Centroid for System of Particles | Engineering Mechanics - Civil Engineering (CE) PDF Download

Centroid: point defines the geometric center

If the material composing a body is uniform or homogeneous, the density or specific weight will be constant throughout the body, then the centroid is the same as the center of gravity or center of mass

Find Centroid of area?

Centroids: Using Single Integration

1) DRAW a differential element on the graph.

2) Label the centroid  of the differential element.

3) Label the point where the element intersects the curve (x, y)

4) Write down the appropriate general equation to use.

5) Express each term in the general equation using the coordinates describing the curve or function.

6) Determine the limits of integration

7) Integrate

Centroids: Using Single Integration

1) DRAW a differential element on the graph.

2) Label the centroid  of the differential element.

3) Label the point where the element intersects the curve (x, y)

4) Write down the appropriate general equation to use.

5) Express each term in the general equation using the coordinates describing the curve or function.

6) Determine the limits of integration

7) Integrate

Determine the center of gravity of a thin homogeneous wire

segment	L (mm)
AB	600	300	0	18x104
BC	650	300	125	19.5x104	8.125x104
CA	250	0	125	0	3.125x104
	1500			37.5x104	11.25x104

Locate the centroid  of the uniform wire bent in the shape shown. 

The given composite line can be divided intofollowing three parts having simpler shapes:

Segment	L (mm)
1	150	0	75	0	11250
2	100	50	150	5000	15000
3	50	75	130	3750	6500
4	130	50	65	6500	8450
5	50	25	0	1250	0
Σ	480			16500	41200

Locate the distanceyto the centroid of the member’scross-sectional area. 

Particle #	A (in2)
1	7.5	4.75	35.625
2	1.875	1.5	2.8125
3	1.875	1.5	2.8125
4	6	0.5	3.0
Σ	ΣA = 17.25		= 44.25

The given composite line can be divided into following three parts having simpler shapes: 

Segment	L (mm)
1	π(60)=188.5	60	-38.2	0	11310	-7200	0
2	40	0	20	0	0	800	0
3	20	0	40	-10	0	800	200
Σ	ΣL= 248.5				ΣxL=11310	ΣyL= -5600	ΣzL= -200

Locate the centroid of the plate area shown in Fig

Segment	A (m2)
1	0.5 * 3 * 3 = 4.5	1	1	4.5	4.5
2	3 * 3 = 9	-1.5	-13.5	13.5	13.5
3	−2 * 1 = −2	-2.5	2	5	-4
Σ	ΣA = 11.5

Segment	A (cm2)
S.circle	π/2*22=6.28	2	6.85	12.56	43.02
Rectangle	6*4=24	2	3	48	72
Triangle	1/2*3*6=9	-1	2	-9	18
Q. circle	-π/4*22 = −3.14	3.15	0.85	-9.89	2.67
Σ	36.14			41.67	130.35

Find: The centroid of the part

Solution: 1. This body can be divided into the following pieces:

rectangle (a) + triangle (b) + quarter circular (c) – semicircular area (d). (Note the negative sign on the hole!)

Steps 2 & 3: Make up and fill the table using parts a, b, c, and d.

Segment	A (m2)
Rectangle	18	3	1.5	54	27
Triangle	4.5	7	1	31.5	4.5
Q. Circle	9π⁄4	−4 * 3⁄3π	4 * 3⁄3π	-9	9
Semi-Circle	−π⁄2	0	−4 * 1⁄3π	0	-2/3
Σ	28.0

For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid.

Solution:

Divide the area into a triangle, rectangle, and semicircle with a circular cutout. Calculate the first moments of each area with respect to the axes. Find the total area and first moments of the triangle, rectangle, and semicircle.  Subtract the area and first moment of the circular cutout. Compute the coordinates of the area centroid by dividing the first moments by the total area.

Segment	A, mm2
Rectangle	120 * 80 = 9.6 * 103	60	40	576 * 103	384 * 103
Triangle	1/2*120 * 60 = 3.6 * 103	40	-20	144 × 103	−72 × 103
Semicircle	1/2 *π * 602 = 5.655 * 103	60	105.46	339.3 × 103	596.4 × 103
Circle	−π * 402 = −5.027* 103	60	80	−301.6* 103	−402.2 * 103
Σ	∑A= 13.828 * 103

Segment	A (mm2)
1	20*60=1200	10	30	12,000	36,000
2	1/2* 30 * 36 = 540	30	36	16,200	19,440
Σ	1740			28,200	55,440