RCC Structures

Given that,
Coarse aggregate = 260 gms
Fine aggregate = 240 gms
G_{CA}=2.6
G_{FA}=2.4
Bulk specific gravity G_m=2.3
Percentage air voids in the mix = ?
G_t (Theoretical specific gravity)

\begin{aligned} &=\frac{\Sigma W}{\Sigma \frac{W}{G}}\\ &=\frac{260+240}{\frac{260}{2.6}+\frac{240}{2.4}}\\ &=2.5 \end{aligned}
\begin{aligned} % \text{ air voids} (V_V)&=\frac{G_t-G_m}{G_t} \times 100\\ &=\frac{2.5-2.3}{2.3}\times 100\\ V_V&=8% \end{aligned}

This is a question of calculation of elastic shortening loss in post tensioned member.
Given data,
length = 15 m
b x D = 450 mm x 450 mm
Numberof tendons = 3
Cross-section area of each tendon (A_s)=200 mm^2.
Prestress = 1500 MPa
Modular ratio (m) = 6

From the given data, eccentricity (e) = 450/2-125 = 100 mm
Force in each cable (P) = 1500 \times 200 \times 10^{-3} = 300 kN
The tendons are tensioned one after another, and hence when tendon (1) is pulled no loss in tenson (1)
When tendon (2) is pulled, loss in tendon (1) but no loss in tendon (2).
When tendon (3) is pulled, loss in tendon (1) and (2), but no loss in tendon (3).
Hence, there will be 2 times losses in tendon (1), time loss in tendon (1) and no loss in tendon (3).
While calculating elastic shortening loss, self weight of the structure is neglected to be on the conservative side.
Consider tensioning of tendon-1
No loss in tendon (1)
Consider tensioning of tendon-2

Stress in concrete at the level of prestressing tendon
\begin{aligned} f_c&= \frac{P}{A}+\frac{Pe}{I}\\ e&=\frac{300 \times 10^3}{450 \times 450}+\frac{300 \times 10^3 \times 100 \times 100}{\frac{450 \times 450^3}{12}} \\ &= 2.36MPa \end{aligned}

I = moment of inertial of the section about the centroidal axis.
As the tendons are horizontal and at the same level f_{c,avg} = f_c.
Loss due to elastic deformation = mfc = 6 \times 2.36 = 14.16 MPa.
Considering tensioning of tendon 3
Loss due to elastic deformation in (1) = mfc = 6 \times 2.36 = 14.16 MPa.
Loss due to elastic deformation in (2) = mfc = 6 \times 2.36 = 14.16 MPa.
Total loss in tendon (1) = 2 x 14.16 = 28.32 MPa
Total loss in tendon (1) = 2 x 14.16 = 28.32 MPa
In tendon (2) = 14.16 MPa
In tendon (3) = 0
Average loss of pre-stress, considering all three tendons is =\frac{28.32+14.16+0}{3}=14.16MPa

C_3S- Responsible for early age strength
C_2S - Responsible for later age strength and lowest heat of hydration
C_3A- Flash setting and highest heat of hydration

The question is based on LSM design principle as it is describing different conditions related to strain
Depending on amount of reinforcement in a cross- section, here ca be three types of sections viz. balanced, under reinforced and over reinforced.
Balanced section is a section that is expected to result in a balanced failure. It means at the ultimate limit state in flexure, the concrete will attain a limiting compressive strain of 0.0035 and steel will attain minimum specified tensile strain of 0.002+\frac{0.87f_y}{E_s}
Under reinforced section is a section in which steel yield before collapse. Over reinforced section is a section in which crushing of concrete in compression i.e. attainment of compressive strain of 0.0035 occurs prior to yielding of steel.
In case of over reinforced section the deflection, crack width remain relatively low and failure occurs without any sign of warning and hence over reinforced flexural members are not recommended by IS code.
Based on the above information:
Statement I is true.
Statement II is true.
Statement III is false.
Statement IV is false.

