Chapter 5

5.0 Bending Stress

5.1 IntroductionWhen some external load acts on a beam, the shear force and bending moments are set up at all sections of the beam. Due to the shear force and bending moment, the beam undergoes certain deformation. The material of the beam will offer resistance or stresses against these deformations. These stresses with certain assumptions can be calculated. The stresses introduced by bending moment are known as bending stresses. In this chapter, the theory of pure bending, expression for bending stresses, bending stress in symmetrical and unsymmetrical sections, strength of a beam and composite beams will be discussed.

5.2 Theory of Simple Bending with Assumptions MadeBefore discussing the theory of simple bending, let us see the assumptions made in the theory of simple bending. The following are the important assumptions:

1. The material of the beam is homogeneous and isotropic.2. The value of Young’s modulus of elasticity is the same in tension and

compression.3. The transverse sections, which were plane before bending, remain plain after

bending also.4. The beam is initially straight and all longitudinal filaments bend into circular arcs

with a common centre of curvature.5. The radius of curvature is large compared with the dimensions of the cross-

section.6. Each layer of the beam is free to expand or contrast, independently of the layer,

above or below it.

Theory of Simple Bending

Figure1(a) shows a part of a beam subjected to simple bending. Consider a small length x of this part of beam. Consider two sections AB and CD which are normal to the axis

of the beam N – N. Due to the action of the bending moment, the part of length x will be deformed as shown in Figure1(b). From this figure, it is clear that all the layers of the beam, which were originally of the same length, do not remain of the same length any more.

The top layer such as AC has deformed to the shape A’C’. This layer has been shortened in its length. The bottom layer BD has deformed to the shape B’D’.

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Figure 1(a) Before bending Figure 1(b) After bending

This layer has been elongated. From Figure1(b), it is clear that some of the layers have been shortened while some of them are elongated. At a level between the top and bottom of the beam, there will be a layer, which is neither shortened nor elongated. This layer is known as neutral layer or neutral surface. This layer in Figure1(b) is shown by N’ – N’ and in figure1 (a) by N – N. The line of intersection of the neutral layer on a cross-section of a beam is known as neutral axis (written as N.A.).

The layers above N-N (or N’-N’) have been shortened and those below have been elongated. Due to the decrease in lengths of the layers above N-N, these layers will be subjected to compressive stresses.

We also see that the top layer has been shortened maximum. As we proceed towards the layer N-N, the decrease in length of the layers decreases. At the layer N-N, there is no change in length. This means the compressive stress will be maximum at the top layer. Similarly, the increase in length will be maximum at the bottom layer. As we proceed from bottom layer N-N, the increase in length of layers decreases. Hence, the amount by which a layer increases or decreases in length, depends upon the position of the layer with respect to N-N. This theory of bending is known as theory of simple bending.

5.3 Expression For Bending Stress

Figure 2(a) shows a small length x of a beam subjected to a simple bending. Due to the action of bending, the part of length x will be deformed as shown in Figure2(b). Let A’B’ and C’D’ meet at O.

Let R = Radius of neutral layer N’N’

=Angle subtended at O by A’B’ and C’D’ produced.

Consider a layer EF at a distance y below the neutral layer NN. After bending, this layer will be elongated to E’F’.

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Figure 2(a) Before bending

Figure 2(b) After bending

Original length of layer EF = x Also length of neutral layer NN = x

After bending, the length of neutral layer N’N’ will remain unchanged. But length of layer E’F’ will increase. Hence

N’N’ = NN = x

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Now from Figure 2(b),

N’N’ = Rand E’F’ = (R+y) ( Radius of E’F’ = R + y)But N’N’ = NN

= x Hence x = R

Increase in the length of layer EF = E’F’- EF

= (R+y) R ( EF x = R )= y

Strain in the layer EF =

=

= ( EF = x = R )

=

As R is constant, hence the strain in a layer is proportional to its distance from the neutral axis.

Let f = Stress in the layer EF E = Young’s modules of the beam

Then

E =

= ( Strain in EF = )

f = E

= --------------(5.1)

Since E and R are constant, therefore stress in any layer is directly proportional to the distance of the layer from the neutral layer.

In the above case, all layers below neutral layer are subjected to tensile stresses whereas the layers above neutral layer are subjected to compressive stresses. The equation (5.1) can also be written as

= --------------(5.2)

5.4 Neutral Axis and Moment of ResistanceThe neutral axis of any transverse section of a beam is defined as the line of intersection of the neutral layer with the transverse section. It is written as N.A.

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In Figure 3, we have seen that if a section of a beam is subjected to pure sagging moment, then the stresses will be compressive at any point above the neutral axis and tensile below the neutral axis. There is no stress at the neutral axis. The stress at a

distance y from the neutral axis is given by equation 5.1 as

F =

Figure 3 Stress distribution across a section

Figure 3 shows the cross-section of a beam. Let N.A. be the neutral axis of the section. Consider a small layer at a distance y from the neutral axis.

Let dA = Area of the layerNow the force on the layer = Stress on layer Area of layer

= f dA

= --------------(i)

f =

Total force on the beam section is obtained by integrating the above equation.

Total force on the beam section =

= ( E and R is

constant)

But for pure bending, there is no force on the section of the beam (or force is zero).

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= 0

or = 0 (as cannot be zero)

Now y dA represents the moment of area dA about neutral axis. Hence

represents the moment of entire area of the section about neutral axis.Due to pure bending, the layers above N.A. are subjected to compressive stresses whereas the layers below the N.A. are subjected to tensile stresses. Due to these stresses the forces will be acting on the layers. These forces will have moment about the N.A. The total moment of these forces about the N.A. for a section is known as moment of resistance of that section.

The Force on the layer at a distance y from neutral axis in Figure 3 is given by equation (i), as Stress Area.

Force on layer =

Moment of this force about N.A. = Force on layer y

=

=

Total moment of the forces on the section of the beam (or moment of resistance).

=

=

Let M = External moment applied on the beam section. For equilibrium, the moment of resistance offered by the section should be equal to the external bending moment.

M =

represents the moment of inertia of the area of the section about the neutral axis. Let this moment of inertia be I.

M =

or = --------------(5.3)

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But from equation (5.2), we have

=

= = --------------(5.4)

The equation (5.4) is known as bending equation.In equation (5.4), the different quantities are expressed in consistent units as given below:M is expressed in N mm; I in mm4

f is expressed in N / mm2; y in mm and E is expressed N / mm2; R in mm.

Worked out Examples

5.4.1 A steel plate of width 120mm and of thickness 20mm is bent into a circular arc of radius 10m. Determine the maximum stress induced and the bending moment, which will produce the maximum stress. Take E = 2 105 N/mm2.

