon the beam, especially since the major deflection will take place when the concrete is wet, before it dries.
(8) V) Design the turbine gate case support beams:
A) Sketch the design:
Plan View
B) Sketch the beam loading:
C) Determine the beam size: 1) Size the beam at 80 % of the AISC Codes maximum allowable stress or
0.8 * 21,600 psi= 17,280 psi 2) Maximum moment is 8' * 3500 # = 28,000 ft-lbs
3) Z= M/sigma= (28,000 ft-lbs * 12 in/ft)/17,280 psi= 19.5 in^3
D) From the AISC Handbook, 7th ed., P. 1-42, tentatively choose a W8"x24 lbs/ft
E) Beam Properties:
height= 7.93" weight= 24 lbs/ft-width length= 18 feet section modulus= 20.8 in^3 moment of inertia= 82.5 in^4 depth= 7.93" width= 6.50" flange thickness= 0.398" web= 0.245" (9)
F) Check the deflection:
Sigma max= P*L^3/(48*E*I)= 3500# * (15.5'*(12"/ft)^3)/(48 * 29,000,000 psi * 82.5 in^3= 0.00006 inch <<<< okay
The choosen beam will meet the design requirements.
VI) Design the generator support beams:
A) Sketch the design:
Plan View
B) Sketch the beam loading:
C) Determine the beam size:
1) Size the beam at 80 % of the AISC Codes maximum allowable stress or
0.8 * 21,600 psi= 17,280 psi
2) Maximum moment is 9.0' * (21,636 #+ 17,600 #)/2 = 176,562 ft-lbs
3) Z= M/sigma= (176,562 ft-lbs * 12 in/ft)/17,280 psi= 123 in^3
D) From the AISC Handbook, 7th ed., P. 1-42, tentatively choose a W12"x92 lbs/ft
(10) E) Beam Properties:
height= 12.62" weight= 92 lbs/ft-width length= 18 feet section modulus= 125 in^3 moment of inertia= 789 in^4 depth= 12.62" width= 12.155" flange thickness= 0.856" web= 0.545"
F) Check the deflection:
Sigma max= P*L^3/(48*E*I)= 19618# * (18'*(12"/ft)^3)/(48 * 29,000,000 psi * 125 in^3= 1.1 inch <<<< okay
The chosen beam will meet the design requirements. The deflection is high, but in reality, the beams will be supported along their edges by the rigid generating room floor
VII) Design the generator floor:
A) Dimensions- - from Fay Engineering Services layout dated 12/27/89 entitled "Powerhouse Longitudinal Section". Assume these dimensions are correct.
1) External dimensions:
a) length = 40.0'
b) width = 18'
2) Internal dimensions:
a) length = 36.0'
b) width = 16.0'
B) Loading:
1) Concrete, total load = 150 #/ft3 * 16' * 36'*
(12"/(12"/ft)) = 86,400 lbf
2) Concrete, load per foot width of box = 86,400 lbf/36' = 2400 lbf/ft width of box
3) Concrete, load per foot width of box per foot width of beam =
2400 lbf/ft-width/16' = 150 lbf/ ft
(11) C) Sketch of typical section through proposed generator floor:
D) Design Calculations: - reference, "Design of Concrete Structures", 9th ed., Winter & Nilson.
1) Assume the yield strength of the steel reinforcing is 30,000 psi and that the compressive strength of the concrete is 3000 psi.
2) Select the trial thickness of the slab, use L/20 from Table 5.1, p.206 in Winter & Nilson.
