Powerhouse Design- Miniwatt Hydro Web Page
Design of Powerhouse Substructure W. Fay P.E.
Mini-Watt Electric Expansion 10/15/88
Orange, Massachusetts
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 considst 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
250 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 - 476.3 msl
2) Mean tailwater elevation - 490.0 msl
3) Waterbox floor - 493.3 msl
4) Mean headwater elevation - 501.3 msl
5) Top of generating room floor slab - 522.0 msl
Assume main beams forming the discharge pit/tailrace and water box inlet
arches are surplus from the Route 2 bridge rebuild (WF-30"x 124 #/ft).
(2)
III) Design of the Waterbox Floor Slab:
A) Dimensions- from Bruce Dexter layout dated 8/22/88 entitled "Lower
Walls etc". Assume these dimensions are correct.
1) External dimensions:
a) length = 173" + 173" + 3*(24") = 418" = 34' 10" = 34.83'
b) width = 18'
2) Internal dimensions:
a) length = 30.83'
b) width = 14.0'
B) Loading:
1) Water, total load = 62.4 #/ft3 * (14'* (501.3-493.3)
= 215,465 lbs
2) Water, load per foot width of box = 1575 lbs/ft-width of flume
3) Water, load per foot width of box per foot width of beam = 499 lb/ft
4) Concrete, total load = 150 #/ft3 * 14' * 30.83'*
(9"/(12"/ft)) = 48,557 lbf
5) Concrete, load per foot width of box = 48,557 lbf/30.83' = 1575
lbf/ft width of box
6) Concrete, load per foot width of box per foot width of beam =
1575 lbf/ft-width/14' = 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 * 14')/20 = 8.4" approximately = 9"
3) The slab weight is 150 #/ft3 * (9/12) = 113 PSF
4) Apply the ACI load multipliers and obtain the factored load:
Dead Load = 113 PSF * 1.4 = 158.2 PSF
Live Load = 499 PSF * 1.7 = 848.3 PSF
Total Factored Load = 1007 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*1.01 KSF *14'^2 = 18.0 ft-kips
e) At the upstream end: -M = 1/14*1.01 KSF *14'^2 = 14.0 ft-kips
d) At the midspan: -M = 1/16*1.01 KSF *14'^2 = 12.4 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
= (18 ft-kips*(12"\ft))/(0.9*0.04*30*12*(1-.59*0.04*(30,000/3000)))
= 21.8 in^2
Therefore, d = 4.7 inchs
7) Determine the minimum effective depth using code restrictions:
dm= 9"- 1" = 8"
8) Since the calculated value of 4.7 inches is less then the coded
effective depth, use d= 9 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))= (18 ft-kips*12"/ft)/0.9*30*(8-1/2)= 1.06 in^2
10) Check the assumed depth:
a= As*fy/(0.85*fc'*b)= 1.06 in^2 *30,000/(0.85*3000*12"/ft)= 1.04 in^2
11) The assumed area of steel and the calculated area of steel are
reasonably close so use 1.06 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= 12.4 ksi*12/(0.9*30*(8-1/2))= 0.73 in^2
b) at the upstream wall: As= 14 ksi*12/(0.9*30*(8-1/2))= 0.83 in^2
13) The minimum reinforcement required to control shrinkage is: see p.
207, W&N.
As= 0.002*12*9= 0.216 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 * (1007*14/2)-1007*(8.5/12)= 8106-713 = 7393 lbs
15) The nominal shear strength of the slab is:
Vn= Vc= 2*b*d*fc'^0.5
= 2*12*8.5*(3000 psi^0.5)
= 11,173 lbs
16) The design shear strength is:
phi*Vc= 0.85*11,173 lbs= 9,498 lbs
17) Since the design shear strength is above the required shear strength
by 30 %, no additional steel is necessary to resist the internal shear
forces.
