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AMS LECTURE-20

Dr. C. R. DEO

CURVED BEAM 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. Again for simple bending, the bending equation = = is suitable for

beam which is initially straight before the application of bending moment.

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. It has also

been found that the stresses in the fibres of a curved beam are not proportional to the

distances of the fibres from the neutral surfaces, as is assumed for a straight beam.

Differences between Straight and Curved Beams

Sl. no Straight beam Curved beam

1 The neutral axis of beam coincides with

centroidal axis.

The neutral axis is shifted towards

the centre of curvature by a distance

called eccentricity i.e. the neutral axis

lies between centroidal axis and

centre of curvature

2 The variation of normal stress due to

bending is linear, tensile at the inner fibre

and compressive at the outer fibre with

zero value at the centroidal axis.

The variation of normal stress due to

bending across section is non-linear

and is hyperbolic.

Stresses in Curved Beam (WINKLER-BATCH THEORY)

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:

AMS LECTURE-20

Dr. C. R. DEO

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.

vii) The radial strain is negligible.

NOTATIONS AND SYMBOLS USED 흈 = 푺풕풓풆풔풔 흈풅 = 푫풊풓풆풄풕 풔풕풓풆풔풔, 풕풆풏풔풊풍풆 풐풓 풄풐풎풑풓풆풔풔풊풗풆 흈풊 = 푺풕풓풆풔풔 풂풕 풊풏풏풆풓 풇풊풃풓풆 흈풐 = 푺풕풓풆풔풔 풂풕 풐풖풕풆풓 풇풊풃풓풆

흈풃풊 = 푵풐풓풎풂풍 풔풕풓풆풔풔 풅풖풆 풕풐 풃풆풏풅풊풏품 풂풕 풊풏풏풆풓 풇풊풃풓풆

흈풃풐 = 푵풐풓풎풂풍 풔풕풓풆풔풔 풅풖풆 풕풐 풃풆풏풅풊풏품 풂풕 풐풖풕풆풓 풇풊풃풓풆 푴풃 = 푩풆풏풅풊풏품 풎풐풎풆풏풕 풇풐풓 풄풓풊풕풊풄풂풍 풔풆풄풕풊풐풏, 풊. 풆 풎풐풎풆풏풕 풂풃풐풖풕 풄풆풏풕풓풐풊풅풂풍 풂풙풊풔

풓풄 = 푫풊풔풕풂풏풄풆 풐풇 풄풆풏풕풓풐풊풅풂풍 풂풙풊풔 풇풓풐풎 풄풆풏풕풓풆 풐풇 풄풖풓풗풂풕풖풓풆 풓풏 = 푫풊풔풕풂풏풄풆 풐풇 풏풆풖풕풓풂풍 풂풙풊풔 풇풓풐풎 풄풆풏풕풓풆 풐풇 풄풖풓풗풂풕풖풓풆

풆 = 푬풄풄풆풏풕풓풊풄풊풕풚, 풊. 풆 풅풊풔풕풂풏풄풆 풃풆풕풘풆풆풏풄풆풏풕풓풐풊풅풂풍 풂풙풊풔 & 푛푒푢푡푟푎푙 푎푥푖푠

풚 = 푫풊풔풕풂풏풄풆 풐풇 풇풊풃풆풓 풇풓풐풎 풏풆풕풓풂풍 풂풙풊풔 푨 = 푨풓풆풂 풐풇 풄풓풐풔풔 − 풔풆풄풕풊풐풏 풐풇 풎풆풎풃풆풓, (풄풖풓풗풆풅 풃풆풂풎),

푷 = 푳풐풂풅 풐풏 풎풆풎풃풆풓 흉풎풂풙 = 푴풂풙풊풎풖풎 풔풉풆풂풓 풔풕풓풆풔풔 Derivation of Expression to Determine Stress at any Point on the Fibres of a Curved Beam

AMS LECTURE-20

Dr. C. R. DEO

Consider a curved beam with rc, as the radius of centroidal axis, rn, the radius of neutral

surface, ri, the radius of inner fibre, ro, the radius of outer fibre subjected to bending moment

Mb.

Let AB and CD be the two adjacent cross-sections separated from each other by a small

angle d.

Because of Mb the section CD rotates through a small angle dα and point “C” shifted to “C’ “

& point D shifted to D’.

Consider a elemental fiber “PQ” at a distance “r” from center of curvature “o” or at a distance

“y” from neutral axis. Due to application of moment Mb , the point Q shifted to Q’. The unit

deformation of fibre PQ at a distance y from neutral surface is:

Deformation 휀 =

=> 휀 =푦푑훼푟푑휃

=푦푑훼

(푟 − 푦)푑휃 −− −−− −− − − −− − −− −−(1)

The unit stress on this fibre is, Stress = Strain × Young’s modulus of material of beam

휎 = 퐸 × 휖 =퐸푦푑훼

(푟 − 푦)푑휃 − −− − − −− − −− −− − (2)

For equilibrium, the summation of the forces acting on the cross sectional area must be zero.

=> 휎푑퐴 = 0

=> 퐸푦푑훼

(푟 − 푦)푑휃푑퐴 = 0

퐸푑훼푑휃

푦푑퐴

(푟 − 푦) = 0 −− − − −− −−− −− − − −(3)

AMS LECTURE-20

Dr. C. R. DEO

Also the external moment Mb applied is resisted by internal moment. i.e Mb=-M

=> 푦(휎푑퐴) = 푀

=> 푦퐸푦푑훼

(푟 − 푦)푑휃푑퐴 = 푀

=> 퐸푦 푑훼

(푟 − 푦)푑휃푑퐴 = 푀

=> 퐸 푑훼푑휃

푦(푟 − 푦)푑퐴 = 푀

=> 퐸 푑훼푑휃

푦 − 푦푟 + 푦푟(푟 − 푦) 푑퐴 = 푀

=> −퐸 푑훼푑휃

푦푑퐴 + 퐸 푑훼푑휃

푟푦푑퐴

(푟 − 푦) = 푀

퐹푅푂푀 푒푞푢푎푡푖표푛 (3)퐸푑훼푑휃

푦푑퐴

(푟 − 푦) = 0

=> −퐸 푑훼푑휃

푦푑퐴 = 푀 = −푀

=> 푀 = 퐸 푑훼푑휃

푦푑퐴

∫ 푦푑퐴 = The moment of cross sectional area with respect to neutral surface

=> 푀 = 퐸 푑훼푑휃

× 퐴푒

Here ‘e’ represents the distance between the centroidal axis and neutral axis. i.e. 푒 = 푟 − 푟

Rearranging terms in equation

=> 푑훼푑휃

=푀퐸퐴푒

푐표푛푠푖푑푒푟푖푛푔 푡ℎ푒 푒푞푢푎푡푖표푛 (2)휎 = 퐸 × 휖 =퐸푦푑훼

(푟 − 푦)푑휃=푀퐴푒

푦(푟 − 푦)

=> 휎 =푀퐴푒

푦

푟 − 푦

=> 휎 =푀퐴푒

푦푟

Where “r” be the radius of the elemental fiber from center of curvature = 푟 − 푦

On considering the equation (3)

퐸푑훼푑휃

푦푑퐴

(푟 − 푦) = 0

=>푦푑퐴

(푟 − 푦) = 0

From the figure it is found that “y”= rn – r or rn = r + y or r = rn – y

=>푦푑퐴

(푟 − 푦) =(푦 − 푟 + 푟 )푑퐴

(푟 − 푦) = − 푑퐴 + 푟푑퐴푟

= 0

AMS LECTURE-20

Dr. C. R. DEO

=> 푟푑퐴푟

= 푑퐴 = 퐴

=> 푟 =퐴

∫ 푑퐴푟

The bending stress distribution diagram along the depth of the curved beam is shown as

follows:

AMS LECTURE-21

Dr. C. R. DEO Page 1

SIGN CONVENTION:- Consider the simple bending case:-

A cantilever beam subjected to end load as shown in figure. Due to load a bending moment acting about the centroidal axis in clock wise direction which tends to bend the beam & produce a convex surface at the top and a concave surface at the bottom. This implies that the upper fibers / layer at the top are under tension whereas the fibers / layers at the bottom are under compression.

Let consider the cross-section of the beam (plane hatched in red colour), due to moment “M” the cross-section plane (plane hatched in red colour) tends to rotate in clock wise direction (moment direction), i.e the plane hatched in green colour above the neutral axis comes out of the cross-section plane whereas bellow the neutral axis it goes into the cross-section plane. The side of the neutral axis where plane hatched in green colour comes out, that side of the neutral axis the bending stress is tensile in nature and vice-versa.

AMS LECTURE-21


SIGN CONVENTION FOR BENDING MOMENT “M” CONSIDER THE CASE FOR CURVED BEAM Case-1 The applied moment trying to close the curve beam or trying to bend more or trying to reduce the radius of curvature. When the applied moment trying to reduce the radius of curvature, the outer surface is under tension and the inner surface is under compression. The bending moment “M” is considered as negative.

