written 6.2 years ago by | • modified 2.3 years ago |
Subject : Structural Analysis 1
Topic : Deflection of Beams
Difficulty : High
written 6.2 years ago by | • modified 2.3 years ago |
Subject : Structural Analysis 1
Topic : Deflection of Beams
Difficulty : High
written 6.2 years ago by | • modified 6.1 years ago |
$1. Reaction:$
$\sum M_A=0$
$-10+16*5+8*8-V_C*6=0$
$V_C=22.33KN$
$\sum F_Y=0$
$V_A-16+22.33-8=0$
$V_A=1.67KN$
$2. By \hspace{1mm} Macaulay's \int\int \hspace{1mm} integration \hspace{1mm} method:$
$Consider \hspace{1mm} part(xA)$
$B.M_x=EI\frac{d^2y}{dx^2}=1.67*x-10(x-2)^0-8(x-4)\frac{(x-4)}{2}+22.33(x-6)+8(x-6)\frac{x-6}{2}$
$\hspace{10mm} =1.67*x+10(x-2)^0-4(x-4)^2+22.33(x-6)+4(x-6)^2 \hspace{1mm} eqn.1$
$Integrating \hspace{1mm} wrt \hspace{1mm} x$
$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 $