Question 1 |

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 1 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 2 |

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 2 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 3 |

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 3 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}

Question 4 |

A simply supported prismatic concrete beam of rectangular cross-section, having a span
of 8 m, is prestressed with an effective prestressing force of 600 kN. The eccentricity
of the prestressing tendon is zero at supports and varies linearly to a value of e at
the mid-span. In order to balance an external concentrated load of 12 kN applied at
the mid-span, the required value of e (in mm, round off to the nearest integer) of the
tendon, is _______.

25 | |

40 | |

80 | |

120 |

Question 4 Explanation:

P = 600 kN

Simply supported span = L = 8 m

To support a point load applied at mid span (W) = 12 kN

\begin{aligned} \text{Balancing load} &= \text{Point load}\\ 2P \sin \theta &=Q\\ 2P\left ( \frac{e}{L/2} \right )&=2\\ \frac{2Pe \times 2}{L}&=W\\ \frac{4Pe}{L}&=W\\ e&=\frac{WL}{4P}\\ &=\frac{12000 N \times 8000mm}{4 \times 600 \times 1000N}\\ &=40mm \end{aligned}

Question 5 |

A 3mx3m square precast reinforced concrete segments to be installed by pushing them through an existing railway embankment for making an underpass as shown in the figure. A reaction arrangement using precast PCC blocks placed on the ground is to be made for the jacks.

At each stage, the jacks are required to apply a force of 1875 kN to push the segment. The jacks will react against the rigid steel plate placed against the reaction arrangement. The footprint area of reaction arrangement on natural ground is 37.5 m^2. The unit weight of PCC block is 24 kN/m^3. The properties of the natural ground are: c = 17 kPa; \phi =25^{\circ} and \gamma =18 kN/m^3. Assuming that the reaction arrangement has rough interface and has the same properties that of soil, the factor of safety (round off to 1 decimal place) against shear failure is _____

At each stage, the jacks are required to apply a force of 1875 kN to push the segment. The jacks will react against the rigid steel plate placed against the reaction arrangement. The footprint area of reaction arrangement on natural ground is 37.5 m^2. The unit weight of PCC block is 24 kN/m^3. The properties of the natural ground are: c = 17 kPa; \phi =25^{\circ} and \gamma =18 kN/m^3. Assuming that the reaction arrangement has rough interface and has the same properties that of soil, the factor of safety (round off to 1 decimal place) against shear failure is _____

2.0 | |

2.4 | |

1.2 | |

2.8 |

Question 5 Explanation:

Weight of Precast PCC block,

W=volume \times \gamma _c=37.5 \times 7.5 \times 24=6750kN

Shear resistance below the Precast PCC block is,

S = C.A + W\tan \phi

F.O safety against shear failure,

\begin{aligned} F&=\frac{S}{T} \\ F&=\frac{C.A.+W \tan \phi }{T} \\ &= \frac{C.A.+N \tan \phi }{T}\;\;\; (Since, \; N=W)\\ &= \frac{17 \times 37.5+6750 \times \tan 25^{\circ}}{1875}\\ &= 2.0187\approx 2.0 \end{aligned}

W=volume \times \gamma _c=37.5 \times 7.5 \times 24=6750kN

Shear resistance below the Precast PCC block is,

S = C.A + W\tan \phi

F.O safety against shear failure,

\begin{aligned} F&=\frac{S}{T} \\ F&=\frac{C.A.+W \tan \phi }{T} \\ &= \frac{C.A.+N \tan \phi }{T}\;\;\; (Since, \; N=W)\\ &= \frac{17 \times 37.5+6750 \times \tan 25^{\circ}}{1875}\\ &= 2.0187\approx 2.0 \end{aligned}

Question 6 |

A 6 m long simply-supported beam is prestressed as shown in the figure.

