Question

The propped cantilever beam AB of span 8m is shown below. The beam has an internal pin at point 'C'. If the moment of Inertia of span AC is half of that of span CB, the vertical deflection of point 'D' is? Take $I_{AC}=74300c{m}^{4}andE=110GPa$

• A. 0.002 m
• B. 0.016m
• C. 0.029 m
• D. 0.042 m

Answer 1

The correct option is C 0.029 m

Portion	at Origin	range of x	EI	M	m
AE	A	0-2	$E{I}_{AC}$	-650+190x	-2.5+0.5x
CE	C	0-3	$E{I}_{AC}$	-90x	-0.5x
CD	C	0-1.5	$2E{I}_{AC}$	$90x-\frac{60x^2}{2}$	0.5x
BD	B	0-1.5	$2E{I}_{AC}$	$90x-\frac{60x^2}{2}$	0.5 x

${\Delta }_D={\int }_0^2\frac{(-650+190x)(-2.5+0.5x)dx}{E{I}_{AC}}+{\int }_0^3\frac{45x^{2}dx}{E{I}_{AC}}+2{\int }_0^{1.5}\frac{(90x-\frac{60x^2}{2})\left(0.5x\right)dx}{2E{I}_{AC}}$

$=\frac{5710}{3E{I}_{AC}}+\frac{405}{E{I}_{AC}}+\frac{2025}{64E{I}_{AC}}$

$=\frac{2339.974}{110\times {10}^3\times {10}^3\times 74600\times {10}^{-8}}$

${\Delta }_D=0.0286m$