b=300mm
d=580 mm
V_u=440 kN
Concrete used is M30
Raft steel is Fe415
V_{cu}=\tau _c Bd=0.66 \times 300 \times \frac{580}{1000}=114.84 kN
V_{su}=V_u-V_{cu}=440-114.84=325.16 kN
Spacing of 2-legged shear reinforcement
\begin{aligned} s_V&=\frac{A_{SV} \times 0.87 f_y \times d}{V_{su}}\\ &=\frac{2 \times \frac{\pi}{2} \times (10)^2 \times 0.87 \times 415 \times 580}{325.16 \times 1000}\\ &=101.16 mm \end{aligned}

Pozzolanic materials have no cementing properties itself but have the property of combining with lime to produce stable compound.
Pozzolana is considered as siliceous and aluminous materials and when added in cement it have SiO_2 and Al_2O_3 form.
So, pozzolanic reaction :
H_2O + Reactive slilica-di-oxide + H_2O \rightarrow C-S-H gel or tobermonite gel

This is a question of working stress method i.e. elastic theory.
Modular ratio
=\frac{E_s}{E_c}=\frac{\text{Young's modulus of steel}}{\text{Young's modulus of concrete}}

\begin{aligned} m&=\frac{E_{s}}{E_{C}}=\frac{2.1 \times 10^{5}}{2 \times 10^{4}}=10.5 \\ A_{s t}&=3 \times \frac{\pi}{4}(16)^{2}=603.20 \mathrm{~mm}^{2} \\ \bar{y}&=\frac{\left(B \cdot D \cdot \frac{D}{2}+(m-1) \times A_{s t} \times d\right)}{B \cdot D+(m-1) \cdot A_{s t}} \\ &=\frac{\left(200 \times \frac{350^{2}}{2}+(10.5-1) \times 603.2 \times 315\right)}{200 \times 350+(10.5-1) \times 603.2}=185.59 \mathrm{~mm}\\ & \text{Distance of N-A from reinforcement}\\ y_{2} &=d-\bar{y} \\ &=315-185.59=129.41 \mathrm{~mm} \end{aligned}

C.G. of load from P_{1}

\begin{aligned} P_{R} \bar{x} &=P_{1} \times 0+P_{2} \times 5 \\ \bar{x} &=\frac{1500 \times 5}{3500}=2.143 \mathrm{~m} \end{aligned}
Distance of C.G. of footing from face of P_{1}
\bar{y}=\bar{x}+0.25=2.393 \mathrm{~m}

C.G of footing \begin{aligned} \bar{y} &=\left(\frac{B_{1}+2 B_{2}}{B_{1}+B_{2}}\right) \times \frac{L}{3} \\ 2.393 &=\left(\frac{5+2 \times 1.5}{5+1.5}\right) \times \frac{L}{3} \\ L &=5.833 \mathrm{~m} \text { say } 5.83 \mathrm{~m} \end{aligned}

\begin{aligned} \mathrm{e} &=\frac{D}{2}-50=\frac{300}{2}-50=100 \mathrm{~mm} \\ \mathrm{DL} &=0.2 \times 0.3 \times 1.0 \times 25=1.50 \mathrm{kN} / \mathrm{m} \\ P &=50 \mathrm{kN}=50000 \mathrm{~N} \\ W &=5 \mathrm{kN} \\ \text { Maximum BM } &=\frac{w^{2}}{2}+\mathrm{W} \\ &=\frac{1.5 \times 1.5^{2}}{2}+5 \times 1.50=9.1875 \mathrm{kNm} \end{aligned}

\begin{aligned} \frac{P}{A}&=\frac{50000}{200 \times 300}=0.833 \mathrm{~N} / \mathrm{mm}^{2} \\ \frac{P e}{Z}&=\frac{50000 \times 100}{200 \times \frac{300^{2}}{6}}=1.67 \mathrm{~N} / \mathrm{mm}^{2} \\ \frac{M}{Z}&=\frac{9.1875 \times 10^{6}}{200 \times \frac{300^{2}}{6}}=3.0625 \mathrm{~N} / \mathrm{mm}^{2}\\ \text{Stress at Q,}\qquad \qquad\\ \text { Stress at } Q &=\frac{P}{A}+\frac{P e}{Z}-\frac{M}{Z} \\ &=0.833+1.67-3.0625 \\ &=-0.56 \mathrm{~N} / \mathrm{mm}^{2} \text { (Tensile) } \end{aligned}