CE 318
STRUCTURE ANALYSIS AND DESIGN II LAB
LESSON 10
UNDERGROUND WATER RESERVOIR
SEMESTER: SUMMER 2021
COURSE TEACHER: SAURAV BARUA CONTACT NO: +8801715334075
EMAIL: [email protected]
LECTURE PLAN
Underground water reservoir (UGWR) capacity calculation
Base slab design
Cover slab design
CAPACITY OF UGWR
Underground water reservoir is RCC box
Per capita consumption as per BNBC = 235 lpcd
Assume, there are 6 person/apartment. There are one apartment per floor.
So total population in the building = 6 x 6 = 36 persons Consider, 2 day storage of water.
CAPACITY OF UGWR
Capacity of water reservoir = 36 x 2 x 235 = 16920 liter We know, 1 cft = 28.32 liter
So, Volume required for underground water reservoir = 16920/28.32
= 597.46 cft
≈ 600 cft Assume, size of UGWR = 12’ x 6’
Height required for UGWR = 600/(12 x 6) = 8.33’
DIMENSION OF UGWR
Consider 6” free board. (Explain in class!!) 6” = 0.5ft
Height of UGWR = 8.33+0.5 = 8.83’ = 8’-10”
So, dimension of UGWR = 12’ x 6’ x 8’-10”
Assume, base slab = 9”, cover slab = 6” and side wall = 9”.
UNDERGROUND WATER RESERVOIR
Size of base slab
Length = 11+ 1 +1 + (0.75 x 2)
= 14.5’
Width = 6 + 1+ 1+ (0.75 x 2)
= 9.5’
So size of base slab = 14.5’ x 9.5’
BASE SLAB DESIGN
Assume, ground water table (GWT) at GL Consider worst condition for the tank.
Upward water pressure = (8.83 + 0.75) x gw
= 9.58’ x 62.4 = 598 psf = 0.6 ksf (upward) This pressure is exerted at the bottom of the base slab.
BASE SLAB DESIGN
Weight of base slab = 9/12 x 14.5 x 9.5 x 0.15
= 15.5 kip (downward)
Size of cover slab = 12.5’x 7.5’
Weight of cover slab = 6/12 x 12.5 x 7.5 x 0.15
= 7.03 kip (downward)
Side wall: long wall = 2 x 12 = 24 ft, short wall = 2 x 6 = 12 ft
Weight of side wall = 9/12 x 8.33 x (24+12) x 0.15 = 33.74 kip (downward)
BASE SLAB DESIGN
Self weight of UGWR = weight of cover slab + based slab + side wall
= 15.5 + 7.03+ 33.74
= 56.27 kip (downward) self weight of UGWR per sft = 56.27/ (14.5 x 9.5)
= 0.41 ksf (downward)
Net upward force on base = 0.6-0.41 = 0.19 ksf( upward) Factored force or load = 1.4 x 0.19 = 0.27 ksf
MOMENT AT BASE SLAB
All sides are discontinuous and m = A/B = 6/12 = 0.5 And slab is case I.
Short direction moment, Ma+ve = 0.095 x 0.19 x 62 = 0.65 k-ft/ft Long direction moment, Mb+ve = 0.006 x 0.19 x 122 = 0.16 k-ft/ft
MOMENT AT BASE SLAB
dreq = 𝑀𝑢
Øρ𝑓𝑦𝑏(1−0.59ρ𝑓𝑦/𝑓𝑐′
=√( 0.65 x 12 /(0.9 x 0.016 x 60x 12 x(1-0.59 x 0.016 x 60/3)))
= 0.96” < 6”
dprovided = 9-3 = 6”
So, the design is ok.
REINFORCEMENT AT BASE SLAB
Steel for short direction:
As = 0.65 x 12/(0.9 x 60 x (6-1/2))
= 0.026 in2/ft
a = 0.026 x 60/(0.85 x 3 x 12)
= 0.05
Retrial, As = 0.024 in2/ft
Minimum steel for temperature A = 0.0018 bh = 0.0018 x 12 x 9
REINFORCEMENT AT BASE SLAB
Steel for long direction:
As = 0.16 x 12/(0.9 x 60 x (6-1/2))
= 0.007 in2/ft
a = 0.007 x 60/(0.85 x 3 x 12)
= 0.014
Retrial, As = 0.006 in2/ft
Provide temperature reinforcement for both directions.
As = 0.19 in2/ft
REINFORCEMENT AT BASE SLAB
Provide Ø12mm rebar.
Spacing = 0.2 x 12/ 0.19
≈ 12”
So, provide Ø12mm @ 12” c/c both ways of base slab
BASE SLAB
COVER SLAB DESIGN
• Cover slab is simply supported.
• Thickness = 6”
• DL = 6/12 x 0.15 = 0.075 ksf
• LL = 60 psf = 0.06 ksf
So factored load, Wu = 1.4 DL + 1.7 LL
= 1.4 x 0.075 + 1.7 x 0.06 = 0.21 ksf Mlong = +wl2/8 = 0.21 x 12.52 /8 = 4.1 k-ft/ft
Mshort = +wl2/8 = 0.21 x 7.52 /8 = 1.48 k-ft/ft
REINFORCEMENT AT COVER SLAB
Steel for long direction:
d = 6-1 = 5”
As = 4.1 x 12/(0.9 x 60 x (5-1/2))
= 0.02 in2/ft
a = 0.02 x 60/(0.85 x 3 x 12)
= 0.4
Retrial, As = 0.19 in2/ft
Minimum steel for temperature Astemp = 0.0018 bh = 0.0018 x 12 x 6
= 0.13 in2/ft Use Ø10mm bar, spacing = 0.11 x 12/0.13 ≈ 10” c/c
REINFORCEMENT AT COVER SLAB
Steel for short direction:
As = 1.48 x 12/(0.9 x 60 x (5-1/2))
= 0.07 in2/ft
a = 0.07 x 60/(0.85 x 3 x 12)
= 0.14
Retrial, As = 0.066 in2/ft
So, provide, temperature rebar at short direction.
Use Ø10mm bar, spacing = 0.11 x 12/0.13 ≈ 10” c/c
