CURVED BEAMS CONTENT:
WHAT’S A CURVED BEAM?
DIFFERENCE BET CURVED BEAM
WHY STRESS CONC NTRA NTRATI TION ON OCCU OCCUR R AT AT INN INNE ER SIDE OR CONCAVE SIDE OF URVED BEAM?
DERIVATION FOR S RESSES IN CURVED BEAM
PROBLEMS.
EEN
A
STRAIGHT
BEA
AND
A
Theory of Simple Bending Due to bending moment, tensile stress develops in one portion of section and compressive stress in the other portion across the depth. In between these two portions, there is a layer where stresses are zero. Such a layer is called neutral layer. Its trace on the cross section is called neutral axis.
Assumption The material of the beam is perfectly homogeneous and isotropic. The cross section has an axis of symmetry in a plane along the length of the beam. The material of the beam obeys Hooke’s law. The transverse sections which are plane before bending remain plane after bending also. Each layer of the beam is free to expand or contract, independent of the layer above or below it. Young’s modulus is same in tension & compression.
Consider a portion of beam between sections AB and CD as shown in the figure. Let e1f 1 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
RѲ
Also Stress
OR dF = 0
∴
there is no direct force acting on the element considered.
Since Σyδa is first mo ent of area about neutral axis, yδa/a is the dist distan ance ce of cent centro roid id fro fro neutral axis. Thus neutral axis c incides with centroid of the cross section. Cross sectional area coincides with neutral axis.
From (1) and (2)
CURVED BEAM
Curved beams are the parts of machine members f und in C clamps, crane hooks, frames of presses, riveters, punches, s ears, boring machi machines nes,, plane planers rs etc etc.. I straight straight beams beams the neutral neutral axis of the section coincides with its centr idal axis and the stress distribution in the beam is linear. But in the case of curved beams the neutral axis f the section is shifted towards the centre of curvature of the beam ca sing a nonlinear [hyperbolic] distribution of stress. The neutral axis lies between the the cent centro roiidal dal axis axis and and t e cen centr tree of of curv curvat atur uree and and will will alwa alwa s be present within the curved beams.
Stresses in Curved Beam
Consider a curved beam subjected to bending moment M b 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
CA NA F
2 c
1 c
c
o
e c
i
F
F M
b
F
M
b
r i
C L
r n r c ro
Stresses in curved beam
Mb = Applied Bending Mo ent 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 th lines 'ab' and 'cd' represent two such planes before bending. i.e., wh n there are no stresses induced. When a bending moment 'Mb' is applied to the beam the plane cd rotates w th respect to 'ab' through an angle ' θ' to the position 'fg' and the outer fibers are shortened while the inn r fibers are elongated. The original length of a strip at a distance 'y' fro the neutral axis is (y + rn)θ. It is shortened by the amount ydθ and the tress in this fiber is, σ = E.e Where σ = stress, e = strain and E = Young's Modulus
We know, stress σ = E.e We know, stress i.e., σ
=–E
θ
..... (i)
θ
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
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.
θ
∴ As
θ
θ
=0
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 get Mb
= = = θ
θ
θ
θ
θ
θ
Since 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.
∴
A.e θ
Mb =
From equation (i)
θ
=–
θ
. A. e.
σ
Mb = – σ =
σ
Substituting in equation (iii)
∴
..... (iii)
θ
..... (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 = – c i ∴ 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 1
straight beam In Straight beams the neutral axis of the section coincides with its centroidal axis and the stress distribution in the beam is linear.
curved beam 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 axi and centroidal Neutral axis is shifted axis coincides towards the lea t centre of curvature
Location of the neutral axis By considering a rectang lar cross section
Centroidal and Neut al Axis of Typical Section of Cur ed Beams
Why stress concentration occur at inner side or co cave side of curved beam
Consider the elements of the curved beam lying between two axial planes ‘ab’ and ‘cd’ separated by angle θ. Let fg is the fin l position of the plane cd having rotated through an angle dθ about neutral axis. Consider two fibers sy metrically located on either side f the neutral axis. Deformation in both the fibers is same and equal to yd .
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=2ri
e = rc - rn = h – 0.91 h = 0.0898h co = ro - rn=
h – 0.910h = 0590h
ci = rn - ri = 0.910h -
= 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 3 F = 22X10 N b = 20mm h = 150-50 = 100mm 2 A = bh = 20X100 = 2000mm
e = rc - rn = 100 - 91.024 = 8.976mm Section A-B will be s bjected to a combination of dir ct load and bending, due to the ecce tricity of the force. Stress due to direct load will be,
y = rn – r = 91.024 – r 3
6
Mb = 22X10 X1 0 = 2.2X10 N-mm
Problem no.2 Determine the value of “t” in the cross section of a cur ed beam as shown in fig such that the normal stress due to bending at the extreme fibers are numerically e ual.