Case-2 The applied moment trying to open the curve beam or trying to straight the beam or trying to increase the radius of curvature. When the applied moment trying to increase the radius of curvature, the outer surface is under compression and the inner surface is under tension. The bending moment “M” is considered as positive.

AMS LECTURE-21


SIGN CONVENTION FOR “Y” “y” considered as positive when it is measure in the direction towards the center of curvature & negative when it is measure in the direction outwards from the center of curvature.

LOCATION OF THE NEUTRAL AXIS A. Rectangular section:-

푅푎푑푖푢푠 표푓 푛푒푢푡푟푎푙 푎푥푖푠 " 푟 " =퐴

∫푑퐴푟

Area of the rectangular section “A”= bxh

Consider a elemental section at distance “y” from neutral axis of thickness “dy”

Area of the elemental section “dA”= bxdy

=> 푑퐴푟

=푏 × 푑푦푟

As”r” = rn + y

Differentiating both the side w.r.t “y” = 1 => 푑푟 = 푑푦

AMS LECTURE-21


=> 푑퐴푟

=푏 × 푑푟푟

= 푏푑푟푟

= 푏[푙푛푟] = 푏[푙푛푟 − 푙푛푟 ] = 푏 푙푛푟푟

Hence rn :

rn =퐴

∫ 푑퐴푟=

푏 × ℎ

푏 푙푛 푟푟

=ℎ

푙푛 푟푟

Determination of C.G. of the section:- 퐶.퐺 표푓 푡ℎ푒 푠푒푐푡푖표푛 푠푖푡푢푎푡푒푑 푎푡 푎 푑푖푠푡푎푛푐푒 푦 푓푟표푚 푡ℎ푒 푏표푡푡표푚 표푓 푡ℎ푒 푠푒푐푡푖표푛.

푦 푓푟표푚 푡ℎ푒 푏표푡푡표푚 표푓 푡ℎ푒 푠푒푐푡푖표푛 =ℎ2

푅푎푑푖푢푠 표푓 푐푢푟푣푎푡푢푟푒 표푓 푐푒푛푡푟푖푑푎푙 푎푥푖푠 푟 = 푟 +ℎ2

B. Trapezoidal Section:-

Consider an elemental section at a distance “y” from neutral axis of thickness “dy”. Let width of the elemental section “b’ “, where :

풃 = 풃풐 +풃풊 − 풃풐

풉× (풓풐 − 풓)

AMS LECTURE-21


Area of the elemental section “dA”

풅푨 = 풃 × 풅풚 = 풃풐 +풃풊 − 풃풐

풉× (풓풐 − 풓) 풅풚

푨풔 풓 = 풓풏 + 풚 풅풓풅풚

= ퟏ => 푑푟 = 푑푦

푯풆풏풄풆 풅푨 = 풃 × 풅풚 = 풃 × 풅풓 = 풃풐 +풃풊 −풃풐

풉× (풓풐 − 풓) 풅풓

Where “h” =ro – ri

푻풐풕풂풍 풂풓풆풂 푨 = 풅푨 = 풃 × 풅풓 = 풃풐 +풃풊 − 풃풐

풉× (풓풐 − 풓)

풓풐

풓풊 풅풓

= 풃풐풓풐

풓풊 풅풓+

풃풊 − 풃풐풉

(풓풐 − 풓) 풓풐

풓풊 풅풓

= 풃풐[풓]풓풊풓풐 +

풃풊 −풃풐풉

풓풐[풓]풓풊풓풐 −

풓ퟐ

ퟐ 풓풊

풓풐

= 풃풐[풓풐 − 풓풊] +풃풊 −풃풐

풉 풓풐[풓풐 − 풓풊]−

풓풐ퟐ

ퟐ−풓풊ퟐ

ퟐ

= 풃풐[풉] +풃풊 − 풃풐

풉 풓풐[풉] −

ퟏퟐ

[(풓풐 − 풓풊)(풓풐 + 풓풊)]


풉 풓풐[풉] −

ퟏퟐ

[풉(풓풐 + 풓풊)]


풉 ퟏퟐ

[풉(풓풐 − 풓풊)]


풉 ퟏퟐ

[풉(풉)]

= 풃풐[풉] + (풃풊 − 풃풐) ퟏퟐ풉 =

ퟏퟐ

(풃풊 + 풃풐)풉

푅푎푑푖푢푠 표푓 푛푒푢푡푟푎푙 푎푥푖푠 " 푟 " =퐴

∫푑퐴푟

=> 푑퐴푟

=푏′ × 푑푟푟

=풃풐 + 풃풊 − 풃풐

풉 × (풓풐 − 풓) × 푑푟

푟

=> 푑퐴푟

=풃풐 + 풃풊 − 풃풐

풉 × (풓풐 − 풓)

풓

풓풐

풓풊푑푟 =

풃풐풓

풓풐

풓풊 풅풓 +

풃풊 − 풃풐풉

풓풐풓

풅풓 − 풓풐

풓풊풅풓

풓풐

풓풊

= 풃풐[풍풏풓]풓풊풓풐 +

풃풊 −풃풐풉

[풓풐풍풏풓 − 풓]풓풊풓풐

= 풃풐 풍풏풓풐풓풊

+ 풃풊 −풃풐

풉 풓풐 풍풏

풓풐풓풊

− (풓풐 − 풓풊)

AMS LECTURE-21


= 풃풐 풍풏풓풐풓풊

+ 풃풊 −풃풐

풉 풓풐 풍풏

풓풐풓풊

− 풉

푯풆풏풄풆 푅푎푑푖푢푠 표푓 푛푒푢푡푟푎푙 푎푥푖푠 " 푟 " =퐴

∫푑퐴푟

푟 =ퟏퟐ (풃풊 + 풃풐)풉

풃풐 풍풏 풓풐풓풊

+ 풃풊 − 풃풐풉 풓풐 풍풏 풓풐

풓풊− 풉


푦 푓푟표푚 푡ℎ푒 푏표푡푡표푚 표푓 푡ℎ푒 푠푒푐푡푖표푛 =ℎ3푏 + 2푏푏 + 푏

푅푎푑푖푢푠 표푓 푐푢푟푣푎푡푢푟푒 표푓 푐푒푛푡푟푖푑푎푙 푎푥푖푠 푟 = 푟 + 푦

C. Triangular Section:

On putting the boundary condition in formulation of trapezoidal section, the radius of curvature of

neutral axis may be calculated.

푏 = 푏 & 푏 = 0

AMS LECTURE-21





풓풊− 풉

=ퟏퟐ (풃)풉

풃풉 풓풐 풍풏 풓풐

풓풊− 풉

=풉ퟐ

ퟐ 풓풐 풍풏 풓풐풓풊

− 풉



=ℎ3푏 + 0푏 + 0

=ℎ3

푅푎푑푖푢푠 표푓 푐푢푟푣푎푡푢푟푒 표푓 푐푒푛푡푟푖푑푎푙 푎푥푖푠 "푟 " = 푟 + 푦

AMS LECTURE-22


A. Composite section: (I-section)

Total area of the I-section is “A”

퐴 = 퐴 = 푏 × ℎ + 푏 × ℎ + 푏 × ℎ

푑퐴푟

=푑퐴푟

= 푏 푙푛푟푟

+ 푏 푙푛푟푟

+ 푏 푙푛푟푟

푅푎푑푖푢푠 표푓 푛푒푢푡푟푎푙 푎푥푖푠 " rn" =퐴

∫ 푑퐴푟=

푏 × ℎ + 푏 × ℎ + 푏 × ℎ

푏 푙푛 푟푟 + 푏 푙푛 푟

푟 + 푏 푙푛 푟푟

Where :

푟 = 푟 + ℎ

푟 = 푟 + ℎ = 푟 + ℎ + ℎ + ℎ


푅푎푑푖푢푠 표푓 푐푢푟푣푎푡푢푟푒 표푓 푐푒푛푡푟푖푑푎푙 푎푥푖푠 "푟 " = 푟 + 푦

AMS LECTURE-22


B. Circular section:-

푅푎푑푖푢푠 표푓 푛푒푢푡푟푎푙 푎푥푖푠 " rn" =퐴

∫푑퐴푟=

푟 + 푟4


푅푎푑푖푢푠 표푓 푐푢푟푣푎푡푢푟푒 표푓 푐푒푛푡푟푖푑푎푙 푎푥푖푠 rc = 푟 + 푦 = 푟 + 푅

CRANE HOOK PROBLEM:-

AMS LECTURE-22


Consider a section “PQ” at an angel from horizontal.

Bending moment about the centroid “C” at the section “PQ” is :

푀 = 푃 × 푂 퐷 = 푃 × 푂 퐶 cos 휃 = 푃 × 푟 cos 휃

Normal load acting on the cross- section “PQ” is :

퐹 = 푃 cos 휃

Direct-stress acting at the section is :

휎 =퐹퐴

=푃 cos 휃

퐴

Where “A” is the cross-sectional area of the beam.

Note:- when =0, the bending moment and direct stress are maximum & plane is known as critical plane.

