Minimum grade of concrete to be used in the design of prestressed concrete structure as per IS 1343 is as below:

1. For Post tensioning minimum grade of concrete used is M-30.

2. For Pre-tensioning minimum grade of concrete used is M-40.

Cover to be used in the design of prestressed concrete structure as per IS 1343 is as below:

1. For Posttensioning minimum cover to be used is 30 mm.

Concept:

Post-Tensioned member: If all the cables are stressed simultaneously and anchored then there will not by any elastic shortening loss because the recording of prestress is done just before anchoring and cables are stretched by taking reaction from the concrete number. Hence at the time of recording of prestress, elastic shortening in concrete would have already occurred.

Hence there will be no loss in stress in the post-tensioned member when all the wires are stretched simultaneously.

Important Point:

Elastic shortening loss:

• Pre-tensioning member: When cables are stretched prestressing force is transferred to the member and concrete undergoes immediate elastic shortening due to the prestressed force. Due to the bond between steel and concrete, elastic shortening will also occur in steel leading to strain loss. Strain loss leads to stress loss.

As per clause 11.1.7.1 of IS 1343,

In case of single wires used in pre-tensioning system, the minimum clear spacing shall not be lesser than greater of the following:

1) 3 times the diameter of wire, and

2) \(1\frac{1}{3}\) times the maximum size of aggregate.

Concept:

Loss due to anchorage slip:

•  In a post-tensioned member in which prestress is transferred through concrete, the anchorage slip loss occurs before the tendon wedge assembly is properly sealed. The anchor slip reads to reduction in the strain in the cable (ε_s)

          Stress loss in cable due to anchorage slip = ε_s E_s

          Also, \({\varepsilon _s} = \frac{{{\rm{\Delta }}\ell }}{\ell }\)                     

where, Δℓ = Anchorage slip

Calculation:

ℓ = 9m = 9000 mm, Δℓ = 9 mm

\(\therefore {\varepsilon _s} = \frac{{{\rm{\Delta }}\ell }}{\ell } = \frac{9}{{9000}} = \frac{1}{{1000}}\)

∴ Stress loss in the cable due to anchorage slip = ε_s E_{s \(= \frac{1}{{1000}} \times 2 \times {10^5}\)} = 200 MPa    

Important Point:

Note that there is no loss due to anchorage slip in a pre-tensioned beam.

Since the applied load is uniformly distributed the tendon will follows parabolic profile for complete load balancing as shown below.

For complete load balancing

Intensity of applied U.D.L. = Intensity of U.D.L. produced by eccentric prestressing

\(\therefore w = \frac{{8P \times e}}{{{L^2}}}\)

\(\therefore 30 = \frac{{8 \times 1400 \times e}}{{{8^2}}}\)

∴ e = 0.17143 m = 171.43 mm.

Calculation:

When load is placed eccentrically, the stress is calculated as:

Thus the stress distribution is

e = 150 – 50 = 100 mm

\(P = Stress \times Area = 1200 \times \left( {4 \times \frac{\pi }{4} \times {5^2}} \right)\) = 94.25 kN

\(\therefore {\sigma _A} = \frac{P}{A} + \frac{{Pe}}{Z} = \frac{{94.25 \times {{10}^3}}}{{150 \times 300}} + \frac{{94.25 \times {{10}^3} \times 100}}{{\left( {\frac{{150 \times {{300}^2}}}{6}} \right)}}\) = 6.28 MPa

Concept:

Loss due to friction = P_x – P₀

P_X = P_o [1 – (μα + kx)]

Where

P_x = Initial prestress

P_o = stress in wire while jacking

μ = coefficient of friction

k = wobbling constant

α = cumulative angle of inclination for prestress bar

x = distance for a particular section of the beam taken from face end

Calculation:

Given:

Coefficient of friction (μ) = 0.55

Friction coefficient of wave effect = 0.0015/m

\({\rm{Slope\;of\;angle\;}} = \left( {\frac{{4{\rm{e}}}}{{\rm{L}}}} \right) = \left( {\frac{{4\left( {50 + 75} \right)}}{{10 \times {{10}^3}}}} \right) = 0.05\)

Cumulative angle (α) = 2 × 0.05 = 0.1

P_o = stress in wires at jacking end

P_x = 1000 MPa

∴ P_x = P_o [1 – (μα + kx)]

1000 = P_o [1 – (0.55 × 0.1 + 0.0015 × 10)]

\(\therefore {{\rm{P}}_{\rm{o}}} = \frac{{1000}}{{0.93}} = 1075.268{\rm{\;MPa}}\)

Loss of stress in cable = 1075.268 – 1000 = 75.268 MPa

Concept:

The unit weight of PCC is 24 kN/m.

Under the limiting condition for no tension at the soffit of the beam, the stress should be zero.