Given data:

Width of the plate, b = 120 mmThickness of the plate, t = 20 mm

Radius of curvature, R = 10 m = 10 103 mm

Young’s modulus, E = 2 105 N/mm2

Solution

Moment of Inertia, I =

=

= mm4

Let fmax = Maximum stress induced and

M = Bending moment.

Using equation (5.2),

=

f = y --------------(i)

Equation (i) gives the stress at a distance y from N.A. Stress will be maximum, when y is maximum. But y will be maximum at the top layer or bottom layer.

ymax =

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= = 10 mm.

Now equation (i) can written as

fmax = ymax

=

fmax = 200 N/mm2

From equation (5.4), we have

=

M =

=

= N mm

M = 1.6 kNm

5.4.2 Calculate the maximum stress induced in cast iron pipe of external diameter 40 mm, of internal diameter 20 mm and of length 4 meter when the pipe is supported at its ends and carries a point load of 80 N at its center.

Given data:

External diameter, D = 40 mm Internal diameter, d = 20 mm

Length, L = 4 m = 4 1000 = 4000 mm

Point Load, W = 80 N

In case of simply supported beam carrying a point load at the center, the maximum bending moment is at the center of the beam.

Figure 4

Solution

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Maximum B.M. =

Maximum B.M. = = N mm

M = N mm

Figure 4(a) shows the cross section of the pipe.

Figure 4(a)

Moment of inertia of hollow pipe, I =

=

=

= 117809.7 mm4

Now using equation (5.4),

= --------------(i)

When y is maximum, stress will be maximum. But y is maximum at the top layer from the N.A.

ymax =

=

ymax = 20 mm

The above equation (i) can be written as

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=

fmax = ymax

=

fmax = 13.58 N/mm2

5.5 Section ModulusSection modulus is defined as the ratio of moment of inertia of a section about the neutral axis to the distance of the outermost layer from the neutral axis. It is denoted by the symbol Z. Hence mathematically section modulus is given by

Z = --------------(5.5)

where I = M.O.I. about neutral axis

and ymax =Distance of the outermost layer from the neutral axis.

From equation (5.4), we have

=

The stress f will be maximum, when y is maximum. Hence above equation can be written as

=

M = fmax

But = Z

M = fmax Z ----------------(5.6)

In the above equation, M is the maximum bending moment (or moment of resistance offered by the section). Hence, moment of resistance offered the section is maximum when section modulus Z is maximum. Hence section modulus represents the strength of the section.

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5.6 Section Modulus for various shapes of Beam Sections

Rectangular SectionWorked out Examples

5.6.1

Figure 5

Moment of inertia of a rectangular section about an axis through its C.G. (or through N.A.) is given by,

I =

Distance of outermost layer from N.A. is given by,

ymax =

Section modulus is given by,

Z =

=

=

= --------------(5.7)

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Hollow Rectangular Section

5.6.2

Figure 6

Here, I =

=

ymax =

Z =

=

= --------------(5.8)

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Circular Section

5.6.3

For a circular section,

I =

and ymax =

Z =

=

= --------------(5.9)

Hollow Circular Section

5.6.4

Figure 7

Here, I =

and ymax =

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Z =

=

= --------------(5.10)

5.6.5 A cantilever of length 2 meter fails when a load of 2 kN is applied at the free end. If the section of the beam is 40 mm 60mm, find the stress at the failure.

Figure 8

Given data:

Length, L = 2 m= 2 103 mm

Load, W = 2 kN = 2000 N

Section of the beam is = 40 mm 60 mm

Width of the beam, b = 40 mm

Depth of the beam, d = 60 mm

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Figure 8(a)

Figure 8(a) shows the section of the beam.Section modulus of a rectangular section is given by equation (5.7).

Solution

Z =

=

= 24000 mm3

Maximum bending moment for a cantilever shown in figure 8 is at the fixed end.

M = W L= 2000 2 103

= 4 106 Nmm

Let fmax = Stress at the failure

Using equation (5.6), we get

M = fmax Z

fmax =

=

fmax = 166.67 N/mm2.

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5.6.6 A rectangular beam 200 mm deep and 300 mm wide is simply supported over a span of 8 m. What uniformly distributed load per meter the beam may carry, if the bending stress is not to exceed 120 N/mm2.

Figure 9

Given data:Depth of beam, d = 200 mmWidth of beam, b = 300 mm

Length of beam, L = 8 mMax. bending stress, fmax = 120 N/mm2

Figure 9(a)

Figure 9(a) shows the section of the beam.

Solution

Let w =Uniformly distributed load per m length over the beam

Section modulus for a rectangular section is given by equation (5.7).

Z =

=

= 2000000 mm3

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Max. B.M. for a simply supported beam carrying uniformly distributed load as shown in figure 9 is at the center of the beam. It is given by

M =

= ( L = 8 m)

= 8w Nm = 8 w 1000 Nmm= 8000w Nmm 1 m = 1000 mm)

Now using equation (5.6), we getM = fmax Z

or 8000w = 120 2000000

w =

= 30 1000 N/mw = 30 kN/m

5.6.7 A square beam 20 mm 20 mm in section and 2 m long is supported at the ends. The beam fails when a point load of 400 N is applied at the center of the beam. What uniformly distributed load per meter length will break a cantilever of the same material 40 mm wide, 60 mm deep and 3 m long?

Figure 10

Given data:

Depth of beam, d = 20 mm

Width of beam, b = 20 mm

Length of beam, L = 2 m = 2000 mm

Point Load, W = 400 N

In this problem, the maximum stress for the simply supported beam is to be calculated first. As the material of the cantilever is same as that of simply supported beam, hence maximum stress for the cantilever will also be same as that of simply supported beam.

Solution

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Figure 10(a) The section modulus for the rectangular section of simply supported beam is given by equation (5.7).

Z =

= = mm3

Max. B.M. for a simply supported beam carrying a point load at the center (as shown in Figure 10(a) is given by,

M =

= = 200 Nm

= 200 1000 = 200000 NmmLet fmax = Max. stress induced

Now using equation (5.6), we get

M = fmax Z

200000 =

fmax = = 150 N/mm2

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Figure 11Now let us consider the cantilever as shown in Figure11(a).

Let w = Uniformly distributed load per m run.

Maximum stress will be same as in case of simply supported beam.

fmax = 150 N/mm2

Width of cantilever, b = 40 mm

Depth of cantilever, d = 60 mm

Length of cantilever, L = 3 m

Figure 11(a)

Figure 11(a) shows the section of cantilever beam.Section modulus of rectangular

section of cantilever=

Z = = 24000 mm3

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Maximum B.M. for a cantilever

=

=

= 4.5 w Nm

M = 4.5 1000w Nmm

Now using equation (5.6), we get

M = fmax Z

or 4.5 1000w = 150 24000

w =

w = 800 N/m

5.6.8 A beam is simply supported and carries a uniformly distributed load of 40 kN/m run over the whole span. The section of the beam is rectangular having depth as 500 mm. If the maximum stress in the material of the beam is 120 N/mm2 and moment of inertia of the section is 7 108 mm4, find the span of the beam.

