Centroid Theorem

The centroid theorem states that the centroid of the triangle is at 2/3 of the distance from the vertex to the mid-point of the sides.

Fig 1

Suppose PQR is a triangle having a centroid V. S, T and U are the midpoints of the sides of the triangle PQ, QR and PR, respectively. Hence as per the theorem;

QV = 2/3 QU,

PV = 2/3 PT and RV = 2/3 RS

Centroid of a Right Angle Triangle

The centroid of a right angle triangle is the point of intersection of three medians, drawn from the vertices of the triangle to the midpoint of the opposite sides.

Fig 2

Centroid of a Square

The point where the diagonals of the square intersect each other is the centroid of the square. As we all know, the square has all its sides equal. Hence it is easy to locate the centroid in it. See the below figure, where O is the centroid of the square.

Fig 3

Properties of centroid

The properties of the centroid are as follows:

• The centroid is the centre of the object.
• It is the centre of gravity.
• It should always lie inside the object.
• It is the point of concurrency of the medians.

Centroid Formula

Let’s consider a triangle. If the three vertices of the triangle are A(x1, y1), B(x2, y2), C(x3, y3), then the centroid of a triangle can be calculated by taking the average of X and Y coordinate points of all three vertices. Therefore, the centroid of a triangle can be written as:

Centroid of a triangle = ((x1+x2+x3)/3, (y1+y2+y3)/3)

Fig 4

Centroid Formula for Different Shapes

Here, the list of centroid formula is given for different geometrical shapes.

1)     Centroid of a triangle = ((x1+x2+x3)/3, (y1+y2+y3)/3)

2)     The centroid is the centre point of the object. The point in which the three medians of the triangle intersect is known as the centroid of a triangle

Taking the simple case first, we aim to find the centroid for the area defined by a function f(x), and the vertical lines x = a and x = b as indicated in the following figure.

Fig 5

To find the centroid, we use the same basic idea that we were using for the straight-sided case above. The "typical" rectangle indicated is x units from the y-axis, and it has width Δx (which becomes dx when we integrate) and height y = f(x)

Generalizing from the above rectangular areas case, we multiply these 3 values x,  f(x) and Δx, which will give us the area of each thin rectangle times its distance from the x-axis), then add them. If we do this for infinitesimally small strips, we get the x-coordinates of the centroid using the total moments in the x-direction, given by:

And, considering the moments in the y-direction about the x-axis and re-expressing the function in terms of y, we have:

Fig 6

Notice this time the integration is with respect to y, and the distance of the "typical" rectangle from the x-axis is y units. Also note the lower and upper limits of the integral are c and d, which are on the  y-axis.

Of course, there may be rectangular portions we need to consider separately. (I've used a different curve for the y​ case for simplification.)

Alternate method: Depending on the function, it may be easier to use the following alternative formula for the y-coordinate, which is derived from considering moments in the x-direction (Note the "dx" in the integral, and the upper and lower limits are along the x-axis for this alternate method).

This is true since for our thin strip (width dx) the centroid will be half the distance from the top to the bottom of the strip. Another advantage of this second formula is there is no need to re-express the function in terms of y.

Centroids for Areas Bounded by 2 Curves

Fig 7

We extend the simple case given above. The "typical" rectangle indicated has width Δx and height y2 − y1, so the total moments in the x-direction over the total area is given by:

For the y coordinate, we have 2 different ways we can go about it.

Method 1: We take moments about the y-axis and so we'll need to re-express the expressions x2 and x1 as functions of y.

Method 2: We can also keep everything in terms of x by extending the "Alternate Method" given above:

Centroids of Composite sections ( 1D, 2D)

First Moments of Areas and Lines

•An area is symmetric with respect to an axis BB’ if for every point P there exists a point P’ such that PP’ is perpendicular to BB’ and is divided into two equal parts by BB’.

•The first moment of an area with respect to a line of symmetry is zero.

•If an area possesses a line of symmetry, its centroid lies on that axis

•If an area possesses two lines of symmetry, its centroid lies at their intersection.

•An area is symmetric with respect to a center O if for every element dA at (x,y) there exists an area dA’ of equal area at (-x,-y).

•The centroid of the area coincides with the center of symmetry.

Fig 8

Key Takeaways

Centroid formula:

Second moment of area

The integral ∫ y2 dA defines the second moment of area I about an axis and can be obtained by considering a segment of area δA some distance y from the neutral axis, writing down an expression for its second moment of area and then summing all such strips that make up the section concerned, i.e. integrating. As indicated in the discussion of the general bending equation, the second moment of area is needed if we are to relate the stress produced in a beam to the applied bending moment.

