Cec 106 Theory- Strength of Materials

8/12/2019 Cec 106 Theory- Strength of Materials

1/77

1

UNESCO-NIGERIA TECHNICAL &VOCATIONAL EDUCATION

REVITALISATION PROJECT-PHASE II

YEAR I- SE MESTER 2THEORY

Version 1 December 2008

NATIONAL DIPLOMA INCIVIL ENGINEERING TECHNOLOGY


2/77

2

CIVIL ENGINEERING TECHNOLOGY

STRENGTH OF MATERIALS CEC 106

COURSE INDEX

WEEK 1 1.0 Tensile Stress and Compressive Stress------------------ 1

1.1 Tensile Strain and Compressive Strain ------------------ 1

1.3 Stress-Strain Curves for Brittle Materials and DuctileMaterials ------------------------------------------------- 2-3

1.4 Stress Strain Curve for mild Steel ---------------- 3-4

1.5 Limit of Proportionality --------------------------------- 4

1.6 Elastic Limit ------------------------------------------ 4- 5

WEEK 2 2.0 Yield Point ------------------------------------------------- 6

2.1 Elastic Range ---------------------------------------------- 6

2.2 Proof Stress ------------------------------------------------ 7

2.3 Working Stress ------------------------------------------- 7

2.4 Lateral Strain----------------------------------------- 8

2.4 Elasticity ------------------------------------------------ 8-12

WEEK 3 3.0 Define and Compute the Centroid of a Section ---- 13-14

WEEK 4 4.0 Define and Compute Neutral Axis ------------------ 15-20

WEEK 5 5.0 Parallel Axis Theorem -------------------------------- 21-235.1 Section Modulus ------------------------------------------- 24-25

WEEK 6 6.0 Shearing Force and Bending Moment -------------------- 26


3/77

3

6.1 Bending Moment ------------------------------------------- 27

WEEK 7 7.0 Relationship Between Bending Moment and Shearing Force

--------------------------------------------------------- 28-29

WEEK 8 8.0 Expressions for Shear Force and Bending Moment at aSection of loaded beam --------------------------- 30-32

WEEK 9 9.0 Point of Contraflecture ---------------------------- 33-34

WEEK 10 10.0 Torsion of Shafts --------------------------------- 35-39

WEEK 11 11.0 Torsion on Circular Sections ----------------- 40-42

WEEK 12 12.0 Torsional Stiffness of a shaft ------------------ 43-46

WEEK 13 13.0 Solid Steel Shaft Problems and Solution ----47-49

WEEK 14 14.0 Analysis of Complex Stress ------------- 50 55

WEEK 15 15.0 Describe Mohrs circle of Stress ------------ 56 -58


4/77

4

WEEK ONE ( 1 )

1.0 Tensile stress : When equal and opposite forces are axially applied on a

body

Such that the length of the body increases, then the stress produced is called

Tensile stress.

Fig.1

Compressive stress : When equal and opposite forces are axially applied on a

Such that the body is compressed , the stress produced is called compressive

Stress.Fig.2

1.1 Tensile strain : When the stress induced is tensile in nature the corresponding

Strain is called tensile strain.

Tensile strain = _Increase in lengthOriginal length

1.2 Compressive strain : When the body is compressed and shortening in length


5/77

5

Takes place due to compressive stress, the corresponding strain is called

Compressive strain.

Compressive strain = _Decrease in lenghOriginal length

Since strain is a ratio of two dimensions hence it is a pure

Member. It is a dimensionless quantity.

Modulus of Elasticity : The quantity E is the ratio of unit stress to unit strain;

it indicates how much stress accompanies a given strain.

E is measured in GN\ mm2

1.3 Stress-strain curves for brittle materials and ductile materials.

Materials which fracture when the strains are small are known as brittle,

Whilst materials which have an appreciable deformation before failure are

Said to be ductile.

Experimenting stress\strain diagram differ considerably for different

materials. However , broadly speaking two types of diagrams shown

below in figure 2.0 brittle material like concrete, copper, cast iron,

Glass, stone e.t.c.


6/77

6

Figure 2: Ductile and brittle material behavior

Figure 2.0: stress\strain curve of a brittle materials subjected to a tensile test , undergothe following:

1.A Brittle e.g. .cast-ion resisted very little to

Rupture such materials can not undergo deformation.

2. The modulus of elasticity E is not well defined. point A could be

Assumed to be limit of proportionality (i.e. hooks law is being obeyed).

3. Small extension leads to fracture. This occurs without any noticeable

Change in the rate of elongation thus B is the fracture or rupture point of

The brittle material.

4. We can conclude from these observation that normal stresses are the


7/77

7

Primary responsible for this failure of brittle materials.

1.4 Stress- Strain curve for mild steel : When a specimen of mild steel

Is gradually loaded in a tension testing machine and a graph is plot

Between the stress and the corresponding strain a curve is obtained as

Shown in fig.3. Below

U

E B

p Y 1 Y2

Figure 3.

1.5 Limit of proportionality :

It is observed that with a gradual increase in loading there is a proportional

Increase in strain as well. The maximum stress value up to which this

Relationship is maintained is called the limit of proportionality, point p on


8/77

8

The curve shows the limit of proportionality.

1.6 Elastic Limit

It is the maximum stress up to which the material behaves as an elastic

Material. There is no permanent or residual deformation left when load is

Entirely removed. point E on the curve represents elastic limit. These two

Points are very close to each other. But in most cases elastic limit is higher

Than limit of proportionality.

= G / L --------- ( 2 )


9/77

9

WEEK TWO ( 2 ) 2.0 Yield Point

It is the point Y on the stress-strain curve. It will be observed that at a

Point just above the limit of proportionality a considerable increase in

Strain takes place in ductile materials with little increase in stress. The

Stress value at which this large increase in strain takes place is termed as

yield point of the material. There are two yield points on the stress-

Strain curve of which there is an increase in strain without an increase in

Stress . these are known as upper and lower yield points. Stress at yield

Point is called yield stress.

2.1 Elastic range

In the early stages of the test it can be seen that the plot rises steeply in a

Straight line . This is the linear elastic range. Linear simply means that

Stress is directly proportional to strain . Elastic means that the extension

Is reversible- i.e. if the load was removed at this stage the specimen would

Return to its original length. The energy put into the steel specimen by the

Testing machine is stored in the specimen like in the spring of a clock.

The tangent of the slope of initial straight portion of the plot :

Tan = StressStrain

This is the modulus of elasticity .E . 0f the material, which is of course

The measure of the material, so the steeper the slope of the line. the stiffer


10/77


11/77

11

Cases, yield stress to the working stress . This is explained above. Some

Decades later an alternative approach was introduced in terms of the loadFactor based on the ratio of loads rather than stresses, i.e.

Ultimate loadLoad Factor =

Working load

2.4 Lateral strain

A member having a rectangular cross section and a bar of circular

Cross section, both being subject to direct tensile forces, we know that

The bars will extend in length under the action of the external forces p . the

Lateral dimensions will decrease. Common experience, perhaps the

Stretching of a piece of elastic with a rectangular cross section, should

Have led you to make the above general observation. Lateral strain (El) is

Directly proportional to the longitudinal stress (d).

2.5 Elasticity

One of the most significant properties of a structural material is elasticity.

You will observed that when a steel is suspended and gradually loaded along

Its axis up to a certain maximum load, the length of wire increases and when

The applied load are gradually removed, the wire comes back to its original

Length.