Given data:

U.D.L., w = 40 kN/m = 40 1000 N/mDepth, d = 500 mm

Max. stress, fmax = 120 N/mm2

M.O.I of section, I = 7 108mm4

Solution

Let L = Span of simply supported beam.

Section modulus of the section is given by equation (5.5), as

Z =

Where ymax =

= = 250 mm

Z =

= 28 105 mm3

The maximum B.M. for a simple supported beam, carrying a U.D.L. over the whole span

is at the center of the beam and is equal to .

M =

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=

= 5000L2 1000 Nmm

Now using equation (5.6), we get

M = fmax Zor 5000 1000 L2 = 120 28 105

or L2 =

= 2.4 28 L =

= 8.197 m L = 8.20 m

5.6.9 A timber beam of rectangular section is to support a load of 20 kN uniformly distributed over a span of 3.6 m when beam is simply supported. If the depth of section is to be twice the breadth, and the stress in the timber is not to exceed 7 N/mm2, find the dimensions of the cross-section.

How would you modify the cross-section of the beam, if it carries a concentrated load of 20 kN placed at the center with the same ratio of breadth to depth.

Given data:

Total load, W = 20 kN = 20 1000 NSpan, L = 3.6 m

Max. stress, fmax = 7 N/mm2

Solution

Let b = Breadth of beam in mm Then depth, d = 2b mm

Section modulus of rectangular beam

=

Z =

= mm3

Maximum B.M., when the simply supported beam carries a U.D.L. over the entire span,

is at the center of the beam and is equal to or

M =

=

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= 9000 Nm = 9000 1000 NmmNow using equation (5.6), we get

M = fmax Z

or 9000 1000 =

or b3 =

= 1.92857 106

b =

= 124.47 mm b = 124.5 mm

and d = 2b = 2 124.5

d = 249 mm

Dimension of the section when the beam carries a point load at the center

B.M is maximum at the center and it is equal to when the beam carries a point

load at the center.

M =

=

= 18000 Nm = 18000 1000 Nmmfmax = 7 N/mm2

and Z = ( In this case also d = 2b)

Using equation (5.6), we get M = fmaxZ

or 18000 1000 =

b3 =

= 3.85714 106

b =

Breadth of the beam, b = 156.82 mmd = 2 156.82

Depth of the beam, d = 313.64 mm

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5.6.10 A rolled steel joist of I - section has the dimensions: as shown in Figure12. This beam of I section carries a u.d.l. of 40 kN/m run on a span of 10 m, calculate the maximum stress produced due to bending.

Figure 12

Given data:

u.d.l., w = 40 kN/m = 40000 N/mSpan, L = 10 m

Solution

Moment of inertia about the neutral axis

=

== 327946666 mm4

Maximum B.M. is given by, M =

=

= 500000 Nm= 500000 1000 Nmm= 5 108 Nmm

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Now using the relation,

=

f =

or fmax =

= ( ymax = 200 mm)

fmax = 304.92 N/mm2

5.6.11 An I–section shown in figure 13, is simply supported over a span of 12 m. If the maximum permissible bending stress is 80 N/mm2, what concentrated load can be carried at a distance of 4 m from one support?

Figure 13(a)

Given data:

Bending stress, fmax = 80 N/mm2

Span = 12 m

Solution

Let W =Concentrated load carried at a distance of 4 m from support B in Newton

To find the maximum bending moment (which will be at a point C where concentrated load is acting), first calculate the reactions RA and RB. Taking moments about point A, we get

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RB 12 = W 8

RB =

= W

and RA =

=

=

B.M. at point C = RA 8

=

=

But B.M. at C is maximum

Maximum B.M.,

Mmax =

= Nmm

= W Nmm

Now find the moment of inertia of the given I -section about the N.A.

I =

=

= 63535227.55= 31386647.45 mm4

Now using the relation,

=

or =

where ymax =

= 112.5 mm

Now substituting the known values, we get

=

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W =

W = 8369.77 N

5.6.12 A water main of 500 mm internal diameter and 20mm thick is running full. The water main is of cast iron and is supported at two points 10 m apart. Find the maximum stress in the metal. The cast iron and water weigh 72000 N/m3 and 10000 N/m3 respectively.

Figure 13(b)

Given data:

Internal diameter, Di = 500 mm = 0.5 mThickness of pipe, t = 20 mmWeight of cast iron = 72000 N/m3

Weight of water = 10000 N/m3

Solution

Outer Diameter, Do = Di+2 t= 500+2 20= 540mm = 0.54 m

Internal area of pipe =

=

= 0.1960 m2

This is also equal to the area of water section.

Area of water section = 0.196 m2

Outer area of pipe =

= m2

Area of pipe section =

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=

=

= 0.0327 m2

Moment of inertia of the pipe section about neutral axis, I

=

=

= 1.105 109 mm4

Let us now find the weight of pipe and weight of water for one meter length.

Weight of the pipe for one meter run =Weight density of cast iron Volume of pipe

=72000

= 72000 0.0327 1 ( Length = 1 m)= 2354 N

Weight of the water for one meter run

=Weight density of water Volume of water

== 10000 0.196 1= 1960 N

Total weight on the pipe for one meter run

= 2354+1960

= 4314 N

Hence the above weight is the U.D.L. (uniformly distributed load) on the pipe. The

maximum bending moment due to U.D.L. is w

Where w = Rate of U.D.L.= 4314 N per meter run

Maximum bending moment due to U.D.L, M

=

= ( L = 10

m)= 53925 Nm= 53925 103 Nm

Now using =

f =

The stress will be maximum, when y is maximum.

But maximum value of y =

=

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ymax = 270 mm

Maximum stress, fmax =

=

fmax = 13.18 N/mm2

5.7 Strength of a Section The strength of a section means the moment of resistance offered by the section and moment of resistance is given by

M = f Z

[M = or M = = f Z

where Z = ]where M = Moment of resistance

f = Bending stress andZ = Section modulus.

For a given value of allowable stress, the moment of resistance depends upon the section modulus. The section modulus, therefore, represents the strength of the section. Greater the value of Z, stronger will be the section.

The bending stress at any point in any beam section is proportional to its distance from the neutral axis. Hence the maximum tensile and compressive stresses in beam section are proportional to the distances of the most distant tensile and compressive fibres from the neutral axis. Hence for the purposes of economy and weight reduction the material should be concentrated as much as possible at the greatest distance from the neutral axis. This idea is put into practice, by providing beams of I–section, where the flanges alone withstand almost all the bending stress.

