PLAIN AND CIVIL: EXAMPLE 4.4.2. EQUATION METHOD: BEAM WITH RECTANGULAR LOADS

Determine the shear and moment equations for the beam shown. Also draw the shear and moment diagrams.

Example 4.3.7. shows the computation for the support reactions. By cutting the beam into three segments, clearer analysis can follow.

For Segment AB [cutting plane a-a']: 0<x<4m

$P_{1}=\omega x=5x$

$\sum F_{Y}=0]\uparrow$

$18.75kN-P1+V_{AB}=0$

${\color{Red} \mathbf{V_{AB}=5x-18.75kN}}$

$\sum M_{cut}=0]$

$18.75x-P_{1}\left ( \frac{x}{2} \right )-M_{AB}=0$

$M_{AB}=18.75x-5x\left ( \frac{x}{2} \right )$

${\color{Red} \mathbf{M_{AB}=18.75x-2.5x^{2}}}$

For Segment BC [cutting plane b-b']: 4<x<7m

$P_{2}=5\frac{kN}{m}\left ( 4m \right )=20kN$

$\sum F_{Y}=0]\uparrow$

$18.75kN-P_{2}+V_{BC}=0$

${\color{Red} \mathbf{V_{BC}=1.25kN\uparrow }}$

$\sum M_{cut}=0]$

$18.75\left ( x \right )-P_{2}\left ( x-2 \right )+M_{BC}=0$

${\color{Red} \mathbf{M_{BC}=1.25x-40}}$

For Segment CD [cutting plane c-c']: 7<x<12m

$P_{3}=2\frac{kN}{m}\left ( x-7 \right )=2x-14$

$\sum F_{Y}=0]\uparrow$

$18.75-P_{2}-P_{3}+V_{CD}=0$

$V_{CD}=-18.75+20+\left ( 2x-14 \right )$

${\color{Red} \boldsymbol{V_{CD}=2x-12.75}}$

$\sum M_{cut}=0]$

$18.75x-P_{2}\left ( x-2 \right )-P_{3}\left ( \frac{x-7}{2} \right )-M_{CD}=0$

$M_{CD}=18.75x-20\left ( x-2 \right )-\left ( 2x-14 \right )\left ( \frac{x-7}{2} \right )$

$M_{CD}=18.75x-20x+40-2\left ( x-7 \right )\left ( \frac{x-7}{2} \right )$

$M_{CD}=40-1.25x-\left ( x^{2}-9x+14 \right )$

${\color{Red} \mathbf{M_{CD}=-9+12.75x-x^{2}}}$

SHEAR AND MOMENT DIAGRAMS

Substitute values of x in the equations for shear and moment to come up with the diagrams.

RELATED TOPIC:

EQUATION METHOD FOR SHEAR AND MOMENT IN BEAMS

PLAIN AND CIVIL

Pages

Sunday 31 March 2019

EXAMPLE 4.4.2. EQUATION METHOD: BEAM WITH RECTANGULAR LOADS

SHEAR AND MOMENT DIAGRAMS

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