curved beams engineering
TRANSCRIPT
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CURVED BEAMS
CONTENT:
WHAT’S A CURVED
DIFFERENCE BET
CURVED BEAM
WHY STRESS CONC
CONCAVE SIDE OF
DERIVATION FOR S
PROBLEMS.
BEAM?
EEN A STRAIGHT BEA
NTRATION OCCUR AT INNE
URVED BEAM?
RESSES IN CURVED BEAM
AND A
R SIDE OR
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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 layeris 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 remainplane 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 e1f 1 be the neutral axis and g
1h
1 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
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G1H1 = (R+Y) Ѳ
GH
Also Stress
OR
dF = 0
∴∴∴∴ there is no direct force
RѲ
acting on the element considered.
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Since Σyδa is first mo
distance of centroid fro
centroid of the cross sec
axis.
From (1) and (2)
CURVED BEAM
Curved beams are
clamps, crane hooks, fra
machines, planers etc. I
coincides with its centr
is linear. But in the case
is shifted towards the c
linear [hyperbolic] distr
the centroidal axis and t
within the curved beams
ent of area about neutral axis,
neutral axis. Thus neutral axis c
tion. Cross sectional area coincides
the parts of machine members f
mes of presses, riveters, punches, s
straight beams the neutral axis o
idal axis and the stress distribution
of curved beams the neutral axis
entre of curvature of the beam ca
ibution of stress. The neutral axis
e centre of curvature and will alwa
.
yδa/a is the
incides with
with neutral
und in C -
ears, boring
f the section
in the beam
f the section
sing a non-
lies between
s be present
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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 fiberσbo = Bending stress at the outer fiber
σri = Combined stress at the inner fiber
σro = Combined stress at the outer fiber
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Stresses in curved beam
Mb = Applied Bending Mori = Inner radius of curved
ro = Outer radius of curved
rc = Radius of centroidal ax
rn = Radius of neutral axis
CL = Center of curvature
In the above figure th
before bending. i.e., wh
moment 'Mb' is applied
'ab' through an angle '
shortened while the inn
strip at a distance 'y' frothe amount ydθ and the
Where σ = stress, e = str
F
FM
b
entbeam
beam
is
lines 'ab' and 'cd' represent two
n there are no stresses induced. Wh
to the beam the plane cd rotates w
θ' to the position 'fg' and the out
r fibers are elongated. The origina
the neutral axis is (y + rn)θ. It istress in this fiber is,
σ = E.e
ain and E = Young's Modulus
CA
NA
c 2
c 1
c
i
c o
e
F
F
Mb
r i r n
rc
ro
CL
such planes
en a bending
th respect to
er fibers are
l length of a
shortened by
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We know, stress σ = E.e
We know, stress i.e., σ = – E
θθ ..... (i)
Since the fiber is shortened, the stress induced in this fiber iscompressive 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 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
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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. ∴
θθ
=0As
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 = – Substituting the value of dF, we getM
b =
θθ
=
θθ
= θ
θ Since represents the statical moment of area, it may be replacedby A.e., the product of total area A and the distance 'e' from the
centroidal axis to the neutral axis.
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∴ Mb =
θθ
A.e ..... (iii)
From equation (i) θ
θ = –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.
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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 thestress distribution in the beam
is linear.
In case of curved beams the
neutral axis of the section is
shifted towards the center ofcurvature of the beam causing
a non-linear stress
distribution.
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2
3 Neutral axi
axis coincides
Location of the neut
and centroidal Neutral axis
towards the lea
curvature
ral axis By considering a rectang
section
is shifted
t centre of
lar cross
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Centroidal and Neut al Axis of Typical Section of Cur ed Beams
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Why stress concent
curved beam
Consider the elemen
planes ‘ab’ and ‘cd’ sep
the plane cd having ro
Consider two fibers sy
axis. Deformation in bot
ration occur at inner side or co
ts of the curved beam lying betwe
arated by angle θ. Let fg is the fin
tated through an angle dθ about
metrically located on either side
h the fibers is same and equal to yd
cave side of
en two axial
l position of
neutral axis.
f the neutral
.
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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 thecentroidal axis. This results in a better stress distribution.