A body which returns to its original shape and size and all traces of

Deformation disappear when the loads are removed is called an elastic body.

This behavior and the property by virtue of which it returns to its original

Dimensions are called elasticity. Perfectly elastic body shows 100% recovery


12/77

12

i.e. deformation completely disappear . But in practice no material has been

Found to be perfectly elastic.

Ductility

If a material can undergo deformation without rupture, then it is called a

Ductile material. It is due to this property that a material may be drawn into a

Wire. Copper is an example of ductile material.

Brittleness

Brittle material possesses very little resistance to rupture. Such materials

Not undergo deformation when external forces are applied and fail by rupture

Cast iron is an example of brittle material.

Malleability

The property of a material by virtue of which it can be rolled into plates is

Known as malleability. Wrought iron is an example of malleable material.

Homogenous material

A homogenous material is one which has the same modulus of elasticity

(E) And poisons ratio u at all points in the body . The material has the same

Physical and chemical composition throughout.

Isotropic material

The second assumption usually made is Isotropic i.e. it possess the same

Elastic properties.

PlasticitYA material is said to be plastic when the deformation produced by the

Application of an external force does not disappear even after the removal


13/77

13

Of the external force. Lead is an example of plastic material.

Example 1.0

Determine the elongation of a steel rod 2 meters long and 40mm

In diameter, when subjected to an axial force of 6KN. The modulus of

Elasticity of steel may be taken as 200GN\m2.

Solution

Axial load on the rod = 6KN = 6000Newtons

Area of cross- section of the rod = 3.14/4(40)2=1256.6 mm2

Axial loadTensile stress = Area of cross section

= 60001256.6

= 4. 77 N / mm 2

Stress

Strain = --------Modulus of elasticity and E = 200GN/m2

= 4. 77 = 0.0238x10-3200 GN/m2

Strain = change in lengthOriginal length

= dl or dl = xl = 0.0238x10-3x2x103L= 0.0478mm

Elongation = 0.0478mm


14/77

14

Example 2

A straight bar of uniform cross- section is subjected to an axial

Tensile force of 40 KN . The cross-section al area of the bar is 500mm2

And its length is 5 meters. Find the modulus of elasticity of the material

If the total elongation of the bar is 2mm.

SolutionSectional area of the bar = 500mm2

Applied load = 40 KN

Tensile stress = LoadCross- sectional area

= 40x103

500 = 80 Mpa

Strain e = dl = 2 = 4x10 -3 L 5x10 3

Modulus of elasticity = E= d /e = 80\4x10 -3 E = 200KN/mm2


15/77

15

EXERCISES 1. Determine the change in the length of the rod AB as shown in fig. below. the length ofthe rod is 4 meters and diameter 30mm , Take E = 210 000 N/MM.

B C D

1.6m 2.4m

4m 30 KN

A /////////////////

R


16/77

16

WEEK THREE ( 3 )

3.0 Define and compute the centroid of section e.g. rectangular, I section

T section on, channel section

The centre of gravity of a body is a point in or near the body through

Which the resultant attraction of the earth, i.e. the weight of the body

Acts for all positions of the body.

It should be noted. However, that the section of a beam is a plane

Figure without weight , and therefore the term centre of area or centroid

Is more appropriate and is frequently used in this case. The determination

Of a body or centroid of a section is equivalent to determining the resultant

Of a number of like, parallel forces. These forces are vertical and they can be

Replaced by a single vertical resultant of magnitude equal to the total weight

Of the body and acting at the centre of gravity of the body. The position of

The centre of gravity of a body,

Suppose we now wish to locate the centre of gravity (G) of an irregular

Shaped plate of material of uniform thickness and uniform weight w N/mm2

As shown in the figure below. Consider the plate lying horizontally. The

Plan view is as shown with the weight forces acting vertically downwards at

Right angles to the page.

X and Y are to the horizontal plane and intersect at a convenient origin

O. The small element shown has an area dA and is located at a distance x

From the Y axis and distance y from the X Axis.


17/77

17

Let us consider a few common shapes. It should be obvious that for a

Square, rectangle and circle, the centroid will be central as shown below.

Now look at the triangle shown in figure (d) above. Taking moments of

Area of the approximately rectangular shaped elements about an X-axis

Passing through the apex , then __ Axy = E ydA

That is .5BHx y = Ebydy since b/B = y/H_ = B/H y2 dy =B/H{y3/3}0-H =BH2/3

Thus y =2/3H from apex for i/2H above base


18/77

18

WEEK FOUR ( 4 )

4.0 Define and compute neutral axis

It has been seen that, in general, most beams tend to bend in the form

Shown in figure below, and if X- X is the neutral axis of such a beam

The fibres above X-X, are stressed in compression and those, below X-X

In tension .Furthermore, fibres far away from the neutral axis are stressed more

Heavily than those near to the neutral axis, fibres lying on the neutral axis

Are neither in tension or compression.

The location of the neutral axis of a plane area is an important geometric property of

the area. To determine the coordinates of centroid (point of intersection of two or more

neutral axis in different directions) of the plane area, let us refer to figure 2.2.1.

A differential element of area dA, with coordinates x and y, is shown in the figure. The

total area is defined as the following integral.

= dAA \In addition, first moment of the area about the x and y axes, respectively , are defined as

= ydAQ x And

= xdAQ y The coordinates x and y of the centroid C (figure2.2.1) are equal to the first moments

divided by the area itself:

==

dA

xdA

A

Q x y And


19/77

19

==

dA

ydA

AQ

y x

If the boundaries of the area are defined by simple mathematical expressions, we can

evaluate the integrals and thereby obtain formulae for yandx .

If an area is symmetric about an axis the centroid must lie in that axis because the first

moment about an axis of symmetry equals zero. For example, the centroid of single

symmetric area shown in figure 2.2.2 must lie on the x-axis, which is the axis of

symmetry; hence only one distance must be calculated in order to locate C.

If an area has two axes of symmetry, as in figure 2.2.3, the position of the centroid can be

determined by inspection of the axes of symmetry.

x

y

Figure 2.2.2 Area with one axis of symmetry

C


20/77

20

An area of the type shown in figure 2.2.4 is symmetric about a point. It has no axis of

symmetry, but there is a point (called center of symmetry) such that every line in area

drawn through the point is symmetric about that point. The center coincides with the

center of symmetry and can be located by inspection.

x

y

C

Figure 2.2.3 Area with two axes of symmetry

Figure 2.2.4 Area that is symmetric about a point

y

xC


21/77

21

If the boundaries of the area are irregular curves not defined by mathematical expression,

then we evaluate approximately the integrals by numerical methods. The simplest

procedure is to divide the area into small elements of area Ai and replace theintegrations with summations:

=

=n

1i

AiA

=

=n

1ix AiQ yi

=

=n

1iy AiQ xi

In which n is the total number of elements of the area, yi is the y coordinate of thecentroid of area Ai, and xi is the x coordinate of the Ai. The accuracy of the

calculation for yandx depends upon how close the selected elements fit the actual area.

Neutral Axis for Composite Areas

In engineering work, we frequently nee to locate the centroid of an area composed of

several parts, each part having a familiar geometric shape such as a rectangle or a

triangle. Examples of such composite areas are the cross-sections beam, which often are

composed of rectangular areas (for example figure 2.2.2, figure 2.2.3 and figure 2.2.4).The area and first moments of a composite area may be calculated by summing the

corresponding properties of its parts.

