Curved Beam

CURVED BEAMS

CONTENT:

� WHAT’S A CURVED BEAM

� DIFFERENCE BETWEEN A

CURVED BEAM

� WHY STRESS CONCENTRA

CONCAVE SIDE OF CURV

� DERIVATION FOR STRES

� PROBLEMS.

WHAT’S A CURVED BEAM?

DIFFERENCE BETWEEN A STRAIGHT BEAM AND A

WHY STRESS CONCENTRATION OCCUR AT INNER

CONCAVE SIDE OF CURVED BEAM?

DERIVATION FOR STRESSES IN CURVED BEAM

STRAIGHT BEAM AND A

TION OCCUR AT INNER SIDE OR

Theory of Simple Bending

Due to bending moment, tensile stress develops in one portion of section

and compressive stress in the other portion across the depth. In between

these two portions, there is a layer where stresses are zero. Such a layer

is called neutral layer. Its trace on the cross section is called neutral axis.

Assumption

� The material of the beam is perfectly homogeneous and isotropic.

� The cross section has an axis of symmetry in a plane along the

length of the beam.

� The material of the beam obeys Hooke’s law.

� The transverse sections which are plane before bending remain

plane after bending also.

� Each layer of the beam is free to expand or contract, independent of

the layer above or below it.

� Young’s modulus is same in tension & compression.

Consider a portion of beam between sections AB and CD as shown in

the figure.

Let e1f1 be the neutral axis and g1h1 an

element at a distance y from neutral

axis. Figure shows the same portion

after bending. Let r be the

radius of curvature and ѳ is the angle

subtended by a1b1 and c1d1at centre of

radius of curvature. Since it is a neutral

axis, there is no change in its length (at

neutral axis stresses are zero.)

EF = E1F1 = RѲ

G1H1 = (R+Y) Ѳ

GH = R

Also Stress

OR

dF = 0

∴∴∴∴ there is no direct force acting on the element considered.

GH = RѲ

is no direct force acting on the element considered.

Since Σyδa is first moment of area about neutral axis,

distance of centroid from neutral

centroid of the cross section. Cross sec

axis.

From (1) and (2)

CURVED BEAM

Curved beams are the parts of machine members found in C

clamps, crane hooks, frames

machines, planers etc. In straight beams the neutral axis of the section

coincides with its centroidal axis and the stress distribution in the beam

is linear. But in the case of curved beams the neutral axis of

is shifted towards the centre of curvature of the beam causing a non

linear [hyperbolic] distribution of stress. The neutral axis lies between

the centroidal axis and the centre of curvature and will always be present

within the curved beams.

a is first moment of area about neutral axis, Σ

distance of centroid from neutral axis. Thus neutral axis coincides with

centroid of the cross section. Cross sectional area coincides with neutral

Curved beams are the parts of machine members found in C

clamps, crane hooks, frames of presses, riveters, punches, shears, boring



is linear. But in the case of curved beams the neutral axis of

is shifted towards the centre of curvature of the beam causing a non



within the curved beams.

Σyδa/a is the

Thus neutral axis coincides with

tional area coincides with neutral

Curved beams are the parts of machine members found in C -

of presses, riveters, punches, shears, boring



is linear. But in the case of curved beams the neutral axis of the section

is shifted towards the centre of curvature of the beam causing a non-



Stresses in Curved Beam

Consider a curved beam subjected to bending moment Mb as shown

in the figure. The distribution of stress in curved flexural member is

determined by using the following assumptions:

i) The material of the beam is perfectly homogeneous [i.e., same

material throughout] and isotropic [i.e., equal elastic properties in all

directions]

ii) The cross section has an axis of symmetry in a plane along the length

of the beam.

iii) The material of the beam obeys Hooke's law.

iv) The transverse sections which are plane before bending remain plane

after bending also.

v) Each layer of the beam is free to expand or contract, independent of

the layer above or below it.

vi) The Young's modulus is same both in tension and compression.

Derivation for stresses in curved beam

Nomenclature used in curved beam

Ci =Distance from neutral axis to inner radius of curved beam

Co=Distance from neutral axis to outer radius of curved beam

C1=Distance from centroidal axis to inner radius of curved beam

C2= Distance from centroidal axis to outer radius of curved beam

F = Applied load or Force

A = Area of cross section

L = Distance from force to centroidal axis at critical section

σd= Direct stress

σbi = Bending stress at the inner fiber

σbo = Bending stress at the outer fiber

σri = Combined stress at the inner fiber

σro = Combined stress at the outer fiber

Stresses in curved beam

Mb = Applied Bending Moment

ri = Inner radius of curved beam

ro = Outer radius of curved beam

rc = Radius of centroidal axis

rn = Radius of neutral axis

CL = Center of curvature

In the above figure the lines 'ab' and 'cd' represent two such planes

before bending. i.e., when there are no stresses induced. When a bending

moment 'Mb' is applied

'ab' through an angle 'd

shortened while the inner

strip at a distance 'y' from the neutral axis is (y + r

the amount ydθ and the stress in this

Where σ = stress, e = strain and E = Young's Modulus

F

F

Mb

= Applied Bending Moment

= Inner radius of curved beam

= Outer radius of curved beam

= Radius of centroidal axis

the lines 'ab' and 'cd' represent two such planes


' is applied to the beam the plane cd rotates with respect to

'ab' through an angle 'dθ' to the position 'fg' and the outer

shortened while the inner fibers are elongated. The original length of a

strip at a distance 'y' from the neutral axis is (y + rn)θ. It is shortened by

and the stress in this fiber is,

σ = E.e

stress, e = strain and E = Young's Modulus

CA

NA

c 2c 1

c i

c o

e

F

F

Mb

r i

rn

rc

roCL

the lines 'ab' and 'cd' represent two such planes


to the beam the plane cd rotates with respect to

' to the position 'fg' and the outer fibers are

s are elongated. The original length of a

It is shortened by

We know, stress σ = E.e

We know, stress e � �� Ѳ

��Ѳ

i.e., σ = – E ��θ

��θ ..... (i)

Since the fiber is shortened, the stress induced in this fiber is

compressive stress and hence negative sign.

The load on the strip having thickness dy and cross sectional area dA is

'dF'

i.e., dF = σdA = – ��θ

��θ dA

From the condition of equilibrium, the summation of forces over the

whole cross-section is zero and the summation of the moments due to

these forces is equal to the applied bending moment.

Let

Mb = Applied Bending Moment

ri = Inner radius of curved beam

ro = Outer radius of curved beam

rc = Radius of centroidal axis

rn = Radius of neutral axis

CL= Centre line of curvature

Summation of forces over the whole cross section

i.e. �dF � 0

∴ ��θθ� ��

��=0

As ��θ θ

is not equal to zero,

∴ � �� = 0 ..... (ii)

The neutral axis radius 'rn' can be determined from the above equation.

If the moments are taken about the neutral axis,

Mb = – � ydF

Substituting the value of dF, we get

Mb =

��θθ� ��

�� dA

= ��θθ� y ! ��

��" dA

= ��θθ� ydA #$ � ��

�� 0%

Since � ydA represents the statical moment of area, it may be replaced

by A.e., the product of total area A and the distance 'e' from the

centroidal axis to the neutral axis.

∴ Mb =

��θθ

A.e ..... (iii)

From equation (i) ��θθ

= – σ��

�

Substituting in equation (iii)

Mb = – σ�� . A. e.