The beam carries a uniformly distributed load of 6 kN/m over its entire span. If the effective flexural rigidity EI=2\times 10^{4}kN/m^{2} and the effective prestressing force is 200 kN, the net increase in length of the prestressing cable (in mm, up to two decimal places) is ______

The beam carries a uniformly distributed load of 6 kN/m over its entire span. If the effective flexural rigidity EI=2\times 10^{4}kN/m^{2} and the effective prestressing force is 200 kN, the net increase in length of the prestressing cable (in mm, up to two decimal places) is ______

0 | |

0.12 | |

0.25 | |

0.48 |

Question 6 Explanation:

\begin{aligned} \text { Span of PSC beam }&=6 \mathrm{m} \\ E I&=2 \times 10^{4} \mathrm{kNm}^{2} \\ &=2 \times 10^{13} \mathrm{N}-\mathrm{mm}^{2} \\ P&=200 \mathrm{kN} \\ \text { Total UDL }&=6 \mathrm{kN} / \mathrm{m} \\ \text { eccentricity }&=e=50 \mathrm{mm} \end{aligned}

(a) Slope of beam due to P-force

\theta_{1}=\frac{P \cdot e \cdot L}{2 E I}

\begin{array}{l} =\frac{200 \times 10^{3} \times 50 \times 6000}{2 \times 2 \times 10^{13}} \\ =1.5 \times 10^{-3} \text {(upward) } \end{array}

(b) Slope of beam due to UDL

\theta_{2}=\frac{w^{3}}{24 E I}

\begin{array}{l} =\frac{6 \times(6000)^{3}}{24 \times 2 \times 10^{13}} \\ =(+) 2.7 \times 10^{-3} \text {(downward) } \end{array}

(c) Net slope of beam

\begin{aligned} \theta &=\theta_{1}+\theta_{2} \\ &=(-) 1.5 \times 10^{-3}+(+) 2.7 \times 10^{-3} \\ &=1.2 \times 10^{-3} \end{aligned}

Total net increase in length

\begin{aligned} &=2 e \theta \\ &=2 \times 50 \times 1.2 \times 10^{-3} \\ &=0.12 \mathrm{mm} \end{aligned}

(a) Slope of beam due to P-force

\theta_{1}=\frac{P \cdot e \cdot L}{2 E I}

\begin{array}{l} =\frac{200 \times 10^{3} \times 50 \times 6000}{2 \times 2 \times 10^{13}} \\ =1.5 \times 10^{-3} \text {(upward) } \end{array}

(b) Slope of beam due to UDL

\theta_{2}=\frac{w^{3}}{24 E I}

\begin{array}{l} =\frac{6 \times(6000)^{3}}{24 \times 2 \times 10^{13}} \\ =(+) 2.7 \times 10^{-3} \text {(downward) } \end{array}

(c) Net slope of beam

\begin{aligned} \theta &=\theta_{1}+\theta_{2} \\ &=(-) 1.5 \times 10^{-3}+(+) 2.7 \times 10^{-3} \\ &=1.2 \times 10^{-3} \end{aligned}

Total net increase in length

\begin{aligned} &=2 e \theta \\ &=2 \times 50 \times 1.2 \times 10^{-3} \\ &=0.12 \mathrm{mm} \end{aligned}

Question 7 |

A simply supported rectangular concrete beam of span 8m has to be prestressed with a force of 1600 kN. The tendon is of parabolic profile having zero eccentricity at the supports. The beam has to carry an external uniformly distributed load of intensity 30 kN/m. Neglecting the self-weight of the beam, the maximum dip (in meters, up to two decimal places) of the tendon at the mid-span to balance the external load should be_______

0.3 | |

0.45 | |

0.15 | |

0.4 |

Question 7 Explanation:

\text { Central dip, } e=\frac{w^{2}}{8 P}=\frac{30 \times 8^{2}}{8 \times 1600}=0.15 \mathrm{m}

Question 8 |

A pre-tensioned rectangular concrete beam 150 mm wide and 300 mm depth is prestressed with three straight tendons, each having a cross-sectional area of 50 mm^{2}, to an initial stress of 1200 N / mm^{2}. The tendons are located at 100 mm from the soffit of the beam. If the modular ratio is 6, the loss of prestressing force (in kN, up to one decimal place) due to the elastic deformation of concrete only is ______

4 | |

2 | |

2.8 | |

4.8 |

Question 8 Explanation:

Prestressing force,

\begin{aligned} P&=3 \times 50 \times 1200=180000 \mathrm{N} \\ e&=\left(\frac{D}{2}-100\right)=\left(\frac{300}{2}-100\right)=50 \mathrm{mm} \end{aligned}

Stress in concrete at the location of steel

\begin{aligned} &=\frac{P}{A}+\frac{P \cdot e}{I} e\\ &=\frac{180000}{150 \times 300}+\frac{180000 \times 50^{2}}{\left(\frac{150 \times 300^{3}}{12}\right)} \\ &=4.0+1.33=5.333 \end{aligned}