We know the relation:

=

or f = y

=

=

where Z = Section modulusFor a given cross-section, the maximum stress to which the section is subjected due to a given bending moment depends upon the section modulus of the section. If the section modulus is small, then the stress will be more. There are some cases where an increase in the sectional area does not result in a decrease in stress. It may so happen that in some cases a slight increase in the area may result in a decrease in section modulus which result in an increase of stress to resist the same bending moment.

Worked out Examples

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5.7.1 Three beams the same length, same allowable bending stress and the same bending moment. The cross-section of the beams are square, rectangle with depth twice the width of a circle. Find the ratios of weights of the circular and the rectangular beams with respect to square beams.

Figure 14

Given data:

Let x = Side of square beamb = Width of the rectangular beam

2b = Depth of the rectangular beamd = Diameter of a circular section

Solution M = f Z

Where Z = Section modulus

As all the three beams have the same allowable bending stress (f), and same bending moment (M), therefore the section modulus (Z) of the three beams must be equal.

Section modulus of a square beam =

=

= ( b = d =

x)

=

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Section modulus of a rectangular beam

=

= ( d = 2b)

=

=

Section modulus of a circular beam =

=

=

Equating the section modulus of a square beam with that of rectangular beam, we get

=

b3 =

=

= 0.25x3

or b =

= 0.63x --------------(i)Equating the section modulus of a square beam with that of a circular beam, we get

=

d3 =

or d =

= 1.1927xThe weights of the beams are proportional to their cross-sectional areas. Hence

=

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=

=

= 0.7938

and =

=

=

= ( d

=1.1927x)= 1.1172

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5.8 Assignments

1. A mild steel strap 10cm wide and 2cm thick is bent into a circular arc of radius 50 meters. Find the maximum intensity of stress induced in the strap. Take E = 2 x 105 N/mm2.

2. (S.I.Units) A steel flat 4mm wide and 2 mm thick is required to bent in a circular arc of 2m radius. Determine the moment required to bend the flat and the maximum stress induced in the material.

3. A rectangular beam 3.5 cm deep has a moment of inertia of 13 160 cm2. Calculate the radius in which the beam may be bent, if the maximum stress is not exceed 126 N/mm2.

4. A timber cantilever 20 cm wide and 30 cm deep is 3 meters long. It is loaded with a u.d.l. of 3000N/m over the entire span. What concentrated load can be placed at the free end of the cantilever, if the stress is not to exceed 7.2 N/mm2.

5. (S.I.Units). A timber beam 15mm wide, 300 mm deep is simply supported over a span of 4 meters. Find the maximum uniformly distributed load that the beam can carry, if the stress is not to exceed 8 N/mm2.

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Chapter 6

6.0 Torsion of Shaft

6.1 Introduction - Pure TorsionA shaft of circular section is said to be in pure torsion when it is subjected to equal and opposite end couples whose axes coincide with the axis of the shaft. While a beam bends as an effect of bending moment, a shaft twists as an effect of torsion. At any point as a section of the shaft, a shear stress is induced or more exactly, the state of stress at any point in the cross-section of the shaft is one of pure shear. By the principle of complementary shear stresses, we know that in a state of simple shear there are two planes carrying the shear stress of the same intensity. These planes must be perpendicular to each other.

In the case of shaft in torsion, the planes of shear at a point are (i) the cross-section itself and (ii) the plane containing the point and the axis of the shaft.

6.2 Theory of Pure TorsionFigure below shows a solid cylindrical shaft of radius R and length l subjected to a couple or a twisting moment T at one end, while its other end is held or fixed by the application of a balancing couple of the same magnitude.

Figure 1

Let AB be a line on the surface of the shaft and parallel to the axis of the shaft before the deformation of the shaft. As an effect of torsion this line, after the deformation of the shaft, takes the form AC.

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The angle CAB = represents the shear strain of the shaft material at the surface. This angle being small, we have,

BC = l

= --------------(i)

Let the angle BOC be the anguler movement of the radius OB due to the strain in the length of the shaft. Let BOC = . Let fs be the shear stress intensity at the surface of the shaft.

We know, fs = C -------------- (ii)

where C = Modulus of rigidity of the shaft material

fs =

But BC = R

fs =

= -------------- (iii)

The shaft may be taken to consist of an infinite number of elemental hollow shafts, one surrounding the other.

If the deformation of a line on the surface of any such interior cylinder, at a radius r be considered it can be similarly visualized that, the shear stress intensity q at the radius r is given by the relation,

=

=

=

Since C, and l are constants, it follows that at any section of the shaft, the shear stress intensity at any point is proportional to the distance of the point from the axis of the shaft. Hence, the shear stress is maximum at the surface and shear stress is zero at the axis of the shaft.

197 CH- 6

6.3 Torsional Moment of Resistance

Figure 2

Figure above shows the section of the shaft of radius R subjected to pure torsion. Let fs be the maximum shear stress, which occurs at the surface.

Consider an elemental area da at a distance r from the axis of the shaft.

Shear stress offered by the

elemental area q=

Shear resistance offered by the elemental area q da =

Moment of resistance offered by the elemental area =

=

Total moment of resistance offered by the cross-section of the

shaft =

=

But represents the moment of inertia of the shaft section about the axis of

the shaft, i.e., the quantity is the polar moment of inertia Ip of the section of the shaft.

T =

= -------------- (iv)

CH- 6 198

But from equation (iii), =

=

=

= =

6.4 Assumptions in the Theory of Pure TorsionThe theory of pure torsion is based on the following assumptions:

(i) The material of the shaft is uniform throughout.

(ii) The twist along the shaft is uniform.

(iii) The shaft is of uniform circular section throughout.

(iv) Cross-sections of the shaft, which are plane before twist remain plane after twist.

(v) All radii which are straight before twist remain straight after twist.

6.5 Polar ModulesLet T be the torsional moment of resistance of the section of a shaft of radius R and Ip

the polar moment of inertia of the shaft section.

The shear stress intensity q at any point on the section distant r from the axis of the shaft is given by

q =

The maximum shear stress occurs at the greatest radius R

fs=

or T =

or T =

where =

=

This ratio is called the polar modulus of the shaft section. The greatest twisting moment which a given shaft section can resist

=maximum permissible shear stress x polar modulus.

Hence for a shaft of a given material the magnitude of the polar modulus is a measure of its strength in resisting torsion.

199 CH- 6

Given a number of shafts of the same length and material, the shaft, which can resist the greatest twisting moment, is the one whose polar modulus is greatest. Shafts of the same material and length having the same polar modulus have the same strength.

