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

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.)

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 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.

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 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 the section 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

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

σ_bi = Bending stress at the inner fiber σ_bo = Bending stress at the outer fiber σri = Combined stress at the inner 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 before bending. i.e., when there are no stresses induced. When a bending ' 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 c2 c1 ci co e F F Mb ri rn rc ro C_L

the lines 'ab' and 'cd' represent two such planes before bending. i.e., when there are no stresses induced. When a bending 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

θdA

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

M_b = Applied Bending Moment r_{i = Inner radius of curved beam} r_o = Outer radius of curved beam

r_c = Radius of centroidal axis r_n = Radius of neutral axis CL= Centre line of curvature

Summation of forces over the whole cross section i.e.  dF  0 ∴ θ θ   =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,

M_b = –  ydF

Substituting the value of dF, we get M_b = θ θ   _dA = θ θ  y !  _" dA = θ θ  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.

∴ M_{b =} θ

θ A.e ... (iii)

From equation (i) θ

θ = –

σ_  Substituting in equation (iii)

M_b = – 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 σ_bo i.e., σ_bo= !&_'()

) ($ 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

σ_bi= &'(*

 +(*

i.e., σ_{bi =} &'(*

 _*

($ 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

Centroidal and Neutral Axis of

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

Why stress concentration occur at inner side or concave side of

Consider the elements of the curved beam lying between two axial 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. Consider two fibers symmetrically located on either side of the neutral axis. Deformation in both the fibers is same and equal to ydθ

Why stress concentration occur at inner side or concave side of

Consider the elements of the curved beam lying between two axial . Let fg is the final position of about neutral axis. Consider two fibers symmetrically located on either side of the neutral

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 = 22X10 3 X100 = 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

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

= 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 Radius of neutral axis

Applied force

+83.61t = 4000+100t;

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

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

Area of cross section A = π_,d2 = _,π x602 = 2827.433mm2 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 = !

&_'_*  _* = + 56750 8_7/2.015 /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

σ_ri = σd + σbi

= ! 7.0736! 105.00

= - 112.0736N/mm2 (compressive) Combined stress at the outer fiber

σro = σd + σbo

= ! 7.0736+58.016

= 50.9424N/mm2 (tensile) Maximum shear stress

τ_max = 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

Distance of neutral axis to centroidal axis

Distance of neutral axis to inner radius

Distance of neutral axis to outer radius

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 straighten the bar. Find the position of the Neutral axis and draw a diagram to show the variation of stress across the section.

60mm

= 2x106 N-mm 30mm

=130 =(ri+c1+c2)

30= 70mm (rc-c1)

Distance of neutral axis to centroidal axis

e = rc - rn= 100-96.924

=3.075mm Distance of neutral axis to inner radius

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

Distance of neutral axis to outer radius

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

e) = 26.925mm ) = 33.075mm

Bending stress at the outer fiber σbo = &')  ₎ = +/;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=10.8x10 6 N-mm

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! 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 = 2827.433mm2 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

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 + σb = ! 7.0736+58.016

= 50.9424N/mm2 (tensile) Maximum shear stress Gmax = 0.5x σmax = 0.5x112.0736

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 Distance of neutral axis to centroidal axis e = rc - rn = 140-136.1038 = 3.8962mm 80 5 0 R100 175 mm 9000N h = 80mm c2 e c1 b = 5 0 m m Critical Section co ci ro rn rc F r = 100mmi 175mm C_L F CA NA

Distance of neutral axis to inner radius 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

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

= 65.676N/mm2 (tensile) 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

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/mm 2 σri=67.926 N/mm 2 b = 50 mm h =80 mm C A N A Bending stress σbo=-44.362 N/mm 2 Combined stress σro -42.112 N/mm 2 = σd=2.25 N/mm 2 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

Outer radius of curved beam ro = 25+40

= 65mm

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

"

