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Moment of Force

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A beam with a rectangular cross section of size 250 mm wide and 350 mm deep is prestressed by a force of 400 kN using number 7 mm $ϕ$ steel cables located at an eccentricity of 75 mm. Determine the loss of prestress due to creep of concrete. Grade of concrete is M 40; Coefficient of creep is 2; stress at transfer is 80%, Modulus of elasticity of steel ( $E_{s}$ ) is 2.0 x $10^{5}$ MPa.

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Solution

Prestressing force after transfer = 80% of initial
$= 0.8 \times 400$
= 320 kN
Stress in concrete at the level of steel

$f_{c} = \frac{P}{A} + \frac{P e . e}{I}$
$= \frac{320 \times 10^{3}}{250 \times 350} + \frac{320 \times 10^{3} \times 75 \times 75}{\frac{250 \times 350^{3}}{12}}$
$= 5.672 N / m m^{2}$
As per IS : 456 - 2000
Modulus of elasticity of concrete
$E_{c} = 5000 \sqrt{f_{c k}}$
$= 5000 \sqrt{40}$
$= 0.3162 \times 10^{5} N / m m^{2}$
Loss of prestress due to creep $= m ϕ f_{c}$
$= \frac{E_{s}}{E_{c}} ϕ f_{c}$
$= \frac{2 \times 10^{5}}{0.3162 \times 10^{5}} \times 2 \times 5.672$
$= 71.745 N / m m^{2}$

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