01 - MECH3001Y - Unsymmetrical.bending

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Unsymmetrical Bending

K.A. | MECH3001Y | Mechanics of Materials & Machines III Page | 2

Review of Centroids & Moments of InertiaReview of Centroids & Moments of InertiaReview of Centroids & Moments of InertiaReview of Centroids & Moments of Inertia

Centroids and moments of inertia are used repeatedly throughout this chapter, and must therefore be clearly understood to efficiently tackle problems involving unsymmetrical bending of beams.

Essential definitions and formulas must be readily accessible.

Topics covered in this introductory part include: centroids and how to locate them moments of inertia parallel-axis theorem

Only plane surfaces are considered


CENTROID OF PLANE AREAS

A centroid is an important geometric property. The area of the above geometric figure is defined by

A dA=

1st moments of area w.r.t the x and y axes?


First moments represent the sums of the products of the differential areas and their coordinates (Pytel & Kiusalaas, 2010).

First moments may be positive or negative, depending upon the position of the xy axes.

Also, first moments have units of length raised to the third power; for instance, m3.

The coordinates x and y of the centroid C are equal to the 1st

moments divided by the area of the shape.

Centroids of some simple geometric shapes

In engineering work, we rarely need to locate centroids by integration. Centroids of common geometric figures are already


known and tabulated. For objects composed of several parts of familiar geometric shape (rectangle, circle, ...), the integral form is rarely used to find the position of the centroids.

Instead,

1. The overall shape is broken down in simpler ones 2. The centroid of the overall shape is calculated using a

summation

e.g:


Problem 1.1

Locate the centroid of a right angled triangle with baseline a

and height h.

,


Problem 1.2 Centroid of a composite area

Consider the L-shaped area in the given figure. Locate its centroid (find the coordinates) by breaking the L-shape into rectangular elements and applying the summation principle.


Note that absence of an area can be handled by using subtraction.

e.g,


Homework Centroids Locate the centroids of the profiles shown (dimensions in mm).


MOMENTS OF INERTIA OF PLANE AREAS

The moments of inertia of a plane area (see figure) with respect to the x and y axes, respectively, are defined by the integrals:

NB: Moments of inertia are always positive.


Problem 1.3 Moment of Inertia

Calculate the moment of inertia of the given rectangle,

1) w.r.t the x-axis 2) w.r.t the y-axis 3) w.r.t the baseline BB


Note that the moment of inertia with respect to axis BB is larger than the moment of inertia with respect to the centroidal x-axis. In general, the moment of inertia increases as the reference axis is moved parallel to itself farther from the centroid.

The moment of inertia of a composite area with respect to any particular axis is the sum of the moments of inertia of its parts with respect to that same axis.


Problem 1.4 Moment of Inertia

Calculate the moment of inertia of the given box w.r.t. the x-axis.

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PARALLEL AXIS THEOREM

Given the moment of inertia with respect to some arbitrary axis, the parallel axis theorem moment of inertia of the same shape with respect to any parallel axis.

Let I0 denote the moment of inertia (2nd moment of area) of the object shown with respect to the x-axis. The parallel axis theorem states that the moment of inertia Ithe object with respect to X(parallel to X) is given

Show the proof on board

Mechanics of Materials & Machines III

ARALLEL AXIS THEOREM

Given the moment of inertia with respect to some arbitrary axis, parallel axis theorem provides a way to calculate

moment of inertia of the same shape with respect to any

denote the moment of moment of area) of

the object shown with respect to

The parallel axis theorem states that the moment of inertia Ix1 of the object with respect to X1 (parallel to X) is given by

Show the proof on board

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Given the moment of inertia with respect to some arbitrary axis, provides a way to calculate the

moment of inertia of the same shape with respect to any

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Problem 1.5 Parallel axis theoremConsider the built-up beam structure shown in the picture below.

a) Find the centroid of the structure.the baseline of the structure, and the ysymmetry)

b) Find the moment of inertia of the structure about the neutral axis (NA). Note that the NA runs through the centroid.


Parallel axis theorem up beam structure shown in the picture

Find the centroid of the structure.(position the xthe baseline of the structure, and the y-axis along the line of

t of inertia of the structure about the neutral axis (NA). Note that the NA runs through the centroid.

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up beam structure shown in the picture

(position the x-axis along axis along the line of

t of inertia of the structure about the neutral axis (NA). Note that the NA runs through the centroid.