= ,0 4 51/76 516 " = 16.667mm h = 40mm c2 e c1 b = 6 m m o co ci ro rn rc F r = 25mmi 100mm C_L F b = 1 8 m m i C A N A

∴ 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 7,0516 JKN<8L<N8 O? <88+ 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 Applied force F = 5000N

Area of cross section A = 5

/b Q b = 5

/

x4018 Q 6

= 480mm

2

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

Direct stress σd = . = :000_,10 = 10.417N/mm 2

(tensile) Bending stress at the inner fiber σbi =

&_'_*  _* =

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

= 286.232N/mm2 (tensile) Bending stress at the outer fiber σbo = &_')

) =

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

= !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 = 10.417!286.232

= ! 198.189N/mm2 (compressive) Maximum shear stress Gmax = 0.5x σmax = 0.5x296.649

= 148.3245 N/mm2, at the inner fiber

σ_bi=286.232 N/mm2 σ_ri=296.649 N/mm2 b = 18 mmi h =40 mm C A N A Bending stress σ_bo=-208.606 N/mm2 Combined stress σ_ro N/mm2 =-198.189 b = 6 mmo σ_d=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:

Draw the critical section as shown in the figure.

Inner radius of curved beam ri = 250mm

Outer radius of curved beam ro = 550mm

Radius of neutral axis rn =  >_*WI*XY*_I* Z >[I)LY) I)XY*\>) I)LY)I) " ai = 75mm; bi = 300mm; b2 = 75mm; ao = 0; bo = 0 A=a1+a2=75x300+75x225 =39375mm 2 ∴ rn = 4942: 400 8?X]8_8? "2: _8?X]888?L?"0 = 333.217mm Let AB be the ref. line

550 75 300 _{85 kN} 250 750 mm 75 B a = 7 5 m m i 225 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 750 r = 250 mmi F F a1

x^ 

J7J7 _J =

2:7400]82:7//: 2:8 "

4942: = 101.785mm Radius of centroidal axis rc = ri +x^

= 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 l = 750+ rc

= 750+351.785 = 1101.785mm Applied force F = 85kN

Bending moment about centroidal axis Mb = FI

= 85000x1101.785 = 93651725N-mm Direct stress σd = .  = 1:000 4942: = 2.16N/mm 2 (tensile) Bending stress at the inner fiber σbi =

&_'_*

 _* =

946:52/:714./52

4942:751.:617/:0

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

!

946:52/:7/56.214 4942:751.:617::0 = ! 50.49N/mm2 (compressive) Combined stress at the inner fiber σri = σd + σbi = 2.16+42.64

= 44.8N/mm2 (tensile) Combined stress at the outer fiber σro = σd + σbo = 2.16! 50.49

= ! 48.33N/mm2 (compressive) Maximum shear stress Gmax = 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/mm 2 b = 3 0 0 m m i 225 C A N A Bending stress σ_bo=-50.49 N/mm2 Combined stress σro N/mm 2 =-48.33 a =75mmi b = 75 mm2 a2 a1 σd=2.16 N/mm 2 Direct stress (σ_d)

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. Inner radius of curved beam ri = 100mm

Outer radius of curved beam ro = 100+90 =

190mm

Distance of centroidal axis from inner fiber c1 =  4 >_*/>₎ >_*>₎ " = 90 4 x 60/740 6040 " = 40mm 90mm 30mm F = 25 kN 25mm 60mm 10 0 m m ri rc rn ro e ci co N A C A c2 c1 h = 90 mm l F C_L

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

= 140 mm Radius of neutral axis rn =

J >*>) '*I)L')I* M I)_I* + >*+>) = J 79076040 <?NJ_? L `?NJ?? _? J_?J?? + 60+40 = 135.42mm Distance of neutral axis to centroidal axis e = rc! rn

= 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_/x90x60 Q 30 = 4050mm2 Bending moment about centroidal axis Mb = FI = 25,000x115

= 2875000 N-mm Direct stress σd = . = /:000_,0:0 = 6.173N/mm

2

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

* =

/12:00074:.,/ ,0:07,.:17500 = 54.9 N/mm2 (tensile) Bending stress at the outer fiber σbo =! &_')

) = !

