Curved Beams Engineering

8/18/2019 Curved Beams Engineering

1/85

CURVED BEAMS

CONTENT:

WHAT’S A CURVED

DIFFERENCE BET

CURVED BEAM

WHY STRESS CONC

CONCAVE SIDE OF

DERIVATION FOR S

PROBLEMS.

BEAM?

EEN A STRAIGHT BEA

NTRATION OCCUR AT INNE

URVED BEAM?

RESSES IN CURVED BEAM

AND A

R SIDE OR


2/85

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 layeris 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 remainplane 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 e1f 1 be the neutral axis and g

1h

1 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


3/85

G1H1 = (R+Y) Ѳ

GH

Also Stress

OR

dF = 0

∴∴∴∴ there is no direct force

RѲ

acting on the element considered.


4/85

Since Σyδa is first mo

distance of centroid fro

centroid of the cross sec

axis.

From (1) and (2)

CURVED BEAM

Curved beams are

clamps, crane hooks, fra

machines, planers etc. I

coincides with its centr

is linear. But in the case

is shifted towards the c

linear [hyperbolic] distr

the centroidal axis and t

within the curved beams

ent of area about neutral axis,

neutral axis. Thus neutral axis c

tion. Cross sectional area coincides

the parts of machine members f

mes of presses, riveters, punches, s

straight beams the neutral axis o

idal axis and the stress distribution

of curved beams the neutral axis

entre of curvature of the beam ca

ibution of stress. The neutral axis

e centre of curvature and will alwa

.

yδa/a is the

incides with

with neutral

und in C -

ears, boring

f the section

in the beam

f the section

sing a non-

lies between

s be present


5/85

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 fiberσbo = Bending stress at the outer fiber

σri = Combined stress at the inner fiber

σro = Combined stress at the outer fiber


6/85

Stresses in curved beam

Mb = Applied Bending Mori = Inner radius of curved

ro = Outer radius of curved

rc = Radius of centroidal ax

rn = Radius of neutral axis

CL = Center of curvature

In the above figure th

before bending. i.e., wh

moment 'Mb' is applied

'ab' through an angle '

shortened while the inn

strip at a distance 'y' frothe amount ydθ and the

Where σ = stress, e = str

F

FM

b

entbeam

beam

is

lines 'ab' and 'cd' represent two

n there are no stresses induced. Wh

to the beam the plane cd rotates w

θ' to the position 'fg' and the out

r fibers are elongated. The origina

the neutral axis is (y + rn)θ. It istress in this fiber is,

σ = E.e

ain and E = Young's Modulus

CA

NA

c 2

c 1

c

i

c o

e

F

F

Mb

r i r n

rc

ro

CL

such planes

en a bending

th respect to

er fibers are

l length of a

shortened by


7/85

We know, stress σ = E.e

We know, stress i.e., σ = – E

θθ ..... (i)

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


8/85

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. ∴

θθ

=0As

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 = – Substituting the value of dF, we getM

b =

θθ

=

θθ

= θ

θ Since represents the statical moment of area, it may be replacedby A.e., the product of total area A and the distance 'e' from the

centroidal axis to the neutral axis.


9/85

∴ Mb =

θθ

A.e ..... (iii)

From equation (i) θ

θ = –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.


10/85

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 thestress distribution in the beam

is linear.

In case of curved beams the

neutral axis of the section is

shifted towards the center ofcurvature of the beam causing

a non-linear stress

distribution.


11/85

2

3 Neutral axi

axis coincides

Location of the neut

and centroidal Neutral axis

towards the lea

curvature

ral axis By considering a rectang

section

is shifted

t centre of

lar cross


12/85

Centroidal and Neut al Axis of Typical Section of Cur ed Beams


13/85


14/85

Why stress concent

curved beam

Consider the elemen

planes ‘ab’ and ‘cd’ sep

the plane cd having ro

Consider two fibers sy

axis. Deformation in bot

ration occur at inner side or co

ts of the curved beam lying betwe

arated by angle θ. Let fg is the fin

tated through an angle dθ about

metrically located on either side

h the fibers is same and equal to yd

cave side of

en two axial

l position of

neutral axis.

f the neutral

.