Example:
For numerical analysis, consider the depth of the section ass twice
the inner radius.
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For a straight beam
Inner most fiber:
Outer most fiber:
For curved beam:
e = rc - rn = h – 0.91
co = ro - rn= h – 0.9
ci = rn - ri = 0.910h -
:
h=2ri
h = 0.0898h
10h = 0590h
= 0.410h
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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 greaterthan 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 = 22X103Nb = 20mm
h = 150-50 = 100mm
A = bh = 20X100 = 2000mm2
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e = rc - rn = 1
Section A-B will be s
bending, due to the ecce
Stress due to direct load
y = rn – r = 91.0
Mb = 22X103X1
00 - 91.024 = 8.976mm
bjected to a combination of dir
tricity of the force.
will be,
24 – r
0 = 2.2X106 N-mm
ct load and
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Problem no.2Determine the value of
shown in fig such that t
fibers are numerically e
Given data;
Inner radius ri=150m
Outer radius ro=150+4
=290mm
Solution;
From Figure Ci + CO =
= 140mm…
Since the normal stresse
the extreme fiber are nu
have,
i.e Ci=
=
Radius of neutral axis
rn=
rn =197.727 mm
ai = 40mm; bi = 100mm;
ao = 0; bo = 0; ri = 150m
“t” in the cross section of a cur
he normal stress due to bending at
ual.
+100
0 + 100
……… (1)
s due to bending at
merically equal we
.51724Co…………… (2)
b2 =t;
; ro = 290mm;
ed beam as
the extreme
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=
i.e., 4674.069+83.
∴ t = 41.126mm
Problem no.3Determine the stresses
figure.
Solution:
The figure shows the cri
ring.Radius of centroidal axi
Inner radius of curved b
Outer radius of curved b
Radius of neutral axis
Applied force
1t = 4000+100t;
t point A and B of the split ring
tical section of the split
rc = 80mm
am ri = 80
= 50mm
eam ro = 80 +
= 110mm
rn =
=
= 77.081mm
F = 20kN = 20,000N (compress
hown in the
ve)
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Area of cross section A =π
d2=
π
x602 = 2827.433mm
2
Distance from centroidal axis to force = rc = 80mmBending moment about centroidal axis Mb = Fl = 20,000x80
=16x105
N-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=27.081mmDistance of neutral axis to outer radius
co = ro rn= 110 77.081=32.919mm
Direct stress σd = = = 7.0736N/mm
2
(comp.)
Bending stress at the inner fiber σbi = =
=
105N/mm
2(compressive)
Bending stress at the outer fiber σbo = =
= 58.016N/mm2 (tensile)
Combined stress at the inner fiber
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σri = σd + σbi
= 7.0736 105.00= - 112.0736N/mm
2(compressive)
Combined stress at the outer fiberσro = σd + σbo
= 7.0736+58.016= 50.9424N/mm
2 (tensile)
Maximum shear stress
τmax = 0.5x σmax
= 0.5x112.0736
= 56.0368N/mm2, at B
The figure.
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Problem No. 4
Curved bar of rectang
100mm is subjected t
straighten the bar. Findiagram to show the vari
Solution
Given data:
b= 40mm h= 60m
rc=100mm Mb= 2x1
C1=C2= 30m
rn=
ro= rc+h/2=100+30=13
ri= rc- h/2 = 100 - 30=
rn= 96.924mm
Distance of neutral axis
Distance of neutral axis
Distance of neutral axis
Area
A= b
Bending stress at the inn
lar section 40x60mm and a me
a bending moment of 2KN-m
the position of the Neutral axisation of stress across the section.
6N-mm
=(ri+c1+c2)
0mm (rc-c1)
to centroidal axise = rc - rn= 100-96.924
=3.075mm
o inner radius
ci= rn- ri = (c1-e) = 26.925
o outer radius
co=c2+e= (ro-rn) = 33.075m
h = 40x60 = 2400 mm2
er fiber σbi = =
= 104.239 N/mm2
(com
n radius of
tending to
and draw a
m
m
ressive)
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Bending stress at the outer fiber σbo = =
= -68.94 N/mm2 (tensile)
Bending stress at the centroidal axis =
=
= -8.33 N/mm2 (Compressive)
The stress distribution at the inner and outer fiber is as shown in the
figure.