== =

===n

1i

n

1i

n

1iyx xiAiQ yiAiQ AiA

In which Ai is the area of the ith part, xi and yi are both coordinates of the centroid of the

ith part, and n is the number of parts. When using these formulae, it is possible to treat

the absence of an area as a negative area, for instance, this concept is useful when a

hole exist in the figure. After finding A, Q x and Q y, we can determine the coordinates of

the centroid.

To illustrate the procedure, let us consider the several case of a composite area that can

be divided conveniently into only two parts. This L-Shaped area shown in figure 2.2.5 is

of this kind, because it can be divided into two rectangles of area A 1 and A 2. These


22/77

22

rectangles have centroids C 1 and C 2 with known coordinates (x 1, y1) and (x2, y2)

respectively.

Thus we obtain the following expressions:

A = A 1 + A 2

QX = y 1A1 + y 2A2

QY = x 1A= + x 2A2

Therefore, the coordinates of the centroid C are

21

2211Y

AA

xAx

A

Q

x ++

== A

21

2211X

AAyAy

A

Q y

++== A

C

A2

A1

C 2

C 1

x

y

O

Figure 2.2.5 centroid of a composite area consisting of two parts


23/77

23

The distance from the Y- Axis to G will be given by ;

Total weight Xx = E wxdA

Where E is the usual mathematical symbol to denote the

Sum of but total weight = sum of the weight of the small elements = EwdA

Thus EwdA X x- = Ew XdA

Or EdA X x- = E x dA ( since w is uniform )

But E d A = A ( the total surface area )

Thus A = E xdA

From the figure below the position of neutral axis can be compute as follows;

Taking moment of areas about the top flange gives ;

15050 = 7500 25 = 187500

20050 = 10000 150 = 1500000

10050 = 5000 275 = 1375000

22500 y = 3062500

Hence y = 30625/225 = 136.1mm

First moment area of the cross- section;

Consider the general case of a beam with irregular cross-section as shown below;


24/77

24

Consider a thin strip of the cross-section of of width b , thickness y and

At distance y from the neutral axis . Let the stress in this strip =

Then the longitudinal force acting on the strip = stress area= (by)

And hence the longitudinal force acting on the cross- section = y1 bdy-y2

But as there can be no resultant longitudinal force acting on the section , this

Force must be zero if static equilibrium is to be assured .

Thus y1 bdy = 0. y2

By similar triangles in figure (b) you will see that ; y = max

y1thus = max y

y1

and equation becomes ;y1

max - y 2 by dy = 0Y1 y1

But max and y 1 cannot be zero , thus - y2 by dy must be zero.

This integral expression is the first moment area of the cross-section about the

Neutral axis , and we know from programme 1 that if the first moment of area about

An axis is zero then that axis must pass through the centroid of that area


25/77

25

SECOND MOMENT OF AREA

Now refer to the diagram of irregular section above .

The longitudinal force in the thin strip = by

The moment of this force about the neutral axis = force distance (y)= by y

y1 The total moment of the resultant force about the neutral axis = -y2 bydy

But = max yY1

Thus the total moment about the neutral axis = the moment of resistance= M

Y1= max -y2 by 2dy

y

the integral expression in equation above is termed the second moment of area of the

cross-section about the neutral axis , and for convenience is given the symbol I . I has

units of (length) 4 and typically would be calculated in mm 4 . the equation can now be

written more simply as ;

y1M = max I where I = -y2 by2 dy

Y1But max = thus M = I

Y1 y

Or = MY I

Where ; M is the moment of resistance

I is the second moment of area about the neutral axis

is the bending stress in any fibre at a distance y from the neutral axis .


26/77

26

The bending stress at a distance of 50 mm is 10 N/mm 2. If the section is subjected

To a bending moment of 2 KN m, calculate the second moment of area of a beam section.

Solution ;I = M y = 2 10 6 50 = 10 10 6 mm 4

10

Example 1

A parabolic semi-segment OAB is bounded by the x-axis , y-axis and a parabolic curve

having its vertex at A (figure EX.1).

The equation of the curve is

y = f (x) = h (1 x 2 /b2)

In which b is the base and h is the height of the semi-segment. Locate the centroid C of

the semi-segment.

Solution:

yh

0x

b

2 y

x

y

y = f (x)

Figure EX.1

A

B

C

x

y


27/77

27

To carryout the analysis, we will select any element of area dA in the form of a thin strip

of width dx and height y, as shown figure EX.1. The area of this element is

dA = ydx = dxb

2

2x -1h

And therefore, the area of the semi-segment is

A =

= dx

bdA

b

02

2x -1h

A = dxb

b

02

2x -1h

A =b

b0

2

3

3x

-xh

A = 0-3b

-bh0

2

3 b

b

A =3b

-bh

A =3

2bh

The first moment of the element of area about any axis can be obtained by multiplying

the area of the element by the distance from its centroid to the axis.

Since the x and y coordinates of the centroid of the element x and y/2, respectively, the

first moments are:

dxbb

dA

==

b

02

2

2

2

xx

- 1hxx

-12h

Ay

Q

dxb

dA

==

b

0

2

2

22

x

x -1

2h

Ay

Q

+= dx

bb 4

4

2

22

X

x

2x -1

2h

Q

b

bb 04

5

2

32

X5x

32x

-x2

h Q +=


28/77

28

154bh

158b

x2

h Q

232

X ==

154bh

Q2

X =

dxb

dA

==

b

02

2

Yx

-1hxAx

Q

= dx

b 2

3

Y

x -xhQ

b

b 02

42

Y4x

2

x hQ =

4b

h xQ2

Y =

4hb

Q2

X =

Now we can determine the coordinates of the centroid C as follows:

AQ

x Y=

2bh3 x

4hb

32bh

4hb x

22 =

=

83b

x =

AQ

y x=

2bh3

x154bh

32bh

154bh

y

22

=

=

52h

y =

The centroid may also be located by taking the element of area A as a horizontal strip of

height dy and width


29/77

29

=

hy

-1bx

This is obtained by solving the curve equation for x in terms of y.

Example 2:

Find the centroid of a channel section 100mm x 50mm x 15 mm as shown in the figure

below:

Solution:

As the section is symmetrical about X X axis, therefore its centroid will lie on this axis.

Now split up the whole section into three rectangles ABFJ, EGKJ and CDHK as shown

in the figure.