=

R =

=

Moment of resistance =

Moment of resistance =

For a hollow shaft whose external and internal diameters are d1 and d2.

=

R =

Zp =

Torsional Moment of resistance

=

=

Torsional rigidity

Let a twisting moment T produce a twist of radians in a length l.

We know the relation.

=

=

For a given shaft the twist is therefore proportional to the twisting moment T. In a beam the bending moment produces a bend or deflection; in the same manner a torque produces a twist in a shaft. The expression CIp corresponds to a similar quantity EI* in expressions for deflection of beams. The quantity CIp is called Torsional rigidity. Obviously the quantity CIp stands for the torque required to produce a twist of 1 radian per unit length of the shaft.

CH- 6 200

6.6 Power Transmitted by a ShaftLet a shaft turning at N rpm transmit P kilo watts. Let the mean torque to which the shaft is subjected to be T Nm.

Power transmitted = P

=

= Watts

= Kilowatts

P = kW

6.7 Worked out examples

6.7.1 A metal bar of 10 mm diameter when subjected to a pull of 23.55 kN gave an elongation of 0.30 mm on a gauge length of 200 mm. In a torsion test on the same material, a maximum shear stress of 40.71 N/mm2 was measured on a bar of 50 mm diameter, the angle of twist measured over a length of 300 mm being 0o 21’. Determine the Poission’s ratio of the material.

Given data:

Tensile test data, d = 10 mmP = 23.55 kN

= 0.30 mml = 200 mm

Solution

Strain e =

= = 0.0015

Stress f =

= 299.85 N/mm2

Modulus of Elasticity, E =

=

= 1.999 x 105 N/mm2

Torsion test data, d = 50 mm

l = 300 mm

fs = 40.71 N/mm2

= 0o21’

201 CH- 6

= radians

=

C =

=

= 0.79972 x 105 N/mm2

E =

=

=

= 1.25

Poisson’s ratio, = 0.25

In a tensile test, a test piece 25 mm in diameter, 200 mm gauge length stretched 0.0975 mm under a pull of 50,000 N. In a torsion test, the same rod twisted 0.025 radian over a length of 200 mm, when a torque of 400 Nm was applied. Evaluate the Possion’s ratio and the three elastic moduli for the material.

Given data:

d = 25 mm

l = 200 mm

Pull = 50,000 N

Solution

Tensile stress, f =

= 101.86 N/mm2

Tensile strain, e =

Elastic Modules, E =

=

= N/mm2

=

C =

CH- 6 202

=

= N/mm2

We know, E =

=

=

= 1.252

Poisson’s Ratio = 0.252

We know, E =

K =

= N/mm2

K = 1.404 x 105 N/mm2

A 60 mm diameter shaft is subjected to a torque of 4 kNm. Find (i) the maximum shear stress induced in the shaft (ii) Angle of twist per metre length of the shaft (iii) the shear stress at a distance of 10 mm from the axis of the shaft. Take C = 8 x 104 N/mm2.

Solution

Maximum shear stress induced, fs =

=

= 94.31 N/mm2

Angle of twist per metre length, =

=

= 0.0392975 radian

= 2o.2516Shear stress at a distance of 10 mm from the axis of the shaft

q =

=

q = 31.44 N/mm2

203 CH- 6

A hollow steel shaft of external diameter 150 mm and internal diameter 100 mm is 1.5 m long. Find the maximum torque required to produce a twist of 0.5o over the length of the shaft. Take C = 8 x 104 N/mm2.

Given data:

C = 8 104 N/mm2

External diameter, D = 150 mm

Internal diameter, d = 100 mm

Solution

Polar moment of inertia of the shaft section, Ip

=

= 39883500 mm4

= 0.5 degree

=

= 8.72665 x 10-3 radian

=

T =

=N

mm

= 18.5626 x 106 Nmm

T = 18.5626 kNm

A hollow shaft of 20 mm outside diameter and 16 mm inside diameter is subjected to a torque of 40 Nm. Find the shear stress at the outside and inside of the shaft.

Given data:

External diameter, D = 20 mm

Internal diameter, d = 16 mm

CH- 6 204

SolutionPolar moment of inertia of the

section, =

= mm4

=

q =

Shear stress at outer surface of the shaft = N/mm2

= 43.13 N/mm2

Shear stress at inner surface of the shaft =

= 34.50 N/mm2

Determine the diameter of a solid shaft which will transmit 90 kW at 160 rpm if the shear stress in the shaft is limited to 60 N/mm2. Find also the length of the shaft, if the twist must not exceed 1 degree over the entire length. Take C = 8 x 104 N/mm2.

Solution

Power transmitted, P =

T =

= Nm

= 5371.5 Nm

T =

=

=

d = 77 mm

=

= 3451142 mm4

205 CH- 6

=

l =

= mm

l = 897 mm

A solid shaft 125 mm in diameter transmits 120 kW at 160 rpm. Find the maximum shear stress induced in the shaft. Find also the angle of twist in a length of 7.5 m. Take C = 8 x 104 N/mm2.

Solution

Power transmitted, P =

T =

=

=

= 7161.97 Nm

= 7161.97 x 103 Nmm

Max. Shear stress induced, =

=

= 18.676 N/mm2

Angle of twist =

=

= 0.028 radian

= 1o.605

Alternatively, =

=

= 0.028 radian

= 1o.605

CH- 6 206

Find the power transmitted by a 75 mm diameter shaft at 140 rpm at a maximum shear stress of 60 N/mm2.

Given data:

Internal diameter, d = 75 mm

Maximum shear stress, fs = 60 N/mm2

Solution

Torque, T =

= 60

= 4970.1 x 103 Nmm

= 4970.1 Nm

Power transmitted, P =

=

= 72.86 kW

A steel shaft transmits 105 kW at 160 rpm. If the shaft is 100 mm in diameter, find the torque on the shaft and the maximum shearing stress induced. Find also the twist of the shaft in a length of 6 m. Take C = 8 x 104 N/mm2.

Given data:

Power transmitted, P = 105 kW

Diameter, d = 100 mm

Solution

P =

105 =

T =

Nm = 6266 Nm

T =

fs =

=

= 31.91 N/mm2

Twist of the shaft, =

207 CH- 6

= radian

= 0.04786 radian

= 2o 45’

A solid shaft is subjected to a torque of 15 kNm. Find the necessary diameter of the shaft if the allowable shearing stress is 60 N/mm2 and the allowable twist is 1o in a length of 20 diameters of the shaft. Take C = 8 x 104 N/mm2.