Q.1. a crane hook is of trapezoidal cross-section having inner side 80 mm, outer side 30 mm,

and depth 120 mm. the radius of curvature of the inner side is 80 mm. if the load 100 kN is applied

in following two condition, find the maximum tensile and compressive stress across critical cross-

section.

a). when the load pass through center of curvature.

b). when the load shifted by 10 mm towards inside surface from center of curvature.

c). when the load shifted by 10 mm away from center of curvature.

Solution:-

Given data:-

Point “O” be the center of curvature

푟 = 80푚푚, ℎ = 120푚푚, 푏 = 80푚푚, 푏 = 30푚푚, 푃 = 100푘푁 = 10 × 10 푁

푟 = 푟 + ℎ = 200푚푚

AMS LECTURE-22




= 120

380 + 2 × 30

80 + 30= 50.91푚푚

푅푎푑푖푢푠 표푓 푐푢푟푣푎푡푢푟푒 표푓 푐푒푛푡푟푖푑푎푙 푎푥푖푠 푟 = 푟 + 푦 = 80 + 50.91 = 130.91푚푚

푯풆풏풄풆 푅푎푑푖푢푠 표푓 푛푒푢푡푟푎푙 푎푥푖푠 " 푟 " =퐴

∫ 푑퐴푟




풓풊− 풉

=ퟏퟐ (ퟖퟎ + ퟑퟎ)ퟏퟐퟎ

ퟑퟎ 풍풏 ퟐퟎퟎퟖퟎ + ퟖퟎ − ퟑퟎ

ퟏퟐퟎ ퟐퟎퟎ 풍풏 ퟐퟎퟎퟖퟎ − ퟏퟐퟎ

푟 =ퟔퟔퟎퟎ

ퟐퟕ.ퟒퟖퟖ+ ퟎ.ퟒퟏퟕ [ퟏퟖퟑ.ퟐퟓퟖ− ퟏퟐퟎ] =ퟔퟔퟎퟎ

ퟐퟕ.ퟒퟖퟖ+ ퟎ.ퟒퟏퟕ [ퟔퟑ.ퟐퟓퟖ] =ퟔퟔퟎퟎퟓퟑ.ퟖퟔퟔ

= 122.525푚푚

퐴푟푒푎 A =ퟏퟐ

(풃풊 + 풃풐)풉 = ퟔퟔퟎퟎ 풎풎ퟐ

‘e’ represents the distance between the centroidal axis and neutral axis. i.e. 푒 = 푟 − 푟 = 130.91− 122.525 = 8.385푚푚

푤푒 푘푛표푤 푡ℎ푎푡 => 휎 =푀퐴푒

푦

푟 − 푦

A) For case (a) when load passing through center of curvature.

푀 = 푏푒푛푑푖푛푔 푚표푚푒푛푡 푎푏표푢푡 푐푒푛푡푟표푖푑 표푓 푡ℎ푒 푠푒푐푡푖표푛 = 푃 × 푟 = 13.091 × 10 푁푚푚

By applying sign convention:

AMS LECTURE-22


푀 푖푠 푝표푠푖푡푖푣푒, 푎푠 푡ℎ푒 Applied moment trying to increase the radius of curvature, the outer

surface is under compression and the inner surface is under tension. The bending moment “M” is considered as positive. => 푀 = 13.091 × 10 푁푚푚

“y” considered as positive when it is measure in the direction towards the center of curvature & negative when it is measure in the direction outwards from the center of curvature. For point “A” , fiber at inner surface 푦 = 푖푠 푝표푠푖푡푖푣푒

푦 = 푟 − 푟 = 122.525− 80 = 42.525푚푚

=> 푟 = 푟 − 푦 = 푟 = 80푚푚

푆푡푟푒푠푠 푑푢푒 푡표 푏푒푛푑푖푛푔 푎푡 A = 휎 =푀퐴푒

푦

푟 − 푦=푀퐴푒

푦


푦푟

=13.091 × 106600 × 8.385

42.525

80= 125.742 푁/푚푚 (푡푒푛푠푖푙푒)

For point “B” , fiber at outer surface 푦 = 푖푠 푛푒푔푎푡푖푣푒

푦 = 푟 − 푟 = 200− 122.525 = 77.475푚푚

=> 푟 = 푟 − (−푦 ) = 푟 + 푦 = 푟 = 200푚푚


푦


−푦

푟 − (−푦 )= −

푀퐴푒

푦푟

= −13.091 × 106600 × 8.385

77.4752080

= −91.634 푁/푚푚 (푐표푚푝푟푒푠푠푖푣푒)

Direct stress acting at the section “A-B” :-

Direct stress at section “A-B”

휎 = 휎 = 휎 = = = 15.152 푁/푚푚 (푡푒푛푠푖푙푒)

Resultant stress:-

휎 = 휎 + 휎 = 140.894 푁/푚푚 (푡푒푛푠푖푙푒)

휎 = 휎 + 휎 = −76.482 푁/푚푚 (푐표푚푝푟푒푠푠푖푣푒)

AMS LECTURE-22


B. For case (b) when the load shifted by 10 mm towards inside surface from center of curvature.

푀 = 푏푒푛푑푖푛푔 푚표푚푒푛푡 푎푏표푢푡 푐푒푛푡푟표푖푑 표푓 푡ℎ푒 푠푒푐푡푖표푛 푥 = 푟 − 푂푂 = 130.91− 10 = 120.91푀푀 푀 = 푃 × 푥 = 12.091 × 10 푁푚푚


푦

푟 − 푦

=푀퐴푒

푦


푦푟

=12.091 × 106600 × 8.385

42.525

80= 116.137 푁/푚푚 (푡푒푛푠푖푙푒)


푦

푟 − 푦

=푀퐴푒

−푦

푟 − (−푦 )= −

푀퐴푒

푦푟

= −12.091 × 106600 × 8.385

77.4752080


Direct stress at section “A-B” 휎 = 휎 = 휎 = = = 15.152 푁/푚푚 (푡푒푛푠푖푙푒)

Resultant stress:- 휎 = 휎 + 휎 = 131.289 푁/푚푚 (푡푒푛푠푖푙푒) 휎 = 휎 + 휎 = −69.482 푁/푚푚 (푐표푚푝푟푒푠푠푖푣푒)

C. For case (b) when the load shifted by 10 mm away from center of curvature.

푀 = 푏푒푛푑푖푛푔 푚표푚푒푛푡 푎푏표푢푡 푐푒푛푡푟표푖푑 표푓 푡ℎ푒 푠푒푐푡푖표푛 푥 = 푟 + 푂푂 = 130.91 + 10 = 140.91푀푀 푀 = 푃 × 푥 = 12.091 × 10 푁푚푚


푦

푟 − 푦

=푀퐴푒

푦


푦푟

=14.091 × 106600 × 8.385

42.525

80= 135.347 푁/푚푚 (푡푒푛푠푖푙푒)


푦

푟 − 푦

=푀퐴푒

−푦

푟 − (−푦 )= −

푀퐴푒

푦푟

= −14.091 × 106600 × 8.385

77.4752080


AMS LECTURE-22


Direct stress at section “A-B” 휎 = 휎 = 휎 = = = 15.152 푁/푚푚 (푡푒푛푠푖푙푒)

Resultant stress:- 휎 = 휎 + 휎 = 150.499 푁/푚푚 (푡푒푛푠푖푙푒) 휎 = 휎 + 휎 = −83.482 푁/푚푚 (푐표푚푝푟푒푠푠푖푣푒)

AMS LECTURE-23


WORK SHEET

Q.2. Plot the stress distribution about section A-B of the hook as shown in figure.

Given data: ri = 50mm, ro = 150mm, P = 22X103N, b = 20mm, h = 150-50 = 100mm A = bh = 20X100 = 2000mm2

Determination of C.G. of the section:- 퐶.퐺 표푓 푡ℎ푒 푠푒푐푡푖표푛 푠푖푡푢푎푡푒푑 푎푡 푎 푑푖푠푡푎푛푐푒


=100

2= 50푚푚

푅푎푑푖푢푠 표푓 푐푢푟푣푎푡푢푟푒 표푓 푐푒푛푡푟푖푑푎푙 푎푥푖푠 푟

= 푟 +ℎ2

= 50 + 50 = 100푚푚

rn =퐴

∫ 푑퐴푟=

푏 × ℎ

푏 푙푛 푟푟

=ℎ

푙푛 푟푟

=100

푙푛 15050

= 91.024푚푚

푒 = 푟 − 푟 = 100− 91.024 = 8.976푚푚 푀 = 푏푒푛푑푖푛푔 푚표푚푒푛푡 푎푏표푢푡 푐푒푛푡푟표푖푑 표푓 푡ℎ푒 푠푒푐푡푖표푛 AB = 푃 × 푟 = 22 × 10 × 100 = 2.2 × 10 푁푚푚 푀 푖푠 푝표푠푖푡푖푣푒,푎푠 푡ℎ푒 Applied moment trying to increase the radius of curvature, the outer surface is under compression and the inner surface is under tension. The bending moment “M” is considered as positive.