/12:0007:,.:1 ,0:07,.:17590

= ! 44.524N/mm2 (compressive) Combined stress at the inner fiber σri = σd + σbi = 6.173+54.9

= 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 The figure shows the stress distribution in the critical section.

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/mm 2 Also C2 = 4>/>_47/>>) 7 )  --- From DDHB

C2 = 4740/740790_47/74040 = 50mm c1 = 90-50= 40mm Moment of inertia I = C67>67>7>)>)E` 46c/>><sub>)</sub>d = C6740 <sub>674074040E90</sub>` 46c/74040d = 2632500mm4 Direct stress σb = .  = /:000 ,0:0 = 6.173N/mm 2 (tensile) Bending stress at the inner fiber σbi =

&_'_J

e =

/12:0007,0

2632500

= 43.685 N/mm2 (tensile) Bending stress at the outer fiber σbo = - &'_e =2875000x50_/64/:00

= -54.606N/mm2 (compressive) Combined stress at the inner fiber σri = σd + σbi = 6.173+43.685

= 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/mm 2 σri= 49.858 N/mm 2 6 0 m m h =90 mm N A , C A σbo=-54.606 N/mm 2 σro N/mm 2 =-48.433 b = 30 mm c =50mm 2 c =40mm1 b /2 = 15o b /2 = 15o σ_d= 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:

Draw the critical section as shown in the figure.

Inner radius of curved beam ri = 125mm

Outer radius of curved beam ro = 125+60

= 185mm Radius of centroidal axis rc =100+ 60_/

= 130mm Radius of neutral axis rn = ln 

#I)_I* % = ln 60 #JK8_J8% = 153.045mm

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

= 155-153.045 = 1.955mm h=60mm b=30mm F = ? 12 5 m m 100 h = 60mm c₂ e c₁ b = 3 0 m m co ci ro rn rc F r = 125mmi C_L l 100 C A N A L o ad l in e

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 Direct stress σd = .  = . 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

b

o

H

M

=

b

i

Z

r =59.5 mmi r = c l rn ro e ci co N A C A c2 c1 h = 111 mm F C_L b =28o b =71 i

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

Radius of neutral axis rn =

J >*>) '*I)L')I* M I)_I*+ >*>) = J 7555725/1 ]JNJ]?.8LKN8_.8 O? J]?.88_.8+ 25/1 = 98.095mm

Distance of centroidal axis from inner fiber c1 =

 4

>_*/>₎ >_*>₎ "

= 555₄ 25/7/1_25/1 " = 47.465mm Radius 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

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

= 170.5-98.095=72.0405mm

Applied force F = 105N

Area of cross section A = 5_/b Q b_ = 5

/x111x71 Q 28 = 5494.5mm

2

Bending moment about centroidal axis Mb = Fl = 10 5

x141.667 = 106.965x105N-mm Direct stress σd = . = 500000_:,9,.:

= 18.2N/mm2 (tensile) Bending stress at the inner fiber σbi =

&_'_*

 _* = 506.96:750

8_741.:9:

:,9,.:71.127:9.: = 142.365/mm2 (tensile) Bending stress at the outer fiber σbo =

&_'₎  ₎ =

506.96:750872/.,0: :,9,.5x8.127520.: = -93.2 N/mm2 (compressive) Combined stress at the inner fiber σri = σd + σbi = 18.2+142.365

= 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 The figure shows the stress distribution in the critical section.

σbi=142,365 N/mm2 σri=160.565 N/mm2 b = 71 mmi C A N A 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. 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 = WDr

o√ri 4 Z 2 = √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 = 7853.98mm2 Mb = Fx = 5000 x

Combined maximum stress at the inner fiber (i.e., at B)

σ

_ri

= Direct stress + bending stress

=

.