∴ σ = &'�

�� ..... (iv)

This is the general equation for the stress in a fiber at a distance 'y' from

neutral axis.

At the outer fiber, y = co

∴ Bending stress at the outer fiber σbo

i.e., σbo

= !&'()��) ($ rn + co = ro) ..... (v)

Where co = Distance from neutral axis to outer fiber. It is compressive

stress and hence negative sign. At the inner fiber, y = – ci

∴ Bending stress at the inner fiber

σbi

= &'(*

��+(*�

i.e., σbi =

&'(*��* ($ rn – ci = ri) ..... (vi)

Where ci = Distance from neutral axis to inner fiber. It is tensile stress

and hence positive sign.

Difference between a straight beam and a curved beam

Sl.no straight beam curved beam

1 In Straight beams the neutral

axis of the section coincides

with its centroidal axis and the

stress distribution in the beam

is linear.

In case of curved beams the

neutral axis of the section is

shifted towards the center of

curvature of the beam causing

a non-linear stress

distribution.

2

3 Neutral axis and centroidal

axis coincides

Location of the neutral axis

Neutral axis and centroidal

Neutral axis is shifted

towards the least centre of

curvature

Location of the neutral axis By considering a rectangular

section

Neutral axis is shifted

towards the least centre of

By considering a rectangular cross

Centroidal and Neutral Axis of

Centroidal and Neutral Axis of Typical Section of Curved Beams

Typical Section of Curved Beams

Why stress concentration occur at inner side or concave side of

curved beam

Consider the elements of the curved beam lying between two axial

planes ‘ab’ and ‘cd’ separated by angle

the plane cd having rotated through an angle d

Consider two fibers symmetrically located on either side of the neutral

axis. Deformation in both the fibers is same and equal to yd



planes ‘ab’ and ‘cd’ separated by angle θ. Let fg is the final position of

the plane cd having rotated through an angle dθ about neutral axis.


axis. Deformation in both the fibers is same and equal to ydθ



. Let fg is the final position of

about neutral axis.


θ.

Since length of inner element is smaller than outer element, the strain

induced and stress developed are higher for inner element than outer

element as shown.

Thus stress concentration occur at inner side or concave side of curved

beam

The actual magnitude of stress in the curved beam would be influenced

by magnitude of curvature However, for a general comparison the stress

distribution for the same section and same bending moment for the

straight beam and the curved beam are shown in figure.

It is observed that the neutral axis shifts inwards for the curved beam.

This results in stress to be zero at this position, rather than at the centre

of gravity.

In cases where holes and discontinuities are provided in the beam, they

should be preferably placed at the neutral axis, rather than that at the

centroidal axis. This results in a better stress distribution.

Example:

For numerical analysis, consider the depth of the section ass twice

the inner radius.

For a straight beam:

Inner most fiber:

Outer most fiber:

For curved beam: h=2r

e = rc - rn = h – 0.910h = 0.0898h

co = ro - rn= h – 0.910h = 0590h

ci = rn - ri = 0.910h -

For a straight beam:

Outer most fiber:

For curved beam: h=2ri

0.910h = 0.0898h

0.910h = 0590h

= 0.410h

Comparing the stresses at the inner most fiber based on (1) and (3), we

observe that the stress at the inner most fiber in this case is:

σbci = 1.522σBSi

Thus the stress at the inner most fiber for this case is 1.522 times greater

than that for a straight beam.

From the stress distribution it is observed that the maximum stress in a

curved beam is higher than the straight beam.

Comparing the stresses at the outer most fiber based on (2) and (4), we

observe that the stress at the outer most fiber in this case is:

σbco = 1.522σBSi

Thus the stress at the inner most fiber for this case is 0.730 times that for

a straight beam.

The curvatures thus introduce a non linear stress distribution.

This is due to the change in force flow lines, resulting in stress

concentration on the inner side.

To achieve a better stress distribution, section where the centroidal axis

is the shifted towards the insides must be chosen, this tends to equalize

the stress variation on the inside and outside fibers for a curved beam.

Such sections are trapeziums, non symmetrical I section, and T sections.

It should be noted that these sections should always be placed in a

manner such that the centroidal axis is inwards.

Problem no.1

Plot the stress distribution about section A-B of the hook as shown in

figure.

Given data:

ri = 50mm

ro = 150mm

F = 22X103N

b = 20mm

h = 150-50 = 100mm

A = bh = 20X100 = 2000mm2

e = rc - rn = 100

Section A-B will be subjected to a combination of direct load and

bending, due to the eccentricity of the force.

Stress due to direct load will be,

y = rn – r = 91.024

Mb = 22X103X100 = 2.2X10

= 100 - 91.024 = 8.976mm

B will be subjected to a combination of direct load and

bending, due to the eccentricity of the force.

Stress due to direct load will be,

r = 91.024 – r

X100 = 2.2X106 N-mm

B will be subjected to a combination of direct load and

Problem no.2

Determine the value of “t” in the cross

shown in fig such that the normal stress due to bending at the extreme

fibers are numerically equal.

Given data;

Inner radius ri=150mm

Outer radius ro=150+40+100

=290mm

Solution;

From Figure Ci + CO = 40 + 100

= 140mm…………… (1)

Since the normal stresses due to bending at

the extreme fiber are numerically equal

have,

i.e Ci=

= 0.51724C

Radius of neutral axis

rn=

rn =197.727 mm

ai = 40mm; bi = 100mm; b

ao = 0; bo = 0; ri = 150mm; r

Determine the value of “t” in the cross section of a curved beam as


fibers are numerically equal.

=150mm

=150+40+100

= 40 + 100

= 140mm…………… (1)

Since the normal stresses due to bending at

the extreme fiber are numerically equal we

0.51724Co…………… (2)

= 100mm; b2 =t;

= 150mm; ro = 290mm;

section of a curved beam as


=

i.e., 4674.069+83.61t = 4000+100t;

∴ t = 41.126mm

Problem no.3

Determine the stresses at point A and B of the split ring shown in

figure.

Solution:

The figure shows the critical section of the split

ring.

Radius of centroidal axis

Inner radius of curved beam

Outer radius of curved beam


Applied force

+83.61t = 4000+100t;


The figure shows the critical section of the split

Radius of centroidal axis rc = 80mm

Inner radius of curved beam ri = 80

= 50mm

Outer radius of curved beam ro = 80 +

= 110mm

rn =

=

= 77.081mm

F = 20kN = 20,000N (compressive)

Determine the stresses at point A and B of the split ring shown in the

F = 20kN = 20,000N (compressive)

Area of cross section A = π

,d2 =

π

, x602 = 2827.433mm

2

Distance from centroidal axis to force l = rc = 80mm

Bending moment about centroidal axis Mb = Fl = 20,000x80

=16x105N-mm

Distance of neutral axis to centroidal axis

e = rc ! rn

= 80! 77.081=2.919mm

Distance of neutral axis to inner radius

ci = rn ! ri

= 77.081! 50=27.081mm

Distance of neutral axis to outer radius

co = ro ! rn

= 110 ! 77.081=32.919mm

Direct stress σd = ! .� = /0000

/1/2.,44

=! 7.0736N/mm2 (comp.)