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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 theextreme 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 numericallyequal 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
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b=150.34mm
To find the centroidal axis, (C2)
bo= 125.84mm; b=25mm; h=120mm
= 74.313mm.
But C1=C2rc= ro-c2 =240 - 74.313 =165.687mm
e=rc- rn = 165.687 - 160 = 5.6869 mm
Bending stress in the outer fiber,
σ
A= = 1050.4mm
60 =
Mb=10.8x106 N-mm
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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
= 50mm
Outer radius of curved beam ro = 80 + = 110mmRadius of neutral axis rn =
=
=77.081mmApplied force F = 20kN = 20,000N (compressive)
Area of cross section A =π
d2=
π
x602 = 2827.433mm
2
Distance from centroidal axis to force = rc = 80mmBending 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
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Distance of neutral axis to inner radius
ci = rn ri = 77.081 50 = 27.081mmDistance of neutral axis to outer radius
co = ro rn = 110 77.081 = 32.919mmDirect stress σd = =
= 7.0736N/mm2 (comp.)Bending stress at the inner fiber σbi =
=
= 105N/mm2 (compressive)Bending stress at the outer fiber σbo =
=
= 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/mm
2 (tensile)
Maximum shear stress
max = 0.5x σmax = 0.5x112.0736= 56.0368N/mm
2, at B
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The figure shows the stress distribution in the critical section.
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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+ = 140mm
Radius of neutral axis rn = =
= 136.1038mm
Distance of neutral axis to centroidal axis
e = rc - rn = 140-136.1038 = 3.8962mm
80
R 1 0 0
175 mm
9000N
h = 80mm
c2 e c1
b = 5 0 m m
CriticalSection
co ciro
rn
rc
r = 100mmi
1
CL
C A
N A
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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
= 175+ rc = 175+140 = 315mmApplied force F = 9000N
Area of cross section A = 50x80 = 4000mm
2
Bending moment about centroidal axis Mb = FI = 9000x315
= 2835000 N-mm
Direct stress σd = =
= 2.25N/mm2 (tensile)
Bending stress at the inner fiber σbi =
=
= 65.676N/mm2(tensile)
Bending stress at the outer fiber σbo = = = 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)
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Maximum shear stress max = 0.5x σmax = 0.5x67.926= 33.963 N/mm
2, 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
C A
N A
Bending stress =-44.362 N/mmσbo2
Combined stress -42.112 N/mmσro2
=
σd=2.25 N/mm2
Direct stress ( )σd
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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 = =
= 16.667mm
h = 40mm
c2e
c1
b = 6 m m
o
co ci
rorn
rc
r = 25mmi 100m
CL
b = 1 8 m m
i
C A
N A
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∴ Radius of centroidal axis rc = ri c1= 25+16.667 = 41.667 mm
Radius of neutral axis rn =
=
=38.8175mm
Distance of neutral axis to centroidal axis e = rc rn= 1.66738.8175=2.8495mm
Distance of neutral axis to inner radius ci = rn ri= 38.817525=13.8175mm
Distance of neutral axis to outer radius co = ro rn= 65-38.8175=26.1825mm
Distance from centroidal axis to force = 100+ rc = 100+41.667= 141.667mm
Applied force F = 5000N
Area of cross section A = = = 480mm2 Bending moment about centroidal axis Mb = FI = 5000x141.667
= 708335 N-mm
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Direct stress σd = =
= 10.417N/mm
2 (tensile)
Bending stress at the inner fiber σbi =
=
= 286.232N/mm2
(tensile)
Bending stress at the outer fiber σbo = =
= 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 = 10.417286.232= 198.189N/mm2 (compressive)
Maximum shear stress max = 0.5x σmax = 0.5x296.649= 148.3245 N/mm
2, at the inner fiber
The figure shows the stress distribution in the critical section.