PRODUCT OF INERTIA

The product of inertia of a plane area is defined with respect to a set of perpendicular axes lying in the plane of the area.

Thus, referring to the figure, the product of inertia (or product moment of area) is defined by the integral:

xyI can be positive, negative, or zero, depending upon the

position of the xy axes with respect to the area.

Area lies entirely in 1st quadrant: Ixy positive

Area lies entirely in 2nd quadrant: Ixy negative

Area lies entirely in 3rd quadrant: Ixy positive

Area lies entirely in 4th quadrant: Ixy negative

Product of inertia: Symmetry The product of inertia of an area is zero with respect to any pair of axes in which at least one axis is an axis of symmetry of the area.


Parallel axis theorem


Problem 1.6 Product of inertia

Determine the product of inertia of a rectangle with respect to xy axes having their origin at point O at the lower left-hand corner of the rectangle.


Problem 1.7 Product of inertia

Determine the product of inertia

xyI of the Z-section shown. The

section has width b, height h and constant thickness t.


Unsymmetrical BendingUnsymmetrical BendingUnsymmetrical BendingUnsymmetrical Bending

Frequently, a beam undergoes a simultaneous deflection in two different perpendicular directions. In this case, the bending is referred to as unsymmetrical bending.

This type of bending occurs when the beam:

1. carries loads along two perpendicular directions 2. cross section is not symmetrical

Unsymmetrical bending can be considered as the superposition of 2 simple bending problems.

For a beam loaded in the y-direction, there is a moment about the z-axis (cf. diagram).

If we now consider loads being applied simultaneously to the beam in the z and y directions resp., moments about the y and the z-axis resp. will apply to the beam.

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Each direction can be solved separately for bending stress, and the results added together to obtain the

ASSUMPTIONS

a) We consider that the beam cross section has an arbitrary shape, and therefore, no particular symmetry is present.

b) We assume that the beam cross-section, i.e loads act through

NB: Observe that we use a right

c) The axis of the beam bends but does not stretch.d) Plane sections of the beam remain plane (do not warp) and

perpendicular to the deformed axis of the beam.section of the beam rotates as a rigid entity about a line called the neutral axis of the cross section

e) Changes in the crossnegligible.


Each direction can be solved separately for bending stress, and the results added together to obtain the overall bending stress.

We consider that the beam cross section has an arbitrary shape, and therefore, no particular symmetry is present.

that loading is such that there is no twisting of section, i.e loads act through the centroid.

Observe that we use a right-handed coordinate system

The axis of the beam bends but does not stretch.

Plane sections of the beam remain plane (do not warp) and to the deformed axis of the beam.

beam rotates as a rigid entity about a line called the of the cross section).

Changes in the cross-sectional dimensions of the beam are

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Each direction can be solved separately for bending stress, and overall bending stress.

We consider that the beam cross section has an arbitrary shape, and therefore, no particular symmetry is present.

loading is such that there is no twisting of the centroid.

handed coordinate system

The axis of the beam bends but does not stretch.

Plane sections of the beam remain plane (do not warp) and to the deformed axis of the beam. (Each cross

beam rotates as a rigid entity about a line called the

sectional dimensions of the beam are

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The deflection of the beam is caused by a momentperpendicular to the load.


The deflection of the beam is caused by a momentperpendicular to the load.

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The deflection of the beam is caused by a moment which is

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SOLVING THE PROBLEM

a) Resolve the load into 2 componentsb) Find the corresponding moments and stressesc) Use the principle of superposition, i.e., add the stresses

Recall from the simple bending theory

:

:

: . . .

. .

= =

z

z

z

z

bending stressM bending moment about ZI moment of inertia wr t Z

M y M yI I


OLVING THE PROBLEM

esolve the load into 2 components

Find the corresponding moments and stresses

Use the principle of superposition, i.e., add the stresses

Recall from the simple bending theory

:

:

: . . .

. .

= = z

z

bending stressM bending moment about ZI moment of inertia wr t Z

M y M yI I

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Use the principle of superposition, i.e., add the stresses

K.A. | MECH3001Y |

The bending stress in the xmoment about the Z

Similarly, the bending stress due to bending moment around the y-axis only is given by

:

: Y: Y

.

=

y

y

y

yy

ybending stress

M bending moment aboutI moment of inertia about

M zI


The bending stress in the x-direction is caused by a bending -axis.