Let the face AC be the axis of reference.

i. Rectangle ABFJ

a1 = 50 x 15 = 750 mm2

x1 = 50/2 = 25mm

ii. Rectangle EGKJ

50 mm

15 mm

1 0 0 m m

XX

B

C

A

D

E F

G H

J

K


30/77

30

a2 = (100 30) x 15 = 1050 mm2

x2 = 15/2 = 7.5 mm

iii. Rectangle CDHK

a3 = 50 x 15 = 750 mm2x3 = 50/2 = 25mm

We know that distance between the center of gravity of the section and left face of the

section AC,

321

332211

aaaxaxaxa

x++++=

( ) ( ) ( )7501050750

25x7507.5x105025x750 x ++

++=

mm17.8 x =


31/77

31

WEEK FIVE ( 5 )

5.0 PARALLEL AXIS THEOREM

Consider the rectangular area shown . Let Icc be the second moment of area of the

section about an axis through its centroid and let Ixx be the second moment of area about

a parallel axis at a distance h from the centroidal axis . In subsequent frames this axis (

X- X) will be the neutral axis of the composite section of which this rectangle is part.

bThe second moment of area of the elemental

Strip about the X- X axis = b yy12

Hence the

Second moment of area of the whole section C - ----------------------------C

About the X- X axis Ixx = by 12dy d hy1

But y 1= y + h Ixx = b(y + h) 2 dy = b( y 2 + 2yh + h2 ) dy

= by2 dy + b2yhdy + bh 2 dy

= by 2 dy + 2h bydy + h2 b dy X-----------------------------------X

But ; by 2 dy = the second moment of area ( Icc) about the centroidal axis

by dy = the first moment of area about an axis through the centroid

= 0( by definition of centroid )

And b dy = the areaof rectangle ( A = b d )

Thus ; generally Ixx = Icc + Ah 2 , or for a solid rectangle ; Ixx = bd 3 + bdh 2 ,12

On to the next frame ,21

Now we will use the theorem of parallel axes to help us calculate the second moment of


32/77

32

area of a beam with an I shaped cross- section , The second moment of area is calculated

about the neutral axis which , as we have shown , passes through the centroid of the

section . Many I beams are symmetrical . In this example however we have selected an

asymmeteical beam to emphasise that the first step is to determine the position of the

centroid . In a symmetrical section , of course , the centroid is located by inspection .

The section shown below is treated as if made of four rectangular parts A , B , C, D.

First we determine the position of the centroid by taking the sum of the first moments of

Area of each part about a convenient axis ( we will take the bottom face of the beam )

and equating to the total area times__ where __ is the height of the centroid above they ybottom face.thus A y = ( A part y part )

200 Y

25 AB 20

C---------------------------------------- 20

X X

D ---------150

_22

{( 200 25 ) + ( 150 20 ) + ( 20 200 ) + ( 150 20 ) }y


33/77

33

= ( 200 25 252.5 ) + ( 150 20 230 ) + ( 20 200 120 ) + ( 150 20 10 )__

Thus y = 164.2 mmNow Ixx = Icc + Ah 2

= bd 312 + bdh 2 Thus for part A; Ixx = ( 200 25

3

)/ 12 +( 200 25 ) ( 252.5 164.2 )2

= 39.24 10 6 mm 4

And for part B ; Ixx = ( 150 20 3 ) / 12 + ( 150 20 ) ( 230 164.2 ) 2 = 13.09 10 4 mm 4

And for part C ; Ixx = ( 20 200 3 )/ 12 + ( 20 200 ) ( 120 .0 164.2 ) 2 = 21.15 106 mm 4

And part D ; Ixx = ( 150 20 3)/ 12 + ( 150 20 ) ( 10 164.2 ) 2= 71.43 10 6 mm 4

For the complete section Ixx = ( 39.24 + 13.09 + 21.15 + 71.43 ) 106 = 144.91 10 6 mm 4

5.1 SECTION MODULUS


34/77

34

It is the property of a section and is determined by dividing

Moment of inertia or the area about an axis passing through the centroid of the

Section by the distance of extreme fibre of the section from the axis. It is

Denoted by the letter Z . M. I. about centriodal axisZ =

Distance of extreme fibre of section from the axis

Through the centroid . mm 3

Example :

Determine, for the plane area shown in figure below(a) Moment of inertia about x x axis and about the base AB.

( b ) Moment of inertia about y y axis and about the side AD.

Section modulus.

Solution ;(a) Moment of inertia about x x .

Ixx = bd 3

12= 60 (100) 3

12 = 5 10 6 mm 4 Moment of inertia about the base AB.

IAB = I xx +Ay 2 = 5 10 6 + (60) (100)(50) 2 = 20 10 6 mm 4

(b) Moment of inertia about Y Y axis

Iyy = db 3 = 100(60) 3

12 12= 18 10 5 mm 4 Moment of inertia about side AD

IAD = IYY + Ax 2

= 18 10 5 + ( 60 ) ( 100) ( 30) 2 = 72 105 mm 4


35/77

35

( c) Section Modulus

Zxx = Ix x = 5 10 6 = 10 5 mm 3y 50

Zyy = Iyy = 18 105

mm3

X 30= 60 10 3 mm 3


36/77

36

WEEK SIX( 6 )

6.0 Shearing forces and bending moments with sign convections

Shear force at a section of a loaded beam may be defined as the algebraic

Sum of all vertical forces acting on any one side of the section

X

W1 W 2 W 3 W 4 W 5 W6

A B

RA RB

X FIG. 6

The shear force at section x x of the beam shown in figure above when forces

To the left of x x are considered.

SF X X = R A W1 W2 W3When the forces on the right hand side of the section are considered,

SFx x = R B W4 W 5 W6 .


37/77

37

6.1 Bending Moment ;

Bending moment at a section of a loaded beam is the algebraic sum of

The moments of all the force on any one side of the section .

XW1 W 2 W 3 W 4 a

RA RB b

X

XC

Bending moment at section x x of the beam shown in the figure above can be written as;

M x x = R Ax W 1( x a ) W 2 ( x b )

Similarly at a section after x x or before it can be treat the same way.


38/77

38

WEEK SEVEN ( 7 )

7.0 Relation Between Bending Moment and Shear force

F F + F w/unit length

A A

M F F

Bx B

F F +FM F + F

M +M

M + MFigure 8

Consider a small length x of a simply supported beam carrying

uniformly distributed load w/ unit length . Let M and F be the B.M and S.F at AB

And ( M + m ) and ( F + F ) be the bending moment and shearing force at CD .

Since the element ABCD is in equilibrium, the sum of all vertical forces on it must

Be zero . Hence F + wx = F + FOr df = w ------------------------ ( i )

dx

Thus the rate of change of shear force is equal to the intensity of

Loading on the beam, similarly equating all moments at AB to Zero.

M +( F + F )x w( x ) 2 ( M + M ) = 02


39/77

39

Neglecting the products and squares of small quantities, we get F x M = 0

Or dm = F --------------------------------(ii)dx

That is the rate of change of bending moment is equal to the shearing force.

xNow integrating equation ( i ) we get F = wdx-------------------------------- (iii)

0

Integrating equation ( ii ) we get x x xM = Fdx = wdx --------------- ( iv )

0 0 0

Hence the change of bending moment from 0 to x is proportional to the area of shear

force diagram from 0 to x.

For bending moment to be maximum dM = 0dx

but dM = F from equation ( ii )dx

Thus bending moment is maximum where shear force is Zero or changes sign.


40/77

40

WEEK EIGHT ( 8 )

8.0 Expressions for shear force and bending moment at a section ofloaded beam.

A simply supported beam of intensity w kn\metre run (fig. a) below. The span

Consists of L m of load, and thus the total load is WL KN, and, the beam is symmetrical,

each reaction will be half of the total load. R L = R R = WL KN

The shear (up to the left) at a point just in the span and very near to R L is quite

simply the L . H . S. reaction of wL .

When the point concerned is say from R L however, then shear to the left

(summation of loads to the left of the point) is then wl upwards + 1m of load w downwards= wl w

Similarly when the point concerned is 2m fromRL, the shear up to the left is

wl upwards 2w (2m of load) downwards= wl 2w

Putting this in general terms , the shear at any point C on the span at distance x from

RL is wl x meters of uniform load = wl wx.

This will give a positive result wherever x is less than L , and a negative result

Where x exceeds L . So the shear will change sign where x = L ( at the point

of mid span ) , and the SFD will be as shown in fig. b below.