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σbi=286.232 N/mm2
σri=296.649 N/mm2
b = 18 mmi
h =40 mm
C A
N A
Bending stress =-208.606 N/mmσbo2
Combined stress N/mmσro2
=-198.189
b = 6 mmo
σd=10.417 N/mm2
Direct stress (σd)
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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 =
ai = 75mm; bi = 300mm; b2 = 75mm; ao = 0; bo = 0
A=a1+a2=75x300+75x225 =39375mm2
∴ rn = = 333.217mmLet AB be the ref. line
5 5 0
7585 kN
300
2 5 0
750 mm75
B a = 7 5 m
i225 mm
a2
b =75mm2
b = 3 0 0 m m
i
ci
co
A
X
e rn
rc
r =550 mmo
C A
N A
CL
7
r = 250 mmi
a1
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=
= 101.785mm
Radius of centroidal axis rc = ri +
= 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 = 750+ rc = 750+351.785 = 1101.785mm
Applied force F = 85kNBending moment about centroidal axis Mb = FI
= 85000x1101.785
= 93651725N-mm
Direct stress σd = =
= 2.16N/mm
2 (tensile)
Bending stress at the inner fiber σbi = = = 42.64N/mm
2 (tensile)
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Bending stress at the outer fiber σbo = = = 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
max = 0.5x σmax = 0.5x48.33
= 24.165N/mm2, at the outer fiber
The below figure shows the stress distribution.
σ
bi
=42.64 N/mm2
σri=44.8 N/mm2
b = 3 0 0 m m
i
225
C A
N A
Bending stress =-50.49 N/mmσbo2
Combined stress N/mmσro2
=-48.33
a =75mmi
b = 75 mm2 a2
a1
σd=2.16 N/mm2
Direct stress ( )σd
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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 = =
= 40mm
90mm
30mm
F = 25 kN
25mm
60mm
1 0 0 m m
ri
rc
rn
ro
e
cico
N C A
c2
c1
h = 90 mm
l
F
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Radius of centroidal axis rc = ri + c1 = 100+40
= 140 mm
Radius of neutral axis rn =
=
= 135.42mm
Distance of neutral axis to centroidal axis e = rc rn= 140 135.42=4.58mmDistance 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 = rc 25= 140 25= 115mm
Applied force F = 25,000N = 25kN
Area of cross section A =
=
= 4050mm2 Bending moment about centroidal axis Mb = FI = 25,000x115
= 2875000 N-mm
Direct stress σd = =
= 6.173N/mm
2 (tensile)
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Bending stress at the inner fiber σbi = =
= 54.9 N/mm2 (tensile)
Bending stress at the outer fiber σbo = = = 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.
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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 = ---------------------- From DDHB
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C2 =
= 50mm
c1 = 90-50= 40mm
Moment of inertia I =
=
= 2632500mm
4
Direct stress σb = =
= 6.173N/mm
2 (tensile)
Bending stress at the inner fiber σbi =
=
= 43.685 N/mm
2(tensile)
Bending stress at the outer fiber σbo = - = = -54.606N/mm
2 (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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σbi= 43.685 N/mm
2
σri= 49.858 N/mm
2
6 0 m m
h =90 mm
N A , C A
σbo=-54.606 N/mm2
σro
N/mm2
=-48.433
b = 30 mm
c =50mm2 c =40mm1
b /2 = 15o
b /2 = 15o
σd= 6.173 N/mm
2
σd
b
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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 100mmfrom 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+
= 130mm
Radius of neutral axis rn =
=
= 153.045mm
Distance of neutral axis to centroidal axis e = rc - rn
= 155-153.045 = 1.955mm
h=60mm
b=30mm
F = ?
1 2 5 m m
100
h = 60mm
c2
ec
1
b
= 3 0 m m
co
ci
ro
rn
rc
F
r = 125mmi
CL
l
100
C A
N A
L o a d l i n e
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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 = =
Bending stress at the inner fiber σbi = +
Combined stress at the inner fiber σri = σ
d + σ
bi
i.e., 75 =
+
F = 8480.4N =Capacity of the hook.