Direct stress at section “A-B”

AMS LECTURE-23


휎 = 휎 = 휎 = = = 11 푁/푚푚 (푡푒푛푠푖푙푒)

For point “A” , fiber at inner surface 푦 = 푖푠 푝표푠푖푡푖푣푒 푦 = 푟 − 푟 = 91.024− 50 = 41.024푚푚

=> 푟 = 푟 − 푦 = 푟 = 50푚푚


푦


푦


푦푟

=2.2 × 10

2000 × 8.976

41.02450

= 100.55 푁/푚푚 (푡푒푛푠푖푙푒) For point “B” , fiber at outer surface 푦 = 푖푠 푛푒푔푎푡푖푣푒

푦 = 푟 − 푟 = 150− 91.024 = 58.976푚푚 푎푠 푦 = 푖푠 푛푒푔푎푡푖푣푒, 푦 = −58.976푚푚

=> 푟 = 푟 − (−푦 ) = 푟 + 푦 = 푟 = 150푚푚


푦


−푦

푟 − (−푦 )= −

푀퐴푒

푦푟

= −2.2 × 10

2000 × 8.976

58.976150

= −48.183 푁/푚푚 (푐표푚푝푟푒푠푠푖푣푒) 푅푒푠푢푙푡푎푛푡 푠푡푟푒푠푠푒푠 푎푟푒 ∶ 휎 = 휎 + 휎 = 111.55 푁/푚푚 (푡푒푛푠푖푙푒) 휎 = 휎 + 휎 = −37.183 푁/푚푚 (푐표푚푝푟푒푠푠푖푣푒) Bending stress at any radius “r” is 휎 = where y=푟 − 푟 Bending stress at neutral axis i.e at “y” =0, 휎 = 표 But Resultant stress at neutral axis=휎 = 휎 + 휎 =11 푁/푚푚 (푡푒푛푠푖푙푒) Bending stress at centroidal axis i.e at “y” = -e = -8.976mm, 휎 = = − (푐표푚푝푟푒푠푠푖푣푒) But Resultant stress at centroidal axis =휎 = 휎 + 휎 = 0 Stress distribution across the cross-section “AB”

AMS LECTURE-23


Problem on “U” fork

Consider a section “PQ” at an angel from horizontal.


푀 = 푃 × (푀퐷 + 푂 퐾) = 푃 × (푀퐷 + 푂 퐶 cos 휃 ) = 푃 × (퐿 + 푟 cos 휃)

Normal load acting on the cross- section “PQ” is :

퐹 = 푃 cos 휃


휎 =퐹퐴

=푃 cos 휃

퐴

Where “A” is the cross-sectional area of the beam.

The “U” section consists of two portions:-

A. Straight portion B. Curve portion

Straight portion

For calculating stress in straight portion (MD) general pure bending formula can be applied:

푩풆풏풅풊풏품 풔풕풓풆풔풔 흈풃(풔풕풓풂풊품풉풕) =푴풔풕풓풂풊품풉풕

푰× 풚

푾풉풆풓 푴풔풕풓풂풊품풉풕 = 푷 × 풙

AMS LECTURE-23


풙 = 풅풊풔풕풂풏풄풆 풐풇 풕풉풆 풔풆풄풕풊풐풏 풇풓풐풎 풍풐풂풅 푷

푾풉풆풓풆 풎풂풙풊풎풖풎 풃풆풏풅풊풏품 풎풐풎풆풏풕 푴풔풕풓풂풊품풉풕(풎풂풙) = 푷 × 푳 & 푖푡 푖푠 표푐푐푢푟푠 푎푡 푡ℎ푒 푠푒푐푡푖표푛 퐶푫

푶풏 풑풖풕풕풊풏품 풕풉풆 풗풂풍풖풆 휽 = ퟗퟎ 푴푪푫 = 푷 × (푳 + 풓풄 퐜퐨퐬ퟗퟎ) = 푷 × 푳

푫풊풓풆풄풕 풔풕풓풆풔풔 흈풅(풔풕풓풂풊품풉풕) =퐹퐴

=푃 cos 90

퐴= 0

푯풆풏풄풆 풓풆풔풖풍풕풂풏풕 풔풕풓풆풔풔 흈(풔풕풓풂풊품풉풕) = 흈풃(풔풕풓풂풊품풉풕) ± 흈풅(풔풕풓풂풊품풉풕) = 흈풃(풔풕풓풂풊품풉풕)

Curve portion

For calculating stress in curved portion Winkler-batch theory can be applied:


푴풃 = 푷 × (푴푫 +푶ퟏ푲) = 푷 × (푴푫 + 푶ퟏ푪 퐜퐨퐬휽 ) = 푷× (푳 + 풓풄 퐜퐨퐬휽)

흈풃(풄풖풓풗풆풅) =푴풃

푨풆

풚풓풏 − 풚


흈풅(풄풖풓풗풆풅) =푭푨

=푷 퐜퐨퐬 휽

푨

푯풆풏풄풆 풓풆풔풖풍풕풂풏풕 풔풕풓풆풔풔 흈(풄풖풓풗풆풅) = 흈풃(풄풖풓풗풆풅) ± 흈풅(풄풖풓풗풆풅)

Q.3. 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.

AMS LECTURE-23


Given data:-

L=750mm,

ri = 250mm, ro = 550mm, P = 85X103N 푏 = 300푚푚, 푏 = 75푚푚, ℎ = 75푚푚,

ℎ = 225푚푚, 푟 = 250푚푚, 푟 = 250 + 75 = 325푚푚, 푟 = 325 + 225 = 550푚푚

푨 = (ퟑퟎퟎ× ퟕퟓ) + (ퟐퟐퟓ× ퟕퟓ) = ퟑퟗퟑퟕퟓ 풎풎ퟐ

풚 =(ퟑퟎퟎ × ퟕퟓ) × ퟕퟓ

ퟐ + (ퟐퟐퟓ × ퟕퟓ) × ퟕퟓ + ퟐퟐퟓퟐ

(ퟑퟎퟎ× ퟕퟓ) + (ퟐퟐퟓ × ퟕퟓ) = ퟏퟎퟏ.ퟕퟖퟓ풎풎

rc = 퐫ퟏ + 풚 = ퟐퟓퟎ + ퟏퟎퟏ.ퟕퟖퟓ = ퟑퟓퟏ.ퟕퟖퟓ풎풎

푹풂풅풊풖풔 풐풇 풏풆풖풕풓풂풍 풂풙풊풔 " rn" =푨

∫풅푨풓=

풃ퟏ × 풉ퟏ + 풃ퟐ × 풉ퟐ풃ퟏ풍풏

풓ퟐ풓ퟏ

+ 풃ퟐ풍풏풓ퟑ풓ퟐ

rn =(ퟑퟎퟎ × ퟕퟓ) + (ퟐퟐퟓ× ퟕퟓ)

ퟑퟎퟎ풍풏 ퟑퟐퟓퟐퟓퟎ + ퟕퟓ풍풏 ퟓퟓퟎ

ퟑퟐퟓ= ퟑퟑퟑ.ퟐퟏퟕ풎풎

Distance of neutral axis to centroidal axis “e” = rc- rn = 18.568mm Bending moment about the centroid “C” at the section is : 푴풃 = 푷 × (푳 + 풓풄 퐜퐨퐬휽)풘풉풆풏 휽 = ퟎ,푻풉풆 풃풆풏풅풊풏품 풎풐풎풆풏풕 풊풔 풎풂풙풊풎풖풎 & 푡ℎ푒 푠푒푐푡푖표푛 푖푠 푎푡 ℎ표푟푖푧표푛푡푎푙 푖푛 푡ℎ푒 푐푢푟푣푒푑 푝표푟푡푖표푛 푴풃 = 푷 × (푳 + 풓풄 퐜퐨퐬ퟎ) = 85 × 10 × 1101.785 = 93.652 × 10 Nmm 푀 푖푠 푝표푠푖푡푖푣푒,푎푠 푡ℎ푒 Applied moment trying to increase the radius of curvature, the outer surface is under compression and the inner surface is under tension. The bending moment “M” is considered as positive. Direct-stress acting at the section is :

흈풅 =푭푨

=푷 퐜퐨퐬휽

푨=

85 × 10 퐜퐨퐬 ퟎ ퟑퟗퟑퟕퟓ

= ퟐ.ퟏퟔ 푵 풎풎ퟐ (풕풆풏풔풊풍풆) Bending stress at “A”

AMS LECTURE-23



=> 푟 = 푟 − 푦 = 푟 = 250푚푚


푦


푦


푦푟

=93.652 × 10

ퟑퟗퟑퟕퟓ× ퟏퟖ.ퟓퟔퟖ

83.217250

= 42.64 푁/푚푚 (푡푒푛푠푖푙푒) For point “B” , fiber at outer surface 푦 = 푖푠 푛푒푔푎푡푖푣푒

푦 = 푟 − 푟 = 550 − 333.217 = 216.783푚푚 푎푠 푦 = 푖푠 푛푒푔푎푡푖푣푒, 푦 = −216.783푚푚

=> 푟 = 푟 − (−푦 ) = 푟 + 푦 = 푟 = 550푚푚


푦


−푦

푟 − (−푦 )= −

푀퐴푒

푦푟

= −93.652 × 10

ퟑퟗퟑퟕퟓ× ퟏퟖ.ퟓퟔퟖ

216.783550


푅푒푠푢푙푡푎푛푡 푠푡푟푒푠푠푒푠 푎푟푒 ∶ 휎 = 휎 + 휎 = 44.8 푁/푚푚 (푡푒푛푠푖푙푒) 휎 = 휎 + 휎 = −48.33 푁/푚푚 (푐표푚푝푟푒푠푠푖푣푒)

푀푎푥푖푚푢푚 푠ℎ푒푎푟 푠푡푟푒푠푠 휏 = 0.5 휎 = −24.165 푁/푚푚

The below figure shows the stress distribution.