IV) Design the Steel Main Support Beam at the Tailrace
A) Data for Bridge 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
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) Dead Loads:
a) Powerhouse backwall:
((31'*(522.0-506.0)*0.75) + ((506.0-492.0)*1.0))*150 lbs/ft^3= 93,000 #
b) Waterbox floor:
(18'*31'*1')*150 lbs/ft^3= 83,700 #
c) Generating room floor:
(18'*31'*1')*150 lbs/ft^3= 83,700 #
d) Water:
18'*31'*(504.0-493.0)*62.4 lbs/ft^3= 383,011 lbs
4) Total Distributed Load:
(93,000 # + 83,700 #/2 + 383,011 #/2)/31'= 18,000 lbs/ft-width beam
5) Total point load is:
(15,800 # + 1400 # + 14,000 # )/2 + 5000 #= 20,600 lbs
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:
6.8' * 20,600 lbs + 18,000 lb/ft * (13.5'^2)/8 = 550,142 ft-lbs
3) The extreme fiber stress is:
Sigma = M/Z = (550,142 ft-lbs * 12"/ft)/355 in^3 = 18,596 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:
delta max= 5*w*l^4/(384*E*I)
= 5* 18,000 lb/ft * (13.5'*12")^4/(384*29,000,000 psi* 5360
in^4*12)= 0.09 inch
b) For the point load:
delta max= p*l^3/(48*E*I)
= 20,600 lbs * 13.5'*12"/ft)^3/(48*29,000,00 psi*5360 in^4)
= 0.01 inch
c) Total defltion is the superposition of the deflections due to the
two different types of loads and is
0.09" + 0.01" =0.1"
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 7' * 3500 # = 24,500 ft-lbs
3) Z= M/sigma= (24,500 ft-lbs * 12 in/ft)/17,280 psi= 17 in^3
D) From the AISC Handbook, 7th ed., P. 1-42, tentatively choose a
W8"x20 lbs/ft
E) Beam Properties:
height= 8.14"
weight= 20 lbs/ft-width
length= 18 feet
section modulus= 17 in^3
moment of inertia= 69.4 in^4
depth= 8.14"
width= 5.268"
flange thickness= 0.378"
web= 0.248"
(9)
F) Check the deflection:
Sigma max= P*L^3/(48*E*I)= 3500# * (14'*(12"/ft)^3)/(48 * 29,000,000 psi
* 69.4 in^3= 0.17 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 7' * 17,600 # = 123,200 ft-lbs
3) Z= M/sigma= (123,200 ft-lbs * 12 in/ft)/17,280 psi= 86 in^3
D) From the AISC Handbook, 7th ed., P. 1-42, tentatively choose a
W10"x77 lbs/ft
(10)
E) Beam Properties:
height= 10.62"
weight= 77 lbs/ft-width
length= 18 feet
section modulus= 86.1 in^3
moment of inertia= 457 in^4
depth= 10.62"
width= 10.2"
flange thickness= 0.868"
web= 0.535"
F) Check the deflection:
Sigma max= P*L^3/(48*E*I)= 17,600# * (14'*(12"/ft)^3)/(48 * 29,000,000
psi * 86.1 in^3= 0.7 inch <<<< okay
The choosen beam will meet the design requirements.
VII) Design the generator floor:
A) Dimensions- from Bruce Dexter layout dated 8/22/88 entitled "Lower
Walls etc". Assume these dimensions are correct. Assume the T/G set is
supported by 10" WF beams. Design the floor for 300 lbs/ft live load,
plus the dead load of the concrete.
