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Question: Using Macaulay's double integration method, Find slope at band deflection at C for the beam shown.
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enter image description here

Subject : Structural Analysis 1

Topic : Deflection of Beams

Difficulty : High

sa1(74) • 242 views
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modified 8 months ago by gravatar for awari.swati831 awari.swati831140 written 9 months ago by gravatar for Gyanendra pal Gyanendra pal10
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enter image description here

$1. Reaction:$

$\sum M_A=0$

$16*1+12*2-15-V_B*6=0$

$V_B=4.167KN$

$\sum F_Y=0$

$V_A-16-12+4.167=0$

$V_A=23.53KN$


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

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

$B.M_x=EI\frac{d^2y}{dx^2}=4.167*x+15(x-2)^0-12(x-4)-8(x-4)\frac{(x-4)}{2}$

$\hspace{10mm} =4.167*x+15(x-2)^0-12(x-4)-4(x-4)^2 \hspace{1mm} eqn.1$

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

$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 $


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

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

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

$C_2=0$

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

$0=4.167*\frac{6^3}{6}+\frac{15(6-2)^2}{2}-\frac{12(6-4)^3}{6}-\frac{4(6-4)^4}{12}+C_1*6+0$

$C_1=-41.446$

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

$EI\frac{dy}{dx}=4.167*\frac{x^2}{2}+15(x-2)-\frac{12(x-4)^2}{2}-\frac{4(x-4)^3}{3}-41.446 \hspace{1mm} eqn.A$

$EIy=4.167*\frac{x^3}{3}+\frac{15(x-2)^2}{2}-\frac{12(x-4)^3}{6}-\frac{4(x-4)^4}{12}-41.446x+0 \hspace{1mm} eqn.B$


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

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

$EI\frac{dy}{dx}=EIQ_B=0+0-0-0-41.446$

$Q_B=\frac{-41.446}{EI} radians$


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

$EIY_c=4.167*\frac{4^3}{6}+\frac{15(4-2)^2}{2}-0-41.446*4-0$

$Y_c=\frac{-91.336}{EI}$ mm

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modified 7 months ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 220 written 9 months ago by gravatar for Gyanendra pal Gyanendra pal10
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