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Problem no.12Design 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 =10tSelection the standard crane hook dimensions from table 25.3 when safe
load =10t and steel (MS)
∴ c =11933; Z = 14mm; M = 71mm andh = 111mm
bi= M = 7133
bo = 2xZ = 2x14 = 28 mm
r1 = = = 59.5mmh = 111mm
o
H
M
= b
i
Z
r =59.5 mmi
r =c l
rn
ro
e
cico
N A
C A
c2 c1
h = 111 mm
CLb =28o b=71i
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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 =
=
= 98.095mm
Distance of centroidal axis from inner fiber c1 =
=
= 47.465mmRadius 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
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Applied force F = 105N
Area of cross section A =
= = 5494.5mm2 Bending moment about centroidal axis Mb = Fl = 10
5x141.667
= 106.965x105N-mm
Direct stress σd = =
= 18.2N/mm2 (tensile)
Bending stress at the inner fiber σbi = =
= 142.365/mm2
(tensile)
Bending stress at the outer fiber σbo = =
= -93.2 N/mm
2
(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.
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σbi=142,365 N/mm
2
σri=160.565 N/mm
2
b = 71 mmi
C A
N A
σbo=-93.2 N/mm
2
σro
N/mm2
=-75
b = 28 mmo
h = 111 mm
σd=18.2 N/mm
2
σd
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Problem no.13The 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 = r o r i4
2 = 150 50
42=
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 = x d
2 =
x 100
2 = 7853.98mm
2
Mb = F = 5000 Combined maximum stress at the inner fiber
(i.e., at B)
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σri = Direct stress + bending stress
= ∴ x= 599.9 = Maximum offset distance.
Problem no.14An 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
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Draw the critical section
Radius of centroidal axi
Inner radius ri =100
Outer radius ro = 100+
Radius of neutral axis
rn = =
= 99.6mm
Distance of neutral axis
Distance of neutral axis
at P-Q as shown in the figure.
rc =100mm
= 87.5mm
= 112.5mm
rom centroidal axis e =rc - rn
=100 - 99.6 =
o inner fiber ci = rn – ri
= 99.6 – 87.5 =12.
0.4mm
1 mm
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Distance of neutral axis to outer fiber co = ro -rn
=112.5 – 99.6 = 12.9 mm
Area of cross-section A =π
4
d
2 =
π
4 x25
2= 490.87 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 -
M bCo
Aeo =
1000
49087 –100000 X129
49087 X 04 X 1125
= - 56.36 N/mm2 (compressive)
Combined stress at inner fibre (i.e., at p)
σri= Direct stress + bending stress
=F
A +
M bci
Aer i =
1000 49087 +
100000 X 12149087 X 04 X 875
= 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
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ro = 75 +25
2 = 87.5 mm
A =π
4
d2 =
π
4
X 252 = 490.87 mm2
rn = r o r i4
2 = 87 5 625
42 =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
–M b coAer o
=1000
49087 - 75000 X13024549087 X 05245 X 625 = - 41.324 N / mm
2
(compressive)
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Combined stress at the inner fiber (at S) = Direct stress + Bending stress
σri =F
A +
M bco
Aer o =
1000
49087 +75000 X 119755
49087 X 05245 X 625 = 55.816 N / mm2 (tensile)
∴ Maximum tensile stress = 72.466 N / mm2 at P Maximum compressive stress = 56.36 N / mm
2 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.
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The ring is symmetri
horizontal and vertical d
Consider the horizontal
A and B, the vertical for
No horizontal forces w
proved by understandin
symmetrical, the reactio
Assume that two horizo
half, as shown in the fi
half must have forces H
This however, results i
hence H must be zero.
of equal magnitude M0
noted that these mo
condition of symmetry.
be treated as that shown
quantity is M0. Againconclude that the tang
vertical and must rema
does not rotate. By Cast
derivative of the strai
displacement of the load
al and is loaded symmetrically
irections.
section as shown in the figure. At
es would be F/2.
uld be there at A and B. this argu
that since the ring and the extern
s too must be symmetrical.
tal inward forces H, act at A and B
gure. In this case, the lower
acting outwards as shown.
violation of symmetry and
esides the forces, moments
ct at A and B. It should be
ents do not violate the
hus loads on the section can
in the figure. The unknown
Considering symmetry, Weents at A and B must be
n so after deflection or M0
igliano’s theorem, the partial
energy with respect to the lo
. In this case, this would be zero.
……………………….(1)
in both the
the two ends
ment can be
al forces are
in the upper
d gives the
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The bending moment at
figure.
