Comparing the stresses at the inner most fiber based on … · 2016-01-13 · Comparing the...

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

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

σbci = 1.522σBSi

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

than that for a straight beam.

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

curved beam is higher than the straight beam.

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

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

σbco = 1.522σBSi

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

a straight beam.

The curvatures thus introduce a non linear stress distribution.

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

concentration on the inner side.

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

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

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

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

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

manner such that the centroidal axis is inwards.

Problem no.1

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

figure.

Given data:

ri = 50mm

ro = 150mm

F = 22X103N

b = 20mm

h = 150-50 = 100mm

A = bh = 20X100 = 2000mm2

e = rc - rn = 100

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

bending, due to the eccentricity of the force.

Stress due to direct load will be,

y = rn – r = 91.024

Mb = 22X103X100 = 2.2X10

= 100 - 91.024 = 8.976mm

B will be subjected to a combination of direct load and

bending, due to the eccentricity of the force.

Stress due to direct load will be,

r = 91.024 – r

X100 = 2.2X106 N-mm

B will be subjected to a combination of direct load and

Problem no.2

Determine the value of “t” in the cross

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

fibers are numerically equal.

Given data;

Inner radius ri=150mm

Outer radius ro=150+40+100

=290mm

Solution;

From Figure Ci + CO = 40 + 100

= 140mm…………… (1)

Since the normal stresses due to bending at

the extreme fiber are numerically equal

have,

i.e Ci=

= 0.51724C

Radius of neutral axis

rn=

rn =197.727 mm

ai = 40mm; bi = 100mm; b

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

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


fibers are numerically equal.

=150mm

=150+40+100

= 40 + 100

= 140mm…………… (1)

Since the normal stresses due to bending at

the extreme fiber are numerically equal we

0.51724Co…………… (2)

= 100mm; b2 =t;

= 150mm; ro = 290mm;

section of a curved beam as


=

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

∴ t = 41.126mm

Problem no.3

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

figure.

Solution:

The figure shows the critical section of the split

ring.

Radius of centroidal axis

Inner radius of curved beam

Outer radius of curved beam


Applied force

+83.61t = 4000+100t;


The figure shows the critical section of the split

Radius of centroidal axis rc = 80mm

Inner radius of curved beam ri = 80

= 50mm

Outer radius of curved beam ro = 80 +

= 110mm

rn =

=

= 77.081mm

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

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

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

Area of cross section A = π

,d2 =

π

, x602 = 2827.433mm

2

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

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

=16x105N-mm

Distance of neutral axis to centroidal axis

e = rc ! rn

= 80! 77.081=2.919mm

Distance of neutral axis to inner radius

ci = rn ! ri

= 77.081! 50=27.081mm

Distance of neutral axis to outer radius

co = ro ! rn

= 110 ! 77.081=32.919mm

Direct stress σd = ! .� = /0000

/1/2.,44

=! 7.0736N/mm2 (comp.)

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

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

= ! 105N/mm2 (compressive)

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

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

= 58.016N/mm2 (tensile)

Combined stress at the inner fiber

σri = σd + σbi

= ! 7.0736! 105.00

= - 112.0736N/mm2

(compressive)

Combined stress at the outer fiber

σro = σd + σbo

= ! 7.0736+58.016


Maximum shear stress

τmax = 0.5x σmax

= 0.5x112.0736

= 56.0368N/mm2, at B

The figure.

Problem No. 4

Curved bar of rectangular section 40x60mm and a mean

100mm is subjected to a bending moment of 2KN

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

diagram to show the variation of stress across the section.

Solution

Given data:

b= 40mm h= 60mm

rc=100mm Mb= 2x10

C1=C2= 30mm

rn=

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

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

rn= 96.924mm




Area

A= bxh = 40x60 = 2400 mm

Bending stress at the inner fiber

Curved bar of rectangular section 40x60mm and a mean

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


diagram to show the variation of stress across the section.

60mm

= 2x106 N-mm

30mm

=130 =(ri+c1+c2)

30= 70mm (rc-c1)


e = rc - rn= 100-96.924

=3.075mm


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


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

A= bxh = 40x60 = 2400 mm2

Bending stress at the inner fiber σbi = =

= 104.239 N/mm2

(compressive)

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

m tending to


e) = 26.925mm

) = 33.075mm

(compressive)


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

= -68.94 N/mm2 (tensile)

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

=+/;50<

/,00;500

= -8.33 N/mm2 (Compressive)

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

figure.