As an illustration of the derivation of a second moment of area from first principles, consider a rectangular cross-section of breadth b and depth d . For a layer of thickness δy a distance y from the neutral axis, which passes through the centroid, the second moment of area for the layer is:

Second moment of area of strip=y2δA=y2b δy

The total second moment of area for the section is thus:

Second moment of area=∫limits from −d/2 to d/2) y2bdy=bd3/12

Fig 9

Determine the second moment of area about the neutral axis of the I-section shown in Figure 10.

Fig 10

We can determine the second moment of area for such a section by determining the second moment of area for the entire rectangle containing the section and then subtracting the second moments of area for the rectangular pieces ‘missing’

Fig11

Thus for the rectangle containing the entire section, the second moment of area is given by I = bd3/12 = (50 × 703)/12 = 1.43 × 106mm4. Each of the ‘missing’ rectangles will have a second moment of area of (20 × 503)/12 = 0.21 × 106 mm4. Thus the second moment of area of the I-section is 1.43 × 106 − 2 × 0.21 × 106 = 1.01 × 106mm4

The product of inertia is defined as

           The x and y terms inside the integral denote the centroidal position of the differential area measured from the y and x axes, respectively. Similar to moments of inertia discussed previously, the value of product of inertia depends on the position and orientation of selected axes. It is possible for the product of inertia to have a positive, negative, or even a zero value.

Fig 12

            For example, either x or y represents an axis of symmetry, then the product of inertia Ixy would be zero. To see why this is the case, take a look at the figure to the right. Consider the small area A1 to the right of y axis at the distance of x1. Then consider a similar area to the left of this axis of symmetry at the distance of -x1. Since both areas are at the same vertical position from the x-axis, they have the same value of y. The contribution from the left area is -x1yA1 and that from the right is x1yA1 which add up to zero. Since every point on one side of the axis of symmetry has an equal counterpart on the other side, the total value of the integral would be zero.

            However, if we were to consider the product of inertia with respect to the x' and y' axes, then Ix'y' would not be zero. We will have more discussion about the product of inertia in the section on principal axes.

For a general three-dimensional body, it is always possible to find 3 mutually orthogonal axes (an x, y, z coordinate system) for which the products of inertia are zero, and the inertia matrix takes a diagonal form. In most problems, this would be the preferred system in which to formulate a problem. For a rotation about only one of these axis, the angular momentum vector is parallel to the angular velocity vector. For symmetric bodies, it may be obvious which axis is principle axis. However, for an irregular-shaped body this coordinate system may be difficult to determine by inspection; we will present a general method to determine these axes in the next section. But, if the body has symmetries with respect to some of the axis, then some of the products of inertia become zero and we can identify the principal axes.

For instance, if the body is symmetric with respect to the plane then, we will have and will be a principal axis. This can be shown by looking at the definition of the products of inertia.

Fig 13

The integral for, say, can be decomposed into two integrals for the two halves of the body at either side of the plane. The integrand on one half will be equal in magnitude and opposite in sign to the integrand on the other half (because will change sign). Therefore, the integrals over the two halves will cancel each other and the product of inertia will be zero. (As will the product of inertia ) Also, if the body is symmetric with respect to two planes passing through the center of mass which are orthogonal to the coordinate axis, then the tensor of inertia is diagonal, with

Fig 14

Another case of practical importance is when we consider axisymmetric bodies of revolution. In this case, if one of the axes coincides with the axis of symmetry, the tensor of inertia has a simple diagonal form. For an axisymmetric body, the moments of inertia about the two axis in the plane will be equal. Therefore, the moment about any axis in this plane is equal to one of these. And therefore, any axis in the plane is a principal axis. One can extend this to show that if the moment of inertia is equal about two axis in the plane (IP P = Ixx), whether or not they are orthogonal, then all axes in the plane are principal axes and the moment of inertia is the same about all of them. In its inertial properties, the body behaves like a circular cylinder.

Fig 15

The tensor of inertia will take different forms when expressed in different axes. When the axes are such that the tensor of inertia is diagonal, then these axes are called the principal axes of inertia.

1)     Second moment of area of strip=y2δA=y2b δy

2)     Thus for the rectangle containing the entire section, the second moment of area is given by I = bd3/12

In addition to the moment of inertia, the product of inertia is commonly used. Here only the product of the area is defined and discussed. The product of inertia defined as

Ixixj=∫A xixjdA

For example, the product of inertia for xx and yy axes is

Ixy=∫AxydA

Product of inertia can be positive or negative value as oppose the moment of inertia. The calculation of the product of inertia isn't different much for the calculation of the moment of inertia. The units of the product of inertia are the same as for moment of inertia.

