[4] Properties of Geometry

Transcription

1 [4] Properties of Geometr Page 1 of 6 [4] Properties of Geometr [4.1] Center of Gravit and Centroid [4.] Composite Bodies [4.3] Moments of Inertia [4.4] Composite reas and Products of Inertia

2 [4] Properties of Geometr Page of 6 [4.1] Center of Gravit and Centroid The center of gravit (denoted as: G) is a point which locates the resultant weight of a sstem of particles: a) The weight of the sstem of n particles comprise a sstem of parallel forces which can be replaced b a single resultant weight having the defined point G of application. b) This means that the resultant weight be equal to the total weight of all n particles as: WR = W c) The sum of the moments of the weights of all the particles about the x and z axes is then equal to the moment of the resultant weight about these axes: xw = xw + x W + + x W R 1 1 n n W = W + W + + W R 1 1 n n zwr = z1w 1 + zw + + znwn where x = coordinates of the center of gravit G of the sstem of particles = coordinates of each particle in the sstem hence xw x W = W W zw z = W x z z The center of mass is given b assuming that the acceleration due to gravit g for ever particle is constant (note: W = mg). x xm = m m = m zm z = m

3 [4] Properties of Geometr Page 3 of 6 rigid bod is composed of an infinite number of particles and the same principle can be applied: x xdw = dw dw = dw z zdw = dw a) In order to appl these equations properl the differential weight dw must be expressed in terms of its associated volume d. b) The specific weight (weight per unit volume) is defined as: dw = d hence x d x = d = d d z = z d d c) Note that: = g where = densit (mass per unit volume) the center of mass (under the constant gravit) can be determined as: x = xd d = d d z = zd d

4 [4] Properties of Geometr Page 4 of 6 The centroid C is a point which defines the geometric center of an object: a) The volume centroid: x = xd d = d d z = zd d b) The area centroid: x = xd d = d d z = zd d x = c) The line centroid: L L xdl dl = L L dl dl z = L L zdl dl The centroids can be specified b the conditions of smmetr: a) In cases where the shape has an axis of smmetr the centroid of the shape will lie along the axis: xdl = 0. L

5 [4] Properties of Geometr Page 5 of 6 b) In cases where a shape has two or three axes of smmetr it follows that the centroid lies at the intersection of these axes:

6 [4] Properties of Geometr Page 6 of 6 CLSS EXMPLE Determine the distance x to the center of mass of the homogeneous rod bent into the shape shown. If the rod has a mass per unit length of 0.5 kg/m determine the reactions at the fixed support O.

7 [4] Properties of Geometr Page 7 of 6 Name: Student ID: HOMEWORK Determine the location ( ) of the centroid of the wire. x

8 [4] Properties of Geometr Page 8 of 6 CLSS EXMPLE 4.1. Locate the centroid x of the shaded area.

9 [4] Properties of Geometr Page 9 of 6 Name: Student ID: HOMEWORK Locate the centroid x of the shaded area.

10 [4] Properties of Geometr Page 10 of 6 [4.] Composite Bodies composite bod can be divided into several simple geometries. a) The same principle can be applied to find center of gravit and centroid: xw x = W W W x x x = = x = = zw z = W = z z = z z = (center of gravit) (volume centroid) (area centroid) x xl = L L = L z zl = L (line centroid) b) It is often convenient to calculate the center of gravit and centroid of composite bodies in a table.

11 [4] Properties of Geometr Page 11 of 6 CLSS EXMPLE 4..1 Locate the centroid of the cross-sectional area of the channel.

12 [4] Properties of Geometr Page 1 of 6 Name: Student ID: HOMEWORK Locate the centroid of the cross-sectional area of the beam.

13 [4] Properties of Geometr Page 13 of 6 [4.3] Moments of Inertia The moment of inertia is a propert of a cross-sectional area of a bod: area moment of inertia. a) Recall that the centroid for an area was determined b considering the first moment of the area about an axis:. xd b) n integral of the second moment of an area such as x d are referred to as the moment of inertia. c) The moment of inertia has units of (length) 4. d) The term moment of inertia has been adopted because of the similarit with integrals of the same form related to mass ( mass moment of inertia). The moments of inertia can be defined for the area which lies in the x- plane. The moments of inertia of the differential are d about the x and axes are defined as: di x = di = x For the entire area the moments of inertia are: I x = d I = x d We can also formulate the second moment of d moment of inertia. about the z axis (or pole ): called the polar a) The polar moment of inertia of d about the z axis (or pole ) is defined as: djo = where r = perpendicular distance from the pole r d b) Since r = x + the polar moment of inertia for the area can be given as: J = r d = I + I O x