AMS LECTURE-24


Q.4. 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.

Given data:

휎 = 75

푁푚푚

(푡푒푛푠푖표푛) Point “O” be the center of curvature 푟 = 125푚푚, ℎ = 60푚푚, b= 30푚푚, 푃 =?

푟 = 푟 + ℎ = 185푚푚 Determination of C.G. of the section:- 퐶.퐺 표푓 푡ℎ푒 푠푒푐푡푖표푛 푠푖푡푢푎푡푒푑 푎푡 푎 푑푖푠푡푎푛푐푒 푦 푓푟표푚 푡ℎ푒 푏표푡푡표푚


=602

= 30푚푚

푅푎푑푖푢푠 표푓 푐푢푟푣푎푡푢푟푒 표푓 푐푒푛푡푟푖푑푎푙 푎푥푖푠 푟 = 푟 +ℎ2

= 125 + 30 = 155푚푚

rn =퐴

∫ 푑퐴푟=

푏 × ℎ

푏 푙푛 푟푟

=ℎ

푙푛 푟푟

=60

푙푛 185125

= 153.045푚푚

푒 = 푟 − 푟 = 155 − 153.045 = 1.955푚푚 퐴 = 30 × 60 = 1800 푚푚

푀= 푏푒푛푑푖푛푔 푚표푚푒푛푡 푎푏표푢푡 푐푒푛푡푟표푖푑 표푓 푡ℎ푒 푠푒푐푡푖표푛 AB= 푃 × 푟 = 푃 × 155 = 155푃 푁푚푚 푀 푖푠 푝표푠푖푡푖푣푒,푎푠 푡ℎ푒 Applied moment trying to increase the radius of curvature, the outer surface is under compression and the inner surface is under tension. The bending moment “M” is considered as positive.

Direct stress at section “A-B” 휎 = 휎 = 휎 = = = 0.555 × 10 푃 푁/푚푚 (푡푒푛푠푖푙푒)


=> 푟 = 푟 − 푦 = 푟 = 125푚푚

AMS LECTURE-24



푦


푦


푦푟

=155푃

1800 × 1.955

28.045125

= 9.882 × 10 푃 푁/푚푚 (푡푒푛푠푖푙푒) For point “B” , fiber at outer surface 푦 = 푖푠 푛푒푔푎푡푖푣푒

푦 = 푟 − 푟 = 185− 153.045 = 31.955푚푚 푎푠 푦 = 푖푠 푛푒푔푎푡푖푣푒, 푦 = −31.955푚푚

=> 푟 = 푟 − (−푦 ) = 푟 + 푦 = 푟 = 185푚푚


푦


−푦

푟 − (−푦 )= −

푀퐴푒

푦푟

= −155푃

1800 × 1.955

31.955185

= −7.608 × 10 푃 푁/푚푚 (푐표푚푝푟푒푠푠푖푣푒) 푅푒푠푢푙푡푎푛푡 푠푡푟푒푠푠푒푠 푎푟푒 ∶ 휎 = 휎 + 휎 = 10.437 × 10 푃 푁/푚푚 (푡푒푛푠푖푙푒) 휎 = 휎 + 휎 = −7.053 × 10 푃 푁/푚푚 (푐표푚푝푟푒푠푠푖푣푒) Maximum tensile stress is :

휎 = 75푁

푚푚 => 75 = 10.437 × 10 푃

P=7185.972 N

Q.5. The figure shows a loaded offset bar. What is the maximum off-set distance ’x’ if the

allowable stress in tension is limited to 50N/mm2.

Solution:-

AMS LECTURE-24


Given data:- 풓풄 = ퟏퟎퟎ풎풎

풓풊 = 풓풄 − 푹 = ퟓퟎ풎풎, 푹 = ퟓퟎ풎풎, 푷 = ퟓ × ퟏퟎퟑ푵

풓풐 = 풓풄 +푹 = ퟏퟓퟎ풎풎 Maximum tensile stress is :

흈풎풂풙 = ퟓퟎ푵

풎풎ퟐ

Determination of C.G. of the section:- 풚 = 푹 = ퟓퟎ풎풎


∫풅푨풓=

풓풐 + 풓풊ퟐ

ퟒ

= √ퟏퟓퟎ + √ퟓퟎ

ퟐ

ퟒ= ퟗퟑ.ퟑퟎퟏ풎풎

푨 =흅ퟒ푫ퟐ = 흅푹ퟐ = ퟕퟖퟓퟑ.ퟗퟖퟐ풎풎ퟐ

풆 = 풓풄 − 풓풏 = ퟏퟎퟎ − ퟗퟑ.ퟑퟎퟏ = ퟔ.ퟔퟗퟗ풎풎 푴풃

= 풃풆풏풅풊풏품 풎풐풎풆풏풕 풂풃풐풖풕 풄풆풏풕풓풐풊풅 풐풇 풕풉풆 풔풆풄풕풊풐풏 AB

= 푷 × 풙 = ퟓ× ퟏퟎퟑ × 풙 = ퟓ × ퟏퟎퟑ풙 푵풎풎 푴풃풊풔 풑풐풔풊풕풊풗풆,풂풔 풕풉풆 Applied moment trying to

increase the radius of curvature, the outer surface is under compression and the inner surface is under tension. The bending moment “Mb” is considered as positive. Direct stress at section “A-B”

흈풅푨 = 흈풅푩 = 흈풅 =푷푨

=ퟓ × ퟏퟎퟑ

ퟕퟖퟓퟑ.ퟗퟖퟐ

= ퟎ.ퟔퟑퟕ푵/풎풎ퟐ(풄풐풎풑풓풆풔풔풊풗풆)

For point “A” , fiber at inner surface 풚푨 = 풊풔 풑풐풔풊풕풊풗풆

풚푨 = 풓풏 − 풓풊 = ퟗퟑ.ퟑퟎퟏ− ퟓퟎ = ퟒퟑ.ퟑퟎퟏ풎풎 => 풓푨 = 풓풏 − 풚푨 = 풓풊 = ퟓퟎ풎풎

푺풕풓풆풔풔 풅풖풆 풕풐 풃풆풏풅풊풏품 풂풕 A = 흈풃푨 =푴풃

푨풆

풚풓풏 − 풚

=푴풃

푨풆

풚푨풓풏 − 풚푨

=푴풃

푨풆 풚푨풓푨

=ퟓ × ퟏퟎퟑ풙

ퟕퟖퟓퟑ.ퟗퟖퟐ × ퟔ.ퟔퟗퟗ ퟒퟑ.ퟑퟎퟏퟓퟎ

= ퟎ.ퟎퟖퟐퟑ풙 푵/풎풎ퟐ(풕풆풏풔풊풍풆)

AMS LECTURE-24


For point “B” , fiber at outer surface 풚푩 = 풊풔 풏풆품풂풕풊풗풆

풚푩 = 풓푶 − 풓푵 = ퟏퟓퟎ − ퟗퟑ.ퟑퟎퟏ = ퟓퟔ.ퟔퟗퟗ풎풎 풂풔 풚푩 = 풊풔 풏풆품풂풕풊풗풆, 풚푩 = −ퟓퟔ.ퟔퟗퟗ풎풎

=> 풓푩 = 풓풏 − (−풚푩) = 풓풏 + 풚푩 = 풓푶 = ퟏퟓퟎ풎풎


푨풆

풚풓풏 − 풚

=푴풃

푨풆

−풚푩풓풏 − (−풚푩)

= −푴풃

푨풆 풚푩풓푩

= −ퟓ × ퟏퟎퟑ풙

ퟕퟖퟓퟑ.ퟗퟖퟐ× ퟔ.ퟔퟗퟗ ퟓퟔ.ퟔퟗퟗퟏퟓퟎ

= −ퟎ.ퟎퟑퟔ풙 푵/풎풎ퟐ(풄풐풎풑풓풆풔풔풊풗풆)

푹풆풔풖풍풕풂풏풕 풔풕풓풆풔풔풆풔 풂풓풆 ∶

흈푨 = 흈풃푨 −흈풅푨 = (ퟎ.ퟎퟖퟐퟑ풙 − ퟎ.ퟔퟑퟕ)푵/풎풎ퟐ(풕풆풏풔풊풍풆)

흈푩 = 흈풃푩 + 흈풅푩 = −(ퟎ.ퟎퟑퟔ풙 + ퟎ.ퟔퟑퟕ) 푵/풎풎ퟐ(풄풐풎풑풓풆풔풔풊풗풆) Maximum tensile stress is :

흈풎풂풙 = ퟓퟎ푵

풎풎ퟐ => 50 = ퟎ.ퟎퟖퟐퟑ풙− ퟎ.ퟔퟑퟕ

x=615.273 mm

Q.6. Determine the stresses at point A and B of the split ring shown in figure.