Bending stress at the inner fiber σbi = ! &'�*��*

= + 5675087/2.015/1/2.,447/.9597:0

= ! 105N/mm2 (compressive)

Bending stress at the outer fiber σbo = &'�)��) =

56750874/.959/1/2.,447/.9597550

= 58.016N/mm2 (tensile)

Combined stress at the inner fiber

σri = σd + σbi

= ! 7.0736! 105.00

= - 112.0736N/mm2

(compressive)

Combined stress at the outer fiber

σro = σd + σbo

= ! 7.0736+58.016


Maximum shear stress

τmax = 0.5x σmax

= 0.5x112.0736

= 56.0368N/mm2, at B

The figure.

Problem No. 4

Curved bar of rectangular section 40x60mm and a mean

100mm is subjected to a bending moment of 2KN

straighten the bar. Find the position of the Neutral axis and draw a

diagram to show the variation of stress across the section.

Solution

Given data:

b= 40mm h= 60mm

rc=100mm Mb= 2x10

C1=C2= 30mm

rn=

ro= rc+h/2=100+30=130 =(r

ri= rc- h/2 = 100 - 30= 70mm

rn= 96.924mm




Area

A= bxh = 40x60 = 2400 mm

Bending stress at the inner fiber

Curved bar of rectangular section 40x60mm and a mean

100mm is subjected to a bending moment of 2KN-m tending to


diagram to show the variation of stress across the section.

60mm

= 2x106 N-mm

30mm

=130 =(ri+c1+c2)

30= 70mm (rc-c1)


e = rc - rn= 100-96.924

=3.075mm


ci= rn- ri = (c1-e) = 26.925mm


co=c2+e= (ro-rn) = 33.075mm

A= bxh = 40x60 = 2400 mm2

Bending stress at the inner fiber σbi = =

= 104.239 N/mm2

(compressive)

Curved bar of rectangular section 40x60mm and a mean radius of

m tending to


e) = 26.925mm

) = 33.075mm

(compressive)


+/;50<;44.02:/,00;4.02:;540

= -68.94 N/mm2 (tensile)

Bending stress at the centroidal axis = +&'��=

=+/;50<

/,00;500

= -8.33 N/mm2 (Compressive)

The stress distribution at the inner and outer fiber is as shown in the

figure.

Problem No. 5

The section of a crane hook is a trapezium; the inner face is b and is at a

distance of 120mm from the centre line of curvature. The outer face is

25mm and depth of trapezium =120mm.Find the proper value of b, if the

extreme fiber stresses due to pure bending are numerically equal, if the

section is subjected to a couple which develop a maximum fiber stress of

60Mpa.Determine the magnitude of the couple.

Solution

ri = 120mm; bi = b; bo= 25mm; h = 120mm

σbi = σbo = 60MPa

Since the extreme fibers stresses due to pure bending are numerically

equal we have,

&'�*��* =

&'�)��)

We have,

Ci/ri =co/ro =ci/co =120/240

2ci=co

But h= ci + co

120 = ci+2ci

Ci=40mm; co=80mm

rn= ri + ci = 120+40 =160 mm

b=150.34mm

To find the centroidal axis, (C2)

bo= 125.84mm; b=25mm; h=120mm

= 74.313mm.

But C1=C2

rc= ro-c2 =240 - 74.313 =165.687mm

e=rc- rn = 165.687 - 160 = 5.6869 mm

Bending stress in the outer fiber,

σ>? � M>c�Aer�

A= �5:0.1,�/:�5/0/

= 1050.4mm

60 = &';10

50::0.,;:.612;/,0

Mb=10.8x106 N-mm

Problem no.6


fig.1.9a

Solution:

Redraw the critical section as shown in the figure.


Inner radius of curved beam ri = 80! 60/ = 50mm

Outer radius of curved beam ro = 80 + 60/ = 110mm

Radius of neutral axis rn = CD�)�D�*E�

,

= C√550�√:0E�

, =77.081mm

Applied force F = 20kN = 20,000N (compressive)

Area of cross section A = π

,d2 =

π

, x602 = 2827.433mm

2

Distance from centroidal axis to force l = rc = 80mm

Bending moment about centroidal axis Mb = FI = 20,000x80

=16x105N-mm

Distance of neutral axis to centroidal axis e = rc ! rn

= 80! 77.081

=2.919mm


ci = rn ! ri = 77.081! 50 = 27.081mm


co = ro ! rn = 110! 77.081 = 32.919mm

Direct stress σd =! .� = /0000

/1/2.,44

=! 7.0736N/mm2 (comp.)

Bending stress at the inner fiber σbi = ! &'�*��* =

+ 5675087/2.015/1/2.,447/.9597:0

= ! 105N/mm2 (compressive)


56750874/.959/1/2.,447/.9597550


Combined stress at the inner fiber

σri = σd + σbi =! 7.0736! 105.00

=! 112.0736N/mm2

(compressive)

Combined stress at the outer fiber

σro = σd + σb = ! 7.0736+58.016


Maximum shear stress

Gmax = 0.5x σmax = 0.5x112.0736

= 56.0368N/mm2, at B

The figure shows the stress distribution in the critical section.

Problem no.7

Determine the maximum tensile, compressive and shear stress induced

in a ‘c’ frame of a hydraulic portable riveter shown in fig.1.6a

Solution:

Draw the critical section as shown in

the figure.

Inner radius of curved beam ri =

100mm

Outer radius of curved beam ro = 100+80

= 180mm

Radius of centroidal axis rc = 100+ 10/

= 140mm

Radius of neutral axis rn = ln �#I)I* %

= ln 10#JK?J�?%

= 136.1038mm


e = rc - rn = 140-136.1038 = 3.8962mm

805

0

R100

175 mm

9000N

h = 80mm

c2

ec1

b =

50

mm

CriticalSection

co ci

ro

rn

rc

F

r = 100mmi 175mm

CL

F

CA

NA


ci = rn - ri = 136.1038-100 = 36.1038mm


co = ro - rn = 180-136.1038 = 43.8962mm

Distance from centroidal axis to force

l = 175+ rc = 175+140 = 315mm

Applied force F = 9000N

Area of cross section A = 50x80 = 4000mm2

Bending moment about centroidal axis Mb = FI = 9000x315

= 2835000 N-mm

Direct stress σd = .� =

9000,000 = 2.25N/mm

2 (tensile)

Bending stress at the inner fiber σbi = &'�*��* =

/14:000746.5041,00074.196/7500


Bending stress at the outer fiber σbo = !&'�)��) = ! /14:0007,4.196/

,00074.196/7510

= ! 44.326N/mm2 (compressive)

Combined stress at the inner fiber σri = σd + σbi = 2.25+65.676

= 67.926N/mm2

(tensile)

Combined stress at the outer fiber σro = σd + σbo = 2.25! 44.362

= !42.112 N/mm2 (compressive)

Maximum shear stress Gmax = 0.5x σmax = 0.5x67.926

= 33.963 N/mm2, at the inner fiber

The stress distribution on the critical section is as shown in the figure.

σbi=65.676 N/mm2

σri=67.926 N/mm2

b = 50 mm

h =80 mm

CA

NA

Bending stress =-44.362 N/mmσbo

2

Combined stress -42.112 N/mmσro

2=

σd=2.25 N/mm2

Direct stress ( )σd

Problem no.8

The frame punch press is shown in fig. 1.7s. Find the stress in inner and

outer surface at section A-B the frame if F = 5000N

Solution:

Draw the critical section as shown in the

figure.

