Question

A simply supported prestressed beam of span $8m$ is subjected to an uniformly distributed loading of $8kN/m$ on its entire span. The prestressing force of $200kN$ is applied to tendons at an uniform eccentricity of $50mm$. The flexural rigidity of beam is $2.5\times {10}^{4}kN/m^2$. The net increase in length (in mm, rounded off to two decimal place) of the tendons is ____.

• A. 0.18

Answer 1

The correct option is A 0.18
Given, $P=200kN,e=50mm$
$EI=2.5\times {10}^{4}kN/m^2$
$L=8m,W=6kN/m$Moment due to prestress, $M=Pe=200\times \frac{50}{1000}=10kN-m$
Rotation due to prestress, ${\theta }_1=\frac{ML}{2EI}$
$=\frac{10\times 8}{2.5\times {10}^4}=3.2\times {10}^{-3}rad$
Rotation due to $udl=\frac{w{L}^3}{24EI}$
${\theta }_2=\frac{6\times 8^3}{24\times 2.5\times {10}^4}=5.12\times {10}^{-3}rad$
Net rotation $={\theta }_2-{\theta }_1$
${\theta }_{net}=1.92\times {10}^{-3}$
Elongation in tendon $=2e{\theta }_{net}$
$=2\times 50\times 1.92\times {10}^{-3}$
$=0.192mm$