Solution:-

AMS LECTURE-24


Given data:- 풓풄 = ퟖퟎ풎풎

풓풊 = 풓풄 − 푹 = ퟓퟎ풎풎, 푹 = ퟑퟎ풎풎, 푷 = ퟐퟎ × ퟏퟎퟑ푵

풓풐 = 풓풄 + 푹 = ퟏퟏퟎ풎풎 Determination of C.G. of the section:- 풚 = 푹 = ퟑퟎ풎풎


∫풅푨풓

= 풓풐 + 풓풊

ퟐ

ퟒ

= √ퟏퟏퟎ + √ퟓퟎ

ퟐ

ퟒ= ퟕퟕ.ퟎퟖퟏ풎풎

푨 =흅ퟒ푫ퟐ = 흅푹ퟐ = ퟐퟖퟐퟕ.ퟒퟑퟑ풎풎ퟐ

풆 = 풓풄 − 풓풏 = ퟖퟎ − ퟕퟕ.ퟎퟖퟏ = ퟐ.ퟗퟏퟗ풎풎

푴풃

= 풃풆풏풅풊풏품 풎풐풎풆풏풕 풂풃풐풖풕 풄풆풏풕풓풐풊풅 풐풇 풕풉풆 풔풆풄풕풊풐풏

= 푷 × 풓풄 = ퟐퟎ × ퟏퟎퟑ × ퟖퟎ = ퟏퟔ × ퟏퟎퟓ 푵풎풎 푴풃풊풔 풏풆품풂풕풊풗풆,풂풔 풕풉풆 Applied moment trying to

decrease the radius of curvature, the outer surface is under tension and the inner surface is under compression. The bending moment “Mb” is considered as negative. Direct stress at section “A-B”

흈풅푨 = 흈풅푩 = 흈풅 =푷푨

=ퟐퟎ × ퟏퟎퟑ

ퟐퟖퟐퟕ.ퟒퟑퟑ

= ퟕ.ퟎퟕퟒ 푵/풎풎ퟐ(풄풐풎풑풓풆풔풔풊풗풆)

For point “B” , fiber at inner surface 풚푩 = 풊풔 풑풐풔풊풕풊풗풆

풚푩 = 풓풏 − 풓풊 = ퟕퟕ.ퟎퟖퟏ− ퟓퟎ = ퟐퟕ.ퟎퟖퟏ풎풎

=> 풓푩 = 풓풏 − 풚푨 = 풓풊 = ퟓퟎ풎풎

푺풕풓풆풔풔 풅풖풆 풕풐 풃풆풏풅풊풏품 풂풕 B = 흈풃푩 =푴풃

푨풆

풚풓풏 − 풚

=푴풃

푨풆

풚푩풓풏 − 풚푩

=푴풃

푨풆 풚푩풓푩

=−ퟏퟔ× ퟏퟎퟓ

ퟐퟖퟐퟕ.ퟒퟑퟑ × ퟐ.ퟗퟏퟗ ퟐퟕ.ퟎퟖퟏퟓퟎ

= −ퟏퟎퟓ 푵/풎풎ퟐ(풄풐풎풑풓풆풔풔풊풗풆)

For point “A” , fiber at outer surface 풚푨 = 풊풔 풏풆품풂풕풊풗풆

풚푨 = 풓푶 − 풓푵 = ퟏퟏퟎ− ퟕퟕ.ퟎퟖퟏ = ퟑퟐ.ퟗퟏퟗ 풎풎

AMS LECTURE-24


풂풔 풚푨 = 풊풔 풏풆품풂풕풊풗풆, 풚푨 = −ퟑퟐ.ퟗퟏퟗ풎풎

=> 풓푨 = 풓풏 − (−풚푨) = 풓풏 + 풚푩 = 풓푶 = ퟏퟏퟎ풎풎


푨풆

풚풓풏 − 풚

=푴풃

푨풆

−풚푨풓풏 − (−풚푨)

= −푴풃

푨풆 풚푨풓푨

= −−ퟏퟔ× ퟏퟎퟓ

ퟐퟖퟐퟕ.ퟒퟑퟑ × ퟐ.ퟗퟏퟗ ퟑퟐ.ퟗퟏퟗퟏퟏퟎ

= ퟓퟖ.ퟎퟏퟔ 푵/풎풎ퟐ(푻푬푵푺푰푳푬)

푹풆풔풖풍풕풂풏풕 풔풕풓풆풔풔풆풔 풂풓풆 ∶

흈푨 = 흈풃푨 −흈풅푨 = ퟓퟎ.ퟗퟒퟐ 푵/풎풎ퟐ(풕풆풏풔풊풍풆)

흈푩 = 흈풃푩 + 흈풅푩 = −ퟏퟏퟐ.ퟎퟕퟒ 푵/풎풎ퟐ(풄풐풎풑풓풆풔풔풊풗풆)

AMS LECTURE-25


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 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 i.e H=0. this argument can be proved by understanding that since the ring and the external forces are symmetrical, the reactions too must be symmetrical. 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 to maintain equilibrium. This however, results in violation of symmetry and hence H must be zero. Besides the forces, moments of equal magnitude M0 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 quantity is M0. Again Considering symmetry, we conclude that the tangents at A and B must be vertical and must remain so after deflection or M0 does not rotate. By Castigliano’s theorem, the partial derivative of the strain energy with respect to the load gives the displacement of the load. In this would be zero.

휕푈휕푀

= 0 − −− − −− − − −− − −(1)

The bending moment at any point C, located at angle “”, as shown in figure. Will be

AMS LECTURE-25


푀 = 푀 −퐹2

(푅 − 푅 cos 휃)

푀 = 푀 −퐹푅2

(1− cos 휃)−− − − − (2) As per Castigliano’s theorem

휕푈휕푀

= 0

=> 휕푈휕푀

=푀퐸퐼

휕푀휕푀

푑푠 = 0

휕푀휕푀

=휕[푀 − 퐹푅

2 (1− cos 휃)]휕푀

= 1

& 푑푠 = 푅푑휃

=> 휕푈휕푀

=푀퐸퐼

휕푀휕푀

푑푠 = 4푀 −퐹푅

2 (1 − cos 휃)

퐸퐼 × 1 × 푅푑휃

=4푅퐸퐼

푀 −퐹푅2

(1− cos 휃) 푑휃 = 0

=> 푀 −퐹푅2

+퐹푅2

cos 휃 푑휃 = 0

=> 푀 휃 −퐹푅2휃 +

퐹푅2

sin 휃 = 0

=> 푀휋2−퐹푅2휋2

+퐹푅2

= 0

=> 푀 =퐹푅휋

휋2− 1 − − −− −−− −(3)

As this quantity (M0) is positive the direction assumed for M0 is correct and it produces tension in the inner fibers and compression on the outer. It should be noted that these equations are valid in the region, = 0 to = 900. The bending moment Mb at any angle from equation (2) will be:-

푀 =퐹푅휋

휋2− 1 −

퐹푅2

(1− cos 휃) =퐹푅2

cos 휃 −2휋

− − −− − −− (4) Bending moment will be Zero when,

cos 휃 =2휋

=> 휃 = 50.46 − −− −−− −− − (5) At load point, i.e. at = 휋

2 , the bending moment is maximum :

푀 =퐹푅2

−2휋

= −퐹푅휋

− − −− −−− −− − − −− (6) It is seen that numerically, Mb-max is greater than M0. The stress at any angle can be found out by considering the forces as shown in the figure.

AMS LECTURE-25


The vertical force F/2 can be resolved into two components (creates normal direct stresses) (creates shear stresses).

푁 =퐹2

cos 휃

푆 =퐹2

sin 휃

The combined normal stress across any section will be: 휎 = ±( ± )

+ cos 휃

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.

AMS LECTURE-25


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

M0.

Again considering symmetry, we conclude that the tangents at C and D must be vertical and must

remain so after deflection or M0 does not rotate.

