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Transcript of 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
TION OCCUR AT INNER SIDE OR
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.)
EF = E1F1 = RѲ
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
σbi = Bending stress at the inner fiber
σbo = Bending stress at the outer fiber
σri = Combined stress at the inner fiber
σro = Combined stress at the outer 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
c 2c 1
c i
c o
e
F
F
Mb
r i
rn
rc
roCL
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
������θ dA
From the condition of equilibrium, the summation of forces over the
whole cross-section is zero and the summation of the moments due to
these forces is equal to the applied bending moment.
Let
Mb = Applied Bending Moment
ri = Inner radius of curved beam
ro = Outer radius of curved beam
rc = Radius of centroidal axis
rn = Radius of neutral axis
CL= Centre line of curvature
Summation of forces over the whole cross section
i.e. �dF � 0
∴ ��θθ� ���
������=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,
Mb = – � ydF
Substituting the value of dF, we get
Mb =
��θθ� ��
������ dA
= ��θθ� y ! ���
������" dA
= ��θθ� 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.
∴ Mb =
��θθ
A.e ..... (iii)
From equation (i) ��θθ
= – σ������
�
Substituting in equation (iii)
Mb = – σ������� . A. e.
∴ σ = &'�
������� ..... (iv)
This is the general equation for the stress in a fiber at a distance 'y' from
neutral axis.
At the outer fiber, y = co
∴ Bending stress at the outer fiber σbo
i.e., σbo
= !&'()��) ($ rn + co = ro) ..... (v)
Where co = Distance from neutral axis to outer fiber. It is compressive
stress and hence negative sign. At the inner fiber, y = – ci
∴ Bending stress at the inner fiber
σbi
= &'(*
����+(*�
i.e., σbi =
&'(*��* ($ rn – ci = ri) ..... (vi)
Where ci = Distance from neutral axis to inner fiber. It is tensile stress
and hence positive sign.
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
By considering a rectangular cross
Centroidal and Neutral Axis of
Centroidal and Neutral Axis of Typical Section of Curved Beams
Typical Section of Curved Beams
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
θ.
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 = 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
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
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.
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
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#JK8J�8%
= 153.045mm
Distance of neutral axis to centroidal axis e = rc - rn
= 155-153.045 = 1.955mm
h=60mm
b=30mm
F = ?
125 mm
100
h = 60mm
c2
ec1
b =
30
mm
co ci
ro
rn
rc
F
r = 125mmi
CL
l
100
CA
NA
Lo
ad l
ine
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
bo
H
M=
bi
Z
r =59.5 mmi
r = c l
rn
ro
e
cico
NA
CA
c2 c1
h = 111 mm
F
CLb =28o b =71i
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�75557�25�/1�]JNJ]?.8L�KN8_.8
O? ��J]?.88_.8 + �25�/1�
= 98.095mm
Distance of centroidal axis from inner fiber c1 = �4 >*�/>)>*�>) "
= 5554 25�/7/125�/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
Distance from centroidal axis to force l = rc -106.965
Applied force F = 105N
Area of cross section A = 5/ �b� Q b��
= 5/ x111x�71 Q 28� = 5494.5mm
2
Bending moment about centroidal axis Mb = Fl = 105x141.667
= 106.965x105N-mm
Direct stress σd = .� =
500000:,9,.:
= 18.2N/mm2 (tensile)
Bending stress at the inner fiber σbi = &'�*��* =
506.96:7508741.: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
CA
NA
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 = WDro�√ri4
Z2 = √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 = 7853.98mm
2
Mb = Fx = 5000 x
Combined maximum stress at the inner fiber
(i.e., at B)
σri= Direct stress + bending stress
= .� Q &'�*
��*
50 � :00021:4.91 Q �:0007��,4.4�
21:4.91j6.2j:0 ∴ 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:
(I) Consider the section P-Q
Draw the critical section at P
Radius of centroidal axis
Inner radius ri =100
Outer radius ro = 100+
Radius of neutral axis
rn = =
= 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 rc =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
87.5 =12.1 mm
Distance of neutral axis to outer fiber co = ro -rn
=112.5 – 99.6 = 12.9 mm
Area of cross-section A = π
4 d2 = π
4 x25
2 = 490.87 mm
2
Distance from centroidal axis I = rc = 100mm
Bending moment about centroidal axis Mb = F.l = 100 x 100
= 100000Nmm
Combined stress at the outer fiber (i.e., at Q) =Direct stress +bending
stress
σro= F
A - MbCo
Aeo =
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)