1) External dimensions:
a) length = 173" + 173" + 3*(24") = 418" = 34' 10" = 34.83'
b) width = 18'
2) Internal dimensions:
a) length = 30.83'
b) width = 14.0'
B) Loading:
1) Concrete, total load = 150 #/ft3 * 14' * 30.83'*
(9"/(12"/ft)) = 48,557 lbf
2) Concrete, load per foot width of box = 48,557 lbf/30.83' = 1575
lbf/ft width of box
3) Concrete, load per foot width of box per foot width of beam =
1575 lbf/ft-width/14' = 113 lbf/ ft
(11)
C) Sketch of typical section through proposed waterbox 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 * 14')/20 = 8.4" approximately = 9"
3) The slab weight is 150 #/ft3 * (9/12) = 113 PSF
4) Apply the ACI load multipliers and obtain the factored load:
Dead Load = 113 PSF * 1.4 = 158.2 PSF
Live Load = 300 PSF * 1.7 = 510.0 PSF
Total Factored Load = 668 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*0.668 KSF *14'^2 = 11.9 ft-kips
e) At the downstream end: -M = 1/14*0.668 KSF *14'^2 = 9.35 ft-kips
f) At the midspan: -M = 1/16*0.668 KSF *14'^2 = 8.18 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
= (11.9ft-kips*(12"\ft))/(0.9*0.04*30*12*(1-.59*0.04*(30,000/3000)))
= 14.43 in^2
Therefore, d = 3.8 inchs
7) Determine the minimum effective depth using code restrictions:
dm= 9"- 1" = 8"
8) Since the calculated value of 4.7 inches is less then the coded
effective depth, use d= 9 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))= (11.9 ft-kips*12"/ft)/0.9*30*(8-1/2)= 0.62 in^2
10) Check the assumed depth:
a= As*fy/(0.85*fc'*b)= 0.62 in^2 *30,000/(0.85*3000*12"/ft)= 0.61 in
11) Reiterate assuming a=0.61 in:
As=Mu/(phi*f*(d-a/2))=(11.9 ft-kips*12"/ft)/0.9*30*(8-0.61/2)= 0.69 in^2
12) Reiterate assuming As=0.69 in^2:
a= As*fy/(0.85*fc'*b)= 0.69 in^2 *30,000/(0.85*3000*12"/ft)= 0.68 in
13) Reiterate assuming a=0.68 in:
As=Mu/(phi*f*(d-a/2))=(11.9 ft-kips*12"/ft)/0.9*30*(8-0.68/2)= 0.69 in^2
14) The assumed area of steel and the calculated area of steel are
reasonably close so use 0.69 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= 8.18 ksi*12/(0.9*30*(8-1/2))= 0.48 in^2
b) at the upstream wall: As= 9.35 ksi*12/(0.9*30*(8-1/2))= 0.55 in^2
16) The minimum reinforcement required to control shrinkage is: see p.
207, W&N.
As= 0.002*12*9= 0.216 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= .668 * (1007*14/2)-1007*(8.5/12)= 8106-713 = 4294 lbs
15) The nominal shear strength of the slab is:
Vn= Vc= 2*b*d*fc'^0.5
= 2*12*8.5*(3000 psi^0.5)
= 11,173 lbs
(14)
16) The design shear strength is:
phi*Vc= 0.85*11,173 lbs= 9,498 lbs
17) Since the design shear strength is above the required shear strength
by 60 %, no additional steel is necessary to resist the internal shear
forces.
VIII Design Waterbox Walls:
A) Dimensions- from Bruce Dexter layout dated 8/22/88 entitled "Lower
Walls etc". Assume these dimensions are correct. Note, these rear
walls are not designed as axial bearing members. They are only designed
as panels to take the water pressure.
1) External dimensions:
a) height= 14'
b) width = 14'
2) Internal dimensions:
a) length = 14'
b) width = 14.0'
B) Loading:
1) Concrete, total load = 150 #/ft3 * 14' * 14'* (9"/(12"/ft)) =
22,050 lbf
2) Concrete, load per sq. ft. = 48,557 lbf/30.83' = 113 lbf/ft^2
3) Water Load= 62.4 lbf/ft^3 * 14' wide * 14' deep * 1' thick/ 14'wide
= 873 PSF. This is the maximum load at the base of the
hydrostatic load.