T= (12 "/ft * 16')/20 = 9.6" approximately = 12"
3) The slab weight is 150 #/ft3 * (12/12) = 150 PSF
4) Apply the ACI load multipliers and obtain the factored load:
Dead Load = 150 PSF * 1.4 = 210.0 PSF
Live Load = 300 PSF * 1.7 = 510.0 PSF
Total Factored Load = 720 PSF
5) Use the ACI moment coefficients to determine the design moments at the critical sections:
a) Since the floor slab is being designed as a one-way slab in the short direction (ie: from the inlet end to the tailrace end), the slab will be resting on the main 30" beam which acts as the arch at the front of the powerhouse, over the inlets and will be built into the top of the rear (downstream wall) of the water boxes. At the inlet end, the floor slab is simply supported and the beam is free to twist and cannot be assumed to be rigid, so use:
(1/11)*Wu*ln^2 >>>>>>>>from Table 8.1 W&N
Sketch:
(12)
b) At the upstream end, the slab is to be built into the discharge pit wall and can be assumed to be rigid, so use:
(1/14)*Wu*ln^2>>>>>>>>from Table 8.1 W&N
Sketch:
c) At the interior span use:
(1/16)*Wu*ln^2>>>>>>>>from Table 8.1 W&N
d) At the inlets: -M = 1/11*0.720 KSF *14'^2 = 12.8 ft-kips
e) At the downstream end: -M = 1/14*0.720 KSF *14'^2 = 10.1 ft-kips
f) At the midspan: -M = 1/16*0.720 KSF *14'^2 = 8.8 ft-kips
5) Determine the maximum steel ratio permitted by the ACI Code:
Pmax = 0.75*Pbalanced = 0.75*0.85*B1*(fc'/fy)*(87,000/(87,000+fc'))
this formula for fc'<4000 psi and B1=0.85
Pmax=0.75*0.85*0.85*(3000 psi/30,000 psi)*(87,000/(87,000 + 30,000))
= 0.04
6) Determine the minimum required effective depth: (This is controlled by the largest moment at the inlet end)
d^2 = Mu/(phi*p*fy*b*(1-(0.59*p*fy/fc))) note: phi=0.9 for bending
= (12.8ft-kips*(12"\ft))/(0.9*0.04*30*12*(1-.59*0.04*(30,000/3000)))
= 15.51 in^2
Therefore, d = 3.9 inchs
7) Determine the minimum effective depth using code restrictions:
dm= 12"- 1" = 11"
8) Since the calculated value of 4.7 inches is less then the coded effective depth, use d= 11 inches
(13) 9) At the tailrace end, assume the stress block depth a = 1.00 inch. Then the area of steel required per foot width in the top of the slab is:
As= Mu/(phi*fy*(d-a/2))= (12.8 ft-kips*12"/ft)/.9*30*(11-1/2)= 0.54 in^2
10) Check the assumed depth:
a= As*fy/(0.85*fc'*b)= 0.54 in^2 *30,000/(0.85*3000*12"/ft)= 0.53 in
11) Reiterate assuming a=0.53 in:
As=Mu/(phi*f*(d-a/2))=(12.8 ft-kips*12"/ft)/0.9*30*(11-.53/2)= 0.52in^2
12) Reiterate assuming As=0.52 in^2:
a= As*fy/(0.85*fc'*b)= 0.52 in^2 *30,000/(0.85*3000*12"/ft)= 0.51 in
13) Reiterate assuming a=0.51 in:
As=Mu/(phi*f*(d-a/2))=(12.8 ft-kips*12"/ft)/0.9*30*(11-0.51/2)= 0.53 in^2
14) The assumed area of steel and the calculated area of steel are reasonably close so use 0.53 in^2 of rebar per foot width of floor slab.
15) At the other critical sections use the same lever arm to determine the required cross sectional areas of steel rebar:
a) at the midspan: As= 10.1 ksi*12/(0.9*30*(11-0.51/2))= 0.42 in^2
b) at the upstream wall: As= 8.8 ksi*12/(0.9*30*(11-0.51/2))= 0.36 in^2
16) The minimum reinforcement required to control shrinkage is: see p. 207, W&N.
As= 0.002*12*12= 0.288 in^2/ 12" with strip
The required steel necessary for shrinkage is met by the steel required to meet the externally applied loads.
17) Determine the factored shear force:
Vu= 1.15*(720*18/2)-720*(12/12)= 7452-720 = 6372 lbs
15) The nominal shear strength of the slab is:
Vn= Vc= 2*b*d*fc'^0.5
= 2*12*12*(3000 psi^0.5)
= 15,774 lbs
(14)
16) The design shear strength is:
phi*Vc= 0.85*15,774 lbs= 13,408 lbs
17) Since the design shear strength is above the required shear strength by 15 %, no additional steel is necessary to resist the internal shear forces.
VIII Design Waterbox Walls:
A) Dimensions- - from Fay Engineering Services layout dated 12/27/89 entitled "Powerhouse Longitudinal Section". Assume these dimensions are correct.
1) External dimensions:
a) height= 14'
b) width = 18'
2) Internal dimensions:
a) length = 14'
b) width = 18'
B) Loading:
1) Concrete, total load = 150 #/ft3 * 14' * 18'* (12"/(12"/ft)) = 37,800 lbf
2) Concrete, load per sq. ft. = 37,800 lbf/40' = 150 lbf/ft^2
3) Water Load= 62.4 lbf/ft^3 * 14' wide * 18' deep * 1' thick/ 14'wide = 1123 PSF. This is the maximum load at the base of the hydrostatic load.
C) Sketch of typical section through proposed waterbox wall:
(15) D) Design Calculations: - reference, "Design of Concrete Structures", 9th ed., Winter & Nilson.
1) Assume the yield strength of the steel reinforcing is 30,000 psi and that the compressive strength of the concrete is 3000 psi.
