Question

A 6 m long simply- supported beam is prestressed as shown in the figure.


The beam carries a uniformly distributed load of 6 kN/m over its entire span. If the effective flexural rigidity EI = 2 $\times {10}^{4}kN{m}^2$ and the effective prestressing force is 200 kN, the net increase in length of the prestressing cable (in mm, up to two decimal places) is _______.

• A. 0.12

Answer 1

The correct option is A 0.12
Downward UDL, w = 6 kN/m
Eccentricity, e = 50 mm
Prestressing force, P = 200 kN
$EI=2\times {10}^{4}kN{m}^2$
L = 6 m
Rotation due to Prestress

$=\frac{PeL}{2EI}$

$=\frac{200\times 50\times {10}^{-3}\times 6}{2\times 2\times {10}^4}=1.5\times {10}^{-3}$

Rotation due to UDL

$=\frac{w{L}^3}{24EI}$

$=\frac{6\times (6{)}^3}{24\times 2\times {10}^4}=2.7\times {10}^{-3}$

Net rotation = $2.7\times {10}^{-3}-1.5\times {10}^{-3}$
$=1.2\times {10}^{-3}$ radian

Elongation of the cable
$=2\times 50\times 1.2\times {10}^{-3}$ mm
= 0.12 mm