41/77

41

Referring again to fig.(a) , the bending moment at the L.H.S. end is again Zero , and

at a point 1m from R L , the bending moment ( summation of moment to the left ) is

simply the algebraic sum of ;

( a ) the clock wise moment of the L . hand reaction ( wL 1 ) ;

( b ) the anti clockwise moment from 1m of downward load ( w 1= w .

if the bending moment is required at a point C , at x m from R L , then the

bending moment is the algebraic sum again of;

( a ) the clock wise moment of the l. hand reaction ( wl x )

( b ) the anticlockwise moment of xm of load ( wx ) x . Thus , thebending

moment = wlx wx 2. This bending moment will be positive for any

value of x , and will reach a maximum value of ;( wl l ) w( l) 2 = wl 2 1/8 wl 2

=1/8 wl2

It should be most carefully noted that the maximum bending moment

(Mmax) is 1/8wl 2 , where w is the amount of uniform load per metre.

Sometimes it is more convenient to think in terms of the total load W (

i.e capital W ) , and in case W = wl , and Mmax in terms of the total ioad

will then be 1/8 wl l = 1/8 Wl

If the values of this bending moment at point along the span are plotted

as a graph , the resulting BMD will be a parabola with a maximum


42/77

42

ordinate of 1/8wl 2 or 1/8 Wl as shown in fig. c . Where the diagram has

to be drawn , it will be necessary only to draw a parabola having a

central height of 1/8 Wl , and any other ordinates at points

away from the centre may be scaled or calculated as desired.

C Total load = wl

X wl L wl

SFD

BMD


43/77

43

WEEK NINE ( 9 )WEEK NINE ( 9 )WEEK NINE ( 9 )WEEK NINE ( 9 )

9.0 Point of Contra flexure is the point at which the negative bending

moment changes to positive ( and vice versa ) . the value of the bending

moment at that point is zero

Example ;

A beam 4m long carrying a uniformly distributed load of 60kn/m cantilever overboth supports as shown in figure below . sketch the shear force and bending momentdiagrams and determine the position of the point of contraflexure.

60 kn/m0.5m 2.5m 1.0m

RL RR

Solution;

The shear force and bending moment diagrams are given., check thatreactions are 96kn and 144kn respectively . the maximum positivebending moment occurs at 1.6m from the left end of the beam and is

Mmax = ( 66 1.1 30 0.5 )= ( 72.6 15 )= 28 .8 knm

The maximum negative bending moment occurs over the right handsupport and is 60 0.5 =30 knm.

By calculating several other bending moment values the diagram can beconstructed as shown in the figure below Bending moment at pont of contraflexure is96 ( x 0.5 ) 60 x 1/2x = 05x 2 16x + 8 =o


44/77

44

X = 16 ( 256 160 )10

X 1 = 0.62m x 2 = 2.58m

60 kn/m0.5m 2.5m 1.0m

RL = 96KN RR = 144KN

66KN 60KN

S.F.D.

30KN 84KN

1.1m

X 1 28.8 knm

B.M.D.

30KNM7

X2

34


45/77

45

WEEK TEN ( 10 )

( 10.0 ) Torsion of shafts

Consider a shaft , with circular cross- section , having radius R and length L , rigidly

fixed at one end while twisting moment T is applied to the other end.

Shear stress / Shear strain = modulus of rigidity

/ = G l/R = G /R = G /L --------------------- (i)

Obviously , this equation , while developed considering the outside material of shaft ,

might have been developed considering material at any radius r

Hence /R = q/r = G /L ------------------------------------- ( 2 )

Where is the maximum shear stress ( occurring at radius R ) and q is the shear stress in

39the shaft at radius r . This , of course , assumes that the Straight line OB remains straight, becoming OB , consider the element indicated .

Area = 2 rdr , shear force = 2 qrdrThe moment of this force about the shaft axis = 2 qr2 dr.

Since q = r/R , this moment may be expressed as 2 ( /R ) r 3 dr


46/77

46

drR R

D

Shaft for solid steel

The applied torque = the total resisting moment

T = 2 ( /R ) R r3 dr = ( / R ) R2r3 dr0 0

= ( / R ) J

Where J = R 2r3 dr = the polar moment of inertia ( second moment of area ) of the shaft0

about the shaft axis = 2 R4

/ 4 = ( /32 ) D4

Hence T/J = /R ------------------------------------------------ ( 3 )

Combining equations ( 2 ) and ( 3 ) we have ;

T/J = /R = q/r = G /L


47/77

47

Example ;

A Solid Steel shaft , having diameter 75 mm , is subjected to twisting . If the angle of

twist is not exceed 1 0 in 2m length determine ; ( I ) the maximum torque which the shaft

may carry; ( ii ) the maximum shear stress produced by this torque for steel , G =

8010 9N/m 2

Solution ;

(i) T /J = G /L

T = JG / L = ( /32 ) ( 0.075 ) 4 80 10 9 ( / 180 ) /2 Nm

= 2170 Nm

(ii) / R = G / L

= RG / L = 0.0375 80 10 9 ( /180 ) /2 Nm 2

= 26.2 10 6N/m 2

Example ;

A Solid Shaft is required to transmit a torque of 140 KNm. If the shear stress must notexceed 55 MN/m 2 , determine the required shaft diameter.

Solution;

T/J = /R T = J/ R

J/R is termed the twisting modulus of the shaft section ( Z ) . For a solid shaft , Z = J / R

= ( / 32 ) D4

/ D/2 = ( / 16 ) D3

D3

= 16T /


48/77

48

= 16 140 10 3 / ( 55 10 6 ) m 3

The diameter required = 3 ( 12.97 10 -3 ) m = 0.235m

Example ;

A 100 mm diameter shaft transmits 60Kw at 60 rev / min . If the maximum torque is

1 times the mean torque, determine the maximum shear stress produced in the shaft .

Calculate also the maximum angle of twist per metre length , taking G = 80 10 9 N/ m 2

Solution ;

Power , P = WT = ( / 30 ) NT

T = 30P / ( N ) = 30 60 103 / ( 60) Nm

mean Torque = 9560Nm

maximum torque = 14330 Nm

T/ J = /R

= TR /J = 14330 0.05 / [ ( /32) ( 0.1 )4 ] N/m2

The maximum shear stress = 73 106 N /m2

T/J = G

/ L

= TL /JG= 14330 1 / ( 9.82 10-6 80 109 ) rad

Over 1m length , the maximum angle of twist = 18.25 10-3rad = 1.0450


49/77

49

FURTHER PROBLEMS

(I) A solid circular drive shaft has a diameter of 50 mm . If it is subjected to a torque of 2

KN m ; calculate ( I ) the maximum shear stress developed and ( 2 ) the angular twistPer unit length of shaft

( ii ) The shear stresses in a solid circular shaft are limited to a maximum value of 75 N /mm2

and the angular twist permitted is no greater than 0.5 degrees over a length of 1 metre .

If both these limiting conditions are to be satisfied simultaneously , determine the least radius

of the section and the maximum permissible torque

( iii ) A solid circular shaft rotates at 100 rev/min and is to be designed to transmit 200KW , If

the maxmum shear stress is not to exceed 50 N /mm2 . What is the least radius of the shaft..