Problem No. 5

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

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

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

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

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

60Mpa.Determine the magnitude of the couple.

Solution

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

σbi = σbo = 60MPa

Since the extreme fibers stresses due to pure bending are numerically

equal we have,

&'�*��* =

&'�)��)

We have,

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

2ci=co

But h= ci + co

120 = ci+2ci

Ci=40mm; co=80mm

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

b=150.34mm

To find the centroidal axis, (C2)

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

= 74.313mm.

But C1=C2

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

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

Bending stress in the outer fiber,

σ>? � M>c�Aer�

A= �5:0.1,�/:�5/0/

= 1050.4mm

60 = &';10

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

Mb=10.8x106 N-mm

Problem no.6


fig.1.9a

Solution:

Redraw the critical section as shown in the figure.

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 = 2827.433mm

2

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

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

=16x105N-mm

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

= 80! 77.081

=2.919mm


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


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

Direct stress σd =! .� = /0000

/1/2.,44

=! 7.0736N/mm2 (comp.)

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

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

= ! 105N/mm2 (compressive)


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


Combined stress at the inner fiber

σri = σd + σbi =! 7.0736! 105.00

=! 112.0736N/mm2

(compressive)

Combined stress at the outer fiber

σro = σd + σb = ! 7.0736+58.016


Maximum shear stress

Gmax = 0.5x σmax = 0.5x112.0736

= 56.0368N/mm2, at B

The figure shows the stress distribution in the critical section.

Problem no.7

Determine the maximum tensile, compressive and shear stress induced

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

Solution:

Draw the critical section as shown in

the figure.

Inner radius of curved beam ri =

100mm

Outer radius of curved beam ro = 100+80

= 180mm

Radius of centroidal axis rc = 100+ 10/

= 140mm

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

= ln 10#JK?J�?%

= 136.1038mm


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

805

0

R100

175 mm

9000N

h = 80mm

c2

ec1

b =

50

mm

CriticalSection

co ci

ro

rn

rc

F

r = 100mmi 175mm

CL

F

CA

NA


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


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

Distance from centroidal axis to force

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

Applied force F = 9000N

Area of cross section A = 50x80 = 4000mm2

Bending moment about centroidal axis Mb = FI = 9000x315

= 2835000 N-mm

Direct stress σd = .� =

9000,000 = 2.25N/mm

2 (tensile)

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

/14:000746.5041,00074.196/7500


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

,00074.196/7510

= ! 44.326N/mm2 (compressive)

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

= 67.926N/mm2

(tensile)

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

= !42.112 N/mm2 (compressive)

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

= 33.963 N/mm2, at the inner fiber

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

σbi=65.676 N/mm2

σri=67.926 N/mm2

b = 50 mm

h =80 mm

CA

NA

Bending stress =-44.362 N/mmσbo

2

Combined stress -42.112 N/mmσro

2=

σd=2.25 N/mm2

Direct stress ( )σd

Problem no.8

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

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

Solution:

Draw the critical section as shown in the

figure.

Inner radius of curved beam ri = 25mm

Outer radius of curved beam ro = 25+40

= 65mm

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

= ,04 51�/7651�6 " = 16.667mm

h = 40mm

c2

ec1

b =

6 m

mo

co ci

ro

rn

rc

F

r = 25mmi 100mm

CL

F

b =

18 m

mi

CA

NA

∴ Radius of centroidal axis rc = ri ! c1

= 25+16.667 = 41.667 mm

Radius of neutral axis rn =

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

=

J�7,0�51�6� JKN<8L<N�8

O? ��<8�8+ �51+6�

=38.8175mm

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

= 1.667!38.8175

=2.8495mm

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

= 38.8175!25=13.8175mm

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

= 65-38.8175=26.1825mm

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

= 141.667mm

Applied force F = 5000N

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

5/ x40�18 Q 6� = 480mm

2

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

= 708335 N-mm


:000,10 = 10.417N/mm

2 (tensile)


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

= 286.232N/mm2

(tensile)


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

= !208.606N/mm2 (compressive)



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





σbi=286.232 N/mm2

σri=296.649 N/mm2

b = 18 mmi

h =40 mm

CA

NA


2

Combined stress N/mmσro

2=-198.189

b = 6 mmo

σd=10.417 N/mm2

Direct stress (σd)

Problem no.9

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

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

85kN

Solution:

Draw the critical section as shown in the

figure.


