written 6.3 years ago by | • modified 2.3 years ago |
Subject : Structural Analysis 1
Topic : Deflection of Beams
Difficulty : High
written 6.3 years ago by | • modified 2.3 years ago |
Subject : Structural Analysis 1
Topic : Deflection of Beams
Difficulty : High
written 6.3 years ago by | • modified 6.1 years ago |
$1. Reaction:$
$\sum M_A=0$
$16*1+12*2-15-V_B*6=0$
$V_B=4.167KN$
$\sum F_Y=0$
$V_A-16-12+4.167=0$
$V_A=23.53KN$
$2. By \hspace{1mm} Macaulay's \int\int \hspace{1mm} integration \hspace{1mm} method:$
$Consider \hspace{1mm} part(xA)$
$B.M_x=EI\frac{d^2y}{dx^2}=4.167*x+15(x-2)^0-12(x-4)-8(x-4)\frac{(x-4)}{2}$
$\hspace{10mm} =4.167*x+15(x-2)^0-12(x-4)-4(x-4)^2 \hspace{1mm} eqn.1$
$Integrating \hspace{1mm} wrt \hspace{1mm} x$
$EI\frac{dy}{dx}=4.167*\frac{x^2}{2}+15(x-2)^1-\frac{12(x-4)^2}{2}-\frac{4(x-4)^3}{3}+C_1 \hspace{1mm} eqn.2 $
$EIy=4.167*\frac{x^3}{6}+\frac{15(x-2)^2}{2}-\frac{12(x-4)^3}{6}-\frac{4(x-4)^4}{12}+C_1x+C_2 \hspace{1mm} eqn.3 $
$3. Find \hspace{1mm} C_1 \hspace{1mm} and \hspace{1mm} C_2 \hspace{1mm}[Applying \hspace{1mm}boundary\hspace{1mm} …