Inner radius of curved beam ri = 25mm


= 65mm

Distance of centroidal axis from inner fiber c1 = �4 >*�/>)>*�>) "

= ,04 51�/7651�6 " = 16.667mm

h = 40mm

c2

ec1

b =

6 m

mo

co ci

ro

rn

rc

F

r = 25mmi 100mm

CL

F

b =

18 m

mi

CA

NA

∴ Radius of centroidal axis rc = ri ! c1

= 25+16.667 = 41.667 mm

Radius of neutral axis rn =

J��>*�>)�'*I) L ')I*M ��I)I* + �>*+>)�

=

J�7,0�51�6� JKN<8L<N�8

O? ��<8�8+ �51+6�

=38.8175mm

Distance of neutral axis to centroidal axis e = rc! rn

= 1.667!38.8175

=2.8495mm

Distance of neutral axis to inner radius ci = rn! ri

= 38.8175!25=13.8175mm

Distance of neutral axis to outer radius co = ro! rn

= 65-38.8175=26.1825mm

Distance from centroidal axis to force l = 100+ rc = 100+41.667

= 141.667mm


Area of cross section A = 5/ �b� Q b�� =

5/ x40�18 Q 6� = 480mm

2

Bending moment about centroidal axis Mb = FI = 5000x141.667

= 708335 N-mm


:000,10 = 10.417N/mm

2 (tensile)


20144:754.152:,107/.1,9:7/:

= 286.232N/mm2

(tensile)


20144:7/6.51/:,107/.1,9:76:

= !208.606N/mm2 (compressive)



Combined stress at the outer fiber σro = σd + σbo = 10.417!286.232





σbi=286.232 N/mm2

σri=296.649 N/mm2

b = 18 mmi

h =40 mm

CA

NA


2

Combined stress N/mmσro

2=-198.189

b = 6 mmo

σd=10.417 N/mm2

Direct stress (σd)

Problem no.9

Figure shows a frame of a punching machine and its various dimensions.

Determine the maximum stress in the frame, if it has to resist a force of

85kN

Solution:


figure.


Outer radius of curved beam ro = 550mm


rn = �

>*��WI*XY*I* Z� >��[I)LY)I)XY* \�>)�� I)I)LY)"

ai = 75mm; bi = 300mm; b2 = 75mm; ao = 0; bo = 0

A=a1+a2=75x300+75x225 =39375mm2

∴ rn = 4942:

400�� 8?X]8�8? "�2:�� 88?L?

�8?X]8"�0 = 333.217mm

Let AB be the ref. line

550

75 85 kN300

250

750 mm75

Ba = 75mm

i225 mm

a2b =75mm2

b=300mm

i

cico A

X

ern

rc

r =550 mmo

CA

NA

CL

750

r = 250 mmi

F

F

a1

x̂ � �J7J��7��J�� =

�2:7400�]8� ��2:7//:� 2:��8� "

4942: = 101.785mm

Radius of centroidal axis rc = ri +x̂

= 250+101.785=351.785 mm


= 351.785-333.217=18.568mm

Distance of neutral axis to inner radius ci = rn! ri

= 333.217! 250=83.217mm

Distance of neutral axis to outer radius co = ro! rn

= 550! 333.217=216.783mm

Distance from centroidal axis to force l = 750+ rc

= 750+351.785 = 1101.785mm

Applied force F = 85kN

Bending moment about centroidal axis Mb = FI

= 85000x1101.785

= 93651725N-mm


1:0004942: = 2.16N/mm

2 (tensile)


946:52/:714./524942:751.:617/:0


Bending stress at the outer fiber σbo =! &'�)��) = ! 946:52/:7/56.2144942:751.:617::0







= 24.165N/mm2, at the outer fiber

The below figure shows the stress distribution.

σbi=42.64 N/mm2

σri=44.8 N/mm2

b =

300 m

mi

225

CA

NA


2

Combined stress N/mmσro

2=-48.33

a =75mmi

b = 75 mm2a2

a1

σd=2.16 N/mm2

Direct stress ( )σd

Problem no.10

Compute the combined stress at the inner and outer fibers in the critical

cross section of a crane hook is required to lift loads up to 25kN. The

hook has trapezoidal cross section with parallel sides 60mm and 30mm,

the distance between them being 90mm .The inner radius of the hook is

100mm. The load line is nearer to the surface of the hook by 25 mm the

centre of curvature at the critical

section. What will be the stress at inner

and outer fiber, if the beam is treated as

straight beam for the given load?

Solution:

Draw the critical section as shown in the figure.


Outer radius of curved beam ro = 100+90 =

190mm

Distance of centroidal axis from inner fiber

c1 = �4 >*�/>)>*�>) " =

904 x 60�/74060�40 "

= 40mm

90mm

30mm

F = 25 kN

25mm60mm

100 mm

ri

rc

rn

ro

e

cico

NA

CA

c2 c1

h = 90 mm

l

F CL

Radius of centroidal axis rc = ri + c1 = 100+40

= 140 mm


J��>*�>)�'*I)L')I*M ��I)I* + �>*+>)�

=

J�7907�60�40� <?NJ_? L `?NJ??

_? ��J_?J?? + �60+40�

= 135.42mm


= 140! 135.42=4.58mm

Distance of neutral axis to inner radius ci = rn ! ri = 135.42! 100

=35.42mm

Distance of neutral axis to outer radius co = ro ! rn = 190! 135.42

= 54.58mm

Distance from centroidal axis to force l = rc ! 25= 140! 25

= 115mm

Applied force F = 25,000N = 25kN

Area of cross section A = 5/ �b� Q b�� =

5/ x90x�60 Q 30� = 4050mm

2

Bending moment about centroidal axis Mb = FI = 25,000x115

= 2875000 N-mm


/:000,0:0 = 6.173N/mm

2 (tensile)


/12:00074:.,/,0:07,.:17500

= 54.9 N/mm2

(tensile)

Bending stress at the outer fiber σbo =! &'�)��) = ! /12:0007:,.:1,0:07,.:17590



= 61.073N/mm2

(tensile)



Maximum shear stress τmax = 0.5x σmax = 0.5x61.072



b) Beam is treated as straight beam

From DDHB refer table,

b = 30mm

bo = 60-30 = 30mm

h = 90

c1 = 40mm

c2 = 90-50 = 40mm

A = 4050 mm2

Mb = 28750000 N/mm2

Also

C2 = �4>�/>) �7�47�/>�>) � ---------------------- From DDHB

C2 = �4740�/740�79047�/740�40� = 50mm

c1 = 90-50= 40mm

Moment of inertia I = C67>��67>7>)�>)�E�`

46c/>�>)d

= C6740��6740740�40�E90`

46c/740�40d

= 2632500mm4

Direct stress σb = .� =

/:000,0:0 = 6.173N/mm

2 (tensile)

Bending stress at the inner fiber σbi = &'�J

e = /12:0007,02632500

= 43.685 N/mm2 (tensile)

Bending stress at the outer fiber σbo = - &'��

e =2875000x50/64/:00

= -54.606N/mm2 (compressive)


= 49.858N/mm2

(tensile)

Combined stress at the outer fiber σro = σd + σbo = 6.173-54.606

= -48.433N/mm2 (compressive)

The stress distribution on the straight beam is as shown in the figure.