Will be
As per Castigliano’s the
From equation (2)
And, ds = Rdθ
any point C, located at angle θ, as s
………..(2)
rem,
own in the
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As this quantity is positi
produces tension in the i
It should be noted that t
θ = 0 to θ = 900.
The bending moment M
e the direction assumed for Mo is c
ner fibers and compression on the
ese equations are valid in the regio
at any angle θ from equation (2) w
orrect and it
outer.
,
ill be:
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It is seen that numericall
The stress at any angle
forces as shown in the fi
Put θ = 0 in Bending
get,
At A-A Mbi = 0.181
Mbo = - 0.18
And θ = 90,
At B-B Mbi = -
Mbo = 0.3
The vertical force F/
components (creates n
(creates shear stresses).
The combined norm
y, Mb-max is greater than Mo.
can be found by considering the
gure.
moment equation (4) then we will
FR
1FR
.318FR
18FR
can be resolved in two
rmal direct stresses) and S
l stress across any section will be:
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The stress at inner (σ1Ai)
On similar lines, the stre
outer points will be (at
It should be noted that i
that the radius is large
straight beam.
and outer points (σ1Ao) at A-A will
ss at the point of application of loa
= 900)
calculating the bending stresses,
ompared to the depth, or the bea
be (at = 0)
at inner and
t is assumed
is almost a
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A Thin Extended C
Consider a thin closed ri
the figure. At the two en
No horizontal forces w
ring.
The unknown quantity
conclude that the tange
so after deflection or M0 There are two region
The straight porti
Mb
The curved porti
osed Link
ng subjected to symmetrical load F
ds C and D, the vertical forces woul
uld be there at C and D, as disc
is M0. Again considering sy
ts at C and D must be vertical and
does not rotate.s to be considered in this case:
n, (0 < y < L) where
= MO
n, where
as shown in
d be F/2.
ssed earlier
metry, we
must remain
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As per Castigliano’s
\
theorem
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It can be observed that
as obtained for a circul
and Compression on the
The bending moment M
Noting that the equation
At = 0At section B-B bending
At section A-A the loa
bending moment occurs
part of the equation is m
It can be observed tha
expression as obtained f
It is seen that numericall
t L = 0 equation reduces to the sa
r ring. Mo produces tension in th
outer.
at any angle
will be
are valid in the region, = 0 to
oment at inner and outer side of t
point, i.e., at = p/2, the maxi(numerically), as it is observed th
ch greater than the first part.
at L = 0, equation (v) reduces
r a circular ring.
y, Mb-max is greater than Mo.
e expression
inner fibers
p/2,
e fiber is
um value of
t the second
to the same
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The stress at any angl
shown in the figure.
The vertical force F/2
normal direct stresses) a
The combined norm
The stress at inner fiber
be (at = 0):
The stress at inner fi
will be (at the loading p
can be found by considering
can be resolved in two compon
d S (creates shear stresses).
l stress across any section will be
σ1Bi and outer fiber σ1Bo and at sect
ber σ1Ai and outer fiber σ1Ao and at
int = 900):
the force as
nts (creates
ion B-B will
section A-A
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Problem 15
Determine the stress ind
of 25 mm diameter subj
diameter of the ring is 6
Solution: the circular ri
1.29a and 1.29b respecti
Inner radius ri = = 30
Outer radius = 30+25 =
Radius of centroidal axi
Radius of neutral axi
Distance of neutral axis
=
Distance of neutral axis
=
ced in a circular ring of circular cr
cted to a tensile load 6500N. The i
mm.
ng and its critical section are as sho
ely.
m
5mm
rc = 30 + = 42.5mm
rn =
= =42.5mm
o centroidal axis e = rc - rn
42.5 – 41.56 = 0.94mm
o inner radius ci = rn - ri
41.56 – 30 = 11.56mm
ss section
ner
n in fig.