Transfer of Axis Theorem

Same as for moment of inertia there is also similar theorem.

I x′y′ = ∫A x′y′dA = ∫A(x+Δx)(y+Δy)dA

Expanding equation above results in

I x′y′ = ∫AxydA+ ∫AxΔydA + ∫AΔxydA + ∫AΔxΔydA

The final form is

I x′y′ = Ixy + ΔxΔydA

There are several relationships should be mentioned

I xy =I yx

Symmetrical area has zero product of inertia because integration of odd function (asymmmertial function) left part cancel the right part.

Fig. 20 Product of inertia

The principle of virtual work was proposed by the Swiss mathematician Jean Bernoulli in the eighteenth century. It provides an alternative method for solving problems involving the equilibrium of a particle, a rigid body, or a system of connected rigid bodies.

The definitions of the work of a force and a couple can be presented in terms of the actual movements which are expressed by differential displacements having values of and . Consider now an imaginary or virtual movement of a body in static equilibrium, which indicates a displacement or rotation that is assumed and does not actually exist. These movements are first-order differential quantities and will be denoted by the symbols and (delta r and delta ), respectively. The virtual work done by a force having a virtual displacement is

Similarly, when a couple undergoes a virtual rotation in the plane of the couple forces, the virtual work is

According to the principle of virtual work if a body is in equilibrium, then  virtual displacement of the body is zero for all the algebraic sum of the virtual work which is done by all the forces and moments of couple that are acting on the body. Thus,

For example, consider the free-body diagram of the particle (ball) that rests on the floor, Fig. below. If we “imagine” the ball to be displaced downwards a virtual amount then the weight does positive virtual work , and the normal force does negative virtual work , For equilibrium the total virtual work must be zero, so that

Fig 21

As seen from the above example, no added advantage is gained by solving particle and rigid-body equilibrium problems using the principle of virtual work. This is because for each application of the virtual-work equation, the virtual displacement, common to every term, factors out, leaving an equation that could have been obtained in a more direct manner by simply applying an equation of equilibrium.

A particle of mass m is acted upon by a number of forces f1,f2….fN Suppose the particle undergoes a virtual displacement to reiterate, these impressed forces fi do not cause the particle to move, one imagines it to be incorrectly positioned a little away from the true equilibrium position.

Fig 22

If the particle is moving with an acceleration a, the quantity -ma is treated as an inertial force. The total virtual work is then (each term here is the dot product of two vectors)

Now if the particle is in equilibrium by the action of the effective (impressed plus inertial) force then                              

This can be expressed as follows:

The principle of virtual work (or principle of virtual displacements) if a particle is in equilibrium under the action of a number of forces (including the inertial force) the total work done by the forces for a virtual displacement is zero.

Alternatively, one can define the external virtual work    Wext and the virtual kinetic energy    in which case the principle takes the form (compare with the work-energy principle

In the above, the principle of virtual work was derived using Newton’s second law. One could just as well regard the principle of virtual work as the fundamental principle and from it derive the conditions for equilibrium. In this case one can say that The principle of virtual work (or principle of virtual displacements)

a particle is in equilibrium under the action of a system of forces (including the inertial force) if the total work done by the forces is zero for any virtual displacement of the particle. The term kinematically admissible displacement is used to mean one that does not violate the constraints, and hence one arrives at the version of the principle which is often used in practice:

The principle of virtual work (or principle of virtual displacements):a particle is in equilibrium under the action of a system of forces (including the inertial force) if the total work done by the forces (excluding reaction forces) is zero for any kinematic ally admissible virtual displacement of the particle.

Beam –

It is horizontal structural member on which different types of loads can be supported. It takes load from floor or roof & transfers to column.

The beam may be sometimes constructed in the inclined position for architectural point of view.

Types of Beam: -

1. Simply supported beam –

A beam which is just resting on the supports at the end without any connection is known as simply supported beam. It is generally used for vertical landing system.

Fig 23

2.     Overhanging Beam –

A beam which is supported at the intermediate point other than ends is called as overhanging beam. Here portion of beam is extended beyond the support

a)

                        (singly overhanging beam)

b)

(doubly overhanging beam)

Fig 24

3.     Cantilever Beam –

A beam which is fixed at one end is called as cantilever beam

Fig 25

Here, there are three reactions components:

1. Vertical reaction at A (RVA)
2. Horizontal reaction at A (RHA)
3. Fixing moment at A (MA)

We can assume any direction for above components.

4.     Continuous Beam:

A beam having more than two supports is called as continuous beam.

Fig 26

5.     Compound Beam:

When two or more beams are joined together by using internal hinge; or when one beam rests over another beam by using internal roller, then such beam is called as compound beam.

Fig 27