15/85

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 thecentroidal axis. This results in a better stress distribution.

Example:

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

the inner radius.


16/85

For a straight beam

Inner most fiber:

Outer most fiber:

For curved beam:

e = rc - rn = h – 0.91

co = ro - rn= h – 0.9

ci = rn - ri = 0.910h -

:

h=2ri

h = 0.0898h

10h = 0590h

= 0.410h


17/85

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 greaterthan 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 = 22X103Nb = 20mm

h = 150-50 = 100mm

A = bh = 20X100 = 2000mm2


18/85

e = rc - rn = 1

Section A-B will be s

bending, due to the ecce

Stress due to direct load

y = rn – r = 91.0

Mb = 22X103X1

00 - 91.024 = 8.976mm

bjected to a combination of dir

tricity of the force.

will be,

24 – r

0 = 2.2X106 N-mm

ct load and


19/85


20/85

Problem no.2Determine the value of

shown in fig such that t

fibers are numerically e

Given data;

Inner radius ri=150m

Outer radius ro=150+4

=290mm

Solution;

From Figure Ci + CO =

= 140mm…

Since the normal stresse

the extreme fiber are nu

have,

i.e Ci=

=

Radius of neutral axis

rn=

rn =197.727 mm

ai = 40mm; bi = 100mm;

ao = 0; bo = 0; ri = 150m

“t” in the cross section of a cur

he normal stress due to bending at

ual.

+100

0 + 100

……… (1)

s due to bending at

merically equal we

.51724Co…………… (2)

b2 =t;

; ro = 290mm;

ed beam as

the extreme


21/85

=

i.e., 4674.069+83.

∴ t = 41.126mm

Problem no.3Determine the stresses

figure.

Solution:

The figure shows the cri

ring.Radius of centroidal axi

Inner radius of curved b

Outer radius of curved b


Applied force

1t = 4000+100t;

t point A and B of the split ring

tical section of the split

rc = 80mm

am ri = 80

= 50mm

eam ro = 80 +

= 110mm

rn =

=

= 77.081mm

F = 20kN = 20,000N (compress

hown in the

ve)


22/85

Area of cross section A =π

d2=

π

x602 = 2827.433mm

2

Distance from centroidal axis to force = rc = 80mmBending moment about centroidal axis Mb = Fl = 20,000x80

=16x105

N-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=27.081mmDistance of neutral axis to outer radius

co = ro rn= 110 77.081=32.919mm

Direct stress σd = = = 7.0736N/mm

2

(comp.)

Bending stress at the inner fiber σbi = =

=

105N/mm

2(compressive)

Bending stress at the outer fiber σbo = =

= 58.016N/mm2 (tensile)

Combined stress at the inner fiber


23/85

σri = σd + σbi

= 7.0736 105.00= - 112.0736N/mm

2(compressive)

Combined stress at the outer fiberσro = σd + σbo

= 7.0736+58.016= 50.9424N/mm

2 (tensile)

Maximum shear stress

τmax = 0.5x σmax

= 0.5x112.0736

= 56.0368N/mm2, at B

The figure.


24/85

Problem No. 4

Curved bar of rectang

100mm is subjected t

straighten the bar. Findiagram to show the vari

Solution

Given data:

b= 40mm h= 60m

rc=100mm Mb= 2x1

C1=C2= 30m

rn=

ro= rc+h/2=100+30=13

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

rn= 96.924mm

Distance of neutral axis



Area

A= b

Bending stress at the inn

lar section 40x60mm and a me

a bending moment of 2KN-m

the position of the Neutral axisation of stress across the section.

6N-mm

=(ri+c1+c2)

0mm (rc-c1)

to centroidal axise = rc - rn= 100-96.924

=3.075mm

o inner radius

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

o outer radius

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

h = 40x60 = 2400 mm2

er fiber σbi = =

= 104.239 N/mm2

(com

n radius of

tending to

and draw a

m

m

ressive)


25/85


= -68.94 N/mm2 (tensile)

Bending stress at the centroidal axis =

=

= -8.33 N/mm2 (Compressive)

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

figure.