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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 = θ
Consider the cross section A – A
At section A – A, θ = 900 with respect to horizontal
Direct stress σd =
= 0
Bending moment Mb = - 0.318Fr
Where r = rc, negative sign refers to tensile load
= - 0.318x6500x42.5 = -87847.5 N-mmThis couple produces compressive stress at the inner fiber and tensile
stress at the at outer fiber
Maximum stress at the inner fiber σ=Direct stress + Bending stress
= 0 - =
= - 73.36N/mm2(compressive)
Maximum stress at outer fiber σ= Direct stress + Bending stress
=0+ =
= 46.52N/mm2 (tensile)
Consider the cross section B – B
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At section B – B, θ = 00 with respect to horizontal
Direct stress σd =
= = 6.621 N/mm2
Bending moment Mb = 0.182Fr
Where r = rc, positive sign refers to tensile load
= 0.182x6500x42.5 = 50277.5 N-mmThis couple produces compressive stress at the inner fiber and tensile
stress at the at outer fiber
Maximum stress at the inner fiber σ=Direct stress + Bending stress= σd -
= 6.621 +
= 48.6 N/mm2
(tensile)
Maximum stress at outer fiber σ= Direct stress + Bending stress
= σd + =6.621+ = -20 N/mm
2 (compressive)
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Problem 16
Determine the stress ind
of 50 mm diameter rod s
mean diameter of the rin
Solution: the circular ri
1.30a and 1.30b respecti
Inner radius ri = -
Outer radius = +
Radius of centroidal axi
Radius of neutral axi
Distance of neutral axis
= 50 -
Distance of neutral axis
= 46.65
Distance of neutral axis
= 75 - 4
Area of cross section A
ced in a circular ring of circular cr
ubjected to a compressive load of 2
g is 100 mm.
ng and its critical section are as sho
ely.
25mm
75mm
rc = = 50mm
rn =
= = 46.65mm
o centroidal axis e = rc - rn
6.65 = 3.35mm
o inner radius ci = rn - ri
-25 = 21.65 mm
o outer radius co = ro - rn
6.65 = 28.35mm
= x552 = 1963.5mm
2
ss section
kNN. The
n in fig.
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Direct stress at any cross section at an angle θ with horizontal
σd = Consider the cross section A – A
At section A – A, θ = 900 with respect to horizontal
Direct stress σd =
= 0Bending moment Mb = + 0.318Fr
Where r = rc, positive sign refers to tensile load
= + 0.318x20000x50 = 318000 N-mmThis couple produces compressive stress at the inner fiber and tensile
stress at the at outer fiber
Maximum stress at the inner fiber
=Direct stress + Bending stress
= 0 + =
= 41.86 N/mm2
(tensile)
Maximum stress at outer fiber = Direct stress + Bending stress=0 -
= -
= - 18.27 N/mm2 (compressive)
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Consider the cross section B – B
At section B – B, θ = 00 with respect to horizontal
Direct stress σd =
=
= 5.093 N/mm2 (compressive)
Bending moment Mb = -0.1828Fr
Where r = rc, negative sign refers to tensile load
= - 0.182x20000x50 = -182000 N-mmThis couple produces compressive stress at the inner fiber and tensile
stress at the at outer fiber
Maximum stress at the inner fiber σ=Direct stress + Bending stress
= σd - = -5.093 + = - 29.05 N/mm
2(compressive)
Maximum stress at outer fiber σ= Direct stress + Bending stress
= σd + = -5.093 +
= 5.366 N/mm2 (tensile)
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Problem 17
A chain link is made of
mean diameter of which
80mm. The straight sidload of 90kN; estimate
along the section of loa
from the load line
Solution: refer figure
= 80mm; dc = 80mm;
F = 90kN = 90000N
Draw the critical cross s
Inner radius ri = 40 -
Outer radius = + =
Radius of centroidal axi
Radius of neutral axi
Distance of neutral axis
Distance of neutral axis
40 mm diameter rod is circular at
is 80mm. The straight sides of the
s of the link are also 80mm.If thethe tensile and compressive stres
line. Also find the stress at a secti
rc = 40mm;
ction as shown in fig.1.32
20mm
60mm
rc = 40mm
rn =
= = 37.32mm
o centroidal axis e = rc - rn
=40-37.32 = 2.68mm
o inner radius ci = rn - ri
= 37.32-20 = 17.32 mm
each end the
link are also
ink carries ain the link
on 900 away
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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 = 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 =
= 0
Bending moment = - π where r = rc,= π = 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=Direct stress + Bending stress= 0 +
=
π
= - 360 N/mm2(tensile)
Maximum stress at outer fiber σ= Direct stress + Bending stress
= 0 - = - π = 157.14 N/mm
2 (compressive)
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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 = θ
=
π = 35.81 N/mm2
(compressive)
Bending moment = - ππ where r = rc,= ππ = - 399655.7N-mm
This couple produces compressive stress at the inner fiber and tensilestress at the at outer fiber
Maximum stress at the inner fiber σ=Direct stress + Bending stress
= σd - = 35.81 +
π
= 138.578 N/mm2(tensile)
Maximum stress at outer fiber σ= Direct stress + Bending stress
= σd + = 35.81 -
π
= - 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.
