A propped cantilever AB of 8 m span with flexural rigidity of $2.8 \times 10^{6} k N - c m^{2}$ 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

48.20

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15.75

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24.10

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30.25

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Solution

The correct option is C 24.10
$M_{F A B} = \frac{- 16 \times 8}{8} = - 16 k N - m$
$M_{F A B} = \frac{16 \times 8}{8} = + 16 k N - m$

When support B sinks by 1 cm

$\frac{Δ}{l} = \frac{1}{800}$

$M_{B A} = 0$

$M_{B A} = M_{F B A} + \frac{2 E I}{l} (2 θ_{B} + θ_{A} - \frac{3 Δ}{l})$

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

$θ_{B} = - 0.112$
$M_{B A} = - 16 + \frac{2 E I}{l} (2 θ_{A} + θ_{B} - \frac{3 Δ}{l})$

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

$M_{A B} = - 24.1 k N - m$

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A propped cantilever AB of 8 m span with flexural rigidity of 2.8×106 kN−cm2 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 propped cantilever AB of 8 m span with flexural rigidity of $2.8 \times 10^{6} k N - c m^{2}$ 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