26/85

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 theextreme 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 numericallyequal 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


27/85

b=150.34mm

To find the centroidal axis, (C2)

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

= 74.313mm.

But C1=C2rc= ro-c2 =240 - 74.313 =165.687mm

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

Bending stress in the outer fiber,

σ

A= = 1050.4mm

60 =

Mb=10.8x106 N-mm


28/85

Problem no.6

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

fig.1.9a

Solution:

Redraw the critical section as shown in the figure.

Radius of centroidal axis rc = 80mm

Inner radius of curved beam ri = 80

= 50mm

Outer radius of curved beam ro = 80 + = 110mmRadius of neutral axis rn =

=

=77.081mmApplied force F = 20kN = 20,000N (compressive)

Area of cross section A =π

d2=

π

x602 = 2827.433mm

2

Distance from centroidal axis to force = rc = 80mmBending 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


29/85


ci = rn ri = 77.081 50 = 27.081mmDistance of neutral axis to outer radius

co = ro rn = 110 77.081 = 32.919mmDirect stress σd = =

= 7.0736N/mm2 (comp.)Bending stress at the inner fiber σbi =

=

= 105N/mm2 (compressive)Bending stress at the outer fiber σbo =

=


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= 50.9424N/mm

2 (tensile)


max = 0.5x σmax = 0.5x112.0736= 56.0368N/mm

2, at B


30/85

The figure shows the stress distribution in the critical section.


31/85

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+ = 140mm

Radius of neutral axis rn = =

= 136.1038mm

Distance of neutral axis to centroidal axis

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

80

R 1 0 0

175 mm

9000N

h = 80mm

c2 e c1

b = 5 0 m m

CriticalSection

co ciro

rn

rc

r = 100mmi

1

CL

C A

N A


32/85


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

Distance of neutral axis to outer radius

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

Distance from centroidal axis to force

= 175+ rc = 175+140 = 315mmApplied force F = 9000N

Area of cross section A = 50x80 = 4000mm

2

Bending moment about centroidal axis Mb = FI = 9000x315

= 2835000 N-mm

Direct stress σd = =


Bending stress at the inner fiber σbi =

=

= 65.676N/mm2(tensile)

Bending stress at the outer fiber σbo = = = 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)


33/85

Maximum shear stress max = 0.5x σmax = 0.5x67.926= 33.963 N/mm

2, 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

C A

N A

Bending stress =-44.362 N/mmσbo2

Combined stress -42.112 N/mmσro2

=

σd=2.25 N/mm2

Direct stress ( )σd


34/85

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 = =

= 16.667mm

h = 40mm

c2e

c1

b = 6 m m

o

co ci

rorn

rc

r = 25mmi 100m

CL

b = 1 8 m m

i

C A

N A


35/85

∴ Radius of centroidal axis rc = ri c1= 25+16.667 = 41.667 mm

Radius of neutral axis rn =

=

=38.8175mm

Distance of neutral axis to centroidal axis e = rc rn= 1.66738.8175=2.8495mm

Distance of neutral axis to inner radius ci = rn ri= 38.817525=13.8175mm

Distance of neutral axis to outer radius co = ro rn= 65-38.8175=26.1825mm

Distance from centroidal axis to force = 100+ rc = 100+41.667= 141.667mm

Applied force F = 5000N

Area of cross section A = = = 480mm2 Bending moment about centroidal axis Mb = FI = 5000x141.667

= 708335 N-mm


36/85


= 10.417N/mm

2 (tensile)


=

= 286.232N/mm2

(tensile)


= 208.606N/mm2 (compressive)Combined stress at the inner fiber σri = σd + σbi = 10.417+286.232

= 296.649N/mm2

(tensile)

Combined stress at the outer fiber σro = σd + σbo = 10.417286.232= 198.189N/mm2 (compressive)

