Advertisements
Advertisements
рдкреНрд░рд╢реНрди
Find the centroid of the shaded portion of the plate shown in the figure.
рд╕рдВрдХреНрд╖реЗрдк рдореЗрдВ рдЙрддреНрддрд░
рдЙрддреНрддрд░
Y = X is the axis of symmetry
The centroid would lie on this line
| Sr.no. | PART | AREA(in mm2) | X co-ordinate(mm) | Ax(mm3) |
| 1. | RECTANGLE | =1000 X 1000 =1000000 |
`1000/2` = 500 | 500000000 |
| 2. | TRIANGLE (to be removed) | `1/2` X 750 X 750 = -281250 |
1000 –`750/2`= 750 | -210937500 |
| 3. | QUARTER CIRCLE (To be removed) | `(pir^2)/4` = 441786.4669 |
`(4 x 7509)/(3pi)`= 3141.5926 | -140625000 |
| TOTAL | 276963.4669 | 148437500 |
`overlineX =(Σ Ax)/(Σ A)=(148437500)/(276963.5331)=535.946 mm`
ЁЭСж╠Е = ЁЭСЛ╠Е = 535.946 mm
CENTROID IS AT (535.946,535.946)mm
shaalaa.com
Centroid for Plane Laminas
рдХреНрдпрд╛ рдЗрд╕ рдкреНрд░рд╢реНрди рдпрд╛ рдЙрддреНрддрд░ рдореЗрдВ рдХреЛрдИ рддреНрд░реБрдЯрд┐ рд╣реИ?