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Using usual notations
curved beam of initia
uniform bending mom
Consider a curved bea
below. Its transverse se
in its unstressed state; it
xy plane along the arcs
Now apply two equal an
(c). The length of neut
central angles before anof neutral surface remai
R1θ =R2θ'
Consider the arc of circ
surface. Let r1 and r2 b
couples have been appli
From Fig. 1.2 a and
prove that the moment of resist
l radius R1 when bent to a ra
nt is
M = EAeR1
of uniform cross section as sho
tion is symmetric with respect to t
s upper and lower surfaces intersec
of circle AB and EF centered at
d opposite couples M and M' as sho
al surface remains the same. θ a
after applying the moment M. Sins the same
Figure
le JK located at a distance y abov
the radius of this arc before and
d. Now, the deformation of JK,
, r1 = R1 – y; r2 = R2 – y ..
nce M of a
dius R2 by
n in Figure
e y axis and
t the vertical
[Fig. 1(a)].
wn in Fig. 1.
d θ' are the
ce the length
..... (i)
e the neutral
fter bending
.... (ii)
... (iii)
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∴ δ = (R2
=R2θ'
= – y
∴ δ =– y∆ [ θ' − θ = θ + ∆
The normal strain ∈
the deformation δ by the
∴ ∈x =
The normal stress σx
∴ σx
=–E
i.e. σx = – E
Equation (vi) shows
with the distance y fro
obtain an arc of hyperbo
From the condition
entire area is zero and th
is equal to the applied b∴ ∫δF =0
i.e., ∫σxdA =0
and ∫(– yσxdA)=M
– y) θ' – (R1 – y) θ
– θ'y – R1θ + θy
θ' – θ) [ R1θ = R2θ' from equ (
θ − θ = ∆θ] ..
x in the element of JK is obtaine
original length r1θ of arc JK.
.
may be obtained from Hooke's law
..
( r1 = R1 – y) .(vii)
that the normal stress σx does not
the neutral surface. Plotting σx
la as shown in Fig. 1.3.
f equilibrium the summation of fo
e summation of the moments due t
nding moment.
....
..
)]
... (iv)
by dividing
.... (v)
σx = E∈x
... (vi)
vary linearly
ersus y, we
ces over the
these forces
. (viii)
... (ix)
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Substituting the value of the σx from equation (vii) into equation
(viii)
–
dA=0Since is not equal to zero dA = 0i.e. R1 1 =0i.e., . R1 1 =0
∴ R1 = 1
∴ It follows the distance R1 from the centre of curvature O to the
neutral surface is obtained by the relation R1 = 1 ..... (x)
The value of R1 is not equal to the 1 distance from O to the centroidof the cross-section, since 1 is obtained by the relation,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)
y dA =M
i.e.,
dA = M ( r1 = R1 - y from iii)
i.e.,
dA = M
i.e., = Mi.e.,
= M [using equations (x) and (xi)]
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i.e.,
i.e.,
=Mi.e.,
=
1 1 ..... (xii)
i.e., = (e = 1– R1 from Fig. 1.2a)………xiii)
Substituting
into equation (VI)
σx =
1
..... (xiv)
∴ σx=M11
1 ( 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 MFrom equation (i),
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= {
From equation (xiii)
=
}
i.e.
= MEAeR 1 ∴ M =EAe R1
Hence proved
References:
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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
4. Determine of value of t in the cross section of a curved beam shown in Figure such that the normalstresses due to bending at the extreme fibers are numerically equal.
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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 70MPa.
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.
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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 inthe 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
6 2. 5 m m
A B
112.5
2
5
8
7 .
5
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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