Outer radius of curved beam ro = 550mm


rn = �

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

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

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

∴ rn = 4942:

400�� 8?X]8�8? "�2:�� 88?L?

�8?X]8"�0 = 333.217mm

Let AB be the ref. line

550

75 85 kN300

250

750 mm75

Ba = 75mm

i225 mm

a2b =75mm2

b=300mm

i

cico A

X

ern

rc

r =550 mmo

CA

NA

CL

750

r = 250 mmi

F

F

a1

x̂ � �J7J��7��J�� =

�2:7400�]8� ��2:7//:� 2:��8� "

4942: = 101.785mm

Radius of centroidal axis rc = ri +x̂

= 250+101.785=351.785 mm


= 351.785-333.217=18.568mm

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

= 333.217! 250=83.217mm

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

= 550! 333.217=216.783mm

Distance from centroidal axis to force l = 750+ rc

= 750+351.785 = 1101.785mm

Applied force F = 85kN

Bending moment about centroidal axis Mb = FI

= 85000x1101.785

= 93651725N-mm


1:0004942: = 2.16N/mm

2 (tensile)


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


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







= 24.165N/mm2, at the outer fiber

The below figure shows the stress distribution.

σbi=42.64 N/mm2

σri=44.8 N/mm2

b =

300 m

mi

225

CA

NA


2

Combined stress N/mmσro

2=-48.33

a =75mmi

b = 75 mm2a2

a1

σd=2.16 N/mm2

Direct stress ( )σd

Problem no.10

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

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

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

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

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

centre of curvature at the critical

section. What will be the stress at inner

and outer fiber, if the beam is treated as

straight beam for the given load?

Solution:

Draw the critical section as shown in the figure.


Outer radius of curved beam ro = 100+90 =

190mm

Distance of centroidal axis from inner fiber

c1 = �4 >*�/>)>*�>) " =

904 x 60�/74060�40 "

= 40mm

90mm

30mm

F = 25 kN

25mm60mm

100 mm

ri

rc

rn

ro

e

cico

NA

CA

c2 c1

h = 90 mm

l

F CL

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

= 140 mm

Radius of neutral axis rn =

J��>*�>)�'*I)L')I*M ��I)I* + �>*+>)�

=

J�7907�60�40� <?NJ_? L `?NJ??

_? ��J_?J?? + �60+40�

= 135.42mm


= 140! 135.42=4.58mm

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

=35.42mm

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

= 54.58mm

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

= 115mm

Applied force F = 25,000N = 25kN

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

5/ x90x�60 Q 30� = 4050mm

2

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

= 2875000 N-mm


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

2 (tensile)


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

= 54.9 N/mm2

(tensile)

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



= 61.073N/mm2

(tensile)



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



b) Beam is treated as straight beam

From DDHB refer table,

b = 30mm

bo = 60-30 = 30mm

h = 90

c1 = 40mm

c2 = 90-50 = 40mm

A = 4050 mm2

Mb = 28750000 N/mm2

Also

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

C2 = �4740�/740�79047�/740�40� = 50mm

c1 = 90-50= 40mm

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

46c/>�>)d

= C6740��6740740�40�E90`

46c/740�40d

= 2632500mm4

Direct stress σb = .� =

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

2 (tensile)

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

e = /12:0007,02632500

= 43.685 N/mm2 (tensile)

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

e =2875000x50/64/:00

= -54.606N/mm2 (compressive)


= 49.858N/mm2

(tensile)

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

= -48.433N/mm2 (compressive)

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

σbi= 43.685 N/mm2

σri= 49.858 N/mm2

60 m

m

h =90 mm

NA

, CA

σbo=-54.606 N/mm2

σro N/mm2

=-48.433

b = 30 mm

c =50mm 2c =40mm1

b /2 = 15o

b /2 = 15o

σd= 6.173 N/mm2

σd

b

Comparing the stresses at the inner most fiber based on … · 2016-01-13 · Comparing the...

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