outline: centres of mass and centroids

Outline:

Centres of Mass and Centroids ◦ Centre of Mass

◦ Centroids of Lines, Areas and Volumes

◦ Composite Bodies

Beams – External Effects

Beams – Internal Effects

ENGR 1205 Chapter 5 1

Up to now all forces have been concentrated forces (force applied at single point)

But forces are actually applied over some contact area

If contact area is significant compared to the other dimensions we must account for it

Sum the effects of the distributed force over the contact region using mathematical integration


Three types of Force Distribution (see fig. 5/2):

◦ Line Distribution - vertical load supported along a cable

Intensity is in N/m

◦ Area Distribution - hydraulic water pressure against dam face

Intensity is in 𝑁/𝑚2 or Pascal (Pa)

Called pressure for the action or fluid forces

Called stress for internal distribution in solids

◦ Volume Distribution - gravitational attraction

Intensity is in N/m3

For gravity the intensity is the specific weight (density times gravitational field strength ρg)


For external forces a concentrated load can be used to replace a distributed load

Need to find the position for a concentrated load that will give an equivalent system

Typically, we are dealing with weight

Need to find Centre of Gravity/Mass for rigid bodies (for location of load)

Also, we need to be able to analyze the internal forces caused by distributed loads


Integration is the inverse of differentiation

The integral of a function is a measure of the area between it and the x-axis

The symbol of integration is ∫ , an elongated S (stands for "sum").

The integral is written as

and is read "the integral of f-of-x with respect to x."


b

a

dxxf )(

Example:

If we know that (dy/dx)=3x2 and we need to know the function this derivative came from, then we "undo" the differentiation process.

y = x3 is ONE antiderivative of 3x2 (since the derivative of any constant is zero)

In general: y = x3 + C, is the indefinite integral, C is called the constant of integration

Indefinite integrals deal with algebraic quantities (define an entire function)


Integrals with a definite bound are definite integrals

The definite integral is written as

and is read "the integral from a to b of f-of-x with respect to x."

a and b are the lower limit and upper limit, respectively, of integration, defining the domain of integration;

f is the integrand, to be evaluated as x varies over the interval [a,b];

and dx is the variable of integration.


b

a

dxxf )(


Back to the example:

to evaluate the integral of 3x2 between x=1 and x=5:

Some integration rules and properties

124)1()5(3

5

1

335

1

32 xdxx

Cn

xdxx

dvdxdvdx

dxaadx

Cxdx

nn

1

1

Image from http://24hours7days.com/Toys/Circus_Sam_Balancing_Man_1972.html

Accessed March 2009


How does it balance?

http://24hours7days.com/Toys/Circus_Sam_Balancing_Man_1972.html







Rigid Body

◦ The center of mass is a unique point whose location is a function of the distribution of mass throughout the body

◦ It is the point through which the total weight vector acts


G G

A B B A

Consider a 2D body: Divide the plate into n small elements each with weight DW and coordinates x & y


W

zdWz

W

ydWy

W

xdWx

dWW


co-ordinates for the

center of gravity:

For a Wire:

gA

LgAW

gALW

DD


W

zdWz

W

ydWy

W

xdWx

L

zdLz

L

ydLy

L

xdLx

ρ (density), g, and A (the cross-

sectional area) are constant over the

length (L)

For an Area:

gt

AgtW

gtAW

DD


W

zdWz

W

ydWy

W

xdWx

A

zdAz

A

ydAy

A

xdAx

ρ (density), g, and t (the thickness)

are constant over the surface area (A)

For a Volume:

g

VgW

gVW

DD


V

zdVz

V

ydVy

V

xdVx

ρ (density), and g are constant over

the volume (V)

W

zdWz

W

ydWy

W

xdWx

Steps to solve:

Choose a differential element for integration

Express dA, dL or dV in terms of 1 variable

Express x, y or z (location of centroid of element) in terms of 1 variable

Substitute into equations and integrate

Use axes of symmetry whenever possible


zdLzL

ydLyL

xdLxL

zdAzA

ydAyA

xdAxA

zdVzV

ydVyV

xdVxV

Find the centroid of the area under y=x2 ?



Expression for differential element

(tiny rectangle)

Integrate to get expression for A


ANSWER: The centroid of the area is 3

4 ,3

10 m

Find the centroid of the cone, with radius “a” and height “h” ?


An irregularly shaped area may be divided into the smaller shapes with known G’s

Summing moments as before:

For plates and wires with uniform thickness and homogeneous material:


ii

ii

ii

AzAZ

AyAY

AxAX

ii

ii

ii

LzLZ

LyLY

LxLX

332211321 )( xWxWxWXWWW

An irregularly shaped body may be divided into smaller shapes with known G’s (see Appendix D for centroids of common shapes)

As before:


ii

ii

ii

WzWZ

WyWY

WxWX

ii

ii

ii

VzVZ

VyVY

VxVX

Steps to solve a composite body:

Choose a reference frame

Divide the composite area/volume into parts whose centroids you know (APPENDIX D) or can easily determine

Determine the coordinates of the centroids and the volume/ area/ length for each part

Watch for cases of symmetry that can simplify calculations


Find the centroid of the L-shaped bar of negligible thickness.


