Project 2

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 = M – T . x

Ast1 = (0.5 fck / fy) {1-√[1-(4.6 B.M./(fck.b.d²))]}

Ast2 = T / 0.87 fy

Total Ast, req = Ast1 + Ast2

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.
  

Procedure unit weight as per the code

Safe Bearing capacity = 18 KN/m^3

Plain cement concrete = 24 KN/m^3

Reinforced cement concrete = 25KN/m^3

Cement concrete screed = 20KN / m^3

Cement masonry = 22 KN/m63

Structural steel = 78.5 KN/m^3

Loads of a building

Slab weight = 3.75kN/m^2(0.15mx25kN/m^3)

Partitions = 1.5kN/m^2

Floor finish = 2.5kN/m^2

, Others = 1kN/m^2

Total floor load = 8.75kN/m^2

Slab unit weights

Slab weight = 3.75kN/m^2(0.15mx25kN/m^3)

Insulations and water proofing = 0.5kN/m^2

MEP services = 0.5kN/m^2

Others = 0.5kN/m^2

Total roof load = 5.25kN/m^2

Common corridor = 3kN/m^2

Total floor load =3kN/m^2

Accessible roof = 1.5kN/m^2

Total roof load = 1.5kN/m^2

Design of slab(IS-456)

Span shorter direction(clear) Ly = 3.125m

Span longer direction(clear) Lx = 5.5m

Live load on the slab(LL) = 3kN/m

Compressive strength of concrete(Fck) = 25N/mm^2

Yield strength of steel(Fy) = 415N/mm^2

Unit weight of concrete(Yc) = 25kN/m^3

Unit weight of floor finish 100mm(Y) = 22kN/m^3

Clear concrete cover = 25mm

Bearing slab(B) =250mm

Step - 1

Span to effective depth ratio L/d = 32

Minimum effective depth(d) = 97.67mm

Overall depth(D) = 127.66mm

Dia of bars in shorter direction = 10mm

Dia of bars in longer direction = 10mm

Effective depth(d) = 120mm

Loading on the slab

Dead load of the slab(DL) = 3.75kN/m^2

Super dead load = 5kN/m^2

Live load on the slab = 3kN/m^2

Total load on the slan = 11.75kN/m^2

Design load = Total load x load factor

= 17.63kN/m^2

Effective span Lx = 3.25m

Ly = 5.62m

Lx/Ly = 1.732 (two way)

	1.5	Alpha x	1.75	Alpha y
For negative moments (at top)	0.075	0.104683	0.107	0.084
For positive moments (at bottom)	0.056	0.103307	0.107	0.063

Step - 2

BM per unit width of slab

	Mx=αxwLx^2	My=αywLy^2
For negative moments (at top)	19.43	15.59kN-m/m
For positive moments (at bottom)	19.17	11.69kN-m/m

Shear force V = 0.5WLx =28.6 kN/m

Step - 3

To check the effective depth of slab

Mu,lim = 0.138Fckbd^2

d = 74.55mm

Step - 4

Depth of slab for shear force

Tc,max = 3.1N/mm^2

Tc = 0.3

K = 1.2

D = 200mm

Tc = 0.3N/mm^2

Tv = Vu/bd

= 0.24N/mm^2

Tv<Tc<Tc,nax

0.24<0.36<3.1

SAFE

Step - 5

Determination of area of steel

Mu = 0.87FyAstd(1-AstFy/Fck bd)

For negative moment

Mu top = 19.43kN/m

Ast = 561.65mm^2/m

For positive moment

Mu bottom = 19.17kN/m

Ast = 554.27mm^2/m

Determination of distribution steel

Astmin = 0.12bd

= 114mm^2

Selection of steel reinforcing bars

Area of bars(Top steel) = 78.5mm^2

Spacing = 140mm

= 125mm

provide 16mm bars at a spacing 125mm

Area of bars(Bottom steel) = 78.5mm^2

Spacing = 142mm

=125mm

Provide 16mm bars at a spacing 125mm

Calculation of wind load (IS875-Part 3)

Design wind speed Vz = Vbk1k2k3k4 m/s

Basic wind speed(Bangalore) Vb = 33m/s

Probability factor k1 = 1

Terrain factor k2 = Category 4B

Topography factor k3 = 1

Design wind pressure 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

Calculation of Seismic Load (IS1893-Part I)

Design base shear VB = Ah x W

Where,

Ah = horizontal seismic coefficient as calculated as per 6.4.2

W = Total weight of the structure

Zone factor Z = 0.1 (Bangalore)

Soil condition factor Sa/g

Height of building h = 24.5m

Base dimension of building

Along-X = 25m

Along-Z = 16m

T along-X = 0.5513sec

T along-Z = 0.4410sec

Sa/g for hard soil = 1/T = 1.814 Along X

= 2.268 Along Z

Response reduction factor R = 3

Importance factor I = 1.2

Horizontal acceleration co-efficient Ah = 0.0363

= 3.6812%

Seismic weight

Area of each floor A = 400m2

Due to dead loads Typical floor DL = 12.5kN/m2

W DL = 5000kN

Due to dead loads, roof

DL roof = 9kN/m2

W DL roof = 3600kN

Due to live loads, Typical floor LL = 3kN/m2

W LL = 300kN/m2

Due to live loads, Roof LL = 1.5kN/m2

W LL roof = 0

Total seismic weight W = 35400kN

Design base shear VB = Ah x W

= 1605.44kN

The code specifies the minimum and maximum flexural reinforcement for beams or 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 utilized.

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:

As, min: Steel minimum area, mm^2
fc': concrete's compressive strength in MPa
fy: MPa, the steel yield stress
bw: both the beam's and the web's width in millimeters for rectangular and T-beams
D: Effective depth, measured in millimeters, from the center of the steel bars to the extreme concrete compression fiber.

Figure 1: Reinforcement bars, both longitudinal and shear
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.

Equation 2: 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

Figure 2: Reinforcement bars in a one-way concrete slab distributed according to temperature and shrinkage

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.
A concrete beam that is over-reinforced fails in compression before the steel bars reach their maximum strength. Equation 3 can be used to determine the maximum reinforcement ratio for beams.

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) Formula 4
Equation 5: Av,min=0.35*(bw*s/fyt)

Where:
s: stirrup center-to-center spacing, millimeters
fyt: stirrup steel bar yield stress, MPa

2. Minimal Transverse and Longitudinal Reinforcement for 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 from Equation 6.
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 from Equation 6.

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.

Development length for HYSD base in compression.

ld=0.87fyϕ4τbd(1.25×1.6)

=0.87×415×164×1.4×(1.25×1.6)=515.79mm

Outcome:
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 = M – T . x

Ast1 = (0.5 fck / fy) {1-√[1-(4.6 B.M./(fck.b.d²))]}

Ast2 = T / 0.87 fy

Total Ast, req = Ast1 + Ast2

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.