Solution

Shear stress consideration

T =

d3 =

d = 108.4 mm

Twist consideration =

=

d4 =

=

Minimum required diameter, d = 129.8 mm

What must the length of a 5 mm diameter aluminum wire be so that it can be twisted through one complete revolution without exceeding a shearing stress of 40 N/mm2. Take C = 2.7 x 104 N/mm2.

Given data:

d = 5 mm

r = 2.5 mm

= 2 radians

fs = 40 N/mm2

C = N/mm2

CH- 6 208

Solution

=

l =

= mm

= 10603 mm

= 10.603 m

Find the power that can be transmitted by a shaft 60 mm diameter, at 180 r.p.m. if the permissible shear stress is 85 N/mm2.

Solution

T =

= Nmm

= 3605000 Nmm

= 3605 Nm

Power transmitted =

=

= 67.95 kW

A solid circular shaft transmits 75 kW at 200 r.p.m. Calculate the shaft diameter if the twist in the shaft is not to exceed 1o in 2 meters of shaft and the shearing stress is limited to 50 N/mm2. Take C = 1 x 105 N/mm2.

Solution

Power transmitted =

= 75 kW

T =

= 3581 Nm

(i) Twist consideration

=

Ip =

209 CH- 6

=

d4 =

=

d = 80.40 mm

(ii) Shear stress consideration T =

=

d3 =

d = 71.4 mmHence we should provide at least a diameter of 80.40 mm

A hollow shaft is to have an outside diameter d and inside diameter d/2. Calculate the minimum value of d if it is to transmit 375 kW at 105 rpm with a working stress of 40 N/mm2. Determine the twist in a length equal to 10 times the external diameter. Take C = 8 x 104 N/mm2.

SolutionPolar moment of inertia of the shaft section

Ip =

= mm4

= 0.092 d4 mm4

Zp =

= 0.184 d3 mm3

P =

T =

= 34104.631 Nm

T =

=

=d = 167.02 mm

l = 1670.2 mm

=

=

CH- 6 210

=

= 0.01 radian

A shaft has to transmit 105 kW at 160 r.p.m. If the shear stress is not to exceed 65 N/mm2 and the twist in a length of 3.50 m must not exceed 1”, find a suitable diameter. Take C = 8 x 104 N/mm4.

Solution

Power transmitted, P =

105 =

T =

= 6266 Nm

From shear stress consideration T =

=

=

D = 78.9 mm

From stiffness consideration, =

Ip =

=

D4 =

D4 =

Hence the required diameter, D = 112.5 mm

211 CH- 6

A solid shaft is 100 mm in diameter. It transmits 120 kW at 200 rpm. Find the maximum intensity of shear stress induced and the angle of twist for a length of 6 m. Take C = 8 x 104 N/mm2.

Solution

Power transmitted, P =

120 =

T =

= 5729 Nm

= Nm

Polar moment of inertia, Ip =

R =

=

=

=

=

= 29.17 N/mm2

=

=

= radian

= 0.04375 radian

= 2o30’

CH- 6 212

Find the diameter of the shaft required to transmit 60 kW at 150 r.p.m. if the maximum torque is likely to exceed the mean torque by 25% for a maximum permissible shear stress of 60 N/mm2. Find also the angle of twist of a length of 2.5 m. Take C = 8 x 104

N/mm2.

Solution

P =

60 =

T =

= 3819.7 Nm

= NmmThe torque calculated above is the mean torque

Max. torque, Tmax = meantorque

=

= Nmm

Polar modulus =

=

=

Tmax =

=

d3 =

d = 74 mm

=

=

=

= 0.0507 radian

= 2o54’

213 CH- 6

A hollow circular shaft 20 mm thick transmits 300 kW at 200 rpm. Determine the external diameter of the shaft if the shear strain due to torsion is not to exceed 0.00086. Take modules of rigidity equal to 8 x 104 N/mm2.

Solution

Let fs = Maximum shear stress

= Maximum shear strain

= 0.00086

C = Modulus of rigidity

=

= 0.00086

= N/mm2

= 68.8 N/mm2

P =

T =

= 14323.95 Nm

T =

Zp =

= 208197 mm3

Let the external diameter be D1

Internal diameter, D2 =

Zp =

= 208197 mm3

= 1060339 mm3

Solving by trial and error =107.94 mm

CH- 6 214

6.8 Assignments

1. A solid shaft of 40mm diameter is subjected to a torque of 80000N. Find the maximum shearing stress induced in the shaft.

2. (S.I. Units). A solid steel shaft transmits 560KW at 300 revolutions / min with a maximum shear stress of 60 N /mm2. Find the suitable diameter of the shaft.

3. A circular shaft is required to transmit 80KN at 200 r.p.m. Find a suitable diameter for the shaft, if the permissible shear stress in the shaft is 70 N/mm2. The maximum torque transmitted in a revolution exceeds the mean by 30%.

4. Find the diameter of a solid circular shaft to transmit 200kw at 300 r.p.m. The maximum permissible shear stress is 65 N/ mm2 and the angle of twist is not to exceed 1 in a length of 3 metres. Take C = 1 x 105 N/mm2.

5. A solid shaft of 6 cm diameter is running at 160 r.p.m. Find the power which the shaft can transmit, if the permissible shear stress is 80 N/mm2 and the maximum torque likely to exceed the mean by 30%.

6. Design a suitable diameter for a circular shaft required to transmit 100kw at 180 r.p.m. The shear stress in the shaft is not to exceed 70 N/mm2 and the maximum torque exceeds the mean by 40%. Also calculate the angle of twist in a length of 2 metres. Take C = 0.9 x 105 N/mm2.

7. A solid shaft of 10 cm diameter is to transmit 160KW. at 100r.p.m. Find the maximum intensity of shear stress induced and the angle of twist for a length of 6 metres. Take C= 0.8 x 105 N /mm2.

8. (S.I. Units). A hollow steel shaft has to transmit 6000 kW at 110 r.p.m. If the allowable shear stress is 60 N/mm2 and the inside diameter is 3/5th of the outside diameter, find the dimensions of the shaft. Also find the angle of twist in a 3 m length. Take C as 80 kN/mm2.

215 CH- 6

Chapter 7

7.0 Torsion of Springs

7.1 Introduction

A spring is a device, in which the material is arranged in such a way that it can undergo a considerable change, without getting permanently distorted. A spring is used to absorb energy due to resilience, which may be restored as and when required. The quality of a spring is judged from the energy it can absorb e.g. in a watch the spring is wound to absorb strain energy. This energy is released to run the watch, when the spring regains its original shape. A carriage spring is used to absorb shocks. It is thus obvious, that a spring which is capable of absorbing the greatest amount of energy for the given stress, in known to be the best one.

Types of SpringsWe have already discussed that a spring is used for absorbing energy due to resilience. Thus, in general, the springs of the following two types depending upon the type of resilience.