σbi= 43.685 N/mm2

σri= 49.858 N/mm2

60 m

m

h =90 mm

NA

, CA

σbo=-54.606 N/mm2

σro N/mm2

=-48.433

b = 30 mm

c =50mm 2c =40mm1

b /2 = 15o

b /2 = 15o

σd= 6.173 N/mm2

σd

b

Problem no.11

The section of a crane hook is rectangular in shape whose width is

30mm and depth is 60mm. The centre of curvature of the section is at

distance of 125mm from the inside section and the load line is 100mm

from the same point. Find the capacity of hook if the allowable stress in

tension is 75N/mm2

Solution:


figure.



= 185mm

Radius of centroidal axis rc =100+ 60/

= 130mm

Radius of neutral axis rn = ln �#I)I* %

= ln 60#JK8J�8%

= 153.045mm

Distance of neutral axis to centroidal axis e = rc - rn

= 155-153.045 = 1.955mm

h=60mm

b=30mm

F = ?

125 mm

100

h = 60mm

c2

ec1

b =

30

mm

co ci

ro

rn

rc

F

r = 125mmi

CL

l

100

CA

NA

Lo

ad l

ine

Distance of neutral axis to inner radius ci = rn - ri

= 153.045-125 = 28.045mm

Distance of neutral axis to outer radius co = ro - rn

= 185-153.045 = 31.955mm

Distance from centroidal axis to force l = rc -25 = 155-25 = 130mm

Area of cross section A = bh = 30x60 = 1800mm2

Bending moment about centroidal axis Mb = Fl = Fx130

= 130F


.5100

Bending stress at the inner fiber σbi = &'�*��* +

.5100

Combined stress at the inner fiber σri = σd + σbi

i.e., 75 = 5407.7/1.0,:

510075.9::75/: + .

5100

F = 8480.4N =Capacity of the hook.

Problem no.12

Design of steel crane hook to have a capacity of 100kN. Assume factor

of safety (FS) = 2 and trapezoidal section.

Data: Load capacity F = 100kN = 105N;

Trapezoidal section; FS = 2

Solution: Approximately 1kgf = 10N

∴ 105 = 10,000 kgf =10t

Selection the standard crane hook dimensions from table 25.3 when safe

load =10t and steel (MS)

∴ c =11933; Z = 14mm; M = 71mm and

h = 111mm

bi= M = 7133

bo = 2xZ = 2x14 = 28 mm

r1 = �/ =

559/ = 59.5mm

h = 111mm

bo

H

M=

bi

Z

r =59.5 mmi

r = c l

rn

ro

e

cico

NA

CA

c2 c1

h = 111 mm

F

CLb =28o b =71i

Assume the load line passes through the centre of hook. Draw the

critical section as shown in the figure.

Inner radius of curved beam ri = 59.5mm

Outer radius of curved beam ro = 59.5+111 = 170.5mm


J��>*�>)�'*I)L')I*M ��I)I*+ �>*�>)�

= J�75557�25�/1�]JNJ]?.8L�KN8_.8

O? ��J]?.88_.8 + �25�/1�

= 98.095mm

Distance of centroidal axis from inner fiber c1 = �4 >*�/>)>*�>) "

= 5554 25�/7/125�/1 " = 47.465mm

Radius of centroidal axis rc = ri + c1

= 47.465+59.5= 106.965 mm


=106.965-98.095 =8.87mm


= 98.095-59.5=38.595mm


= 170.5-98.095=72.0405mm

Distance from centroidal axis to force l = rc -106.965


Area of cross section A = 5/ �b� Q b��

= 5/ x111x�71 Q 28� = 5494.5mm

2

Bending moment about centroidal axis Mb = Fl = 105x141.667

= 106.965x105N-mm


500000:,9,.:



506.96:7508741.:9::,9,.:71.127:9.:

= 142.365/mm2

(tensile)


506.96:750872/.,0::,9,.5x8.127520.:

= -93.2 N/mm2 (compressive)


= 160.565N/mm2

(tensile)

Combined stress at the outer fiber σro = σd + σbo = 18.2-93.2

= -75N/mm2 (compressive)

Maximum shear stress τmax = 0.5x σmax = 0. 160.565



σbi=142,365 N/mm2

σri=160.565 N/mm2

b = 71 mmi

CA

NA

σbo=-93.2 N/mm2

σro N/mm2

=-75

b = 28 mmo

h = 111 mm

σd=18.2 N/mm2

σd

Problem no.13

The figure shows a loaded offset bar. What is the maximum offset

distance ’x’ if the allowable stress in tension is limited to 50N/mm2

Solution:

Draw the critical section as shown in the figure.


Inner radius ri = 100 – 100/2 = 50mm

Outer radius ro = 100 + 100/2 = 150mm

Radius of neutral axis rn = WDro�√ri4

Z2 = √150�√50

4"2=

93.3mm

e = rc - rn = 100 - 93.3 = 6.7mm

ci = rn – ri = 93.3 – 50 = 43.3

mm

co = ro - rn = 150 - 93.3 =

56.7mm

A = i, x d

2 =

i, x 100

2 = 7853.98mm

2

Mb = Fx = 5000 x

Combined maximum stress at the inner fiber

(i.e., at B)

σri= Direct stress + bending stress

= .� Q &'�*

��*

50 � :00021:4.91 Q �:0007��,4.4�

21:4.91j6.2j:0 ∴ x= 599.9 = Maximum offset distance.

Problem no.14

An Open ‘S’ made from 25mm diameter

rod as shown in the figure determine the

maximum tensile, compressive and shear

stress

Solution:

(I) Consider the section P-Q

Draw the critical section at P

Radius of centroidal axis

Inner radius ri =100

Outer radius ro = 100+


rn = =

= 99.6mm

Distance of neutral axis from centroidal axis e =r

Distance of neutral axis to inner fiber c

Draw the critical section at P-Q as shown in the figure.

Radius of centroidal axis rc =100mm

= 87.5mm

+ = 112.5mm


Distance of neutral axis from centroidal axis e =rc - rn

=100 - 99.6 =

Distance of neutral axis to inner fiber ci = rn – ri

= 99.6 – 87.5 =12.1 mm

0.4mm

87.5 =12.1 mm

Distance of neutral axis to outer fiber co = ro -rn

=112.5 – 99.6 = 12.9 mm

Area of cross-section A = π

4 d2 = π

4 x25

2 = 490.87 mm

2

Distance from centroidal axis I = rc = 100mm

Bending moment about centroidal axis Mb = F.l = 100 x 100

= 100000Nmm

Combined stress at the outer fiber (i.e., at Q) =Direct stress +bending

stress

σro= F

A - MbCo

Aeo =

1000

490.87 – 100000 X12.9

490.87 X 0.4 X 112.5

= - 56.36 N/mm2 (compressive)

Combined stress at inner fibre (i.e., at p)

σri= Direct stress + bending stress

= F

A +Mbci

Aeri =1000 490.87 +

100000 X 12.1490.87 X 0.4 X 87.5

= 72.466 N/MM2 (tensile)

(ii) Consider the section R -S

Redraw the critical section at R –S as shown in fig.

rc = 75mm

ri = 75 - 25

2 =62.5 mm

ro = 75 +

25

2 = 87.5 mm

A =π

4 d2 =

π

4 X 25

2 = 490.87 mm2

rn = WDro � √ri4

Z2 = √87.5 �√62.5

4"2 =74.4755 mm

e = rc - rn = 75 -74.4755 =0.5254 mm

ci = rn - ri =74.4755 – 62.5 =11.9755 mm

co = ro - rn = 87.5 – 74.4755 = 13.0245 mm

l = rc = 75 mm

Mb = Fl = 1000 X 75 = 75000 Nmm

Combined stress at the outer fibre (at R) = Direct stress + Bending stress

σro = FA

– Mb coAero

= 1000

490.87 - 75000 X13.0245490.87 X 0.5245 X 62.5

= - 41.324 N/mm2

(compressive)

Combined stress at the inner fiber (at S) = Direct stress + Bending stress

σri = F

A + Mbco

Aero =

1000

490.87 + 75000 X 11.9755

490.87 X 0.5245 X 62.5 = 55.816 N/mm

2 (tensile)

∴ Maximum tensile stress = 72.466 N/mm2 at P

Maximum compressive stress = 56.36 N/mm2 at Q

Maximum shear stress τmax=0.5 σmax= 0.5 X 72.466

= 36.233 N/mm2 at P

Stresses in Closed Ring

Consider a thin circular ring subjected to symmetrical load F as shown

in the figure.