Q

& '_**

50 

_21:4.91:000

Q

_{21:4.91j6.2j:0}:0007,4.4



∴ 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:

Draw the critical section at P Radius of centroidal axis Inner radius r_i =100 Outer radius r_o

= 100+

Radius of neutral axis

r

_n

=

=

= 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 r_c

=100mm

= 87.5mm

+ = 112.5mm

Radius of neutral axis

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

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 mm2 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_bC_o Ae_o = 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)

σ

_ri

= Direct stress + bending stress

=

F A

+

M_bc_i Aer_i

=

1000 490.87

+

100000 X 12.1 490.87 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.

r

_c

= 75mm

r

i

= 75 -

25

r

_{o = 75 +} 25 2

= 87.5 mm

A =

π 4

d

2

=

π 4

X 25

2

= 490.87 mm

2

r

n

=

W

Dr o  √ri 4

Z

2

=

√87.5 √62.5 4

"

2

=74.4755 mm

e = r

_c

- r

_n

= 75 -74.4755 =0.5254 mm

c

_i

= r

_n

- r

_i

=74.4755 – 62.5 =11.9755 mm

c

o

= r

o

- r

n

= 87.5 – 74.4755 = 13.0245 mm

l = r

_c

= 75 mm

Mb = Fl = 1000 X 75 = 75000 Nmm

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

σ

_ro

=

F A

–

M_bc_o Aer_o

=

1000 490.87

-75000 X13.0245 490.87 X 0.5245 X 62.5

= - 41.324 N/mm

2 (compressive)

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

σ

_ri

=

F A

+

M_bc_o Aer_o

=

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 τ_max

=0.5 σ

_max

= 0.5 X 72.466

= 36.233 N/mm

2 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

The ring is symmetrical and is loaded symmetrically in both the 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.

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 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. 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

derivative of the strain energy with respect to the load gives the he load. In this case, this would be zero.

……….(1)

The ring is symmetrical and is loaded symmetrically in both the

figure. At the two ends

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

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

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 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

produces tension in the inner fibers and compression on the outer. It should be noted that these equations are valid in the region,

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

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,

It is seen that numerically, M The stress at any angle Ѳ forces 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 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 i Ѳ = 900

)

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 A will be (at Ѳ = 0)

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

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

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

Consider a thin closed ring subjected to symmetrical load F as shown in 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 The unknown quantity is M0. Again considering symmetry, we

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

Consider a thin closed ring subjected to symmetrical load F as shown in 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 Again considering symmetry, we conclude that the tangents at C and D must be vertical and must remain

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

It can be observed that at L = 0 equation reduces to the same expression 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 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, equation (v) reduces obtained for a circular ring.

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

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

= p/2,

the fiber is

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

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

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

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

= 30mm Outer radius = 30+25 = 55mm

Radius of centroidal axis rc = 30 + = 42.5mm

Radius of neutral axis rn =

= =42.5mm

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

= 42.5 – 41.56 = 0.94mm axis to inner radius ci = rn - ri

= 41.56 – 30 = 11.56mm

Determine the stress induced in a circular ring of circular cross section diameter subjected to a tensile load 6500N. The inner

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

= 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:: = 46.52N/mm2 (tensile) Consider the cross section B – B

At section B – B, θ = 00 with respect to horizontal Direct stress σd = . k 0 2A = 6:007cos0 /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

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

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)

Maximum stress at outer fiber σ_l= Direct stress + Bending stress = 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.

Solution: the circular ring and its critical section are as shown in fig. 1.30a and 1.30b respectively.

Inner radius ri = - = 25mm

Outer radius = + = 75mm Radius of centroidal axis r

Radius of neutral axis r

Distance of neutral axis to centroidal axis e = r = 50 - 46.65 = 3.35mm

Distance of neutral axis to inner radius c = 46.65

Distance of neutral axis to

= 75 - 46.65 = 28.35mm Area of cross section A =

Determine the stress induced in a circular ring of circular cross section of 50 mm diameter rod subjected to a compressive load of 20kNN. The mean diameter of the ring is 100 mm.