Loss of stress

=m \times f_{c}=6 \times 5.3333=32 \mathrm{N} / \mathrm{mm}^{2}

Loss of prestressing force

=\frac{3 \times 50 \times 32}{1000} \mathrm{kN}=4.8 \mathrm{kN}

Question 9 |

In a pre-stressed concrete beam section shown in the figure, the net loss is 10% and the final pre-stressing force applied at X is 750 kN. The initial fiber stresses (in N/mm^{2} ) at the top and bottom of the beam were:

4.166 and 20.833 | |

-4.166 and -20.833 | |

4.166 and -20.833 | |

-4.166 and 20.833 |

Question 9 Explanation:

Given, net loss =10 %

Final prestressing force after loss =750 kN

Hence, initial prestressing force

=\frac{750}{0.9}=833.33 \mathrm{kN}

Initial stresses at top and bottom

\begin{array}{l} =\frac{P}{A} \pm \frac{P e}{Z}=\frac{833.33 \times 10^{3}}{250 \times 400} \pm \frac{833.33 \times 10^{3} \times 100 \times 6}{250 \times 400^{2}} \\ =8.33 \pm 12.50 \\ \end{array}

Stress at top =8.33-12.50=-4.17 (tensile)

Stress at bottom =8.33+12.50=20.83 (Comp.)

Final prestressing force after loss =750 kN

Hence, initial prestressing force

=\frac{750}{0.9}=833.33 \mathrm{kN}

Initial stresses at top and bottom

\begin{array}{l} =\frac{P}{A} \pm \frac{P e}{Z}=\frac{833.33 \times 10^{3}}{250 \times 400} \pm \frac{833.33 \times 10^{3} \times 100 \times 6}{250 \times 400^{2}} \\ =8.33 \pm 12.50 \\ \end{array}

Stress at top =8.33-12.50=-4.17 (tensile)

Stress at bottom =8.33+12.50=20.83 (Comp.)

Question 10 |

A rectangular concrete beam 250mm wide and 600mm deep is pre-stressed by means of 16 high
tensile wires, each of 7 mm diameter, located at 200mm from the bottom face of the beam at a
given section. If the effective pre-stress in the wires is 700MPa, what is the maximum sagging
bending moment (in kNm) (correct to 1-decimal place) due to live load that this section of the beam
can with stand without causing tensile stress at the bottom face of the beam? Neglect the effect of dead load of beam.

25.65 | |

46.36 | |

86.21 | |

98.64 |

Question 10 Explanation:

since the tensile stress at bottom face of the beam is zero.

\begin{aligned} \frac{P}{A}+\frac{P e}{I} y-\frac{M}{I} y &=0 \\ P &=700 \times \frac{\pi}{4}(7)^{2} \times 16 \\ &=431.026 \times 10^{3} \mathrm{N} \end{aligned}

since the prestressing force is located at 200 mm from the bottom face of the beam.

\therefore Eccentricity =300-200=100 \mathrm{mm}

\begin{array}{l} \frac{431.03 \times 10^{3}}{250 \times 600}+\frac{431.03 \times 10^{3} \times 100}{250 \times 600^{2}} \times 6 \\ =\frac{M \times 6}{250 \times 600^{2}} \\ \Rightarrow 2.87+2.87=6.66 \times 10^{-8} M \\ \Rightarrow M=86.205 \mathrm{kNm} \end{array}

\begin{aligned} \frac{P}{A}+\frac{P e}{I} y-\frac{M}{I} y &=0 \\ P &=700 \times \frac{\pi}{4}(7)^{2} \times 16 \\ &=431.026 \times 10^{3} \mathrm{N} \end{aligned}

since the prestressing force is located at 200 mm from the bottom face of the beam.

\therefore Eccentricity =300-200=100 \mathrm{mm}

\begin{array}{l} \frac{431.03 \times 10^{3}}{250 \times 600}+\frac{431.03 \times 10^{3} \times 100}{250 \times 600^{2}} \times 6 \\ =\frac{M \times 6}{250 \times 600^{2}} \\ \Rightarrow 2.87+2.87=6.66 \times 10^{-8} M \\ \Rightarrow M=86.205 \mathrm{kNm} \end{array}

There are 10 questions to complete.