Maximum shear stress max = 0.5x σmax = 0.5x296.649= 148.3245 N/mm

2, at the inner fiber



37/85

σbi=286.232 N/mm2

σri=296.649 N/mm2

b = 18 mmi

h =40 mm

C A

N A


Combined stress N/mmσro2

=-198.189

b = 6 mmo

σd=10.417 N/mm2

Direct stress (σd)


38/85

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 =

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

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

∴ rn = = 333.217mmLet AB be the ref. line

5 5 0

7585 kN

300

2 5 0

750 mm75

B a = 7 5 m

i225 mm

a2

b =75mm2

b = 3 0 0 m m

i

ci

co

A

X

e rn

rc

r =550 mmo

C A

N A

CL

7

r = 250 mmi

a1


39/85

=

= 101.785mm

Radius of centroidal axis rc = ri +

= 250+101.785=351.785 mm

Distance of neutral axis to centroidal axis e = rc rn= 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 = 750+ rc = 750+351.785 = 1101.785mm

Applied force F = 85kNBending moment about centroidal axis Mb = FI

= 85000x1101.785

= 93651725N-mm


= 2.16N/mm

2 (tensile)

Bending stress at the inner fiber σbi = = = 42.64N/mm

2 (tensile)


40/85



= 44.8N/mm2

(tensile)

Combined stress at the outer fiber σro = σd + σbo = 2.16 50.49= 48.33N/mm2 (compressive)


max = 0.5x σmax = 0.5x48.33

= 24.165N/mm2, at the outer fiber

The below figure shows the stress distribution.

σ

bi

=42.64 N/mm2

σri=44.8 N/mm2

b = 3 0 0 m m

i

225

C A

N A


Combined stress N/mmσro2

=-48.33

a =75mmi

b = 75 mm2 a2

a1

σd=2.16 N/mm2

Direct stress ( )σd


41/85

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 = =

= 40mm

90mm

30mm

F = 25 kN

25mm

60mm

1 0 0 m m

ri

rc

rn

ro

e

cico

N C A

c2

c1

h = 90 mm

l

F


42/85

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

= 140 mm


=

= 135.42mm

Distance of neutral axis to centroidal axis e = rc rn= 140 135.42=4.58mmDistance 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 = rc 25= 140 25= 115mm

Applied force F = 25,000N = 25kN

Area of cross section A =

=

= 4050mm2 Bending moment about centroidal axis Mb = FI = 25,000x115

= 2875000 N-mm


= 6.173N/mm

2 (tensile)


43/85


= 54.9 N/mm2 (tensile)



= 61.073N/mm2

(tensile)

Combined stress at the outer fiber σro = σd + σbo = 6.173 44.524= 38.351N/mm2 (compressive)

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

= 30.5365 N/mm2, at the inner fiber



44/85

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 = ---------------------- From DDHB


45/85

C2 =

= 50mm

c1 = 90-50= 40mm

Moment of inertia I =

=

= 2632500mm

4

Direct stress σb = =

= 6.173N/mm

2 (tensile)


=

= 43.685 N/mm

2(tensile)

Bending stress at the outer fiber σbo = - = = -54.606N/mm

2 (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.


46/85

σbi= 43.685 N/mm

2

σri= 49.858 N/mm

2

6 0 m m

h =90 mm

N A , C A

σbo=-54.606 N/mm2

σro

N/mm2

=-48.433

b = 30 mm

c =50mm2 c =40mm1

b /2 = 15o

b /2 = 15o

σd= 6.173 N/mm

2

σd

b


47/85

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 100mmfrom 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+

= 130mm


=

= 153.045mm

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

= 155-153.045 = 1.955mm

h=60mm

b=30mm

F = ?

1 2 5 m m

100

h = 60mm

c2

ec

1

b

= 3 0 m m

co

ci

ro

rn

rc

F

r = 125mmi

CL

l

100

C A

N A

L o a d l i n e


48/85

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


Bending stress at the inner fiber σbi = +

Combined stress at the inner fiber σri = σ

d + σ

bi

i.e., 75 =

+

F = 8480.4N =Capacity of the hook.