( iv ) Determine the maximum torque that can be applied at the junction of the two solid circularshafts shown below . Both shafts are firmly restrained at their ends as shown and themaximum shear stress is limited to 75 N/mm2

A 20 B 40 C

300 500


50/77

50

WEEK ELEVEN ( 11 )

11.0 TORSION ON CIRCULAR SECTION

Shear Strain is define as the angle of deformation in radians =

Consider an element of ma T erial, rigidly fixed at one face and subjected to

shearing stress on a parallel face . As shown below .

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Relation between shear stress and shear strain;

Modulus of rigidity

For an elastic material is proportional to . Hence / = G , where G =

a constant for the material , referred to as the modulus of rigidity.

Torsion of a thin tube ;

Consider a thin walled tube subjected to torsion as indicated . Let the

shear stress on the cross section be ; in a thin walled tube may be

assumed constant.

Stress = force / area

= (T/R) /2 Rt where R = the mean radius and t = the wall thickness

/ R = T / 2 R3t --------------------- ( 1 )

Also / = G / ( R / L ) = G / R = G / L --------- ( 2 )


51/77

51

Combining equation ( 1 ) and ( 2 ) T/2 R3t = /R = G /L

L

BA B1

T

B B1

Example ;

A thin walled cylindrical tube , having diameter 25mm and wall thickness 1.6mm .

is subjected to torsion under twisting moment T. If the shear is not exceed 35MN/m 2,

determine the value of T. Taking the modulus of rigidity for the material as 80 10 9 N/m 2

determine the angle of twist per metre length .


52/77

52

Solution;

/R = T/2 R3t T = 2 R3t

T = 2 ( 0.0125 ) 2 1.6 10 -3 35 10 6 Nm = 55Nm

/R = G /L = l/RG

= 35 10 6 1/ ( 0.0125 80 10 9 ) rad

= 0.035 rad

Angle of twist per metre length = 0.035 rad

= 0.035 180/

= 2.005 0


53/77

53

WEEK TWELVE ( 12 )

12.0 Torsional Stiffness of a shaft may be defined as the ratio of the torque to the angleof

twist ; Torsional Stiffness = T / .

Torsional rigidity of a shaft may be defined as the ratio of the torque to the angle of

twist per unit length .; Torsional rigidity = T / ( /L ) = G J .

Torsional strength may be defined as the ratio of the torque to the maximum shear stress

due to torsion. ; Torsional strength = T / = J / R ( sometimes referred to as the

torsional modulus of the shaft section )

Example ;

A Solid steel shaft has a uniform diameter of 60 mm and is 1.50m long . This shaft

is required to transmit 100 Kw when running at 400 rev / min . Determine;

( I ) the maximum shear stress , ( ii ) the torsional stiffness , ( iii ) the torsional rigidity;

( iv ) the torsional strength for steel G = 83 Gpa .

Solution;

J = ( /32 ) ( 60 ) 4 = 127.3 10 4 mm 4 = 127.2 10 -8 m 4

400 rev/min = 41.9 rad / S Hence 41.9 T = 100 000

Torque T = 100 000/ 41.9 Nm = 2386 Nm

( I ) T / J = / R

= TR / J = 2386 0.03 / ( 127.3 10 -8 ) N /m 2

( ii ) T /J = G /L

T/ = GJ / L

= 83 10 9 127.3 10 -8 / 1.5 Nm /rad


54/77

54

= 105800 / 1.5 Nm /rad

( iii ) Torsional rigidity = 105 800 Nm 2

( iv ) Torsional strength = 2386 ( Nm ) / [ 56.3 10 6 ( N/m 2 ) ]

= 42.4 10 -6 m3

Example ;

Determine the necessary diameter of solid shaft to transmit 112 Kw at 270 rev / min.If the maximum shear stress due to torsion must not exceed 70 MN /m 2. Take the

maximum torque as 1.3 times , the mean torque.

Solution ;

Maximum torque = 1.3 112 000 / ( 2.70 /30 ) Nm = 5150 Nm 3

D3 = 16 5150 / ( 70 10 6 ) m 3 = 374 10 -6 m3

D = 0.072 m

Hollow shafts

The general torsion equation ; T/J = / R = q /r = G / L

Is applicable to a hollow shaft , but in the case

J = ( /32 ) ( D 4 d 4 )

Thus the torsional modulus for a hollow sectionZ = ( /32 ) ( D 4 d 4 ) / D = ( /16 ) D 3 [ 1- ( d/D ) 4 ].


55/77

55

Example ;

A hollow shaft is to have internal diameter of 0.6 times , the external diameter , and is

required to transmit 1500 Kw at 2500 rev/ min . The shaft is fitted with a flangedcoupling , having 8 bolts on a pitch circle of diameter equal to twice the external diameterof the shaft . Allowing Shear stresses of 70 MN/m 2 and 55 MN/ m 2 in shaft and boltsrespectively , diameter the necessary diameters;

Solution ;

Torque = 150010 3 / ( 2500 /30 ) Nm

= 5730 Nm

T/J = /R T----------------------- = / 1/2D

( /16 ) ( D 4 d 4 )

T = ( / 16 ) D 3 [ 1- ( 0.6 ) 6 ]

D3 = 16T / ( 0.8704 )

D3 = 16 5730 / ( 0.8704 70 10 6 ) m3

D = 0.0782m and d = 0.0469m

Let the required bolt diameter = x ( m ) , T = b n /4 db 2 Rb

5730 = x2 55 10 6 8 0.0782

= 0.01458m


56/77

56

Example ;

Detemine the dimensions of a hollow steel shaft to transmit 1600 Kw at a speed of

120 rev/ min , the maximum torque being 1.25 times the torque. Take , the internaldiameter to be half the external diameter and allow a maximum shear stress of 45 MN/m2 . Calculate also the maximum angle of twist over a length of 3m. G = 80 10 9 N /m 2

Solution;

TMAX = 1.25 1600 10 3 / ( 120 / 30 ) Nm

= 159 000 Nm

T/J = /R

T / ( /32) ( D4

d4

) - / D

T / ( /32 ) D 4 [ 1 ( d/ D ) 4 ] = / 1/2D

T = ( / 16 ) D 3 [ 1- ( ) 4 ]

D3 = ( 16 1380 000 ) / ( 0.9375 3 2240 ) m 3

D3 = 16 159 000 / ( 0.9375 45 10 6 ) m 3

D = 0.267m

External diameter = 0.267m ; internal diameter = 0.1335m

/R = G /L = L / RG

= 45 10 6 3 / ( 0.1335 80 10 9 ) rad

Over a length of 3m , the maximum angle of twist

= 0.01264 rad = 0.723 0


57/77

57

STRENGHT BETWEEN SOLID AND HOLLOW SHAFT

Comparison of strength between solid and hollow shaft

It is possible to compare the strength of two equally strong shafts (transmitting the sametorque). one is of solid section and the other hollow.the two shafts are assumed to be ofthe same material, having the same lengths and both reached the same permissiblemaximum shear stress.

=S

H

T

T 1

12

2

+=

nn

nshaft Solid theof Strenght shaft Hollowtheof Strenght Where: n = R/r , and R > r, and

n > 1

Example

Compared the strength of two shafts one hollow shaft and the other solid of the samematerial, weight and length, and both reached the same permissible maximum shearstress. If the external diameter of the hollow shaft is D = 75 mm external diameter, andinternal diameter d = 50 mm.

we have:

since : n = R/r = 37.5/25 = 1.5

Therefore, the strength of the hollow shaft is 1.938 times the strength of the solid shaft,for the same maximum shear stress, weight, material and length .Or stated otherwisehollow shaft can take up torque 93.8 % more than the solid shaft under the givenconditions.