There are two regions to be considered in this case:

The straight portion, (0 < y < L) where Mb =M0

The curved portion, where bending moment about point “C” as shown in figure is:

AMS LECTURE-25


푴풃 = 푴ퟎ −푭ퟐ

(푹−푹 퐜퐨퐬휽)

As per Castigliano’s theorem 흏푼흏푴ퟎ

= ퟎ => 흏푼풔풕풓풂풊품풉풕

흏푴ퟎ+

흏푼풄풖풓풗풆풅

흏푴ퟎ= ퟎ

흏푼흏푴ퟎ

= ퟒ 푴ퟎ

푬푰

푳

ퟎ 흏푴ퟎ

흏푴ퟎ풅풔 + ퟒ

푴풃

푬푰

흅ퟐ

ퟎ 흏푴풃

흏푴ퟎ 풅풔 = ퟎ

=> ퟒ 푴ퟎ

푬푰

푳

ퟎ 풅풔 + ퟒ

푴풃

푬푰

흅ퟐ

ퟎ (ퟏ) 풅풔 = ퟎ

=> ퟒ푴ퟎ푳푬푰

+ ퟒ푹푬푰

푴ퟎ −푭ퟐ

(푹 −푹 퐜퐨퐬휽)흅ퟐ

ퟎ 풅휽 = ퟎ 풂풔 풅풔 = 푹풅휽

=> 푴ퟎ푳푬푰

+ 푴ퟎ흅푹ퟐ푬푰

−푭흅푹ퟐ

ퟒ푬푰+ 푭푹ퟐ

ퟐ푬푰= ퟎ

=> 푴ퟎ푳 + 푴ퟎ흅푹ퟐ

−푭흅푹ퟐ

ퟒ+ 푭푹ퟐ

ퟐ= ퟎ

=> 푴ퟎ ퟐ푳 + 흅푹

ퟐ=

푭푹ퟐ

ퟐ 흅− ퟐퟐ

It can be observed that at L = 0 equation reduces to the same expression as obtained for a circular

ring. i.e

=> 푀 =퐹푅휋

휋2− 1

The M0 produces tension in inner fiber and Compression on the outer.

The bending moment Mb at any angle will be:

푀 = 푀 −퐹2

(푅 − 푅 cos 휃) = 퐹푅

2

휋 − 22퐿 + 휋푅

− 퐹푅2

(1 − cos 휃)

Noting that the above equation is valid in the region, = 0 to = π/2

AMS LECTURE-25


At = 0, the bending moment at section BB

푴풃 = 푴ퟎ =푭푹ퟐ

ퟐ

흅 − ퟐퟐ푳+ 흅푹

At = π/2, the bending moment at section

AA

푴풃 =푭푹ퟐ

ퟐ

흅 − ퟐퟐ푳 +흅푹

− 푭푹ퟐ

The stress at any angle can be found by considering the force as shown in the above figure. The

vertical force F/2 can be resolved in two components (creates normal direct stresses) and S

(creates shear stresses).

Normal force “N”

푁 =퐹2

cos 휃

& shear stress “S”

푆 =퐹2

sin 휃

The combined normal stress across any section will be

흈 = ±푴풃풚

푨풆(풓풏 ± 풚) + 푭ퟐ푨

퐜퐨퐬 휽

AMS LECTURE-26


Q.7. 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:

Given data:- 푫풊 = ퟔퟎ풎풎, 풅 = ퟐퟓ풎풎, 풓 = ퟏퟐ.ퟓ풎풎 푭 = ퟔퟓퟎퟎ푵,

풓풊 = ퟑퟎ풎풎, 풓풐 = 풓풊 + 풅 = ퟓퟓ풎풎 Determination of C.G. of the section:- 풚 = 풓 = ퟏퟐ.ퟓ풎풎

풓풄 = 푹 = 풓풊 + 풓 = ퟒퟐ.ퟓ풎풎


∫풅푨풓=

풓풐 + 풓풊ퟐ

ퟒ

= √ퟓퟓ + √ퟑퟎ

ퟐ

ퟒ= ퟒퟏ.ퟓퟔ풎풎

푨 =흅ퟒ풅ퟐ = 흅풓ퟐ = ퟒퟗퟎ.ퟖퟕퟒ 풎풎ퟐ

풆 = 풓풄 − 풓풏 = ퟒퟐ.ퟓ − ퟒퟏ.ퟓퟔ = ퟎ.ퟗퟒ 풎풎

The bending moment Mb at any angle

푀 =퐹푅휋

휋2− 1 −

퐹푅2

(1 − cos 휃) =퐹푅2

cos 휃 −2휋

At section B-B, = 0

푀 = 푀 =퐹푅휋

휋2− 1 −

퐹푅2

(1− cos 0)

푀 =퐹푅2

cos0−2휋

=퐹푅2

1−2휋

=6500 × 42.5

21−

2휋

= 50277.5푁푚푚 At inner fiber 푟 = 푟 = 30푚푚, 푦 = 푟 − 푟 = 11.56푚푚(+푣푒)

휎 =푀 × 푦퐴푒(푟 − 푦 )

=푀 × 푦

퐴푒 푟

=50277.5 × 11.56

ퟒퟗퟎ.ퟖퟕퟒ× ퟎ.ퟗퟒ× ퟑퟎ= 41.987

푁푚푚

(푇푒푛푠푖푙푒) At outer fiber 푟 = 푟 = 55푚푚, 푦 = 푟 − 푟 = 13.44푚푚(−푣푒)

AMS LECTURE-26


휎 =푀 × 푦

퐴푒(푟 − (−푦 ))=푀 × 푦

퐴푒 푟

=50277.5 × (−13.44)ퟒퟗퟎ.ퟖퟕퟒ × ퟎ.ퟗퟒ × ퟓퟓ

= −26.626푁

푚푚 (푐표푚푝푟푒푠푠푖푣푒)

Direct stress at section B-B: At section B-B, = 0

휎 =퐹

2퐴cos 휃 =

65002 × ퟒퟗퟎ.ퟖퟕퟒ

cos 0= 6.621 푁/푚푚 (푇푒푛푠푖푙푒)

Resultant stress at section B-B

휎 = 휎 + 휎 = 48.608 푁/푚푚 (푇푒푛푠푖푙푒) 휎 = 휎 + 휎 = −20.005 (푐표푚푝푟푒푠푠푖푣푒) At section A-A, = π/2

푀 = 푀 =퐹푅휋

휋2− 1 −

퐹푅2

(1 − cos 90) =퐹푅2

cos 90−2휋

=6500 × 42.5

2cos 90 −

2휋

= −87933.106 푁푚푚 At inner fiber

푟 = 푟 = 30푚푚, 푦 = 푟 − 푟 = 11.56푚푚(+푣푒)

휎 =푀 × 푦퐴푒(푟 − 푦 )

=푀 × 푦

퐴푒 푟=

−87933.106 × 11.56ퟒퟗퟎ.ퟖퟕퟒ× ퟎ.ퟗퟒ× ퟑퟎ

= −73.433푁


At outer fiber

푟 = 푟 = 55푚푚, 푦 = 푟 − 푟 = 13.44푚푚(−푣푒)

휎 =푀 × 푦

퐴푒(푟 − (−푦 ))=푀 × 푦

퐴푒 푟=−87933.106 × (−13.44)ퟒퟗퟎ.ퟖퟕퟒ× ퟎ.ퟗퟒ × ퟓퟓ

= 46.568푁

푚푚 (푇퐸푁푆퐼퐿퐸)

Direct stress at section A-A: At section A-A, = π/2

휎 =퐹

2퐴cos 휃 =

65002 × ퟒퟗퟎ.ퟖퟕퟒ

cos흅/ퟐ = 0 Resultant stress at section A-A

휎 = 휎 + 휎 == −73.433푁


휎 = 휎 + 휎 = 46.568푁


AMS LECTURE-26


Q.8. 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. If the link carries a load of 90kN;

estimate the tensile and compressive stress along the section of load line. Also find the stress at a

section 900 from the load line.

Solution:

Solution:- Given data:- 푫풄 = ퟖퟎ풎풎, 풓풄 = ퟒퟎ풎풎 풅 = ퟒퟎ풎풎, 풓 = ퟐퟎ풎풎 푭 = ퟗퟎퟎퟎퟎ푵, 풓풊 = ퟐퟎ풎풎, 풓풐 = 풓풊 + 풅 = ퟔퟎ풎풎

ퟐ푳 = ퟖퟎ풎풎 Determination of C.G. of the section:- 풚 = 풓 = ퟐퟎ풎풎

풓풄 = 푹 = ퟒퟎ풎풎


∫풅푨풓=

풓풐 + 풓풊ퟐ

ퟒ

= √ퟔퟎ + √ퟐퟎ

ퟐ

ퟒ= ퟑퟕ.ퟑퟐ풎풎

푨 =흅ퟒ풅ퟐ = 흅풓ퟐ = ퟏퟐퟓퟔ.ퟔퟑퟕ 풎풎ퟐ

풆 = 풓풄 − 풓풏 = ퟒퟎ − ퟑퟕ.ퟑퟐ = ퟐ.ퟔퟖ 풎풎

The bending moment Mb at any angle will be:

푀 = 푀 −퐹2

(푅 − 푅 cos 휃)

= 퐹푅

2

휋 − 22퐿+ 휋푅

− 퐹푅2

(1− cos 휃)

At section B-B, = 0

AMS LECTURE-26


푀 = 푀 =퐹푅

2

휋 − 22퐿+ 휋푅

−퐹푅2

(1 − cos 0)

푀 =퐹푅

2

휋 − 22퐿+ 휋푅

=90000 × 40

2휋 − 2

2 × 40 + 휋40= 399655.7 푁푚푚

At inner fiber 푟 = 푟 = 20푚푚, 푦 = 푟 − 푟 = 17.32 푚푚(+푣푒)

휎 =푀 × 푦퐴푒(푟 − 푦 )

