Question

A propped cantilever AB of 8 m span with flexural rigidity of $2.8\times {10}^{6}kN-c{m}^2$ is subjected to 16 kN load at mid span. When the support B sinks by 1 cm, the bending moment (in kNm) at support A is

• A. 48.20
• B. 15.75
• C. 24.10
• D. 30.25

Answer 1

The correct option is C 24.10

$M_{FAB}=\frac{-16\times 8}{8}=-16kN-m$
$M_{FAB}=\frac{16\times 8}{8}=+16kN-m$

When support B sinks by 1 cm

$\frac{\Delta }{l}=\frac{1}{800}$

$M_{BA}=0$

$M_{BA}=M_{FBA}+\frac{2EI}{l}(2{\theta }_B+{\theta }_A-\frac{3\Delta }{l})$

$0=16+\frac{2\times 2.8\times {10}^6\times {10}^{-4}}{8}(2{\theta }_B+0-\frac{3\times 1}{800})$

${\theta }_B=-0.112$
$M_{BA}=-16+\frac{2EI}{l}(2{\theta }_A+{\theta }_B-\frac{3\Delta }{l})$

$=-16+\frac{2\times 2.8\times {10}^6\times {10}^{-4}}{8}(0-0.0112-3\times \frac{1}{800})$

$M_{AB}=-24.1kN-m$