58/77

58

- Exercise

A compound shaft with applied torques and dimensions is shown below. Section AB ismade of brass. Section BC is made of brass. Section CD is made of steel. For this shaft:A. Determine the maximum shear stress in each of the sections of the shaft.

B. Determine the resultant angle of twist of a point on end D with respect to a point onend A.The modulus of rigidity for steel = G steel = 8 x 10 4 N/mm 2 The modulus of rigidity for brass = G brass = 4 x 10 4 N/mm 2 .

Exercise

A vice is clamped by applying a force of 50kN at each end of a 1m arm on a shaft. Theshaft is solid 0.35m diameter steel. Determine the maximum shear stress in the shaft.

400N

1400Nm 800N

200N

2m

1m 1.5m

2m

1.5m

0.7

1m

0.35

50kN

50kN


59/77

59

WEEK THITEEN ( 13 )

13.0 SOLID STEEL SHAFT PROBLEMS AND SOLUTION

Example;

A solid steel shaft is required to transmit 1120 kw at 300 rev/ min . If the shaft is not

twist more than 1 0 on a length of ten diameter , and the shear stress due to torsion is not

to exceed 60 MN /m 2 , determine the diameter of shaft required .Modulus of rigid for

steel = 82 10 9 N/ m 2 .

Solution ;

/ R = G / L

If the angle of twist is 1 0 on a length of to 10 diameters

/ ( D ) = ( G ) / ( 10D )

= G / 2D

= 82 10 9 ( /180 ) / 2D = 71.6 10 6 N/m 2

Since on this basis is greater than 60 MN/m2

,

Therefore the stress criterion must be used to obtain the required shaft diameter.

T( /32 ) 4 = / D

T = ( / 16 ) D 3

D3 = 16T/

The required diameter = [ 16 ( 1120 103 / 31.42 ) ]

60 10 6

= 0.1445 m


60/77

60

Problem;

( I ) Compare the weight per unit length of a solid shaft with that of a hollowshaft of the same material , having internal diameter equal to 2/3 of the externaldiameter , given that the maximum shear stress is to be the same in both shafts

when they transmit the same torque ( 2 ) A hollow steel shaft of external

Example ;

A Solid Steel shaft , having diameter 75 mm , is subjected to twisting . If the angle of

twist is not exceed 1 0 in 2m length determine ; ( I ) the maximum torque which the shaft

may carry; ( ii ) the maximum shear stress produced by this torque for steel , G =

8010 9N/m 2

Solution ;

(iii) T /J = G /L

T = JG / L = ( /32 ) ( 0.075 ) 4 80 10 9 ( / 180 ) /2 Nm

= 2170 Nm

(iv) / R = G / L

= RG / L = 0.0375 80 10 9 ( /180 ) /2 Nm 2

= 26.2 10 6N/m 2

Example ;

A Solid Shaft is required to transmit a torque of 140 KNm. If the shear stress must notexceed 55 MN/m 2 , determine the required shaft diameter.

Solution;

T/J = /R T = J/ R


61/77

61

J/R is termed the twisting modulus of the shaft section ( Z ) . For a solid shaft , Z = J / R

= ( / 32 ) D 4 / D/2 = ( / 16 ) D 3 D3 = 16T /

= 16 140 10 3 / ( 55 10 6 ) m 3

The diameter required = 3 ( 12.97 10 -3 ) m = 0.235m


62/77

62

11.9 Torsion: Deformation - Angle of Twist

Another effect of applying an external torque to a shaft is a resulting deformation or twistas the material is stressed. The resulting shaft deformation is expressed as an Angle ofTwist of one end of the shaft with respect to the other.

In Diagram 1 we have shown a section of a solid shaft. An external torque T is applied tothe left end of the shaft, and an equal internal torque T develops inside the shaft.Additionally there is a corresponding deformation (angle of twist) which results from theapplied torque and the resisting internal torque causing the shaft to twist through anangle, phi, shown in Diagram 1.

=G J

LT ; where

T = the internal torque in the shaft L = the length of shaft being "twisted" J = the polar moment of inertia of the shaft G = the Modulus of Rigidity (Shear Modulus) for the material, for example forsteel we have, G steel = 8 x 10

4 N/mm 2

We will now look at an example of determining the angle of twist in a shaft. In Diagram2a we have shown a solid steel circular shaft with an external torque of 1.0kN . beingapplied at each end of the shaft, in opposite directions.

Fig. 1


63/77

63

The shaft has a diameter of 1.5 m. We would like to determine the angle of twist of end Bwith respect to end A.

To find the angle of twist we first determine the internal torque in the shaft. We cut theshaft a distance x feet from the left end, and make a free body diagram of the left sectionof the shaft - shown in Diagram 2b. From the free body diagram, we see that the internaltorque must be 1.0kN. to satisfy rotational equilibrium.

We next apply the Angle of Twist formula: = T L / J G ; whereT = 1.0kN. L = 2 m.J = polar moment of inertia = ( /32) d 4 for a solid shaft = (3.1416/32) (1.5m) 4 = 0.497m 4 = 0.497x10 3mm 4

G steel = 8 x 104 N/mm 2 Then,

= T L / J G = (1000 N.* 2000 m) / (0.497 x 10 3 mm 4 *8 x 10 4 N/mm 2 ) = 0.050radians = 2.75 o.

The angle of twist will have units of radians, and in this problem is clockwise withrespect to end A as shown in Diagram 3.

We also take a moment to calculate the maximum shear stress in the shaft, just out ofinterest in it's value. = T r / J = 1000 N *0 .75 x 10 3mm / 0.497 x 10 3 mm 4. = 1509.05. N/mm 2

1kN 1kN

2 m

1.5

1kN

1kN

X m

1kN

2.75 o

2m

1kN

Fig. 2

Fi . 3


64/77

64

11.10 Torsion - Power Transmission

One important process which involves applying torque to a shaft is power transmission.To transmit power a drive torque is applied to a rotating shaft. A short derivation willgives us a relationship between Torque, rotation rate, and Power transmitted.

In Diagram 1a we have shown a shaft with a drive torque of 1.0KN at end A and an equaland opposite load torque at end B.

In Diagram 1b we have shown the shaft from end on. Notice that to apply a torque to ashaft we must exert a force, F, usually at the outer edge of the shaft. This force may beapplied through use of a belt or gear. The product of the force (F) and the radius (r) is theapplied (or load) torque.The work done as we rotate the shaft will be the product of the force and the distance theforce acts through - which is the circumference. That is, for each revolution of the shaft,the force act through a distance of one circumference, or we may write:Work = F x d = F * (2 r) * (N revolutions) The Power sent down the shaft is then the Work per unit time, or if we divide theequation for Work above by the time we can write:Power = Work/Time = F * (2 r) * (N rev/time) If we now rewrite the above equation slightly, as below:Power = 2 ( F * r) * (N rev/sec) Then we recognize the (F * r) term is the torque in the shaft, and we can rewrite as:Power = 2 n (Watt) where T = Torque in Nm.; n = N rev/secThis is the formula for power transmitted in Watt. It is often more convenient to expressit in kilowatt as:

Power P = 1000602 x

NT kilowatts

Next we look at a simple example of determining transmitted horsepower.