The ring is symmetrical and is loaded symmetrically in both the

horizontal and vertical directions.

Consider the horizontal section as shown in the

A and B, the vertical forces would be F/2.

No horizontal forces would be there at A and B. this argument can be

proved by understanding that since the ring and the external forces are

symmetrical, the reactions too must be symmetri

Assume that two horizontal inward forces H, act at A and B in the upper

half, as shown in the figure. In this case, the lower

half must have forces H acting outwards as shown.

This however, results in violation of symmetry and

hence H must be zero. B

of equal magnitude M0 act at A and

noted that these moments do not violate the

condition of symmetry. Thus loads on the section can

be treated as that shown in the figure.

quantity is M0. Again Consi

conclude that the tangents at A and B must be

vertical and must remain so after deflection or M

does not rotate. By Castigliano’s theorem

derivative of the strain energy with respect to the load gives the

displacement of the load. In this


horizontal and vertical directions.

Consider the horizontal section as shown in the figure. At the two ends

A and B, the vertical forces would be F/2.



symmetrical, the reactions too must be symmetrical.


half, as shown in the figure. In this case, the lower

half must have forces H acting outwards as shown.

This however, results in violation of symmetry and

Besides the forces, moments

act at A and B. It should be

noted that these moments do not violate the

condition of symmetry. Thus loads on the section can

be treated as that shown in the figure. The unknown

. Again Considering symmetry, We

conclude that the tangents at A and B must be

vertical and must remain so after deflection or M0

Castigliano’s theorem, the partial


he load. In this case, this would be zero.

……………………….(1)


figure. At the two ends





The bending moment at any point C, located at angle

figure.

Will be

As per Castigliano’s theorem,

From equation (2)

And, ds = Rdθ

The bending moment at any point C, located at angle θ, as shown in the

………..(2)

As per Castigliano’s theorem,

, as shown in the

As this quantity is positive the direction assumed for M

produces tension in the inner fibers and compression on the outer.

It should be noted that these equations are valid in the region,

θ = 0 to θ = 900.

The bending moment Mb

this quantity is positive the direction assumed for Mo is correct and it



b at any angle θ from equation (2) will be:

is correct and it



2) will be:

It is seen that numerically, M

The stress at any angle Ѳforces as shown in the figure.

Put θ = 0 in Bending moment equation (4) then we will

get,

At A-A Mbi = 0.181FR

Mbo = - 0.181FR

And θ = 90,

At B-B Mbi = - 0.318FR

Mbo = 0.318FR

The vertical force F/2 can

components (creates normal direct stresses)

(creates shear stresses).

The combined normal stress across any section will be:

It is seen that numerically, Mb-max is greater than Mo. Ѳ can be found by considering the

forces as shown in the figure.

= 0 in Bending moment equation (4) then we will

= 0.181FR

0.181FR

0.318FR

= 0.318FR

2 can be resolved in two

(creates normal direct stresses) and S

The combined normal stress across any section will be:

The stress at inner (σ1Ai) and outer points (

On similar lines, the stress at the point of application of load at i

outer points will be (at Ѳ

It should be noted that in calculating the bending stresses, it is assumed

that the radius is large compared to the depth, or the beam is almost a

straight beam.

) and outer points (σ1Ao) at A-A will be (at

lines, the stress at the point of application of load at iѲ = 900)



A will be (at Ѳ = 0)

lines, the stress at the point of application of load at inner and



A Thin Extended Closed Link

Consider a thin closed ring subjected to symmetrical load F as shown in

the figure. At the two ends C and D, the vertical forces would be F/2.

No horizontal forces would be there at C and D, as discussed earlier

ring.

The unknown quantity is M

conclude that the tangents at C and D must be vertical and must remain

so after deflection or M0

There are two regions to be considered in this case:

� The straight portion, (0 < y < L) where

Mb

� The curved portion, where

A Thin Extended Closed Link


the two ends C and D, the vertical forces would be F/2.


The unknown quantity is M0. Again considering symmetry, we


does not rotate.

There are two regions to be considered in this case:

The straight portion, (0 < y < L) where

b = MO

The curved portion, where


the two ends C and D, the vertical forces would be F/2.


Again considering symmetry, we


As per Castigliano’s theorem

\

As per Castigliano’s theorem

It can be observed that at L = 0 equation reduces to the same expression

as obtained for a circular

and Compression on the outer.

The bending moment Mb

Noting that the equation are valid in the region,

At Ѳ = 0

At section B-B bending moment at inner and outer side of

At section A-A the load point, i.e., at

bending moment occurs (numerically), as it is observed that the second

part of the equation is much greater than the first part.

It can be observed that at L = 0,

expression as obtained for

It is seen that numerically, M


as obtained for a circular ring. Mo produces tension in the inner fibers

and Compression on the outer.

b at any angle Ѳ will be

Noting that the equation are valid in the region, Ѳ = 0 to Ѳ = p/2

B bending moment at inner and outer side of the fiber is

A the load point, i.e., at Ѳ = p/2, the maximum value of


part of the equation is much greater than the first part.

It can be observed that at L = 0, equation (v) reduces

obtained for a circular ring.

It is seen that numerically, Mb-max is greater than Mo.


in the inner fibers

= p/2,

the fiber is

the maximum value of


to the same

The stress at any angle

shown in the figure.

The vertical force F/2 can

normal direct stresses) and S

The combined normal stress across any section will be

The stress at inner fiber

be (at Ѳ = 0):

The stress at inner fiber

will be (at the loading point

The stress at any angle Ѳ can be found by considering the force as

2 can be resolved in two components (creates

and S (creates shear stresses).

normal stress across any section will be

fiber σ1Bi and outer fiber σ1Bo and at section B

fiber σ1Ai and outer fiber σ1Ao and at section A

will be (at the loading point Ѳ = 900):

can be found by considering the force as

be resolved in two components (creates

and at section B-B will

and at section A-A

Problem 15

Determine the stress induced in a circular ring of circular cross section

of 25 mm diameter subjected to a tensile load 6500N. The inner

diameter of the ring is 60 mm.

Solution: the circular ring and its critical section are as shown in fig.

1.29a and 1.29b respectively.

Inner radius ri = = 30mm

Outer radius = 30+25 = 55mm

Radius of centroidal axis r

Radius of neutral axis r

Distance of neutral axis to centroidal axis e = r

= 42.5

Distance of neutral axis to inner radius c

= 41.56


diameter subjected to a tensile load 6500N. The inner

diameter of the ring is 60 mm.



