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.

 

C) Use 1.2 in^2 of steel per foot width of wall.

 

 

X) Design main 30" beam support column:

 

A) Determine the load on the column:

 

1) Sketch the freebody diagram:

 

 

 

 

 

 

 

 

 

                                  (18)

Ra=((20,600 lbs + (16,000*18'))/2 = 154,300 lbf

 

2) the factored ACI Code load is:

 

Pu=((1.4*(16,000 lbf * 18')/2) + (1.7 * 20,600/2)

 

 = 201,600 lbf + 17,510 lbf = 219,110 lbf

 

B) Determine the nominal axial load strength of the column, Po,

assuming minimal eccentricity:

 

1) Po= 0.85 * fc' * Aconc + fy * Ast

 

2) By ACI Code, the ratio of the longitudinal steel area to the gross

column area must be:

 

0.01 <= Pg <= 0.08. Try 0.025 to start.

 

3) Assume the column is 12" square, built into the discharge pit walls

and has a steel plate between the WF beam flange and the concrete to

transmit the load.

 

4) The gross area of the column, Ag = 144 sq. in.

 

5) Areas of steel are:

 

Ast = Pg * Ag = 0.025 * 144 in. sq. = 3.6 in. sq.

 

6) Po = 144 in. sq. * (0.85*(1-0.025) * 3000 psi + 30,000 psi * 0.025

 

      = 466,020 lbf

 

C) Determine the ACI factored design strength:

 

Pdesign = phi*Po = 0.80 * 0.70 * Po

 

        = 0.8 * 0.7 * 466,020 lbf

 

        = 260,971 lbf

 

D) Since the factored ACI design strength is greater then the factored

ACI load, this design will work

 

phi * Po > Pu >>>>>>>>>>>> 260,971 lbf > 219,110 lbf

 

E) Determine the number of bars and their size:

 

1) Assume six bar design

 

2) Area of single bar = Ast/# bars

 

                      = 3.6 in^2/6 = 0.60 sq. in.

 

3) From ASTM rebar, table #7 bars are 0.6 in. sq.

 

                                  (19)

F) Sketch design:

 

 

 

 

 

 

 

 

 

 

 

1) minimum spacing is 1.5 * 0.875 = 1.31"

 

2) minimum tie wire size is #4

 

3) concrete cover must be 1 1/2" thick

 

4) every corner to be supported by a tie wire

 

5) tie wire spacing shall be every 14" of column rise

 

 

 

 

 

 

 

 

 

 

 

 

G) Check the slenderness ratio of the column:

 

1) SR = K*Lu/r = K*Lu/0.3*W                K = 1 for unbraced columns

                                           Lu = 39 - 21 = 18'

                                           w = 12"

   SR = 1 * 18'/(0.3*1') = 60

 

Since 60>22 this design is slender

 

2) Determine w for SR= 22 minimum

 

W = KLu/(0.3*SR) = 18'/(0.3*22) = 2.73' = 33"

 

H) Retrofit design so that wall is 33" thick for the 12" length of the

wall which the column is embedded into.

 

 

 

 

 

 

 

                                 (20)

I) Sketch design:

 

 

 

 

 

 

 

 

 

 

 

XI) Check the rock foundation bearing capacity:

 

A) Total weight of the powerhouse and equipment:

 

Water + 2 floors + machinery + back & front wall + side walls =

 

400,982 lbf + 100,000 lbf + 36,200 lbf + 432,000 lbf + 388,000 lbf =

 

1,357,182 lbf

 

B) Total surface bearing area is:

 

(40' + 2*18') * 1' = 76 sq. ft.

 

 

C) The stress on the rock is:

 

1,357,182 lbf/76 sq. ft. = 17,858 lbf/ft^2 = 8.92 tons/sq. ft.

 

D) "American Civil Engineer's Handbook", Merriman & Wiggin, 5th edition,

P.  711, table lists the allowable soil pressures in short tons per sq.

ft. for very hard native bedrock at 9 tons/sq.ft..  A short ton is 2000

lbs. Since the calculated pressure of 4.5 tons/sq.ft. is much less then

15 tons/sq.ft.  This design should be all right if the walls are poured

directly on the rock excavation.  The footing should be chipped square

and level before the forms are set up.