49/85

Problem no.12Design 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 =10tSelection the standard crane hook dimensions from table 25.3 when safe

load =10t and steel (MS)

∴ c =11933; Z = 14mm; M = 71mm andh = 111mm

bi= M = 7133

bo = 2xZ = 2x14 = 28 mm

r1 = = = 59.5mmh = 111mm

o

H

M

= b

i

Z

r =59.5 mmi

r =c l

rn

ro

e

cico

N A

C A

c2 c1

h = 111 mm

CLb =28o b=71i


50/85

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


=

= 98.095mm

Distance of centroidal axis from inner fiber c1 =

=

= 47.465mmRadius of centroidal axis rc = ri + c1

= 47.465+59.5= 106.965 mm

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

=106.965-98.095 =8.87mm

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

= 98.095-59.5=38.595mm


= 170.5-98.095=72.0405mm

Distance from centroidal axis to force l = rc -106.965


51/85

Applied force F = 105N

Area of cross section A =

= = 5494.5mm2 Bending moment about centroidal axis Mb = Fl = 10

5x141.667

= 106.965x105N-mm




= 142.365/mm2

(tensile)


= -93.2 N/mm

2

(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

= 80.2825 N/mm2, at the inner fiber



52/85

σbi=142,365 N/mm

2

σri=160.565 N/mm

2

b = 71 mmi

C A

N A

σbo=-93.2 N/mm

2

σro

N/mm2

=-75

b = 28 mmo

h = 111 mm

σd=18.2 N/mm

2

σd


53/85

Problem no.13The 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.

Radius of centroidal axis rc = 100mm

Inner radius ri = 100 – 100/2 = 50mm

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

Radius of neutral axis rn = r o r i4

2 = 150 50

42=

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 = x d

2 =

x 100

2 = 7853.98mm

2

Mb = F = 5000 Combined maximum stress at the inner fiber

(i.e., at B)


54/85

σri = Direct stress + bending stress

= ∴ x= 599.9 = Maximum offset distance.

Problem no.14An 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


55/85

Draw the critical section

Radius of centroidal axi

Inner radius ri =100

Outer radius ro = 100+


rn = =

= 99.6mm



at P-Q as shown in the figure.

rc =100mm

= 87.5mm

= 112.5mm

rom centroidal axis e =rc - rn

=100 - 99.6 =

o inner fiber ci = rn – ri

= 99.6 – 87.5 =12.

0.4mm

1 mm


56/85

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

=112.5 – 99.6 = 12.9 mm

Area of cross-section A =π

4

d

2 =

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 -

M bCo

Aeo =

1000

49087 –100000 X129

49087 X 04 X 1125

= - 56.36 N/mm2 (compressive)

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

σri= Direct stress + bending stress

=F

A +

M bci

Aer i =

1000 49087 +

100000 X 12149087 X 04 X 875

= 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


57/85

ro = 75 +25

2 = 87.5 mm

A =π

4

d2 =

π

4

X 252 = 490.87 mm2

rn = r o r i4

2 = 87 5 625

42 =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

–M b coAer o

=1000

49087 - 75000 X13024549087 X 05245 X 625 = - 41.324 N / mm

2

(compressive)


58/85

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

σri =F

A +

M bco

Aer o =

1000

49087 +75000 X 119755

49087 X 05245 X 625 = 55.816 N / mm2 (tensile)

∴ Maximum tensile stress = 72.466 N / mm2 at P Maximum compressive stress = 56.36 N / mm

2 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.


59/85

The ring is symmetri

horizontal and vertical d

Consider the horizontal

A and B, the vertical for

No horizontal forces w

proved by understandin

symmetrical, the reactio

Assume that two horizo

half, as shown in the fi

half must have forces H

This however, results i

hence H must be zero.

of equal magnitude M0

noted that these mo

condition of symmetry.

be treated as that shown

quantity is M0. Againconclude that the tang

vertical and must rema

does not rotate. By Cast

derivative of the strai

displacement of the load

al and is loaded symmetrically

irections.

section as shown in the figure. At

es would be F/2.

uld be there at A and B. this argu

that since the ring and the extern

s too must be symmetrical.

tal inward forces H, act at A and B

gure. In this case, the lower

acting outwards as shown.

violation of symmetry and

esides the forces, moments

ct at A and B. It should be

ents do not violate the

hus loads on the section can

in the figure. The unknown

Considering symmetry, Weents at A and B must be

n so after deflection or M0

igliano’s theorem, the partial

energy with respect to the lo

. In this case, this would be zero.