Bending spring and Torsion Spring.

Bending SpringsA spring, which is subjected to bending only, and the resilience is also due to it is known as bending spring. Laminated springs or leaf springs are also called bending springs.

Torsion SpringsA spring, which is subjected to torsion or twisting moment only and the resilience is also due to it is known as a torsion spring. Helical springs are also called torsion springs. Some springs are subjected to bending as well as torsion.

7.2 Forms of SpringsThough there are many forms of springs, which are made by the manufactures, yet the following types of springs are commonly used in Engineering practice.

(1) Carriage springs or leaf springs

(2) Helical springs

In this chapter we discuss only on Helical springs.

Helical Springs:It is a torsion spring and made up of a wire coiled into a helix. Though there are many types of helical springs, yet the following are important from the subject point of view:

1. Closely – coiled helical springs, and Open – coiled helical springs.

CH- 7

7.3 Close Coiled Helical Springs

In a closely coiled helical spring, the spring wire is coiled so close that the each turn is practically a plane at right angles to the axis of the helix and the wire is subjected to torsion. The bending stress is neglible as compared to the torsional stress. A closely coiled helical spring may be subjected to (1) axial loading or (2) axial twist. In this chapter, we shall discuss both the cases, one by one

Closely coiled Helical springs subjected to an Axial load:Consider a closely – coiled helical spring subjected to an axial load as shown in fig

Let

d = Diameter of the spring wire

n = No. of turns or coils

C = Modulus of rigidity for the spring materialW = Axial load on the spring

fs =Maximum shear stress induced in the wire due to twisting

= Angle of twist in the spring wire, and

=Deflection of the spring, as a result of axial load

A little consideration will show, that the load W will cause a twisting moment,

T = W x R

We know that the twisting moment,

T =

W.R =

We also know that the length of the wire,

l = Length of one coil x No. of coils

= 2 R n

We have already discussed in Art. 21.5 that

=

or =

= ( T=WR)

=

217 CH- 7

Deflection of the spring,

= R

= R

=

We know that the energy stored in the spring,

U =

And stiffness of the spring,

s =

=

Open – coiled Helical SpringsIn an open helical spring, the spring wire is coiled in such a way, that there is large gap between the two consecutive turns. As a result of this the spring can take compressive load also. An open helical spring, like a closed helical spring, may be subjected to (1) axial loading or (2) axial twist. In this chapter, we shall discuss only the first case.

Now consider an open coiled helical spring subjected to an axial load as shown in figLet

d = Diameter of the spring wire

R = Mean radius of the spring coil

P = Pitch of the spring coilsn = No. of turns or coils

C = Modulus of rigidity for the spring materials

SpringsW = Axial load on the spring

fs =Maximum shear stress induced in the spring wire due to loading

f =Bending stress induced in the spring wire due to loading

=Deflection of the spring as a result of axial load and,

= Angle of helix

The formula for deflection

= 64 WR3n sec Cos2 + 2 sin2 d4 C E

CH- 7 218

7.4 Close Coiled Helical Springs Subjected to Axial LoadA closely coiled helical spring carrying an axial load W. Let the spring consist of n turns. Let d be the diameter of the rod of the spring and R be the mean radius of the coil. Every section of the rod is subjected to a torque WR.

Figure 1

Maximum shear stress at any section of the rod

= fs =

= =

Length of the rod = l = 2

Strain energy stored by the spring = X volume = 2

= 32

If be the deflection for the spring i.e., the downward movement of the load,

Work done on the spring = W

Equating the work done to the strain energy stored, we have,

W = 32

=

219 CH- 7

Strain energy stored by the spring = work done on the spring

= W

Stiffness of the spring s = load required to produce unit deflection

s = =

Solved Problems

7.4.1 A close coiled helical spring of 100 mm mean diameter is made of 10 mm diameter rod and has 20 turns. The spring carries an axial load of 200 N. Determine the shear this load. Also calculate the stiffness of the spring.

Solution

fs

p =

=

= 50.93 N/mm

=

= mm

= 38.1 mm

Stiffness = = N/mm

= 5.25 N/mm of deflection

7.4.2 A closely coiled helical spring is made out of 10 mm diameter steel rod, the coil consisting of 10 complete turns with a mean diameter of 120 mm. The spring carries an axial pull of 200 N. find the maximum shear stress induced in the section of the rod. If C = 8 x 104 N/mm2, find the deflection of the spring, the stiffness and the strain energy stored by the spring.

Solution

fs =

= = 61.10 N/mm2

= = = 34.56 mm

Stiffness = s = = = 5.80 N per mm of deflection

CH- 7 220

Strain Energy stored = W = x 200 x 34.56 Nmm

Strain Energy stored = 3456 Nmm.

7.4.3 A close coiled helical spring is to carry a load of 500 N. Its mean coil diameter is to be 10 times that of the wire diameter. Calculate these diameters if the maximum shear stress in the material of the spring is 80 N/mm2.

Solution

D = 2R = 10d

R = 5d

W = 500 N

T = W R =

W x 5d =

d2 =

d = 12.62 mm

D = 10 x 12.62

= 126.2 mm

7.4.4 A helical spring is made of 12 mm diameter steel wire by winding it on a 120 mm diameter mandrel. If there are 10 active turns what is the spring constant?Take C = 8.2 x 104 N/mm2. What force must be applied to the spring to elongate it by 40 mm?

Solution

=

R = 60 + 6 = 66 mm

Stiffness = = = N/mm

= 9.24 N/mmForce required to produce a

deflection of 40 mm = 9.24 x 40

= 369.6 N.

221 CH- 7

7.4.5 A close coiled helical spring is to carry a load of 120 N and the mean coil diameter is to be 9 times the wire diameter. Calculate these diameters if the maximum shear stress is 100 N/mm2.

Solution

D = 2R = 9d

R = 4.5d

W = 120 N

fs =

100 =

d2 =

d = 5.24 mm

D = 9 x 5.24

= 47.16

7.4.6 A close coiled helical spring is to have a stiffness of 1 N/mm of compression under a maximum load of 45 N and a maximum shearing stress of 126 N/mm2. The solid length of the spring (when the coils are touching) is to be 45 mm. Find the diameter of the wire, the mean diameter of the coils and the number of coils required. Modulus of rigidity C = 4.2 x 104 N/mm2.

Solution

Stiffness = = N/mm

1 =

d4 = R3n

fs =

126 =

R =

R = 0.55 d3

Solid length of the spring, when the coils are touching

= nd = 45 mm.

n = .

Substituting the values of R and n in equation (i), we get

CH- 7 222

D4 = (0.55 d3)3

D4 =

d = 3.06 mm.

R = 0.55 x 3.063

R = 15.75 mmD = 2 x 15.75D = 31.50 mm.

n =

n = 14.7.