Similarly, the bending stress due to bending moment around axis only is given by

:

: Y: Y

bending stressM bending moment aboutI moment of inertia about

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by a bending

Similarly, the bending stress due to bending moment around


Principle of superposition

The total bending stress is given as

..

= +

=

zyb

y zb

zy

M z M yI I

unsymmetrical bending

equation


Neutral axis: board show how to find.

line of ZERO stress.


INCLINATION OF THE NEUTRAL AXIS (NA) In general, the neutral axis for unsymmetrical bending is not parallel to the bending moment M. Because the neutral axis is the line where the bending stress is zero, its equation can be determined by setting 0 = , which yields

..0 = y z

zy

M z M yI I

The angle between the neutral axis

and the Y-axis is given by

1 1 .tan tan.

= =z y

zy

M Izy M I


Problem 1.8 Unsymmetrical Bending Two distributed line loads act on a cantilever beam as shown in the diagram on the left. Both loads act through the center of the rectangular cross section in the directions shown. What is the maximum absolute bending stress in the wall?


Problem 1.9 Unsymmetrical Bending A cantilevered beam of rectangular cross section supports an inclined load P having its line of action along a diagonal of the cross section (see figure). Show that the neutral axis lies along the other diagonal.


Problem 1.10 Unsymmetrical Bending The W250x32.7 section carries a 32kN.m bending moment inclined at 16.2 to the z-axis as shown. Determine a) the moments of inertia of the section w.r.t. the y and z axes. b) the angle between the neutral axis and the z-axis. c) the largest bending stress acting on the section.


TRANSFORMATION OF SECOND MOMENTS OF AREA

In general, the values of Ix, Iy, and Ixy for a plane area depend on the location of the origin of the coordinate system and the orientation of the xy-axes.

Here we investigate the changes caused by rotating the coordinate axes.

Consider a set of axes ( ),x y and a second set of axes ( ),u v having the same origin as ( ),x y and rotated through an angle w.r.t. ( ),x y .


Given the second moments of area w.r.t. the (x, y) axes, find the moments of area w.r.t. (u,v).


PRINCIPAL AXES

The principal axes are the set of axes ( ),u v for which the product moment of area is EQUAL TO ZERO.

It follows that for a couple of perpendicular axes, for which at least one of the axes is an axis of symmetry of a given plane shape, the pair of axes are also principal axes for the given shape.

Finding the principal axes

0uvI =

0 sin2 cos22

x yxy

I II = +


2 tan2

xy

x y

I

I I =

Note that having found 1 principal direction, the 2nd one is at 90 to the first one.


PRINCIPAL AXES AND BENDING

Recall the unsymmetrical bending equation

Choosing the axes for resolving forces/moments

y and z are chosen such that they are principal axes.

All plane sections, whether they have an axis of symmetry or not, have two perpendicular axes about which the product second moment of area is zero.

..

= y zbzy

M z M yI I


BENDING OF BEAMS HAVING UNSYMMETRICAL SECTIONS

1) Find moments of inertia (including product moment of inertia) of section with respect to a set of known axes.

2) Find the principal axes.

3) Resolve load along principal axis directions.

4) Calculate moments of inertia of section with respect to principal axes.

5) Find bending moments / stresses for each principal direction.

6) Superimpose solutions for each direction to obtain the overall solution of the problem.


Problem 1.11 Unsymmetrical Bending // Principal axes A couple of magnitude M0=1.5 kN.m acting in a vertical plane is applied to a beam having the Z-shaped cross-section shown. Determine a) the moments of inertia of the section w.r.t. y and z axes b) the directions of the principal axes c) the stress at point A d) the angle that the neutral axis forms with the horizontal

plane The moments and product of inertia of the section with respect to the y and z axes have been computed and are as follows: Iy = 3.25x10-6 m4 Iz = 4.18x10-6 m4 Iyz = 2.87x10-6 m4


Problem 1.12 Unsymmetrical Bending // Principal axes The angle section shown below is used as a simply supported beam over a span of 2.4m. The beam carries a central concentrated load of 400N acting along line YG , midway between the two supports. Point G

is the centroid of the section. Taking E=200GPa , a) Calculate the positions of the principal axes. b) Calculate the principal moments of inertia. c) Show that the bending moment at the midpoint between the

supports is given by L4

W and calculate the corresponding

stress at point C.

d) Find the position of the neutral axis.


Solution:

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01 - MECH3001Y - Unsymmetrical.bending