……………………….(1)

in both the

the two ends

ment can be

al forces are

in the upper

d gives the


60/85

The bending moment at

figure.

Will be

As per Castigliano’s the

From equation (2)

And, ds = Rdθ

any point C, located at angle θ, as s

………..(2)

rem,

own in the


61/85

As this quantity is positi

produces tension in the i

It should be noted that t

θ = 0 to θ = 900.

The bending moment M

e the direction assumed for Mo is c

ner fibers and compression on the

ese equations are valid in the regio

at any angle θ from equation (2) w

orrect and it

outer.

,

ill be:


62/85

It is seen that numericall

The stress at any angle

forces as shown in the fi

Put θ = 0 in Bending

get,

At A-A Mbi = 0.181

Mbo = - 0.18

And θ = 90,

At B-B Mbi = -

Mbo = 0.3

The vertical force F/

components (creates n

(creates shear stresses).

The combined norm

y, Mb-max is greater than Mo.

can be found by considering the

gure.

moment equation (4) then we will

FR

1FR

.318FR

18FR

can be resolved in two

rmal direct stresses) and S

l stress across any section will be:


63/85

The stress at inner (σ1Ai)

On similar lines, the stre

outer points will be (at

It should be noted that i

that the radius is large

straight beam.

and outer points (σ1Ao) at A-A will

ss at the point of application of loa

= 900)

calculating the bending stresses,

ompared to the depth, or the bea

be (at = 0)

at inner and

t is assumed

is almost a


64/85

A Thin Extended C

Consider a thin closed ri

the figure. At the two en

No horizontal forces w

ring.

The unknown quantity

conclude that the tange

so after deflection or M0 There are two region

The straight porti

Mb

The curved porti

osed Link

ng subjected to symmetrical load F

ds C and D, the vertical forces woul

uld be there at C and D, as disc

is M0. Again considering sy

ts at C and D must be vertical and

does not rotate.s to be considered in this case:

n, (0 < y < L) where

= MO

n, where

as shown in

d be F/2.

ssed earlier

metry, we

must remain


65/85

As per Castigliano’s

\

theorem


66/85

It can be observed that

as obtained for a circul

and Compression on the

The bending moment M

Noting that the equation

At = 0At section B-B bending

At section A-A the loa

bending moment occurs

part of the equation is m

It can be observed tha

expression as obtained f

It is seen that numericall

t L = 0 equation reduces to the sa

r ring. Mo produces tension in th

outer.

at any angle

will be

are valid in the region, = 0 to

oment at inner and outer side of t

point, i.e., at = p/2, the maxi(numerically), as it is observed th

ch greater than the first part.

at L = 0, equation (v) reduces

r a circular ring.

y, Mb-max is greater than Mo.

e expression

inner fibers

p/2,

e fiber is

um value of

t the second

to the same


67/85

The stress at any angl

shown in the figure.

The vertical force F/2

normal direct stresses) a

The combined norm

The stress at inner fiber

be (at = 0):

The stress at inner fi

will be (at the loading p

can be found by considering

can be resolved in two compon

d S (creates shear stresses).

l stress across any section will be

σ1Bi and outer fiber σ1Bo and at sect

ber σ1Ai and outer fiber σ1Ao and at

int = 900):

the force as

nts (creates

ion B-B will

section A-A


68/85

Problem 15

Determine the stress ind

of 25 mm diameter subj

diameter of the ring is 6

Solution: the circular ri

1.29a and 1.29b respecti

Inner radius ri = = 30

Outer radius = 30+25 =


Radius of neutral axi


=


=

ced in a circular ring of circular cr

cted to a tensile load 6500N. The i

mm.

ng and its critical section are as sho

ely.

m

5mm

rc = 30 + = 42.5mm

rn =

= =42.5mm

o centroidal axis e = rc - rn

42.5 – 41.56 = 0.94mm

o inner radius ci = rn - ri

41.56 – 30 = 11.56mm

ss section

ner

n in fig.


