Question

A prestressed concrete rectangular beam has a width = 400 mm and a depth = 900 mm. It is simply supported over a span length of 12 m. The beam is subjected to a prestressing force of 2000 kN using a parabolic cable profile having an eccentricity of 240 mm at mid span and zero eccentricity at the supports. Calculate the minimum dead load moment including the self weight and the corresponding maximum possible live load moment at the centre span of the beam at service load, if the service load stress in the beam should be never tensile.

Answer 1

At transfer stage

Only dead load will act at transfer. Let moment due to dead load be $M_D$,
Stress in bottom fibre

${\sigma }_b=\frac{P}{A}+\frac{P.e.y_b}{I}-\frac{M_D{y}_b}{I}$
[Assuming compression as positive]
For no tension ,${\sigma }_b=0$
$\therefore \frac{M_D{y}_b}{I}=\frac{P}{A}+\frac{P.e.y_b}{I}$ ...(i)
Similarly, stress in top fibre
${\sigma }_t=\frac{P}{A}-\frac{P.e{y}_t}{I}+\frac{M_D{y}_t}{I}$
$\Rightarrow$ for no tension, ${\sigma }_t=0$
$\frac{M_D{y}_t}{I}=\frac{Pe{y}_t}{I}-\frac{P}{A}$ ...(ii)

As it can be easily seen from equation (i) and equation (ii), that value of $M_D$ coming from equation (ii) will be less. Hence, minimum dead load moment will be.

$\frac{M_D{y}_1}{I}=\frac{P.e{y}_T}{I}-\frac{P}{A}$
$\Rightarrow \frac{M_D\times 450}{400\times \frac{{900}^3}{12}}=\frac{2000\times {10}^3\times 240\times 450}{400\times \frac{{900}^3}{12}}-\frac{2000\times {10}^3}{400\times 900}$

$\frac{M_D\times 450\times 12}{400\times {900}^3}=3.33$
$M_D=180$ kN.m Ans.

After loading :

Let moment due to live load be $M_L$ at top-fibre
${\sigma }_t=\frac{P}{A}-\frac{Pe.y_t}{I}+\frac{M_D{y}_t}{I}+\frac{M_L{y}_t}{I}$

From above equation, it can be concluded that whatever be the value of $M_L,{\sigma }_t$ is always.

Positive (i.e., compressive) . Therefore, value of $M_L$ should be calculated by analyzing bottom fibre
At bottom fibre
${\sigma }_b=\frac{P}{A}+\frac{Pe.y_b}{I}-\frac{M_D{y}_b}{I}-\frac{M_L{y}_b}{I}$
For no tension ${\sigma }_b=0$
$\Rightarrow \frac{M_L{y}_b}{I}=\frac{P}{A}+\frac{P.e{y}_b}{I}-\frac{M_D.y_b}{I}$
$\frac{M_L\times 450}{400\times \frac{{900}^3}{12}}=\frac{2000\times {10}^3}{400\times 900}+\frac{2000\times {10}^3\times 240\times 450}{400\times \frac{{900}^3}{12}}-\frac{180\times {10}^6\times 450}{400\times \frac{{900}^3}{12}}$

$\Rightarrow \frac{M_L\times 450\times 12}{400\times {900}^3}=5.5556+8.889-3.33$

$\Rightarrow ML=600$ kN.m