Prestressed Concrete Beams

In a doubly reinforced beam, at the limit state of collapse. The strain in the extreme corression fibre will be 0.0035. As the compression reinforcement will be below this, the strain in if will be always less than 0.0035.
For Fe250, the yield strain
=\frac{0.87 \times 250}{2 \times 10^{5}}=0.0010875

For Fe500, the yield strain
\begin{aligned} & =0.002+\frac{0.87 \mathrm{f}_{y}}{E_{s}} \\ & =0.002+\frac{0.87 \times 500}{2 \times 10^{5}} \\ & =0.004175>0.0035 \end{aligned}
So, it is possible that \mathrm{Fe} 250 can reach upto its yield strain and hence can reach the design strength 0.87 \mathrm{f}_{\mathrm{y}}.

Given data,
Grade of concrete M-20
Grade of steel Fe-415
Balanced section, singly reinforced beam.
As per Clause No. 38.1, IS 456: 2000,
Maximum strain in concrete at the outermost compression fibre =0.0035
and strain in the tension reinforcement for balanced section at ultimate state under flexure
\begin{aligned} & =0.002+\frac{f_{y}}{1.15 E_{s}} \\ & =0.002+\frac{415}{1.15 \times 2 \times 10^{5}}=0.0038 \end{aligned}
Hence, correct answer is (A).

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

\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}

Jacking from one end
x = L = 15 m
Wobble correction factor, K = 0.0015
Coefficient of friction = 0.35 = \mu
P = Not given
p_0 = Unknown
Change of gradient, \alpha = \tan \alpha =\frac{8h}{L}=\frac{8 \times 120}{15000}=0.064
% loss of stress in steel due to friction
\begin{aligned} &=\frac{p_0(Kx+\mu \alpha )}{p_0} \times 100\\ &=(0.0015 \times 15+0.35 \times 0.064)\times 100 \\ &= 4.49\% \end{aligned}