1 m

1 m

Find the centroid of the irregularly-shaped plate below. All dimensions in cm.


50

40

20

30 40 80

y

x

r=10


PART A x Y xA yA

1

2

3

4

Sum

Locate the center of mass of the composite body made of a conical frustrum and a hemisphere, shown below.


Beams are bars of material that support lateral loads (perpendicular to the axis of the beam)

Beams are probably the most important structural member

Beams can support both concentrated and distributed loads

Analyze load carrying capacity of beams for:

◦ Equilibrium and external reactions

◦ Internal Resistance (strength characteristics)


We will only analyze statically determinate beams in this class


Distributed loads: intensity w of a distributed load is expressed as force per unit length of the beam (can be constant or variable)

The resultant of the distributed load is equal to the area formed by the intensity w and the length

The resultant of the distributed load passes through the centroid of the area


For a more general load distribution:


𝑅 = 𝒘𝑑𝑥

𝑅𝑥 = 𝑥𝒘𝑑𝑥

To Determine External Reactions for Beams:

Reduce each distributed load to an equivalent concentrated load (find total force applied and location)

Use given concentrated loads and equivalent concentrated loads with 2-D equations of equilibrium


A beam is subjected to the loads shown. Determine the reactions at the supports A & B.



TOTAL AREA (RESULTANT FORCE) CENTROID (POINT OF ACTION)

1 12 ∗ 2 m ∗ 300 N/m = 300 N 2

3∗ 2 m =

4

3 m = 1.33 m

2 6 m ∗ 300 N/m = 1800 N 1

2∗ 6 m + 2 m = 5 m

3 12 ∗ (4 m) ∗ (300 N/m) = 600 N 1

3∗ 4 m + 8 m = 9.33 m

Given: See FBD

Find: Ax, Ay, By


For Equilibrium 𝚺𝑴 = 𝟎 & 𝚺𝑭 = 𝟎

𝚺𝑴𝑨 = −𝟑𝟎𝟎 𝟏. 𝟑𝟑 − 𝟏𝟖𝟎𝟎 𝟓 − 𝟔𝟎𝟎 𝟗. 𝟑𝟑 + 𝑩𝒀 𝟏𝟐 = 𝟎 𝑩𝒀 = 𝟏𝟐𝟓𝟎 𝑵

𝚺𝑭𝑿 = 𝟎 𝑨𝑿 = 𝟎 N

𝚺𝑭𝒀 = 𝟎 𝑨𝒀 − 𝟑𝟎𝟎 − 𝟏𝟖𝟎𝟎 − 𝟔𝟎𝟎 + 𝑩𝒀 = 𝟎 𝑨𝒀 = 𝟏𝟒𝟓𝟎 𝑵

ANSWER: The reactions at the supports are 𝐀𝐘 = 𝟏𝟒𝟓𝟎 𝐍 [𝐮𝐩] & 𝐁𝐘 = 𝟏𝟐𝟓𝟎 𝐍 [𝐮𝐩]

Beams can resist:

Tension/ Compression

Shear force (V)

Bending moment (M)

Torsional moment (T)

The amount of each kind of internal load can change throughout the length of the beam, depending on external loads


We can ‘cut’ the beam at any point along its length to analyze the internal forces at that point

It is often difficult to tell the direction of the shear and moment without calculations so represent V and M in their positive directions and let the result tell you (+/-) whether you drew it the right way

CONVENTION


The maximum bending moment is often the primary consideration in the design of a beam

Variations in shear and moment are best shown graphically


Use equilibrium of whole beam to find reaction forces

Then cut beam ___________ concentrated loads, replace any distributed loads with _________________ and solve for unknown shear and moment at _________________.

Plot V (shear vs x) and M (moment vs x) along the beam


Solve for R1 and R2

Isolate left section & solve for V and M between the left support & 4kN load

Isolate right section and solve for V and M between the right support & 4kN load

Isolate a section between every change in an external load, but DO NOT CUT the section at the concentrated load

The area under a shear curve between two points is equal to the change in bending moment between the same two points.



Given: See Diagram

Find: Maximum bending moment

Draw shear force and bending moment diagrams

Solve: For Equilibrium 𝐹 = 0 and 𝑀 = 0

Draw the shear and bending moment diagrams for the beam and loading shown and determine the location and magnitude of the maximum bending moment.


Draw the shear and bending moment diagrams for the beam and loading shown.


outline: centres of mass and centroids

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