    

XII) Rebar size and spacing selection:

 

A) Waterbox Floor:

 

1) To obtain 1.72 in. sq./ft. width slab, use one #8 bar every six

inches.

 

2) Use #6 bar at 18 inch spacing in the longitudinal spacing.

 

B) Generating Room Floor:

 

1) To obtain 0.53 in. sq./ft. width slab, use one #7 bar every 12

inches.

 

 

                                  (21)

 

2) Use #5 bar at 18 inch spacing in the longitudinal spacing.

 

C) Walls:

 

1) To obtain 1.20 in. sq./ft. width slab, use one #7 bar every 6

inches.

 

2) Use #5 bar at 12 inch spacing in the longitudinal spacing.

 

 

 

                                      

ADDITIONAL CALCULATIONS

July 19, 1990

 

 

A)  Lap Splice Lengths -

 

In tension

 

- Bars are not larger than #11

- Minimum length of lap for tension lap splices shall be as

   required: Class A = 1.0 L, Class B = 1.3 L and Class C = 1.7 L

 

where: L > 0.0004*D*Fy*1.25D = nominal diameter of bar

Fy = specified yield strength

     = 30000 psi

  L = 13.1 inches

 

  and L > 0.04*Ab*Fy/(Fc)^.5Ab = Area of individual ba

 

Fc = specified compression

stength of concrete

 = 4000 psi 

 

L  = 14.2 inches, Class C = 24 inches

 

In Compression   

 

L = 0.02*D*Fy/(Fc)^.5

 

  = 9.5 inches

 

Thus, the longest the lap splice would need to be in any condition

would be 24 inches long.

 

 

B) Make sure that the extra circular reinforcement around holes in

   the generator room floor and water box floor is equivalent or

   greater than the reinforcement cut to create the holes.

 

   Maximum area cut = 2 - #7 Bars

                                  = 2[3.14*(0.875)^2]/4

                                  = 1.20 sq. in.

 

                                 (22)

 

   Area added  = 4 - #5 Bars

= 4[3.14*(0.625)^2]/4

= 1.22 sq. in.

 

   In the other case 4 - #8 bars are cut and 4 - #8 bars are added for

reinforcement.

 

Therefore, in each case there is enough reinforcement added to be

equivalent or greater than the cross-sectional area cut to create the

holes.

 

 

C) Local Buckling Calculations - addition of ed stiffeners

 

Midsection of 30" I-beam

 

Nmin = [reaction/(0.75)*Fy*tw] - k;  where Fy = yield strength

        of steel

tw = web thickness

k = flange-to-web toe

fillet distance

Nmin = [(2)(154300 lbs)/(0.75)(30000 psi)(0.563)] - 1.06

= 22.7 inches

 

Two ends of 30" I-beam

 

Nmin = [(154300)/(0.75)(30000 psi)(0.563)] - 1.06

= 11.1 inches

 

Turbine gate case support beams

 

Nmin = [(3500 lbs)/(0.75)(30000)(.25)] - 0.875

= 0 inches

 

Generator support beams

 

Nmin = [(17600)/(0.75)(30000)(0.563)] - 1.563

= 0 inches

 

D.  Check for plate bearings on end of beams in concrete wall

 

Design for Fp = 0.35 fc     (Civil Eng Ref Manual, 4th Ed.)

    = 1050 psi

 

Turbine gate case support beams

 

Fp = 1050 psi > 1750 lbs/(6.5")(24") = 11 psi (no plate needed)

 

Generator support beams

 

    Fp = 1050 psi > 17600/(12.13)(24) = 30 psi (no plate needed)

 

 

 

                                  -23-

Main 30" support beam

 

A = P/F = 310,000 lbs/1050 psi = 298 sq in

 

Need to spread load out onto 298 sq in, at present we have

a 24 inch X 30 inch steel plate

 

Area of steel plate = (24")(30") = 720 sq in

 

So the load is spread out enough

 

F = 310000 lbs/720 sq in = 431 psi < 1050 psi

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

                                 -24-