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
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
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
= 50.9424N/mm2 (tensile)
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
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
(compressive)
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=10.8x106 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 = 2827.433mm
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
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
= 56.0368N/mm2, at B
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
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
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
= !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
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
The figure shows the stress distribution in the critical section.
σbi=286.232 N/mm2
σri=296.649 N/mm2
b = 18 mmi
h =40 mm
CA
NA
Bending stress =-208.606 N/mmσbo
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.
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)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
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 = .� =
1:0004942: = 2.16N/mm
2 (tensile)
Bending stress at the inner fiber σbi = &'�*��* =
946:52/:714./524942:751.:617/:0
= 42.64N/mm2 (tensile)
Bending stress at the outer fiber σbo =! &'�)��) = ! 946:52/:7/56.2144942:751.: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.
σbi=42.64 N/mm2
σri=44.8 N/mm2
b =
300 m
mi
225
CA
NA
Bending stress =-50.49 N/mmσbo
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.
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 >*�/>)>*�>) " =
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
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/ x90x�60 Q 30� = 4050mm
2
Bending moment about centroidal axis Mb = FI = 25,000x115
= 2875000 N-mm
Direct stress σd = .� =
/:000,0:0 = 6.173N/mm
2 (tensile)
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/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)
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.
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