= 30mm

Outer radius = 30+25 = 55mm

Radius of centroidal axis rc = 30 + = 42.5mm


= =42.5mm


= 42.5 – 41.56 = 0.94mm

axis to inner radius ci = rn - ri

= 41.56 – 30 = 11.56mm


diameter subjected to a tensile load 6500N. The inner



= 55 - 41.56 = 13.44mm

Direct stress at any cross section at an angle θ with horizontal

σd = . ��k θ/�

Consider the cross section A – A

At section A – A, θ = 900 with respect to horizontal

Direct stress σd = . ��k 90

/� = 0

Bending moment Mb = - 0.318Fr

Where r = rc, negative sign refers to tensile load

M>l = - 0.318x6500x42.5 = -87847.5 N-mm

This couple produces compressive stress at the inner fiber and tensile

stress at the at outer fiber

Maximum stress at the inner fiber σ��l=Direct stress + Bending stress

= 0 - &'�*��* = ! 121,2.:755.:6

,90.12,70.9,740

= - 73.36N/mm2 (compressive)

Maximum stress at outer fiber σ��l= Direct stress + Bending stress

=0+ &'�)��) =

121,2.:754.,,,90.12,70.9,7::


Consider the cross section B – B

At section B – B, θ = 00 with respect to horizontal

Direct stress σd = . ��k 02A =

6:007 cos0/7,90.12, = 6.621 N/mm

2

Bending moment Mb = 0.182Fr

Where r = rc, positive sign refers to tensile load

M>o = 0.182x6500x42.5 = 50277.5 N-mm



Maximum stress at the inner fiber σ��o=Direct stress + Bending stress

= σd - &'�*��* = 6.621 +

:0/22.:755.:6,90.874x0.9,740

= 48.6 N/mm2

(tensile)


= σd + &'�)��) =6.621+

:0/22.:754.,,,90.12,70.9,7::

= -20 N/mm2 (compressive)

Problem 16

Determine the stress induced in a

of 50 mm diameter rod subjected to a compressive load of 20kNN. The

mean diameter of the ring is 100 mm.



Inner radius ri = - = 25mm

Outer radius = + = 75mm



Distance of neutral axis to centroidal axis e = r

= 50 - 46.65 = 3.35mm


= 46.65

Distance of neutral axis to

= 75 - 46.65 = 28.35mm

Area of cross section A =



mean diameter of the ring is 100 mm.



= 25mm

= 75mm

Radius of centroidal axis rc = = 50mm


= = 46.65mm

axis to centroidal axis e = rc - rn

46.65 = 3.35mm


= 46.65-25 = 21.65 mm


46.65 = 28.35mm

Area of cross section A = x552 = 1963.5mm

2

circular ring of circular cross section




σd = p qrs t

/u

Consider the cross section A – A


Direct stress σd = p qrs 90

/u = 0

Bending moment Mb = + 0.318Fr

Where r = rc, positive sign refers to tensile load

vwx = + 0.318x20000x50 = 318000 N-mm



Maximum stress at the inner fiber yz{x=Direct stress + Bending stress

= 0 + |}q~u�z~ =

318000�21.6:5964.:�4.4:�/:

= 41.86 N/mm2

(tensile)

Maximum stress at outer fiber yzrx= Direct stress + Bending stress

=0 - |}q�u�z� = -

451000�/1.355964.:�4.4:�2:


Consider the cross section B – B


Direct stress σd = . ��k 0/� =

/00007��k 0/75964.:

= 5.093 N/mm2 (compressive)

Bending moment Mb = -0.1828Fr

Where r = rc, negative sign refers to tensile load

M>o = - 0.182x20000x50 = -182000 N-mm




= σd - &'�*Aer* = -5.093 +

51/0007/5.6:5964.5x3.4:7/:

= - 29.05 N/mm2

(compressive)


= σd + &'�)��) = -5.093 +

51/0007/1.4:5964.:74.4:72:


Problem 17

A chain link is made of 40 mm diameter rod is circular at each end the

mean diameter of which is 80mm. The straight sides of the link are also

80mm. The straight sides of the link are also 80mm.If the link carries a

load of 90kN; estimate the tensile and compress

along the section of load line. Also find the stress at a section 90

from the load line

Solution: refer figure

= 80mm; dc = 80mm;

F = 90kN = 90000N

Draw the critical cross section as shown in fig.1.32

Inner radius ri = 40 - = 20mm

Outer radius = + = 60mm



Distance of neutral axis to centroidal axis e =


link is made of 40 mm diameter rod is circular at each end the



estimate the tensile and compressive stress in the link

along the section of load line. Also find the stress at a section 90

rc = 40mm;

Draw the critical cross section as shown in fig.1.32

= 20mm

= 60mm



= = 37.32mm


=40-37.32 = 2.68mm


= 37.32-20 = 17.32 mm

link is made of 40 mm diameter rod is circular at each end the



ive stress in the link

along the section of load line. Also find the stress at a section 900 away


= 60 – 37.32 = 22.68mm


σd = . ��k θ/�

Consider the cross section A – A [i.e., Along the load line]


Direct stress σd = . ��k 90

/� = 0

Bending moment M>l = - .I�/��/�π�� where r = rc,

M>l= 900007,07�/7,0�10�

/�π7,0�10� = 1.4x106N-mm



Maximum stress at the inner fiber σ��l=Direct stress + Bending stress

= 0 + &'�*��* =

5.,750<752.4/π

O7,0�7/.617/0

= - 360 N/mm2 (tensile)


= 0 - &'�)��) = -

5.,750<7//.61π

O7,0�7/.61760

= 157.14 N/mm2 (compressive)

Consider the cross section B – B [i.e., 900 away from the load line]


Direct stress σd = . ��k /θ

/� = 5.,750<7��k 0

/7πO7,0� = 35.81 N/mm2

(compressive)

Bending moment M>l = - .I�/�+π��/�π�� where r = rc,

M>l= 900007,07�/7,0+π710�

/�π7,0�10� = - 399655.7N-mm




= σd - &'�*��* = 35.81 +

4996::.2752.4//7πO7,0�7/.617/0



= σd + &'�)��) = 35.81 -

4996::.27//.61/7πO7,0�7/.61760


Maximum tensile stress occurs at outer fiber of section A –A and

maximum compressive stress occurs at the inner fiber of section A –A.

Using usual notations prove that

curved beam of initial radius R

uniform bending moment is

Consider a curved beam of uniform cross section as shown in Figure

below. Its transverse section is symmetric with respect to the y axis and

in its unstressed state; its upper and lower surfaces intersect the vertical

xy plane along the arcs of circle A

Now apply two equal and opposite couples

(c). The length of neutral surface remains the same.

central angles before and after applying the moment M. Since the length

of neutral surface remains the same

R1θ =R2θ'

Consider the arc of circle JK located at a distance y above the neutral

surface. Let r1 and r2 be the radius of this arc before and after bending

couples have been applied. Now, the deformation of JK,

From Fig. 1.2 a and c, r

Using usual notations prove that the moment of resistance M of a

curved beam of initial radius R1 when bent to a radius R

uniform bending moment is

M = EAeR1


Its transverse section is symmetric with respect to the y axis and

its upper and lower surfaces intersect the vertical

xy plane along the arcs of circle AB and EF centered at O [Fig. 1(

Now apply two equal and opposite couples M and M' as shown in Fig. 1.