7.4.7 A 100 mm diameter safety valve is to be designed to blow of at a gauge pressure of 1.2 N/mm2. The valve is held in position by a 180 mm diameter close coiled compression helical spring whose initial compression is 25 mm, find the diameter of the rod of the spring and the number of turns required if the shear stress is not to exceed 80 N/mm2. Take C = 8.5 x 104 N/mm2.

Solution

Total force on the spring W = (100)2 1.2 = 9424.8 N

fs =

d3 =

=

d = 37.8 mm

=

n =

=

= 9.87 turns, say 10 turns.

7.4.8 A weight of 2600 N is dropped on a closely coiled helical spring consisting of 16 turns. Find the height by which the weight is dropped before striking

223 CH- 7

the spring so that the spring may be compressed by 220 mm. The coils have a mean radius of 120 mm and the diameter of the rod of the spring is 30 mm. Take C = 9 x 104 N/mm2.

Solution

Let P be the gradually applied load, producing the same compression of 220 mm.

= 220

P =

= 9064 NEquating the energy supplied by the falling load to the strain energy stored by the spring,

W (h + ) = P

2600 (h + 200) = x 9064 x 220

h + 220 = 383.5 mmh = 163.50 mm

7.4.9 A close coiled helical spring has a stiffiness of 10 N/mm. Its length when fully compressed, with adjacent coils toughing each other is 400 mm. The modulus of rigidity of the material of the spring is 8 x 104 N/mm2. (i) Determine the wire diameter and the mean coil diameter if their ratio =

.

(ii) If the gap between any two adjacent coils is 2 mm, what maximum load can be applied before the spring becomes solid, i.e., adjacent coils touch?(iii) What is the corresponding maximum shear stress in the spring?

Solution

= 10 N/mm

nd = 400 mm. C = 8 X 104 N/mm2.

= 0.1

Gap between adjacent coils = 2 mm.

=

= = 10

d4 =

CH- 7 224

d4 = R3n

But nd = 400

n =

and d = 0.1D = 0.2R

R = 5d

From (i) and (ii), d4 = x 125 d3 x

d2 = 400 d = 20 mm.

n = = = 20 turns

R = 5d = 5 x 20 = 100 mm.D = 200 mm.

Gap between adjacent coils = 2 mm.Max. deflection = = 2 x 20 = 40 mm.

= 10

W = 10 x 40 = 400 N

T = WR =

fs =

= 25.46 N/mm2.

7.4.10 It is required to design a close coiled helical spring which shall deflect 1 mm under an axial load of 100N at a shear stress of 90 N/mm2. The spring is to be made out of round wire having modulus of rigidity of 8 x 104 N/mm2. The mean diameter of the coils is to be 10 times the diameter of the wire. Find the diameter and length of the wire necessary to form the spring.

Solution

= 1 mmW = 100Nfs = 90 N/mm2

C = 8 x 104 N/mm2

D = 10dR = 5d

T = WR =

225 CH- 7

fs =

90 =

d2 = 28.29d = 5.32 mmR = 5 x 5.32 = 26.60 mm

=

n =

= = 5.32

Length of wire required = 2 x 26.60 x 5.32= 889.15 mm.

7.5. Open - coiled Helical springsIn open coil helical spring, the spring wire is coiled in such a way, that there is large gap between the two consecutive turns. As a result of this the spring can take compressive load also. An open helical spring, like a closed helical spring may be subjected to (1) axial loading or (2) axial twist. In this chapter, we shall discuss only first case.

Figure 2

CH- 7 226

Now consider an open coiled helical spring subjected to an axial load as shown in Figure 2.

Let d = Diameter of the spring wire R = Mean radius of the spring coil p = Pitch of the spring coilsn = No. of turns or coils

C =Modulus of rigidity for the spring materials

SpringsW = Axial load on the spring

fs =Maximum shear stress induced in the spring wire due to loading,

f =Bending stress induced in the spring wire due to bending

=Deflection of the spring as a result of axial load, and

= Angle of helix.

Formula for deflection:

= [ ]

Note. If we substitute = 0 in the above equation, it gives the deflection of a closed coiled spring i.e.,

=

7.6. Worked out Problems

7.6.1 An open coil helical spring made of 10mm diameter and of mean diameter 10 cm has 12 coils, angle of helix being 150. Determine the axial deflection and the intensities of bending and shear stress under a load of 50 kg. Take

C as 0.8 106 kg/cm2 and E as 2.0 106 kg/cm2.

Given data:

Dia of spring wire, d = 10 mm = 1.0 cmMean dia of spring, D = 10 cm

Mean radius, R = 5 cm

No. of coils, n = 12

Angle of helix, =

Load W = 50 kg

Modulus of rigidity, C = 0.8 106 kg/cm2

Solution

Young’s modulus, E = 2.0 106 kg/cm2

Deflection of the springLet = Deflection of the spring.

227 CH- 7

Using the relation, = [ ]

with usual notation

=

= 4800000

= 6.13 cmBending stress

Let f = Bending stress in the sectionWe know that the bending moment in the coil.,

M = WR Sin = 50 5 sin = 250 0.2588 kg – cm= 64.7 kg-cm

And moment of inertia of the spring wire section,

I =

=

=

We also know that the bending stress, f

=

= kg / cm2

= kg / cm2

[ ]f = 659.0 kg/ cm2

Shear StressLet fs = Shear stress induced in the wire.

We know that the twisting moment in the coil,T = WR cos

= 50 5 cos kg-cm = 250 0.9659= 241.48 kg-cm

Now using the relation

T = with usual notations.

241.48 =

=

CH- 7 228

fs =

fs = 1229.8 kg / cm2

7.7 Assignments

1. A closely coiled cylindrical spring of mean diameter 14 cm is made of 12mm diameter steel wire. Compute the direct axial load, the spring can carry, if the maximum stress is not to exceed 100 N/mm2.

2. A closely coiled helical spring is made of 6 mm wire. The maximum shear stress and deflection under a load of 200N is not to exceed 80 N/mm2 and 1.1 cm respectively. Determine the number of coils and their mean radius.

Take C = 8.4 x 104 N /mm2. 3. A closely coiled helical spring made of 10 mm diameter wire has 15 coils of 10cm mean diameter. Find the increase in the number of turns and the bending stress induced in the wire, if it is subjected to a twisting moment of 10 Nm. E = 2 x 105 N/mm2.

4. An open coil helical spring made of 1 cm diameter wire has 15 coils of 5 cm radius with a 20 angle of helix. Determine the deflection of the spring when subjected to an axial load of 300 N. Take E= 2 x 105 N /mm2 and C = 8 x 104 N /mm2.

229 CH- 7