σri= Direct stress + bending stress
= F
A +Mbci
Aeri =1000 490.87 +
100000 X 12.1490.87 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.
rc = 75mm
ri = 75 - 25
2 =62.5 mm
ro = 75 +
25
2 = 87.5 mm
A =π
4 d2 =
π
4 X 25
2 = 490.87 mm2
rn = WDro � √ri4
Z2 = √87.5 �√62.5
4"2 =74.4755 mm
e = rc - rn = 75 -74.4755 =0.5254 mm
ci = rn - ri =74.4755 – 62.5 =11.9755 mm
co = ro - rn = 87.5 – 74.4755 = 13.0245 mm
l = rc = 75 mm
Mb = Fl = 1000 X 75 = 75000 Nmm
Combined stress at the outer fibre (at R) = Direct stress + Bending stress
σro = FA
– Mb coAero
= 1000
490.87 - 75000 X13.0245490.87 X 0.5245 X 62.5
= - 41.324 N/mm2
(compressive)
Combined stress at the inner fiber (at S) = Direct stress + Bending stress
σri = F
A + Mbco
Aero =
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 τmax=0.5 σmax= 0.5 X 72.466
= 36.233 N/mm2 at P
Stresses in Closed Ring
Consider a thin circular ring subjected to symmetrical load F as shown
in the figure.
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
derivative of the strain energy with respect to the load gives the
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 shown in the
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,
2) will be:
It is seen that numerically, M
The stress at any angle Ѳforces 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 combined normal stress across any section will be:
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 iѲ = 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
to the same
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
and at section A-A
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
Solution: the circular ring and its critical section are as shown in fig.
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 02A =
6:007 cos0/7,90.12, = 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 = 1963.5mm
2
circular ring of circular cross section
of 50 mm diameter rod subjected to a compressive load of 20kNN. The
Solution: the circular ring and its critical section are as shown in fig.
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
vwx = + 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 yz{x=Direct stress + Bending stress
= 0 + |}q~u�z~ =
318000�21.6:5964.:�4.4:�/:
= 41.86 N/mm2
(tensile)
Maximum stress at outer fiber yzrx= Direct stress + Bending stress
=0 - |}q�u�z� = -
451000�/1.355964.:�4.4:�2:
= - 18.27 N/mm2 (compressive)
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 +
51/0007/5.6:5964.5x3.4:7/:
= - 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�/����/��� where r = rc,
M>l= 900007,07�/7,0�10�
/�π7,0�10� = 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
O7,0�7/.617/0
= - 360 N/mm2 (tensile)
Maximum stress at outer fiber σ��l= Direct stress + Bending stress
= 0 - &'�)��) = -
5.,750<7//.61π
O7,0�7/.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
/7πO7,0� = 35.81 N/mm2
(compressive)
Bending moment M>l = - .I�/�+π��/�π���� where r = rc,
M>l= 900007,07�/7,0+π710�
/�π7,0�10� = - 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//7πO7,0�7/.617/0
= 138.578 N/mm2 (tensile)
Maximum stress at outer fiber σ��l= Direct stress + Bending stress
= σd + &'�)��) = 35.81 -
4996::.27//.61/7πO7,0�7/.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)
∴ δ = (R2
=R2θ'
= – y (
∴ δ =– y∆θ
[ θ' − θ = θ + ∆θ
The normal strain ∈
the deformation δ by the original length r
∴ ∈x =
The normal stress σx
∴ σx =–E
i.e. σx = – E
Equation (vi) shows that the normal stress
with the distance y from the neutral surface. Plotting
obtain an arc of hyperbola as shown in Fig. 1.3.
From the condition of equilibrium the summation of forces over the
entire area is zero and the summation of the moments due to these forces
is equal to the applied bending moment.
∴ ∫δF =0
i.e., ∫σxdA =0
and ∫(– yσxdA)=M
– y) θ' – (R1 – y) θ
' – θ'y – R1θ + θy
y (θ' – θ) [ R1θ = R2θ' from equ (i)]
∆θ
θ = θ + ∆θ − θ = ∆θ] .....
∈x in the element of JK is obtained by dividing
by the original length r1θ of arc JK.
..... (v)
x may be obtained from Hooke's law
..... (vi)
( r1 = R1 – y) .(vii)
Equation (vi) shows that the normal stress σx does not vary linearly
with the distance y from the neutral surface. Plotting σx versus y, we
obtain an arc of hyperbola as shown in Fig. 1.3.
From the condition of equilibrium the summation of forces over the
entire area is zero and the summation of the moments due to these forces
is equal to the applied bending moment.
..... (viii)
..... (ix)
' from equ (i)]
..... (iv)
in the element of JK is obtained by dividing
..... (v)
may be obtained from Hooke's law σx = E∈x
..... (vi)
does not vary linearly
versus y, we
From the condition of equilibrium the summation of forces over the
entire area is zero and the summation of the moments due to these forces
..... (viii)
..... (ix)
Substituting the value of the σx from equation (vii) into equation
(viii)
– �∆t t � �J+zJ
zJ "dA=0
Since �∆t t is not equal to zero � �J+zJ
zJ "dA = 0
i.e. R1� �uz1 ! ���=0
i.e., . R1� �uz1 ! �=0
∴ R1 =u
��x�1
∴ It follows the distance R1 from the centre of curvature O to the
neutral surface is obtained by the relation R1 = u
��x�1 ..... (x)
The value of R1 is not equal to the �1� distance from O to the centroid
of the cross-section, since �1� is obtained by the relation,
�1� =5u � �1�� ..... (xi)
Hence it is proved that in a curved member the neutral axis of a
transverse section does not pass through the centroid of that section.