14 [4] Properties of Geometr Page 14 of 6 If the moment of inertia for an area is known about an axis passing through its centroid it is convenient to determine the moment of inertia of the area about a corresponding parallel axis (parallel-axis theorem). a) The moment of inertia of d about the x axis is: di = ( ' + d ) x hence I = ( ' + d ) d = ' d + d ' d + d d x where ( ' + d ) d = I x' (the moment of inertia about the centroid) d ' d = 0 ( ' d = d = 0 d d = d since 0 = ) b) In general the parallel-axis theorem for the moment of inertia can be defined as: I = I + d and x x' I = I + d ' x c) For the polar moment inertia about an axis perpendicular to the x- plane and passing through the pole O (z axis): J = J + d O C

15 [4] Properties of Geometr Page 15 of 6 The radius of gration of an area is often used for the design of columns in structural members. a) If the areas and moments of inertia are known the radii of gration are determined from: k x = I x k I z = kz = I b) The radius of gration has units of length. When the boundaries for a planer area are expressed b mathematical functions the integration ma be carried out to determine the moments of inertia: I x = and d I = x d

16 [4] Properties of Geometr Page 16 of 6 CLSS EXMPLE Determine the moment of inertia of the area about the x axis. Solve the problem in two was using rectangular differential elements: (a) having a thickness dx and (b) having a thickness of d.

17 [4] Properties of Geometr Page 17 of 6 Name: Student ID: HOMEWORK Determine the moment of inertia of the area about the x axis. Solve the problem in two was using rectangular differential elements: (a) having a thickness dx and (b) having a thickness of d.

18 [4] Properties of Geometr Page 18 of 6 [4.4] Composite reas and Products of Inertia composite area can be divided into several simple geometries: a) The moments of inertia about the centroid of the simple geometries are usuall known. b) Using the parallel-axis theorem the moments of inertia for the parallel axes can be determined. c) It is often convenient to calculate the moments of inertia of composite bodies in a table. The product of inertia is also a propert of a cross-sectional area of a bod: a) The product of inertia for an element of area d located at point (x ) is defined as: di = xd x b) For the entire area the product of inertia is: I x = xd

19 [4] Properties of Geometr Page 19 of 6 c) The products of inertia has the same units as the moments of inertia. Note) If the element of area chosen has a differential size in two directions a double integration must be performed to evaluate I x. If either the x or axis is an axis of smmetr for the area the product of inertia for the area is zero: a) For ever element d at (x ). located at point (x ) there is a corresponding element d located b) Since the products of inertia for these elements are xd and xd the algebraic sum (or integration) of all the elements that are chosen in this wa will cancel each other. c) The sign of the product of inertia depends upon the quadrant where the area is located:

20 [4] Properties of Geometr Page 0 of 6 The parallel-axis theorem can also be applied for the products of inertia: a) The product of inertia of d with respect to the x and axes is: di = ( x' + d )( ' + d ) d x x therefore I = ( x ' + d )( ' + d ) d x x = x ' ' d + d ' d + d x ' d + d d d x x where x ' ' d = the product of inertia about the centroidal axis d ' d = d x ' d = 0 x since 0 x d = x d = since 0 ( ' d = d = 0 ( ' 0 d d d = d d x x = ) x = ) b) In general the parallel-axis theorem for the products of inertia can be expressed as: I x = I x + dxd

21 [4] Properties of Geometr Page 1 of 6 CLSS EXMPLE Determine the distance x to the centroid of the beam s cross-sectional area: then find the moment of inertia about the axis. I ' '

22 [4] Properties of Geometr Page of 6 Name: Student ID: HOMEWORK Locate the centroid of the cross-sectional area for the angle. Then find the moment of inertia the x ' centroidal axis. I x' about

23 [4] Properties of Geometr Page 3 of 6 CLSS EXMPLE 4.4. Determine the moment of inertia of the beam s cross-sectional area about the x axis.

24 [4] Properties of Geometr Page 4 of 6 Name: Student ID: HOMEWORK Locate the centroid x of the cross-sectional area for the angle. Then find the moment of inertia I about the ' centroidal axis. '

25 [4] Properties of Geometr Page 5 of 6 CLSS EXMPLE Determine the product of inertia of the shaded section of the ellipse with respect to the x and axis.

26 [4] Properties of Geometr Page 6 of 6 Name: Student ID: HOMEWORK Determine the product of inertia of the shaded area with respect to the x and axis.