2) Select the trial thickness of the slab, use L/20 from Table 5.1, p.206 in Winter & Nilson.
T= (12 "/ft * 18')/20 = 10.8" approximately = 11"
3) The slab weight is 150 #/ft3 * (12/12) = 150 PSF
4) Apply the ACI load multipliers and obtain the factored load:
Dead Load = 150 PSF * 1.4 = 210 PSF
Live Load = 1123 PSF * 1.7 = 1910.0 PSF
Total Factored Load = 2120 PSF
5) Use the ACI moment coefficients to determine the design moments at the critical sections:
a)The wall is rigidly built into the floor slabs at the top and bottom, so use:
(1/14)*Wu*ln^2 >>>>>>>>from Table 8.1 W&N
Sketch:
b) At the interior span use:
(1/16)*Wu*ln^2>>>>>>>>from Table 8.1 W&N
c) At the top: -M = 1/14*2.2 KSF *14'^2 = 30.8 ft-kips
d) At the bottom: -M = 1/14*2.2 KSF *14'^2 = 30.8 ft-kips
e) At the midspan: -M = 1/16*2.2 KSF *14'^2 = 27.0 ft-kips
5) Determine the maximum steel ratio permitted by the ACI Code:
Pmax = 0.75*Pbalanced = 0.75*0.85*B1*(fc'/fy)*(87,000/(87,000+fc'))
this formula for fc'<4000 psi and B1=0.85 (16) Pmax=0.75*0.85*0.85*(3000 psi/30,000 psi)*(87,000/(87,000 + 30,000))
= 0.04
6) Determine the minimum required effective depth: (This is controlled by the largest moment at either the top or the bottom)
d^2 = Mu/(phi*p*fy*b*(1-(0.59*p*fy/fc))) note: phi=0.9 for bending
= (30.8ft-kips*(12"\ft))/(0.9*0.04*30*12*(1-.59*0.04*(30,000/3000)))
= 35.2 in^2
Therefore, d = 5.9 inchs
7) Determine the minimum effective depth using code restrictions:
dm= 12"- 1" = 11"
8) Since the calculated value of 5.9 inches is less then the coded effective depth, use d= 11 inches
9) At the tailrace end, assume the stress block depth a = 1.00 inch. Then the area of steel required per foot width in the top of the slab is:
As= Mu/(phi*fy*(d-a/2))= (30.8 ft-kips*12"/ft)/.9*30*(11-1/2)= 1.3 in^2
10) Check the assumed depth:
a= As*fy/(0.85*fc'*b)= 1.3 in^2 *30,000/(0.85*3000*12"/ft)= 1.27 in
11) Reiterate assuming a=1.27 in:
As=Mu/(phi*f*(d-a/2))=(30.8 ft-kips*12"/ft)/0.9*30*(12-1.27/2)= 1.2 in^2
12) Reiterate assuming As=1.20 in^2:
a= As*fy/(0.85*fc'*b)= 1.20 in^2 *30,000/(0.85*3000*12"/ft)= 1.18 in
13) Reiterate assuming a=1.18 in:
As=Mu/(phi*f*(d-a/2))=(30.9 ft-kips*12"/ft)/.9*30*(12-1.18/2)= 1.20in^2
14) The assumed area of steel and the calculated area of steel are reasonably close so use 1.20 in^2 of rebar per foot width of floor slab.
15) At the other critical sections use the same lever arm to determine the required cross sectional areas of steel rebar:
a) at the midspan: As= 30.8 ksi*12/(0.9*30*(12-1.18/2))= 1.20 in^2
b) at the bottom: As= 27 ksi*12/(0.9*30*(12-1.18/2))= 1.05 in^2
(17) 16) The minimum reinforcement required to control shrinkage is: see p. 207, W&N.
As= 0.002*12*12= 0.288 in^2/ 12" with strip
The required steel necessary for shrinkage is met by the steel required to meet the externally applied loads.
17) Determine the factored shear force: Note that the dead weight of the vertical concrete does not add to the shear component.
Vu=1.15*1910*14/2-2120*(12/12)= 15,375-2120 = 13,255 lbs
15) The nominal shear strength of the slab is:
Vn= Vc= 2*b*d*fc'^0.5
= 2*12*12*(3000 psi^0.5)
= 15,774 lbs
16) The design shear strength is:
phi*Vc= 0.85*15,774 lbs= 12,897 lbs
17) The design shear strength is slightly less then the required shear strength. However, the differential is small and no additional steel is necessary to resist the internal shear forces.
18) Use 1.2 in^2 of steel for the vertical reinforcement.
IX) Design the discharge pit walls:
A) Assume the pit is dewatered and drained externally. The height of water is 31.5-20.5=11'
B) This design is similiar to the waterbox walls. < |