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
Geometric Properties of Line and Area Elements |
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 | 18x10^{4} | |
BC | 650 | 300 | 125 | 19.5x10^{4} | 8.125x10^{4} |
CA | 250 | 0 | 125 | 0 | 3.125x10^{4} |
1500 | 37.5x10^{4} | 11.25x10^{4} |
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 (in^{2}) | ||
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 (m^{2}) | ||||
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 (cm^{2}) | ||||
S.circle | π/2*2^{2}=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*2^{2} = −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 (m^{2}) | ||||
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 * 10^{3} | 60 | 40 | 576 * 10^{3} | 384 * 10^{3} |
Triangle | 1/2*120 * 60 = 3.6 * 10^{3} | 40 | -20 | 144 × 10^{3} | −72 × 10^{3} |
Semicircle | 1/2 *π * 60^{2} = 5.655 * 10^{3} | 60 | 105.46 | 339.3 × 10^{3} | 596.4 × 10^{3} |
Circle | −π * 40^{2} = −5.027* 10^{3} | 60 | 80 | −301.6* 10^{3} | −402.2 * 10^{3} |
Σ | ∑A= 13.828 * 10^{3} |
Segment | A (mm^{2}) | ||||
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 |