Project 2
PROJECT 2: Design of Tanks by Finite Element Analysis.
Solution:
Aim: Once the analysis is finished, proceed with the post-processing of the results. Checking the base pressure and settlement beneath the base slab, which should be within the parameters specified in Project-1, is the first duty. In order to control base pressure and settlement, the dimensions must be increased or decreased if certain restrictions are exceeded.
Report the following findings once the foundation has been inspected in order to complete the element's structural design.
- Design Base Slab and Wall Maximum and Minimum Moments (Wood Armer Simplified Method)
- Design Shear Stress Maximum and Minimum for Base Slabs and Walls
- Design Axial Stresses at Maximum and Minimum for Base Slabs and Walls
- After the findings have been post-processed, complete the structural design of the elements by performing the following checks.
- Create an Excel spreadsheet to determine the amount of reinforcement needed for base slabs and both long and short walls. The following approach (covered in Week-3) can be used to compute the contribution due to direct tension, assuming it is significant, in order to determine the required design reinforcement:
M,net = T. x – M
Ast1 = (0.5 fck / fy) {1-√[1-(4.6 B.M./(fck.b.d²))]
Ast2 = T / 0.87 fy
Ast1 + Ast2 equals = Ast, req.
M = Design moment derived using the Wood Armer formulas
T = The section's design tension force (stress x c/s area of the FE element).
x = The tension reinforcement's distance from the section's center line.
Verify that the foundation slab's and the walls' short and long crack widths are less than 0.2 mm.
Using CAD software, draft the cross-sectional reinforcement detail (plan and elevation). If CAD is not available, use any drawing tool to sketch the intricacies and make sure every detail is visible.
How to do it:
Unit weight of materials according to the code
Safe bearing capacity = 18 kN/m^3
Concrete with plain cement = 24 kN/m^3
Concrete with reinforced cement = 25 kN/m^3
Concrete screed with cement = 20 kN/m^3
Masonry made of cement = 22kN/m63
Steel for structure = 78.5 kN/m^3.
Building loads
Slab weight = 3.75kN/m^2(0.15mx25kN/m^3)
Divisions equal 1.5 kN/m^2.
Finish floor: 2.5 kN/m^2
For Others, the value is 1kN/m^2.
Floor load total = 8.75kN/m^2
Weights of slab units
Slab weight = 3.75kN/m^2(0.15mx25kN/m^3)
Waterproofing and insulations = 0.5 kN/m^2.
Services for MEP = 0.5kN/m^2
Others equals 0.5kN/m^2.
Roof load total = 5.25 kN/m^2
Shared corridor: 3 kN/m^2.
Floor load total = 3kN/m^2
Roof accessibility = 1.5 kN/m^2
Roof load total = 1.5 kN/m^2
Slab design (IS-456)
Ly = 3.125m is the span shorter direction (clear).
Longer span in a distinct direction Lx is equivalent to 5.5m.
3kN/m is the live load on the slab (LL).
Concrete's compressive strength (Fck) = 25 N/mm^2
Steel yield strength (Fy) = 415 N/mm^2
Concrete weight in units (Yc) = 25 kN/m^3
Floor finish unit weight: 100mm(Y) = 22kN/m^3
Concrete cover that is clear = 25 mm
Slab (B) of bearings = 250 mm
Step 1
L/d = 32 is the span to effective depth ratio.
Effective depth minimum (d) = 97.67 mm
Depth total (D) = 127.66 mm
Bars' diameter in the shorter direction = 10 mm
Bar diameter in the extended direction = 10 mm
120mm is the effective depth(d).
Filling the slab with load
The slab's dead load (DL) is 3.75kN/m^2.
Extremely low load = 5 kN/m^2
3kN/m^2 is the slab's live load.
The slan's total load is 11.75 kN/m^2.
17.63kN/m^2 = Total load x load factor is the design load.
Span effective Lx = 3.25 m
Ly is 5.62 meters.
Lx/Ly (two-way) = 1.732
For negative moments (at top),
1.5 Alpha x 1.75 Alpha y = 0.075 0.104683 0.107 0.084
In the case of positive moments
(bottom), 0.056 0.103307 0.107 0.063
Step 2:
BM per slab width unitMx=αxwLx^2My=αywLy^2
Regarding negative moments (uppermost): 19.43 ± 15.59 kN/m
For positive situations, 19.17 11.69 kN-m/m (bottom)
Shear force V equals 28.6 kN/m/0.5WLx.
Step 3:
To determine the slab's effective depth = 0.138Fckbd^2 for Mu,lim
D is equal to 74.55 mm.
Step 4:
slab's depth for shear force, Tc, max is equal to 3.1N/mm^2.
Tc equals 0.3.
K is equal to 1.2.
D is equivalent to 200 mm.
Tc is equal to 0.3N/mm^2.
Vu/bd = Tv = 0.24 N/mm^2.
Tv
Step 5:
Estimating the steel's area
For a bad moment, Mutop is 19.43 kN/m.