69/85


= 55 - 41.56 = 13.44mm

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

σd = θ

Consider the cross section A – A

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

Direct stress σd =

= 0

Bending moment Mb = - 0.318Fr

Where r = rc, negative sign refers to tensile load

= - 0.318x6500x42.5 = -87847.5 N-mmThis couple produces compressive stress at the inner fiber and tensile

stress at the at outer fiber

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

= 0 - =

= - 73.36N/mm2(compressive)

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

=0+ =


Consider the cross section B – B


70/85

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

Direct stress σd =

= = 6.621 N/mm2

Bending moment Mb = 0.182Fr

Where r = rc, positive sign refers to tensile load

= 0.182x6500x42.5 = 50277.5 N-mmThis couple produces compressive stress at the inner fiber and tensile


Maximum stress at the inner fiber σ=Direct stress + Bending stress= σd -

= 6.621 +

= 48.6 N/mm2

(tensile)


= σd + =6.621+ = -20 N/mm

2 (compressive)


71/85

Problem 16

Determine the stress ind

of 50 mm diameter rod s

mean diameter of the rin

Solution: the circular ri

1.30a and 1.30b respecti

Inner radius ri = -

Outer radius = +




= 50 -


= 46.65


= 75 - 4

Area of cross section A

ced in a circular ring of circular cr

ubjected to a compressive load of 2

g is 100 mm.

ng and its critical section are as sho

ely.

25mm

75mm

rc = = 50mm

rn =

= = 46.65mm


6.65 = 3.35mm


-25 = 21.65 mm

o outer radius co = ro - rn

6.65 = 28.35mm

= x552 = 1963.5mm

2

ss section

kNN. The

n in fig.


72/85


σd = Consider the cross section A – A


Direct stress σd =

= 0Bending moment Mb = + 0.318Fr

Where r = rc, positive sign refers to tensile load

= + 0.318x20000x50 = 318000 N-mmThis couple produces compressive stress at the inner fiber and tensile


Maximum stress at the inner fiber

=Direct stress + Bending stress

= 0 + =

= 41.86 N/mm2

(tensile)

Maximum stress at outer fiber = Direct stress + Bending stress=0 -

= -



73/85

Consider the cross section B – B


Direct stress σd =

=

= 5.093 N/mm2 (compressive)

Bending moment Mb = -0.1828Fr

Where r = rc, negative sign refers to tensile load

= - 0.182x20000x50 = -182000 N-mmThis couple produces compressive stress at the inner fiber and tensile



= σd - = -5.093 + = - 29.05 N/mm

2(compressive)


= σd + = -5.093 +

= 5.366 N/mm2 (tensile)


74/85

Problem 17

A chain link is made of

mean diameter of which

80mm. The straight sidload of 90kN; estimate

along the section of loa

from the load line

Solution: refer figure

= 80mm; dc = 80mm;

F = 90kN = 90000N

Draw the critical cross s

Inner radius ri = 40 -

Outer radius = + =





40 mm diameter rod is circular at

is 80mm. The straight sides of the

s of the link are also 80mm.If thethe tensile and compressive stres

line. Also find the stress at a secti

rc = 40mm;

ction as shown in fig.1.32

20mm

60mm

rc = 40mm

rn =

= = 37.32mm


=40-37.32 = 2.68mm


= 37.32-20 = 17.32 mm

each end the

link are also

ink carries ain the link

on 900 away


75/85


= 60 – 37.32 = 22.68mm


σd = θ

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


Direct stress σd =

= 0

Bending moment = - π where r = rc,= π = 1.4x106N-mm

This couple produces compressive stress at the inner fiber and tensile


Maximum stress at the inner fiber σ=Direct stress + Bending stress= 0 +

=

π

= - 360 N/mm2(tensile)


