Using conjugate beam method find slope at C & deflection at C for a cantilever beam as shown in figure.

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written 6.7 years ago by

teamques10 ★ 68k

• modified 6.6 years ago

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1. Suppport reaction calculation

$\sum M_A=0(\circlearrowright+ve)[ For\ UDL\ only]\\ -M_A+12\times5\times\frac{5}{2}=0\\ \boxed{M_A=150\ kNm}\\ \\ \sum M_A=0(\circlearrowright+ve)[For\ point\ load]\\ -M_A+24\times2=0\\ \boxed{M_A=48\ kNm} $

2. Draw BMD $ [\circlearrowright| \circlearrowleft+ve]$

$B.M_A=150\ kNm [For\ UDL]\\ B.M_A=48\ kNm[For\ point\ load]\\B.M_B=12\times3\times\frac{3}{2}=\underline{54\ kNm} $

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3.$\frac{M}{EI}$ diagram

$\underline{conjugate\ beam}$

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3. Conjugate beam calculation

$\sum F_Y=0(\uparrow+ve)\\ v_c=Total\ load\\ load\ 1=2\times\frac{-27}{EI}=\frac{-54}{EI}\\ load\ 2=\frac{1}{3}\times\frac{-48}{EI}\times2=\frac{-32}{EI}\\ load\ 3=\frac{1}{2}\times\frac{-54}{EI}\times3=\frac{-54}{EI}\\ load\ 4=\frac{1}{2}\times2\times\frac{-24}{EI}=\frac{-24}{EI} \\v_c=\frac{-54-32-54-24}{EI}\\ \boxed{v_c=\frac{-164}{EI}}\\ \sum M @ c =0(\circlearrowright+ve)\\ M_c=load\times C.G$

$M_c$	load	C.G
1	$\frac{-54}{EI}$	$3+\frac{3}{2}=4.5m$	$\frac{-216}{EI}$
2	$\frac{-32}{EI}$	$3+\left(\frac{3}{4}\times2\right)=4.5m$	$\frac{-144}{EI}$
3	$\frac{-54}{EI}$	$\frac{3}{4}\times3=2.25m$	$\frac{-121.5}{EI}$
4	$\frac{-24}{EI}$	$3+\left(\frac{2}{3}\times2\right)=4.333m$	$\frac{-103.99}{EI}$
		$\sum =$	$\frac{-585.5}{EI}$

$\boxed{M_c=\frac{-585.5}{EI}} $

4. Slope & Deflection

$Q_c$ in real beam = S.F @ c in conjugate beam.

$Q_c=v_c=\frac{-164}{EI}\\ \boxed{Q_c=\frac{164}{EI}rad(\downarrow)} $

$y_c$ in real beam = B.M @ c in conjugate beam.

$ y_c=B.M@c\\ y_c=\frac{-585.5}{EI}\\ \boxed{y_c=\frac{585.5}{EI}}(\downarrow)$

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