Ast = 561.65 mm^2/m
For a happy moment, 19.17 kN/m for Mubottom
Ast = 554.27 mm^2/m
Calculating the steel distribution
Astmin is 0.12bd = 114 square millimeters
Choosing steel reinforcement bars
Bar area (upper steel) = 78.5 mm^2
140 mm is the spacing= 125 millimeters
Area of bars (78.5 mm^2) for bottom steel
Measurement of spacing: 142 mm = 125 mm
Give 16 mm bars 125 mm apart.
Wind load calculation (IS875-Part 3)
Wind speed design Vz = Vbk1k2k3k4 m/s
Bangalore's basic wind speed is Vb = 33 m/s.
Factor of probability k1 = 1.
Ground factor k2 equals Category 4B.
Factor of topography k3 = 1.
Wind pressure design (Pz) = 0.6Vz^2
| Floor | Height | K2 | Vz | Pz | Pz in kN |
|---|---|---|---|---|---|
| 1 | 3.5 | 0.76 | 25 | 377.40 | 0.38 |
| 2 | 7 | 0.76 | 25 | 377.40 | 0.38 |
| 3 | 10.5 | 0.76 | 25 | 377.40 | 0.38 |
| 4 | 14 | 0.76 | 25 | 377.40 | 0.38 |
| 5 | 17.5 | 0.76 | 25 | 377.40 | 0.38 |
| 6 | 21 | 0.76 | 26 | 397.53 | 0.40 |
| R | 24.5 | 0.76 | 26 | 461.04 | 0.46 |
Seismic Load Calculation (IS1893-Part I)
Zone factor (Z) in Bangalore is 0.1.
Factor of soil condition Sa/g
Distance Along-X = 25 m
Along-Z is 16 meters.
T - X = 0.5513 seconds
Z - T = 0.4410sec
For difficult soil, Sa/g = 1/T = 1.814 Along X = 2.268 Parallel to Z
R = 3, the responce reduction factor
Factor of importance I = 1.2
Coefficient of horizontal acceleration Ah = 0.0363
3.6812%
seismic mass
Each level A's area is 400 m^.
Because of dead loads Floor DL on average is 12.5 kN/m2.
W DL equals 5000kN.
Because to dead loads, the roof
DL roof is equivalent to 9kN/m2.
3600kN is the W DL roof.
Live loads cause the typical floor's LL to be 3kN/m2.
W LL equals 300 kN/m^2.
Live loads cause Roof LL to equal 1.5 kN/m2.
W LL roof is equal to 0.
Seismic weight total W = 35400kN
Base shear design (VB) = AhxW = 1605.44 kN
Please find the attachments below;
The guidelines to meet the slab's limit stage of cracking are as follows:- The code's minimum and maximum flexural reinforcing values for beams and slab elements:-
The smallest amount of steel that can be inserted in structural concrete parts to avoid early failure when the tensile strength is lost is known as the minimum reinforcing ratio. The concrete members' propensity to crack is managed by the minimum reinforcement ratio.
The largest steel area that can be included into concrete members such as beams and columns is known as the maximum reinforcing ratio. Extra reinforcement above the maximum reinforcement ratio in a reinforced concrete beam would not be helpful since the concrete would be crushed before the steel's full strength is utilised.
A concrete structure will collapse without warning and without any prior indications. The maximum reinforcing ratio protects against brittle concrete failure and guarantees the economy of concrete members.
Last but not least, the required reinforcement area of the specified concrete member must fall between the minimum and maximum reinforcement ratios. As a result, this requirement needs to be verified for the designed member.
1. Minimum Ratio of Reinforcement in Beams.
When the first flexural crack forms in a reinforced concrete beam, the beam will fail without causing any discomfort if the flexural strength of the cracked portion is less than the moment that caused the cracking of the previously uncracked section.
ACI 318-19 provides an equation that can be used to calculate the minimal reinforcement ratio, which can save a concrete beam from failing too soon. The following expression can be used to calculate the minimal reinforcement required for beams:
2. Slabs' Minimum Reinforcement Ratio.
The temperature and shrinkage reinforcement installed to control cracks resulting from temperature fluctuations and concrete shrinkage is the minimum reinforcement area for a slab. Beyond the temperature and shrinkage reinforcement, a reinforcement area is not necessary.
As= ρbd
As: mm2 of temperature reinforcement and shrinkage
B: The 1-meter-wide slab strip that was taken into consideration for design
d: millimeters of effective depth
3. Minimum Ratio of Reinforcement in Even Footing
For uniform footing, the minimal reinforcement ratio—that is, the temperature and shrinkage reinforcement ratio—is comparable to that of a slab.
4. Columns with the Minimum Reinforcement Ratio
Regardless of the analytical conclusions, columns must have a minimum reinforcement ratio to offer resistance against bending. Additionally, it is required to lessen the impact of the concrete's creep and shrinkage under continuous compressive pressures.