Now substitute the value of σx from equation (vii) into equations (ix)
� �∆t t �J+zJ
zJ "y dA =M
i.e., �∆t t � �J+zJ
zJ "/ dA = M ($ r1 = R1 - y from iii)
i.e., �∆t t � C�J�+/�JzJ�zJ�E
zJ/dA = M
i.e., �∆t t #�5/ � �u
zJ ! 2�5 ��� Q� �1��%= M
i.e., �∆t t #�5/ � u
�J ! 2�5� Q �5��% = M [using equations (x) and (xi)]
i.e., �∆t t ��5� ! 2�5� Q �5��� � v
i.e., �∆t t ��5�� ! �5�� =M
i.e., �∆t t =
|u�z1^̂ ^+�1� ..... (xii)
i.e., �∆t t =
|u� ($e = �1�– R1 from Fig. 1.2a)
………xiii)
Substituting �∆t t into equation (VI)
σx =|�
u���1+�� ..... (xiv)
∴ σx=M�z1+�1�
u�z1 ($ r1 = R1 – y ..... (xv)
An equation (xiv) is the general expression for the normal stress σx in a
curved beam.
To determine the change in curvature of the neutral surface caused by
the bending moment M
From equation (i),
5 �� � 5
�Jt� t
� 5 �J t�∆tt "
� 5 �J 1 Q ∆t
t "=5 �J #1 Q |
�u�%
{$From equation (xiii) �∆t t = |u� }
� 1 �5 Q
v����5
i.e. 5 �� ! 5
�J = M
EAeR1
∴ M =EAe R1 5 �� ! 5
�J"
Hence proved
References:
ASSIGNMENT
1. What are the assumptions made in finding stress distribution for a curved flexural member? Also
give two differences between a straight and curved beam
2. Discuss the stress distribution pattern in curved beams when compared to straight beam with
sketches
3. Derive an expression for stress distribution due to bending moment in a curved beam
EXERCISES
1. Determine the force F that will produce a maximum tensile stress of 60N/mm2 in section
A - B and the corresponding stress at the section C - D
2. A crane hook has a section of trapezoidal. The area at the critical section is 115 mm2. The hook
carries a load of 10kN and the inner radius of curvature is 60 mm. calculate the maximum tensile,
compressive and shear stress.
Hint: bi = 75 mm; b
o = 25 mm; h = 115 mm
3. A closed ring is made of 40 mm diameter rod bent to a mean radius of 85 mm. If the pull along
the diameter is 10,000 N, determine the stresses induced in the section of the ring along which it is
divided into two parts by the direction of pull.
4. Determine of value of t in the cross section of a curved beam shown in Figure such that the normal
stresses due to bending at the extreme fibers are numerically equal.
VTU,Jan/Feb.2005
Fig.1.35
5. Determine a safe value for load P for a machine element loaded as shown in Figure limiting the
maximum normal stress induced on the cross section XX to 120 MPa.
6. The section of a crane hook is trapezoidal, whose inner and outer sides are 90 mm and
25 mm respectively and has a depth of 116 mm. The center of curvature of the section is at a
distance of 65 mm from the inner side of the section and load line passes through the center of
curvature. Find the maximum load the hook can carry, if the maximum stress is not to exceed 70
MPa.
7. a) Differentiate between a straight beam and a curved beam with stress distribution in each of the
beam.
b) Figure shows a 100 kN crane hook with a trapezoidal section. Determine stress in the outer,
inner, Cg and also at the neutral fiber and draw the stress distribution across the section AB.
8. A closed ring is made up of 50 mm diameter steel bar having allowable tensile stress of 200
MPa. The inner diameter of the ring is 100 mm. For load of 30 kN find the maximum stress in
the bar and specify the location. If the ring is cut as shown in part -B of
Fig. 1.40, check whether it is safe to support the applied load.
F = 100kN
62.5
mm
A B
112.5
25
87
.5
REFERENCE BOOKS
• “MECHANICAL ENGINEERING DESIGN” by Farazdak Haideri
• “MACHINE DESIGN” by Maleev and Hartman
• “MACHINE DESIGN” by Schaum’s out lines
• “DESIGN OF MACHINE ELEMENTS” by V.B.Bhandari
• “DESING OF MACHINE ELEMENTS-2” by J.B.K Das and P.L. Srinivasa murthy
• “DESIGN OF SPRINGS” Version 2 ME, IIT Kharagpur, IITM