C) Sketch of typical section through proposed waterbox floor:
(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 * 14')/20 = 8.4" approximately = 9"
3) The slab weight is 150 #/ft3 * (9/12) = 113 PSF
4) Apply the ACI load multipliers and obtain the factored load:
Dead Load = 113 PSF * 1.4 = 158.2 PSF
Live Load = 873 PSF * 1.7 = 1484.0 PSF
Total Factored Load = 1643 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*1.643 KSF *14'^2 = 23.0 ft-kips
d) At the bottom: -M = 1/14*1.643 KSF *14'^2 = 23.0 ft-kips
e) At the midspan: -M = 1/16*1.643 KSF *14'^2 = 23.0 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
= (23 ft-kips*(12"\ft))/(0.9*0.04*30*12*(1-.59*0.04*(30,000/3000)))
= 27.9 in^2
Therefore, d = 5.3 inchs
7) Determine the minimum effective depth using code restrictions:
dm= 9"- 1" = 8"
8) Since the calculated value of 5.3 inches is less then the coded
effective depth, use d= 8 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))= (23 ft-kips*12"/ft)/0.9*30*(8-1/2)= 1.36 in^2
10) Check the assumed depth:
a= As*fy/(0.85*fc'*b)= 1.36 in^2 *30,000/(0.85*3000*12"/ft)= 1.33 in
11) Reiterate assuming a=1.33 in:
As=Mu/(phi*f*(d-a/2))=(23 ft-kips*12"/ft)/0.9*30*(8-1.33/2)= 1.39 in^2
12) Reiterate assuming As=1.39 in^2:
a= As*fy/(0.85*fc'*b)= 1.39 in^2 *30,000/(0.85*3000*12"/ft)= 1.37 in
13) Reiterate assuming a=1.37 in:
As=Mu/(phi*f*(d-a/2))=(23 ft-kips*12"/ft)/0.9*30*(8-1.37/2)= 1.40 in^2
14) The assumed area of steel and the calculated area of steel are
reasonably close so use 0.69 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= 23 ksi*12/(0.9*30*(8-1.37/2))= 1.40 in^2
b) at the bottom: As= 23 ksi*12/(0.9*30*(8-1.37/2))= 1.40 in^2
(17)
16) The minimum reinforcement required to control shrinkage is: see p.
207, W&N.
As= 0.002*12*9= 0.216 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*1484*14/2-1484*(9.0/12)= 11,946-1113 = 10833 lbs
15) The nominal shear strength of the slab is:
Vn= Vc= 2*b*d*fc'^0.5
= 2*12*9.0*(3000 psi^0.5)
= 11,831 lbs
16) The design shear strength is:
phi*Vc= 0.85*11,173 lbs= 9,498 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 1.4 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 490.0-476.0=14'
B) This design is identical to the waterbox walls.
C) Use 1.4 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 + (18,000*15'))/2 = 145,300 lbf
2) the factored ACI Code load is:
Pu=(1.4*(1.4*(18,000 lbf * 15')/2) + (1.7 * 20,600/2)
= 189,000 lbf + 17,510 lbf = 206,510 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 > 206,510 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 = 492.3 - 476.3 = 16'
w = 12"
SR = 1 * 16'/(0.3*1') = 53
Since 53>22 this design is slender
2) Determine w for SR= 22 minimum
W = KLu/(0.3*SR) = 16'/(0.3*22) = 2.42' = 30"
H) Retrofit design so that wall is 30" 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 =
215,465 lbf + 100,000 lbf + 36,200 lbf + 186,000 lbf + 93,000 lbf =
630,665 lbf
B) Total surface bearing area is:
(34.83' + 2*18') * 1' = 71 sq. ft.
C) The stress on the rock is:
630,665 lbf/71 sq. ft. = 8900 lbf/ft^2 = 4.45 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 15 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 1.06 in. sq./ft. width slab, use one #7 bar every six
inches.
2) Use #5 bar at 18 inch spacing in the longitudinal spacing.
B) Generating Room Floor:
1) To obtain 0.48 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 1.40 in. sq./ft. width slab, use one #7 bar every 6
inches.
2) Use #5 bar at 12 inch spacing in the longitudinal spacing.