Under a continuous service load, the minimal reinforcement ratio in the column keeps the steel bars from giving way. The minimum longitudinal reinforcement ratio for a column is 0.01 times the column's gross area, according to ACI 318-19.
5. Minimal Reinforcement for Connections with the Foundation and Cast-in-Place Members
A minimum of 0.005 times the gross area of the supported element should be the minimum reinforcement area that spans the interface between the foundation and a cast-in-place column or pedestal.
Maximum Ratio of Reinforcement
The maximum amount of steel that can be added to concrete members is known as the maximum reinforcing ratio. It is frequently given for a number of reasons, some of which are covered below:
1. Maximum Beam Reinforcement Ratio
The ACI rule forbids concrete crushing, an undesirable form of failure, which is why the maximum reinforcing ratio for beams is offered. Additionally, it stays away from using an excessive amount of steel where there are no practical benefits. As a result, it aids in making concrete beam designs more economical.
A beam is referred to as over-reinforced concrete if its reinforcement ratio is higher than the maximum and it typically collapses in compression.
2. Maximum Ratio of Reinforcement in Columns
According to ACI 318.19, the maximum reinforcement has been determined to guarantee that the concrete can be sufficiently compacted around steel bars and that the intended columns resemble the test specimens.
Columns can have a maximum reinforcement ratio of 0.08 times their gross area. It makes column designs more economical and avoids steel congestion, which would otherwise make it difficult to put concrete properly.
To prevent over-reinforcement at the sites where steel bars splice, it is practically advised to take into account a maximum reinforcement ratio of 0.04 times the column's gross area.
Shear Minimum Reinforcement Ratio
ACI 318-19 establishes the minimum reinforcement ratio for shear in beams, etc., in a manner akin to the minimum flexural reinforcement previously mentioned.
1. The lowest ratio of shear reinforcement in beams
In every area of a beam where applied shear exceeds half of the concrete's intended shear strength, a minimum area of shear reinforcement must be present.
In beams, the minimum shear reinforcement (Av,min) should equal the bigger of the two values:
Av,min=fc'(0.5)* 0.062*(bw*s/fyt)
Av,min=0.35*(bw*s/fyt)
Where:
s: stirrup center-to-center spacing, millimeters
fyt: stirrup steel bar yield stress, MPa
2. Minimum Transverse and Longitudinal Reinforcement in Cast-in-place Walls Use values from Table 1 as a minimum reinforcement for both longitudinal and transverse direction if the in-plane applied shear (Vu) of the cast-in-place wall is equal to or less than the value obtained.
On the other hand, (ρt= 0.0025) and the value of (ρ™) is the greatest of 0.0025 and the result from Equation 7 if the in-plane applied shear (Vu) is greater than the value obtained.
Where:
hw: total wall height measured in millimeters from base to top
lw: total wall length in millimeters
Between the earth and the RCC footing, there is a lean piece of concrete, or PCC:-
Lean concrete: Its functions include giving foundation concrete a consistent surface and preventing it from coming into direct touch with the ground. It is 5 centimeters thick. The mixing ratio is 150 kg/m3. 350 kg/m3 of concrete is mixed for column footings and ground beams.
Naturally, RCC has more strength as the reinforcement increases its capacity to carry loads. Because PCC is weaker, it is only utilized for layering surfaces, such as flooring or plastering, and—most importantly—for layering excavations before casting footings.
Duration of development for HYSD base
Results:
successfully finish the post-processing of the data when the analysis is finished. Checking the base pressure and settlement beneath the base slab, which should be within the parameters specified in Project-1, is the first duty. In order to regulate the base pressure and settlement, the dimensions must be raised or lowered if certain restrictions are exceeded.
Report the following findings once the foundation has been inspected in order to complete the element's structural design.
Design Base Slab and Wall Maximum and Minimum Moments (Wood Armer Simplified Method)
Design Shear Stress Maximum and Minimum for Base Slabs and Walls
Design Axial Stresses at Maximum and Minimum for Base Slabs and Walls
After the results have been post-processed, proceed with the structural design of the elements by performing the subsequent verifications:
Create an Excel spreadsheet to determine the amount of reinforcement needed for base slabs and both long and short walls. The following approach (covered in Week-3) can be used to compute the contribution due to direct tension, assuming it is significant, in order to determine the required design reinforcement:
M,net = T. x – M
(0.5 fck / fy) is Ast1. {1-√[1-(4.6 B.M./(fck.b.d²))]~
T / 0.87 fy = Ast2
Ast1 + Ast2 equals Total Ast, req.
M = Design moment derived using the Wood Armer formulas
T is the section's design tension force (stress x c/s area of the FE element).
x is the tension reinforcement's distance from the section's center line.
Verify that the foundation slab's and the walls' short and long crack widths are less than 0.2 mm.
Using CAD software, draft the cross-sectional reinforcement detail (plan and elevation). If CAD is not available, use any drawing tool to sketch the intricacies and make sure every detail is visible.