ExampleIn Diagram 2, we have shown a solid shaft with an applied driving torque of 1.0KN. and

1kN 1kN

2m

Fig. 4


65/77

65

an equal and opposite load torque, rotating at a speed of 1800 rpm (30 rev/sec). Wewould like to determine the horsepower being transmitted down the shaft.

Solution: First we mentally note that since this is a simple shaft with equal externaldriving and load torque, then the internal torque will be equal in value to the externaltorque of 1.0KN.Next we apply the Horsepower equation:Power

hp = [2 /60 x 1000] ; where

T = internal torque in shaft (in Nm) = 1.0KN .n = the number of revolution per second = 30 rev/sec. So we have: Power hp = [2* 3.1416* 1.0KN* 1800 / 60 x 1000] = 0.188 hp .

11.11 Power Transmission

Example 1 :A solid steel shaft, shown in Diagram 1, has a 1 inch diameter and an allowable shearstress of 12,000 N/m 2. What is the largest amount of power which could safely betransmitted down the shaft if it is to rotate at 2400 rpm?

Solution: Step 1: First, using the maximum allowable shear stress, we determine the largest torquewhich may be applied to the shaft. The formula for the shear stress in a shaft is:

= T r / J ; then solving for the torque: T = J / r ; whereJ = d 4 / 32 = (1.5m) 4 / 32; = 0.497m 4 r = 0.75 m; and = 12000 N/m 2, (We use theallowable shear stress as the maximum stress in the shaft.) Putting values into theequation and solving:T = 12,000 * 0.497 / 0.75 = 7952Nm = 7.952kNm

Step 2 : Now that we have the maximum torque we can safely apply, we can determinethe largest amount of power we can transmit from the:

1kN 1kN

2m


66/77

66

Power P =100060

2 x

NT kilowatts

=100060

795224002 x

x x

= 1998.6 kWatts.

ExampleWhat power can be transmitted at 300 r.p.m. by a hollow steel shaft of 75 mmexternal and 50 mm internal diameters; when the permissible shear stress forthe steel is 70 N/mm 2 and the maximum torque is 1.30 times the mean ?.

Solution:Given: N= 300 r.p.m. D = 75 mm external diameter, d = 50 mm internal diameter;permissible shear stress for the steel = = 70 N/mm 2 , maximum torque T = 1.30 timesthe mean torque.

The maximum shear stress produced in the shaft should in no case , be more than 70N/mm 2 when transmitting the maximum torque.

Tmax . =

x = 4.653 x 10 6 Nmm

Mean torque: 1/1.3 x 4.653 x 10 6 = 3.579 x 10 6 Nmm

Power P =


67/77

67

WEEK FOURTEEN ( 14 )

14.0 ANALYSIS OF COMPLEX STRESSES

Two Dimensional Stresses

When a plane element is separated from a body it will be subjected to normal stresses as

well as shearing stresses..

Stresses on An Inclined Plane

If x and y are the normal stresses acting on two mutually perpendicular planes

accompanied by shearing stress xy as shown in the fig. 1 below , then normal stresses

and shearing stresses on a plane inclined at an angle to the x- axis are given by the

expression.

y xy

xy

x x

xy

xy

y


68/77

68

Y

x

y

xy

y Fig.1

Normal stress = x + y - x - y Cos2 + xy Sin2 -------------- ( 1 )2 2

Shearing stress = x - y Sin 2 + xy Cos 2 --------------------- ( 2 )2

Principal stresses

The maximum and the minimum values of normal stress depends upon the angle

or the direction of the inclined plane.

When normal stress assumes maximum and minimum values at a particular inclination ,

then these stresses are termed as principal stresses.

Principal planes

The plane perpendicular to which the principal stresses act are called Pincipal planes, the

shearing stress at which will be zero.


69/77

69

Sign Convention

(1) Normal stress is considered positive if it is a tensile stress

( 2 ) Normal stress is considered negative if it is a compressive stress

( 3 ) Shearing stresses are consider negative if they tend to rotate the element in a clock

wise direction

( 4 ) Shearing stresses are consider negative if they tend to rotate the element in a

counter clock wise direction

Law of Complementary Shears

y xy

xy

x x

xy

xy

y

EXAMPLE

A plane element n a body is subjected to a normal compressive stress in the x-

direction of 60Mpa and a shearing stress of 15Mpa as shown in fig below .

Determine ;

a ) The normal and shearing stress intensities on a plane inclined at an angle of 30 0 to


70/77

70

the normal stress.

( b ) The maximum and minimum value of normal stress on the inclined planes and the

of these stresses

( c ) The magnitude and direction of the maximum shearing stress on inclined plane.

y 15MPa xy

xy

x x = 60MPa

xy

xy

y

28 MPa

33.48MPa60MPa

300

y15 MPa

15MPaxy

y


71/77

71

( a) x = - 60MPa and xy = - 15MPa

Normal stress on a plane inclined at 30 0

= x x cos 2 + xy sin 2 = ( - 60 ) ( - 60 ) cos60 + ( - 15 ) sin 60 0

= - 27.99 MPa

Shearing Stress on a plane inclined at 30 0

= xsin 2 + xycos2

= ( - 60 ) sin 60 15cos 60

= - 25.98- 7.5

= 33.48MPa

( b ) Maximum normal stress

max = x + [1/2 x]2 + ( xy )2

= - 60/2 + ( - 60/2 ) 2 + ( -15) 2

= 3.45MPa

min = x + [1/2 x]2 + ( xy )2

= - 30 33.54

= - 63.55MPa

Tan 2 p = xy 1/2 x

= -


72/77

72

( c ) Maximum Shear stress

max/min = +/- x + [1/2 x]2 + ( xy )2

= +/- 33.54 MPa


73/77

73

WEEK FIFTEEN ( 15 )

Mohrs Circles

15.0 Describe Mohrs Circle of Stress

The graphical approach to the two dimensional stress problem was first presented by otto

mohr in the year 1882 . In this representation a circle is used , accordingly the

construction is called mohrs circle.

The rules for the construction of mohrs circle are summarized as follows;

y xy

xy

x x

xy

xy

y

( I ) Normal stresses x and y are plotted along x- axis to a suitable scale.

( ii ) Shearing stresses xy are plotted to the the same scale along the vertical axis.

( iii ) Tensile stresses are plotted to the right of the origin O compressive stressed are

considered negative and plotted on the left side of the origin O.

( IV ) Shearing stresses which rotate the element in clockwise direction are considered


74/77

74

positive and negative if rotate the element in an anticlock wise direction.

( v ) Point B and D are thus located and joined mark the mi point of the diameter BD as C

( vi ) Now with C as centre and BC = CD as radius draw a circle . This is the mohrs

circle.

( vii ) Now measure an angle 2 from the diameter BD in a counter clock wise direction

and mark points E and F on the circle

( viii ) The coordinates of point F represent the normal and shearing stresses on the plane

inclined at an angle to the x- axis . In the below diagram ON represent and NF

refresent the shearing stress

Shearing stress

x L

B

XY

F

M


75/77

75

EXAMPLE

A Plane element is subjected to the stresses in figure below . Using

mohrs circle determing ;

( a ) The principle stresses and their directions

( b ) The maximum shearing stresses and the direction of the planes in

which they occur.

80MPa xy

xy

x 20MPa

xy

xy

y


76/77

76


77/77

Documents

Cec 106 Theory- Strength of Materials