= 0 - = - π = 157.14 N/mm

2 (compressive)


76/85

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


Direct stress σd = θ

=

π = 35.81 N/mm2

(compressive)

Bending moment = - ππ where r = rc,= ππ = - 399655.7N-mm

This couple produces compressive stress at the inner fiber and tensilestress at the at outer fiber


= σd - = 35.81 +

π

= 138.578 N/mm2(tensile)


= σd + = 35.81 -

π


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

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


77/85

Using usual notations

curved beam of initia

uniform bending mom

Consider a curved bea

below. Its transverse se

in its unstressed state; it

xy plane along the arcs

Now apply two equal an

(c). The length of neut

central angles before anof neutral surface remai

R1θ =R2θ'

Consider the arc of circ

surface. Let r1 and r2 b

couples have been appli

From Fig. 1.2 a and

prove that the moment of resist

l radius R1 when bent to a ra

nt is

M = EAeR1

of uniform cross section as sho

tion is symmetric with respect to t

s upper and lower surfaces intersec

of circle AB and EF centered at

d opposite couples M and M' as sho

al surface remains the same. θ a

after applying the moment M. Sins the same

Figure

le JK located at a distance y abov

the radius of this arc before and

d. Now, the deformation of JK,

, r1 = R1 – y; r2 = R2 – y ..

nce M of a

dius R2 by

n in Figure

e y axis and

t the vertical

[Fig. 1(a)].

wn in Fig. 1.

d θ' are the

ce the length

..... (i)

e the neutral

fter bending

.... (ii)

... (iii)


78/85

∴ δ = (R2

=R2θ'

= – y

∴ δ =– y∆ [ θ' − θ = θ + ∆

The normal strain ∈

the deformation δ by the

∴ ∈x =

The normal stress σx

∴ σx

=–E

i.e. σx = – E

Equation (vi) shows

with the distance y fro

obtain an arc of hyperbo

From the condition

entire area is zero and th

is equal to the applied b∴ ∫δF =0

i.e., ∫σxdA =0

and ∫(– yσxdA)=M

– y) θ' – (R1 – y) θ

– θ'y – R1θ + θy

θ' – θ) [ R1θ = R2θ' from equ (

θ − θ = ∆θ] ..

x in the element of JK is obtaine

original length r1θ of arc JK.

.

may be obtained from Hooke's law

..

( r1 = R1 – y) .(vii)

that the normal stress σx does not

the neutral surface. Plotting σx

la as shown in Fig. 1.3.

f equilibrium the summation of fo

e summation of the moments due t

nding moment.

....

..

)]

... (iv)

by dividing

.... (v)

σx = E∈x

... (vi)

vary linearly

ersus y, we

ces over the

these forces

. (viii)

... (ix)


79/85

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

(viii)

–

dA=0Since is not equal to zero dA = 0i.e. R1 1 =0i.e., . R1 1 =0

∴ R1 = 1

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

neutral surface is obtained by the relation R1 = 1 ..... (x)

The value of R1 is not equal to the 1 distance from O to the centroidof the cross-section, since 1 is obtained by the relation,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)

y dA =M

i.e.,

dA = M ( r1 = R1 - y from iii)

i.e.,

dA = M

i.e., = Mi.e.,

= M [using equations (x) and (xi)]


80/85

i.e.,

i.e.,

=Mi.e.,

=

1 1 ..... (xii)

i.e., = (e = 1– R1 from Fig. 1.2a)………xiii)

Substituting

into equation (VI)

σx =

1

..... (xiv)

∴ σx=M11

1 ( 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 MFrom equation (i),


81/85

= {

From equation (xiii)

=

}

i.e.

= MEAeR 1 ∴ M =EAe R1

Hence proved

References:


82/85

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

4. Determine of value of t in the cross section of a curved beam shown in Figure such that the normalstresses due to bending at the extreme fibers are numerically equal.


83/85

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 70MPa.

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.


84/85

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 inthe 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

6 2. 5 m m

A B

112.5

2

5

8

7 .

5


85/85

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

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