CE 201- Statics

1. CE 201- Statics Chapter 9 – Lecture 1

2. CENTER OF GRAVITY AND CENTROID The following will be studied • Location of center of gravity (C. G.) and center of mass for discrete particles • Location of C. G. and center of mass for an arbitrary-shaped body • Location of centroid or geometric center

3. w1 w2 w3 w4 wn WR z x y Center of Gravity and Center of Mass for a System of Particles • Consider ( n ) particles • Weight of particles are parallel forces • Weight can be replaced by a single resultant weight • The point of application of the resultant weight is called the center of gravity (C. G.) Resultant weight, WR =  w

4. w1 w2 w3 w4 wn WR z x y Moment of all weights about x, y, and z is equal to the moment of the resultant weight about the same axes. My xRWR = x1w1 + x2w2 + ..+xnwn Mx yRWR = y1w1 + y2w2 + ..+ynwn

5. w2 w1 w3 w1 w2 w3 w4 wn wn w4 WR WR z y x x z y To find z, imagine rotating the system coordinates by 90 with particles are fixed in it. Mx zRWR = z1w1 + z2w2 + ……..+znwn

6. then, x = ( x W) /  WR y = ( y W) /  WR Z = ( z W) /  WR Note: x, y, and z for C. G. of the system x, y, and z for C. G. of each particle

7. Center of Mass W = mg x = ( x m) /  mR y = ( y m) /  mR x = ( z m) /  mR The location of the center of gravity coincides with that of the center of mass

8. dW z x y CENTER OF GRAVITY, CENTER OF MASS AND CETROID FOR A BODY Center of Gravity A rigid body is composed of a system of particles, where each particle has a differential weight (dW). Applying the same principles that were used with discrete particles, the following is obtained: x = ( x dW) / ( dW) y = ( y dW) / ( dW) z = ( z dW) / ( dW)

9. dW z x y here, we use integration rather than summation due to differential weight (dW). If dW =  dV Where  = specific weight (weight / volume) V = volume of body Then, x = ( x dV) / ( dV) y = ( y dV) / ( dV) z = ( z dV) / ( dV)

10. Center of Mass Substitute  = g into the previous equations ( = density (mass/volume), then: x = ( xg dV) / (g dV) y = ( yg dV) / (g dV) z = ( zg dV) / (g dV)

11. Centroid • Centroid is the geometric center of the object • Centroid is independent of the weight • Centroid is dependent of the body's geometry

12. c dV z z x x y y Volume Centroid • Subdivide the object into volume elements (dV) • Compute the moments of the volume elements about the coordinate axes x = ( x dV) / ( dV) y = ( y dV) / ( dV) z = ( z dV) / ( dV)

13. c dV z z x x y y Area Centroid • Subdivide the object into area elements (dA) • Compute the moments of the area elements about the coordinate axes x = ( x dA) / ( dA) y = ( y dA) / ( dA) z = ( z dA) / ( dA)

14. dL c z z x x y y Line Centroid • Subdivide the line into elements (dA) • Compute the moments of the line elements about the coordinate axes x = ( x dL) / ( dL) y = ( y dL) / ( dL) z = ( z dL) / ( dL) • Centroid could be located off the object in space • Centroid of some shapes may be specified by using the conditions of symmetry

15. y dL c x dL Line Centroid • If the shape has an axis of symmetry, then the centroid will be located along that axis • For every element dL having a distance ( y ), there is an element dL having a distance ( -y ) So, y = 0

16. The same thing can apply if the shape has more than one axis of symmetry. The centroid lies at the intersection of the axe.

17. Procedure for Analysis To determine the Center of Gravity or the Centroid: • select an appropriate coordinate axes • select an appropriate differential element for integration (dL, dA, dV) • express he differential element (dL, dA, dV) in terms of the coordinates (x, y, z) • determine the coordinate (x, y, z) or moment arms for the centroid or center of gravity of the element • integrate