Solution: the circular ring and its critical section are as shown in fig. 1.30a and 1.30b respectively.

= 25mm = 75mm

Radius of centroidal axis rc = = 50mm

Radius of neutral axis rn =

= = 46.65mm axis to centroidal axis e = rc - rn

46.65 = 3.35mm

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

= 46.65-25 = 21.65 mm

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

46.65 = 28.35mm

Area of cross section A = x552 = 1963.5mm2

circular ring of circular cross section of 50 mm diameter rod subjected to a compressive load of 20kNN. The

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

p qrs t /u

Consider the cross section A – A

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

p qrs 90 /u = 0

Bending moment Mb = + 0.318Fr

Where r = rc, positive sign refers to tensile load

v_wx = + 0.318x20000x50 = 318000 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 y_z{x=Direct stress + Bending stress = 0 + |}q~

uz~ =

318000€21.6:

5964.:4.4:/:

= 41.86 N/mm2 (tensile)

Maximum stress at outer fiber y_zrx= Direct stress + Bending stress =0 - |}q‚

uz_‚ = -

451000/1.35

5964.:4.4:2:

Consider the cross section B – B

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

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

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

&_'_*

Aer_* = -5.093 + 5964.51/0007/5.6:5x3.4:7/:

= - 29.05 N/mm2 (compressive) Maximum stress at outer fiber σ_l= Direct stress + Bending stress = σd + &_'₎  ₎ = -5.093 + 51/0007/1.4: 5964.:74.4:72: = 5.366 N/mm2 (tensile)

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 Radius of centroidal axis r

Radius of neutral axis r

Distance of neutral axis to centroidal axis e =

Distance of neutral axis to inner radius c

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 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 Radius of centroidal axis rc = 40mm

Radius of neutral axis rn =

= = 37.32mm Distance of neutral axis to centroidal axis e = rc - rn

=40-37.32 = 2.68mm Distance of neutral axis to inner radius ci = rn - ri

= 37.32-20 = 17.32 mm

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

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

= 60 – 37.32 = 22.68mm

Direct stress at any cross section at an angle θ with horizontal σd = . k θ_/

Consider the cross section A – A [i.e., Along the load line] At section A – A, θ = 900 with respect to horizontal

Direct stress σd = . k 90_/ = 0

Bending moment M_>l = - .I/

/π where r = rc,

M>_l= 900007,07/7,010_/π7,010 = 1.4x106N-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 + &_'* * = 5.,750<752.4/ π O7,07/.617/0 = - 360 N/mm2 (tensile)

Maximum stress at outer fiber σ_l= Direct stress + Bending stress = 0 - &_') ) = - 5.,750<7//.61 π O7,07/.61760 = 157.14 N/mm2 (compressive)

Consider the cross section B – B [i.e., 900 away from the load line] At section B – B, θ = 00 with respect to horizontal

Direct stress σd = . k /θ / = 5.,750<7 k 0 /7π O7,0 = 35.81 N/mm2 (compressive)

Bending moment M_>l = - .I_/π/+π where r = rc,

M>_l= 900007,07/7,0+π710_/π7,010 = - 399655.7N-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_o=Direct stress + Bending stress = σd - &_'* * = 35.81 + 4996::.2752.4/ /7π O7,07/.617/0 = 138.578 N/mm2 (tensile)

Maximum stress at outer fiber σ_l= Direct stress + Bending stress = σd + &_') ) = 35.81 - 4996::.27//.61 /7π O7,07/.61760 = - 9.047 N/mm2 (compressive) 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

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

central angles before and after applying the moment M. Since the length of neutral surface remains the same

'

Figure

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

Consider a curved beam of uniform cross section as shown in Figure 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 central angles before and after applying the moment M. Since the length

... (i)

Consider the arc of circle JK located at a distance y above the neutral be the radius of this arc before and after bending

.... (ii) ... (iii)