Calculation:

Dead load = 24 × 0.4 × 0.25 × 1 = 2.4 kN/m

∴ Total load = L.L + D.L = 2.4 + 4 = 6.4 kN/m

The maximum moment in a simply supported beam occurs at the mid-span and it is given as \(\frac{{\omega {\ell ^2}}}{8}\)

∴ M at mid-span \(= \frac{{\omega {\ell ^2}}}{8} = \frac{{6.4\; \times \;{8^2}}}{8} = 51.2\;kN - m\)

Let the eccentricity be e.

For no tension at base, σ_A = 0

\(\Rightarrow {\sigma _A} = \frac{P}{A} + \frac{{Pe}}{z} - \frac{M}{z} = 0\)

\(\Rightarrow \frac{{400\; \times \;{{10}^3}}}{{250\; \times \;400}} + \frac{{400\; \times \;{{10}^3}\; \times \;e}}{{\left( {\frac{{250\; \times\; {{400}^2}}}{6}} \right)}} - \frac{{51.2\; \times\; {{10}^6}}}{{\left( {\frac{{250\; \times \;{{400}^2}}}{6}} \right)}} = 0\)  ⇒ e = 61.33 mm

Mistake point:

In prestress, the design is done using the working stress method, and hence factored load is not considered.

Effective Prestressing force = (1 - 0.11) × 240

P = P_eff = 213.6 kN

Span = 7m

EI = 2 × 10⁴ kN - m²

Slope of beam due to UDL:

\({\theta _1} = \frac{{w{L^3}}}{{24EI}}\)

\({\theta _1} = \frac{{8 \times {{\left( {7000} \right)}^3}}}{{24 \times 2 \times {{10}^{13}}}} = 5.7 \times {10^{ - 3}}\;\left( {down\;wood} \right)\)

Slope of beam due to Prestressing force:

Using area moment method

\({\theta _2} - {\theta _C} = \frac{{Pe}}{{EI}} \times \frac{L}{2}\)

θ_C = 0

\({\theta _2} = \frac{{ 213.6 \times {{10}^3} \times 75 \times 7000}}{{2 \times {{10}^4} \times {{10}^9} \times 2}} = 2.80 \times {10^{ - 3}}\)

Net slope of the beam = θ₁ - θ₂

θ_net = 5.7 × 10^-3 – [2.80 × 10^-3]

θ_net = 2.9 × 10^-3

Now,

Total increase in length = 2eθ

= 2 × 75 × 2.9 × 10^-3

= 0.435 mm

Concept:

Loss in prestress due to creep:

Creep loss = Ultimate creep strain × E_s

Ultimate creep coefficient (θ) \(= \frac{{Ultimate\;creep\;strain}}{{Elastic\;strain}}\)

⇒ Ultimate creep strain = Elastic strain × θ

Also, Elastic strain \(= \frac{{{f_c}}}{{{E_c}}}\)

\(\therefore Creep\;loss = \frac{{{f_c}}}{{{E_c}}} × \theta × {E_s}\) = m × f_c × θ

Where,  f_c is the stress in concrete at the level of prestressing cable.

Generally, the f_c calculated shall be the value of f_c after deducting the initial losses or short term losses. Creep losses are around 2 – 3% of the initial prestress.

\({f_{CA}} = \frac{P}{A}\;\;\;;\;\;{f_{CB}} = \frac{P}{A} + \frac{{P{e^2}}}{I}\)

\({f_{Cavg}} = \frac{{{f_{CA}} × \frac{\ell }{2} + \frac{2}{3} × \frac{\ell }{2}\left( {{f_{CB}} - {f_{CA}}} \right)}}{{\frac{\ell }{2}}}\)

\(\Rightarrow {f_{Cavg}} = {f_{CA}} + \frac{2}{3}\left( {{f_{CB}} - {f_{CA}}} \right)\)

Calculation:

P = 1400 × 50 = 70 kN, b = 150 mm, d = 300 mm

Stress in concrete at the support \(A = {f_{CA}} = \frac{P}{A} = \frac{{\left( {1400\; × \;50} \right)}}{{150\; × \;300}} = 1.56\;MPa\)

Stress at the mid-span at B \(= {f_{CB}} = \frac{P}{A} + \frac{{P{e^2}}}{I}\) = \(\frac{{\left( {1400\; × \;50} \right)}}{{150\; × \;300}} + \frac{{1400\; × \;50\; × \;{{60}^2}\; × \;12}}{{150\; × \;{{300}^3}}}\) = 2.302 Mpa

\({f_{cavg}} = {f_{CA}} + \frac{2}{3}\left( {{f_{CB}} - {f_{CA}}} \right) = 1.56 + \frac{2}{3}\left( {2.302 - 1.56} \right)\) = 2.055 MPa

∴ Loss due to creep = m × f_cAvg × θ = 6 × 2.055 × 1.6 = 19.728 MPa