Question 1 |

Consider a doubly reinforced RCC beam with the option of using either Fe250 plain bars or Fe500 deformed bars in the compression zone. The modulus of elasticity of steel is 2 \times 10^{5} \mathrm{~N} / \mathrm{mm}^{2}. As per IS456-2000, in which type(s) of the bars, the stress in the compression steel \left(f_{s c}\right) can reach the design strength \left(0.87 f_{y}\right) at the limit state of collapse?

Fe250 plain bars only | |

Fe500 deformed bars only | |

Both Fe250 plain bars and Fe500 deformed
bars
| |

Neither Fe250 plain bars nor Fe500 deformed
bars |

Question 1 Explanation:

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}}.

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}}.

Question 2 |

A singly reinforced concrete beam of balanced section is made of M20 grade concrete and \mathrm{Fe} 415 grade steel bars. The magnitudes of the maximum compressive strain in concrete and the tensile strain in the bars at ultimate state under flexure, as per IS 456: 2000 are _______ respectively. (round off to four decimal places)

0.0035 and 0.0038 | |

0.0020 and 0.0018 | |

0.0035 and 0.0041 | |

0.0020 and 0.0031 |

Question 2 Explanation:

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).

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).

Question 3 |

A post-tensioned concrete member of span 15 m and cross-section of
450 mm x 450 mm is prestressed with three steel tendons, each of
cross-sectional area 200 mm^2. The tendons are tensioned one after another
to a stress of 1500 MPa. All the tendons are straight and located at 125 mm
from the bottom of the member. Assume the prestress to be the same in all
tendons and the modular ratio to be 6. The average loss of prestress, due to
elastic deformation of concrete, considering all three tendons is

14.16 MPa | |

7.08 MPa | |

28.32 MPa | |

42.48 MPa |

Question 3 Explanation:

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

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

Question 4 |

A prismatic cantilever prestressed concrete beam of span length, L=1.5 m has one straight tendon placed in the cross-section as shown in the following figure (not to scale). The total prestressing force of 50 kN in the tendon is applied at d_{c}=50 \mathrm{~mm} from the top in the cross-section of width, b=200 mm and depth, d=300 mm.

If the concentrated load, P=5 kN, the resultant stress (in Mpa,in integer) experienced at point 'Q' will be _________

If the concentrated load, P=5 kN, the resultant stress (in Mpa,in integer) experienced at point 'Q' will be _________

3 | |

0 | |

5 | |

6 |

Question 4 Explanation:

\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}

Question 5 |

A concrete beam of span 15 m, 150 mm wide and 350 mm deep is prestressed with
a parabolic cable as shown in the figure (not drawn to the scale). Coefficient of friction
for the cable is 0.35, and coefficient of wave effect is 0.0015 per metre.

If the cable is tensioned from one end only, the percentage loss (round off to one decimal place) in the cable force due to friction, is ________.

If the cable is tensioned from one end only, the percentage loss (round off to one decimal place) in the cable force due to friction, is ________.

4.49 | |

2.45 | |

8.56 | |

9.47 |

Question 5 Explanation:

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}

There are 5 questions to complete.