Subject : Structural Analysis 1

Topic : Deflection of Beams

Difficulty : High

$1. Reaction:$

$\sum M_A=0$

$-10+16*5+8*8-V_C*6=0$

$V_C=22.33KN$

$\sum F_Y=0$

$V_A-16+22.33-8=0$

$V_A=1.67KN$

$2. By \hspace{1mm} Macaulay's \int\int \hspace{1mm} integration \hspace{1mm} method:$

$Consider \hspace{1mm} part(xA)$

$B.M_x=EI\frac{d^2y}{dx^2}=1.67*x-10(x-2)^0-8(x-4)\frac{(x-4)}{2}+22.33(x-6)+8(x-6)\frac{x-6}{2}$

$\hspace{10mm} =1.67*x+10(x-2)^0-4(x-4)^2+22.33(x-6)+4(x-6)^2 \hspace{1mm} eqn.1$

$Integrating \hspace{1mm} wrt \hspace{1mm} x$

$EI\frac{dy}{dx}=1.67*\frac{x^2}{2}-10(x-2)^1-\frac{4(x-4)^3}{3}+22.33\frac{(x-6)^2}{2}+\frac{4(x-6)^3}{3}+C_1 \hspace{1mm} eqn.2 $

$EIy=1.67*\frac{x^3}{6}-\frac{10(x-2)^2}{2}-\frac{4(x-4)^4}{12}+22.33\frac{(x-6)^4}{12}+\frac{4(x-6)^4}{12}+C_1x+C_2 \hspace{1mm} eqn.3 $

1. Find $\hspace{1mm} C_1 \hspace{1mm} and \hspace{1mm} C_2 \hspace{1mm}[Applying \hspace{1mm}boundary\hspace{1mm} condition]$

$Put \hspace{1mm} x=0, y=0 \hspace{1mm} in \hspace{1mm} eqn.3$

$0=0-0-0+0+0+0+C_2$

$C_2=0$

$Now \hspace{1mm} at \hspace{1mm} x=6, y=0 \hspace{1mm} put \hspace{1mm} in \hspace{1mm} eqn.3$

$0=1.67*\frac{6^3}{6}+\frac{10(6-2)^2}{2}-\frac{4(6-4)^4}{12}+0+0+C_1*6+0$

$C_1=4.20$

$Put \hspace{1mm} value\hspace{1mm} of \hspace{1mm} C_1 \hspace{1mm} and \hspace{1mm} C_2 in \hspace{1mm} eqn.2 \hspace{1mm} and \hspace{1mm} eqn.3$

$EI\frac{dy}{dx}=1.67*\frac{x^2}{2}-10(x-2)-\frac{4(x-4)^3}{3}+22.33\frac{(x-6)^2}{2}+\frac{4(x-6)^3}{3}+4.20\hspace{1mm} eqn.A$

$EIy=1.67*\frac{x^3}{6}+\frac{10(x-2)^2}{2}-\frac{4(x-4)^4}{12}+22.33\frac{(x-6)^3}{6}+\frac{4(x-6)^4}{12}+4.20x+0 \hspace{1mm} eqn.B$

$4. To \hspace{1mm} get \hspace{1mm} Q_B:$

$Put \hspace{1mm} x=0 \hspace{1mm} in \hspace{1mm} eqn.A$

$EI\frac{dy}{dx}=EIQ_A=0-0-0+0+0+4.20$

$Q_A=\frac{4.20}{EI} radians$

$5. To \hspace{1mm} get \hspace{1mm} Y_c:$

$Put x=6 \hspace{1mm} in \hspace{1mm} eqn.B$

$EIY_B=1.67*\frac{6^3}{6}-\frac{10(6-2)^2}{2}-\frac{4(6-2)^4}{12}+4.20*6$

$Y_B=\frac{57.32}{EI}$