chapter 5 distributed forces - gantep.edu.trerklig/me108/0_lecture8.pdf · chapter 5 distributed...

Chapter 5

Distributed Forces

Ahmet Erkliğ

ME 108 - Statics

Section A

Center of Mass and Centroids

Applications

To design the structure for supporting a water tank, we will need to know the weights of the tank and water as well as the locations

where the resultant forces representing these distributed loads are acting

How can we determine these weights and their locations?

Center of Gravity (CG) and

Center of Mass (CM)

Consider a three dimensional body

of any size and shape, having a mass m

If we suspend the body, as shown

in the figure, from any point such as

A, B, or C, the body will be in

equilibrium under the action of tension

in the cord and the resultant W of the

gravitational forces acting on all particles

of the body

For each instances we mark the line of action of the resultant force

For all practicle purposes these lines of action will be cuncerrunt at a single point G, which is called the center of gravity of the body

To determine the mathematically the location of the CG

of any body, we apply the principle of moments to the

parallel system of gravitational forces

The moment of the resultant gravitational force W about

any axis equals the sum of the moments about the same

axis of the gravitaional forces dW acting on all particles

treated as infinitesimal elements of the body

The resultant of the gravitational forces acting on all

elements is the weight of the body and is given by

the sum W = ∫dW

For example, if we apply the moment principle about the y-axis, the moment about this axis of the elemental weight is x dW, and the sum of these moments for all elements of the body is ∫x dW

This sum of moments must equal , the moment of the sum

Thus, W x = ∫ xdW

Center of Gravity (CG) and

Center of Mass (CM)

Similarly, we can sum moments about the x- and

z-axes to find the coordinates of G

With the substitution of W = mg and dW = g dm,

the expressions for the coordinates of G become

Center of Gravity (CG) and

Center of Mass (CM)

The coordinates of G may be expressed in vector form as

In which

r = xi + yj + zk and r = xi + yj + zk

Center of Gravity (CG) and

Center of Mass (CM)

It is obvious that the previous equations are independent

of gravitational effects since g no longer appears

They therefore define a unique point in the body which is a

function solely of the distribution of mass

This point is called the center of mass, and clearly it

coincides with the center of gravity as long as the gravity

field is treated as uniform and parallel

Center of Gravity (CG) and

Center of Mass (CM)

The centroid C is a point which defines

the geometric center of an object

The centroid coincides with the center

of mass or the center ofgravity only if

the material of the body is

homogenous (density or specific

weight is constant throughout the

body)

If an object has an axis of symmetry,

then the centroid of object lies on that

axis

Concept of Centroid

(1) Lines. For a slender rod or wire of length L, cross-

sectional area A, and desity ρ, the body approximates a

line segment and dm = ρ A dL. If ρ and A are constant over

the length of the rod, the coordinates of the center of mass

become the coordinates of the centroid C of the line

segment.

Centroid of Lines, Area, and Volume

Centroid of Lines, Area, and Volume

Centroid of Lines, Area, and Volume

• Centroid is simple in concept, but not always simple in

evaluation. When possible, choose an incremental element

that reduces the number of integrations required.

example: Select appropriate area increments to find the

centroid of the shape shown.

x ˜ x dAdA

y ˜ y dAdA

x

y

x, y = distances to area element dA ~ ~

Choosing Element for Integration

___

/ 2 (need two

expressions)

double integral single integral

with messy with nice .....

limits limits two single

integrals

dA dxdy dA wdy dA dx

x x x x w

y y y y

x x

y

possible area increments:

x

y

x

y

x

y

x ̃

y ̃

w

dy y ̃

x ̃

(a) (b) (c)

Show that the centroid of the area is

x = 4a/7 and y = 2b/5

x

y

a

b

x = y 3 a

b 3

Example

1/33

1/33

1

2 2

bdA ydx x dx

a

y by x

a

x x

x ˜ x dAdA

xb3

ax

0

a

1/ 3

dx

b3

ax

1/ 3

dx

0

a

y ˜ y dAdA

1

2

b3

ax

1/ 3b3

ax

1 / 3

dx

0

a

dA

a4

7 = b

2

5

x

y

a

b

x = y 3 a

b 3

dx

y ̃ y

x ̃

Solution 1

3

3

3

3

( )

2 2

1

2

adA a x dy a y dy

b

y y

a x x ax x

ay a

b

x

x̃

x

y

a

b

dy

a-x

x ˜ x dAdA

1

2

a

b3y3 a

0

b

a a

b3y3

dy

a a

b3y3

dy

0

b

y ˜ y dAdA

y a a

b3y3

dy

0

b

dA

a4

7 = b

2

5

Solution 2

Example 1

Centroid of a circular arc. Locate the centroid of

a circular arc as shown in the figure.

Solution

Example 2

Centroid of a triangular area. Determine the distance h

from the base of a triangle of altitude h to the centroid of its

area.

Solution

Example

Solution

Example

Solution

example: Find the centroid of the shape shown.

x

y

1 m

1 m

y = x2

y = x 2

A: x y 9 20 m

example: A

homogeneous

uniform wire

is bent to the

shape shown.

Determine the

location of the

centroid.

x

y

45°

45°

r

A: x

2r 2

3

Example

2 m

Locate the center of gravity x of the

homogeneous rod bent in the form of a

semicircular arc. The rod has a weight

per unit length of 0.5 N/m. Also,

determine the horizantal reaction at the

smooth support B and the x and y

components of reaction at the pin A.

Solution

2 m

Solution

Extra Problem

Extra Problem