The length of neutral surface remains the same. θ and


of neutral surface remains the same

'

Figure


be the radius of this arc before and after bending

couples have been applied. Now, the deformation of JK,

From Fig. 1.2 a and c, r1 = R1 – y; r2 = R2 – y ..... (iii)

the moment of resistance M of a

when bent to a radius R2 by


Its transverse section is symmetric with respect to the y axis and

its upper and lower surfaces intersect the vertical

B and EF centered at O [Fig. 1(a)].

d M' as shown in Fig. 1.

and θ' are the


..... (i)


be the radius of this arc before and after bending

.... (ii)

..... (iii)

∴ δ = (R2

=R2θ'

= – y (

∴ δ =– y∆θ

[ θ' − θ = θ + ∆θ

The normal strain ∈

the deformation δ by the original length r

∴ ∈x =

The normal stress σx

∴ σx =–E

i.e. σx = – E

Equation (vi) shows that the normal stress

with the distance y from the neutral surface. Plotting

obtain an arc of hyperbola as shown in Fig. 1.3.

From the condition of equilibrium the summation of forces over the

entire area is zero and the summation of the moments due to these forces

is equal to the applied bending moment.

∴ ∫δF =0

i.e., ∫σxdA =0

and ∫(– yσxdA)=M

– y) θ' – (R1 – y) θ

' – θ'y – R1θ + θy

y (θ' – θ) [ R1θ = R2θ' from equ (i)]

∆θ

θ = θ + ∆θ − θ = ∆θ] .....

∈x in the element of JK is obtained by dividing

by the original length r1θ of arc JK.

..... (v)

x may be obtained from Hooke's law

..... (vi)

( r1 = R1 – y) .(vii)

Equation (vi) shows that the normal stress σx does not vary linearly

with the distance y from the neutral surface. Plotting σx versus y, we

obtain an arc of hyperbola as shown in Fig. 1.3.



is equal to the applied bending moment.

..... (viii)

..... (ix)

' from equ (i)]

..... (iv)

in the element of JK is obtained by dividing

..... (v)

may be obtained from Hooke's law σx = E∈x

..... (vi)

does not vary linearly

versus y, we



..... (viii)

..... (ix)

Substituting the value of the σx from equation (vii) into equation

(viii)

– �∆t t � �J+zJ

zJ "dA=0

Since �∆t t is not equal to zero � �J+zJ

zJ "dA = 0

i.e. R1� �uz1 ! ��=0

i.e., . R1� �uz1 ! �=0

∴ R1 =u

��x�1

∴ It follows the distance R1 from the centre of curvature O to the

neutral surface is obtained by the relation R1 = u

��x�1 ..... (x)

The value of R1 is not equal to the �1� distance from O to the centroid

of the cross-section, since �1� is obtained by the relation,

�1� =5u � �1�� ..... (xi)

Hence it is proved that in a curved member the neutral axis of a

transverse section does not pass through the centroid of that section.

Now substitute the value of σx from equation (vii) into equations (ix)

� �∆t t �J+zJ

zJ "y dA =M

i.e., �∆t t � �J+zJ

zJ "/ dA = M ($ r1 = R1 - y from iii)

i.e., �∆t t � C�J�+/�JzJ�zJ�E

zJ/dA = M

i.e., �∆t t #�5/ � �u

zJ ! 2�5 �� Q� �1��%= M

i.e., �∆t t #�5/ � u

�J ! 2�5� Q �5��% = M [using equations (x) and (xi)]

i.e., �∆t t ��5� ! 2�5� Q �5�� v

i.e., �∆t t ��5�� ! �5�� =M

i.e., �∆t t =

|u�z1^̂ ^+�1� ..... (xii)

i.e., �∆t t =

|u� ($e = �1�– R1 from Fig. 1.2a)

………xiii)

Substituting �∆t t into equation (VI)

σx =|�

u��1+�� ..... (xiv)

∴ σx=M�z1+�1�

u�z1 ($ r1 = R1 – y ..... (xv)

An equation (xiv) is the general expression for the normal stress σx in a

curved beam.

To determine the change in curvature of the neutral surface caused by

the bending moment M

From equation (i),

5 �� 5

�Jt� t

� 5 �J t�∆tt "

� 5 �J 1 Q ∆t

t "=5 �J #1 Q |

�u�%

{$From equation (xiii) �∆t t = |u� }

� 1 �5 Q

v��5

i.e. 5 �� ! 5

�J = M

EAeR1

∴ M =EAe R1 5 �� ! 5

�J"

Hence proved

References:

ASSIGNMENT

1. What are the assumptions made in finding stress distribution for a curved flexural member? Also

give two differences between a straight and curved beam

2. Discuss the stress distribution pattern in curved beams when compared to straight beam with

sketches

3. Derive an expression for stress distribution due to bending moment in a curved beam

EXERCISES

1. Determine the force F that will produce a maximum tensile stress of 60N/mm2 in section

A - B and the corresponding stress at the section C - D

2. A crane hook has a section of trapezoidal. The area at the critical section is 115 mm2. The hook

carries a load of 10kN and the inner radius of curvature is 60 mm. calculate the maximum tensile,

compressive and shear stress.

Hint: bi = 75 mm; b

o = 25 mm; h = 115 mm

3. A closed ring is made of 40 mm diameter rod bent to a mean radius of 85 mm. If the pull along

the diameter is 10,000 N, determine the stresses induced in the section of the ring along which it is

divided into two parts by the direction of pull.

4. Determine of value of t in the cross section of a curved beam shown in Figure such that the normal

stresses due to bending at the extreme fibers are numerically equal.

VTU,Jan/Feb.2005

Fig.1.35

5. Determine a safe value for load P for a machine element loaded as shown in Figure limiting the

maximum normal stress induced on the cross section XX to 120 MPa.

6. The section of a crane hook is trapezoidal, whose inner and outer sides are 90 mm and

25 mm respectively and has a depth of 116 mm. The center of curvature of the section is at a

distance of 65 mm from the inner side of the section and load line passes through the center of

curvature. Find the maximum load the hook can carry, if the maximum stress is not to exceed 70

MPa.

7. a) Differentiate between a straight beam and a curved beam with stress distribution in each of the

beam.

b) Figure shows a 100 kN crane hook with a trapezoidal section. Determine stress in the outer,

inner, Cg and also at the neutral fiber and draw the stress distribution across the section AB.

8. A closed ring is made up of 50 mm diameter steel bar having allowable tensile stress of 200

MPa. The inner diameter of the ring is 100 mm. For load of 30 kN find the maximum stress in

the bar and specify the location. If the ring is cut as shown in part -B of

Fig. 1.40, check whether it is safe to support the applied load.

F = 100kN

62.5

mm

A B

112.5

25

87

.5

REFERENCE BOOKS

• “MECHANICAL ENGINEERING DESIGN” by Farazdak Haideri

• “MACHINE DESIGN” by Maleev and Hartman

• “MACHINE DESIGN” by Schaum’s out lines

• “DESIGN OF MACHINE ELEMENTS” by V.B.Bhandari

• “DESING OF MACHINE ELEMENTS-2” by J.B.K Das and P.L. Srinivasa murthy

• “DESIGN OF SPRINGS” Version 2 ME, IIT Kharagpur, IITM

Curved Beam

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