=푀 × 푦

퐴푒 푟

=399655.7 × 17.32

ퟏퟐퟓퟔ.ퟔퟑퟕ× ퟐ.ퟔퟖ× ퟐퟎ= 102.768

푁푚푚

(푇푒푛푠푖푙푒) At outer fiber 푟 = 푟 = 60 푚푚, 푦 = 푟 − 푟 = 22.68 푚푚(−푣푒)

휎 =푀 × 푦

퐴푒(푟 − (−푦 ))=푀 × (푦 )

퐴푒 푟

=399655.7 × (−22.68)ퟏퟐퟓퟔ.ퟔퟑퟕ× ퟐ.ퟔퟖ × ퟔퟎ

= −44.857푁


Direct stress at section B-B: At section B-B, = 0

휎 =퐹

2퐴cos 휃 =

900002 × ퟏퟐퟓퟔ.ퟔퟑퟕ

cos 0= 35.81 푁/푚푚 (푇푒푛푠푖푙푒)

Resultant stress at section B-B

휎 = 휎 + 휎 = 138.578 푁/푚푚 (푇푒푛푠푖푙푒) 휎 = 휎 + 휎 = −9.047 (푐표푚푝푟푒푠푠푖푣푒)

At section A-A, = π/2

푀 = 푀 =퐹푅

2

휋 − 22퐿 + 휋푅

− 퐹푅2

(1 − cos 90) =푭푹ퟐ

ퟐ

흅 − ퟐퟐ푳+ 흅푹

− 푭푹ퟐ

=90000 × 40

2휋 − 2

2 × 40 + 휋 × 40− ퟗퟎퟎퟎퟎ× ퟒퟎ

ퟐ= −1.4 × 10 푁푚푚

At inner fiber 푟 = 푟 = 20푚푚, 푦 = 푟 − 푟 = 17.32 푚푚(+푣푒)

휎 =푀 × 푦퐴푒(푟 − 푦 )

=푀 × 푦

퐴푒 푟=

−1.4 × 10 × 17.32ퟏퟐퟓퟔ.ퟔퟑퟕ × ퟐ.ퟔퟖ × ퟐퟎ

= −360푁


AMS LECTURE-26


At outer fiber

푟 = 푟 = 60 푚푚, 푦 = 푟 − 푟 = 13.44푚푚(−푣푒)

휎 =푀 × 푦

퐴푒(푟 − (−푦 ))=푀 × 푦

퐴푒 푟=−1.4 × 10 × (−22.68)ퟏퟐퟓퟔ.ퟔퟑퟕ × ퟐ.ퟔퟖ × ퟔퟎ

= 157.14푁


Direct stress at section A-A: At section A-A, = π/2

휎 =퐹

2퐴cos 휃 =

900002 × ퟏퟐퟓퟔ.ퟔퟑퟕ

cos흅/ퟐ = 0 Resultant stress at section A-A

휎 = 휎 + 휎 == −360푁


휎 = 휎 + 휎 = 157.14푁


Q.9. A steel ring of rectangular section 8 mm width by 5 mm thickness has a mean diameter of 30

cm. A narrow saw cut was made and tangential separating force of 0.5 kg each are applied at the

cut in the plane of the ring. Find the additional separation due to these forces.(E=2x106 kg/cm2.)

Solution:-

Given data:-

R= 15 cm =150mm

b= 8 mm, t= 5mm

P=0.5 kg

E=2x106 kg/cm2

AMS LECTURE-26


퐼 =1

12푏 푡 = 83.333 푚푚

Bending moment at any section X-X at any angle is “M”: 푀 = 푃푅(1 − cos 휃)

As per castgliano’s theorem :

훿 =휕 ∪휕푃

=1퐸퐼

푀 휕푀휕푃

푅푑휃

퐴푆 휕푀휕푃

= 푅(1− cos 휃)

훿 =휕 ∪휕푃

=푃푅퐸퐼

(1 − cos 휃) (1− cos 휃)푑휃

=푃푅퐸퐼

(1− cos 휃) 푑휃 =푃푅퐸퐼

(1 + cos 휃 − 2 cos 휃)푑휃

=푃푅퐸퐼

휃 − 2 sin 휃 +휃2

+sin 2휃

4=

3휋푃푅퐸퐼

= 9.54 푚푚

Q.10. A steel spring ABC of radius R=60mm & AB of length 120mm is firmly fixed at point “c” as

shown in figure. Find the vertical deflection at point “A” neglecting the effect of shear, where take

E=2x106 kg/cm2.

Solution:-

Given data:-

AMS LECTURE-26


R=60mm, L=120mm, E=2x106 kg/cm2.

퐼 =1

12푏 푡 =

112

20 × 3 = 45 푚푚 = 45 × 10 푐푚

푈 = 푈 +푈

푉푒푟푡푖푐푎푙 푑푒푓푙푒푐푡푖표푛 푎푡 "퐴" 푖푠 훿

훿 =휕푈휕푃

= 휕푈휕푃

+ 휕푈휕푃

FOR the portion “AB”: (0<x<L)

Bending moment at any distance “x” from point A is:

푀 = 푃푥 휕푀휕푃

= 푥

휕푈휕푃

= 1퐸퐼

푀 휕푀휕푃

푑푥 =1퐸퐼

푃푥 × 푥 푑푥 =푃퐿3퐸퐼

FOR the portion “BC”: (0<<π)


푀 = 푃(퐿 + 푅푠푖푛휃) 휕푀휕푃

= (퐿 + 푅푠푖푛휃)

휕푈휕푃

= 1퐸퐼

푀 휕푀휕푃

푑푠 =1퐸퐼

푃(퐿 + 푅푠푖푛휃) × (퐿 + 푅푠푖푛휃)푅 푑휃 퐴푆 푑푠 = 푅푑휃

=푃푅퐸퐼

(퐿 + 푅 푠푖푛 휃 + 2푅퐿푠푖푛휃) 푑휃 = 푃푅퐸퐼

퐿 휃 − 2푅퐿푐표푠휃 +푅 휃 − 푅 푠푖푛2휃

22

= 푃푅퐸퐼

퐿 휋 + 4푅퐿 +푅 휋

2

훿 =푃퐿3퐸퐼

+푃푅퐸퐼

퐿 휋 + 4푅퐿 +푅 휋

2=

1퐸퐼

푃퐿

3+ 푃푅(퐿 휋 + 4푅퐿 +

푅 휋2

)

=1퐸퐼

[1152 + 12(452.389 + 288 + 56.548)] =1퐸퐼

[10715.244] =10715.244

2 × 10 × 45 × 10

= 1.1906푐푚

Q.11. A steel spring ABCD made of from a rod of diameter “d”. the semi-circular portion B to C of

radius R & straight portion AB and CD of length L is loaded by end loads P as shown in figure. Find

the increase in distance A&D due to loads.

AMS LECTURE-26


Solution:- 푉푒푟푡푖푐푎푙 푑푒푓푙푒푐푡푖표푛/푠푒푝푎푟푎푡푖표푛 푏푒푡푤푒푒푛 AD 푖푠 훿

Due to symmetric

훿 = 2훿 푤ℎ푒푟푒 훿 푖푠 푡ℎ푒 푣푒푟푖푐푎푙 푑푒푓푙푒푐푡푖표푛 표푓 푡ℎ푒 푝표푖푛푡 퐴 푓표푟 푡ℎ푒 푏푒푎푚 푓푟표푚 퐴 푡표퐸

훿 =휕푈휕푃

= 2 ×휕푈휕푃

= 2 휕푈휕푃

+ 휕푈휕푃

= 2훿

훿 =휕푈휕푃

+ 휕푈휕푃

FOR the portion “AB”: (0<x<L)


푀 = 푃푥 휕푀휕푃

= 푥

휕푈휕푃

= 1퐸퐼

푀 휕푀휕푃

푑푥 =1퐸퐼

푃푥 × 푥 푑푥 =푃퐿3퐸퐼

FOR the portion “BC”: (0<<π/2)


푀 = 푃(퐿 + 푅푠푖푛휃) 휕푀휕푃

= (퐿 + 푅푠푖푛휃)

휕푈휕푃

= 1퐸퐼

푀 휕푀휕푃

푑푠 =1퐸퐼

푃(퐿 + 푅푠푖푛휃) × (퐿 + 푅푠푖푛휃)푅 푑휃 퐴푆 푑푠 = 푅푑휃

AMS LECTURE-26


=푃푅퐸퐼

(퐿 + 푅 푠푖푛 휃 + 2푅퐿푠푖푛휃) 푑휃 = 푃푅퐸퐼

퐿 휃 − 2푅퐿푐표푠휃 +푅 휃 − 푅 푠푖푛2휃

22

= 푃푅퐸퐼

퐿휋2

+ 2푅퐿 +푅 휋

4

훿 =푃퐿3퐸퐼

+푃푅퐸퐼

퐿휋2

+ 2푅퐿 +푅 휋

4=

112퐸퐼

[4푃퐿 + 푃푅(6퐿 휋 + 24푅퐿 + 3푅 휋)]

훿 = 2훿 =1

6퐸퐼 [4푃퐿 + 푃푅(6퐿 휋 + 24푅퐿+ 3푅 휋)]

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