Subject : Structural Analysis 1

Topic : Deflection in Beams

Difficulty : Low

$1. Reaction: $

$\sum M_A=0$

$15*2+10*4-V_B*6=0$

$V_B=11.67KN$

$\sum F_Y=0$

$V_A-15-10+11.67=0$

$V_A=13.33KN$

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

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

$B.M_x=EI \frac{d^2y}{dx^2}=13.33*x- 15(x-2)- 10(x-4)\hspace{1mm} eqn.1$

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

$EI\frac{dy}{dx}=13.33*\frac{x^2}{6}- \frac{15(x-2)^2}{2}-\frac{10(x-4)^2}{2}+C_1 \hspace{1mm} eqn.2$

$Again \hspace{1mm} integrating \hspace{1mm} wrt \hspace{1mm} x,$

$EIy=13.33*\frac{x^3}{6}-\frac{15(x-2)^3}{6}-\frac{10(x-4)^3}{6}+C_1x+C_2 \hspace{1mm} eqn.3$

$3. To \hspace{1mm} find \hspace{1mm} C_1 \hspace{1mm} and \hspace{1mm} C_2[Applying \hspace{1mm} boundary \hspace{1mm} condition]$

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

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

$C_2=0$

Note: In bracket (-) sign there consider whole part as 0 

If one term is zero, consider whole part is zero, Macaulays says.

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

$0=13.33*\frac{6^3}{6}-\frac{15(6-2)^3}{6}-\frac{10(6-4)^3}{6}+c_1(6)+0$

$C_1=-51.09$

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

$EI\frac{dy}{dx}=13.34*\frac{x^2}{2}-\frac{15(x-2)^2}{2}-\frac{10(x-4)^2}{2}-51.09 \hspace{1mm} eqn.A$ (G.S.E.)

$EIy=13.34*\frac{x^3}{6}-\frac{7.5(x-2)^3}{3}-\frac{5(x-4)^3}{3}-51.09x+0\hspace{1mm} eqn.B$ (G.D.E.)

$4. To \hspace{1mm} get Q_A[slope \hspace{1mm} at \hspace{1mm} A]:$

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

$Q_A=\frac{dy}{dx}$

$EI\frac{dy}{dx}=EI\hspace{1mm}Q_A=0-0-0-51.15$

$Q_A=\frac{-51.15}{EI} radians$

$5. To \hspace{1mm} get \hspace{1mm} Y_c[deflection at C]:$

$Put x=2m\hspace{1mm}in \hspace{1mm} eqn.B$

$EIY_c=\frac{13.34}{6}*2^3-0-0-51.15*2$

$Y_c=\frac{-84.5}{EI}mm$