RCC Design Test 4

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 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