Given data; Inner radius ri=150m Outer radius ro=150+4 +100 =290mm Solution; From Figure Ci + CO = 0 + 100 = 140mm…
……… (1)
Since the normal stresses due to bending at the extreme fiber are numerically equal we have,
i.e
Ci= = .51724Co…………… (2)
Radius of neutral axis rn=
rn =197.727 mm ai = 40mm; bi = 100mm; b2 =t; ao = 0; bo = 0; ri = 150m ; ro = 290mm;
=
i.e.,
∴
4674.069+83. 1t = 4000+100t;
t = 41.126mm
Problem no.3 Determine the stresses t point A and B of the split ring hown in the figure.
Solution: The figure shows the critical section of the split ring. Radius of centroidal axi rc = 80mm Inner radius of curved b am
ri = 80 = 50mm
Outer radius of curved beam
ro = 80 + = 110mm
Radius of neutral axis
rn = =
Applied force
= 77.081mm F = 20kN = 20,000N (compress ve)
π 2
Area of cross section
π
2
A = d = x60 = 2827.433mm
2
Distance from centroidal axis to force = rc = 80mm Bending moment about centroidal axis M b = Fl = 20,000x80 5 =16x10 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=27.081mm Distance of neutral axis to outer radius co = ro rn = 110 77.081=32.919mm
Direct stress
σd =
=
=
2
7.0736N/mm (comp.)
= =
Bending stress at the inner fiber σbi
=
2
105N/mm (compressive)
= =
Bending stress at the outer fiber σbo
2
= 58.016N/mm (tensile) Combined stress at the inner fiber
σri = σd + σbi
= 7.0736 105.00 2 = - 112.0736N/mm (compressive) Combined stress at the outer fiber σro = σd + σbo = 7.0736+58.016 2 = 50.9424N/mm (tensile) Maximum shear stress τmax = 0.5x σmax = 0.5x112.0736 2 = 56.0368N/mm , at B
The figure.
Problem No. 4 Curved bar of rectang lar section 40x60mm and a me n radius of 100mm is subjected t a bending moment of 2KN-m tending to straighten the bar. Fin 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= 60m 6 rc=100mm Mb= 2x1 N-mm C1=C2= 30m rn= ro= rc+h/2=100+30=13 =(ri+c1+c2) ri= rc- h/2 = 100 - 30= 0mm (rc-c1) rn= 96.924mm Distance of neutral axis to centroidal axis e = rc - rn= 100-96.924 =3.075mm Distance of neutral axis o inner radius ci= rn- ri = (c1-e) = 26.925 m Distance of neutral axis o outer radius co=c2+e= (ro-rn) = 33.075mm Area 2 A= b h = 40x60 = 2400 mm Bending stress at the inner fiber σbi =
= 2
= 104.239 N/mm (com ressive)
Bending stress at the outer fiber σbo =
=
2
= -68.94 N/mm (tensile)
Bending stress at the centroidal axis =
= 2
= -8.33 N/mm (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,
C /r i i =co /ro =c /c i o =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
A=
= 1050.4mm 60 =
6
Mb=10.8x10 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 r i = 80
Outer radius of curved beam ro = 80 +
= 50mm
= 110mm
Radius of neutral axis r = = =77.081mm n
Applied force
F = 20kN = 20,000N (compressive) π 2
π
2
2
Area of cross section A = d = x60 = 2827.433mm
Distance from centroidal axis to force = rc = 80mm Bending moment about centroidal axis M b = FI = 20,000x80 5
=16x10 N-mm Distance of neutral axis to centroidal axis e = r c
= 80
rn 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 σd =
Direct stress
=
77.081 = 32.919mm
=
2
7.0736N/mm (comp.)
Bending stress at the inner fiber σbi
= = =
2
105N/mm (compressive)
= =
Bending stress at the outer fiber σbo
2
= 58.016N/mm (tensile) Combined stress at the inner fiber
σri = σd + σbi =
=
7.0736
105.00 2
112.0736N/mm (compressive)
Combined stress at the outer fiber σro = σd + σb =
7.0736+58.016 2
= 50.9424N/mm (tensile) Maximum shear stress
max =
0.5x σmax = 0.5x112.0736 2
= 56.0368N/mm , 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: R 1 0 0
Draw the critical section as shown the figure.
9000N
in
80
Inner radius of curved beam ri = 100mm
175 mm
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
F
h = 80mm c2
e
c1
m m 0 5 = Critical b
Section co
ri = 100mm
ci
175mm
ro A A C N
rn rc C L
F
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
Applied force
= 175+ rc = 175+140 = 315mm F = 9000N 2
Area of cross section A = 50x80 = 4000mm
Bending moment about centroidal axis M b = FI = 9000x315 = 2835000 N-mm
= 2.25N/mm (tensile) = Bending stress at the inner fiber σ = Direct stress
2
σd = =
bi
2
= 65.676N/mm (tensile) Bending stress at the outer fiber σbo = =
=
2
44.326N/mm (compressive)
Combined stress at the inner fiber σri = σd + σbi = 2.25+65.676 2
= 67.926N/mm (tensile)
Combined stress at the outer fiber σro = σd + σbo = 2.25 =
2
44.362
42.112 N/mm (compressive)
Maximum shear stress
max =
0.5x σmax = 0.5x67.926 2
= 33.963 N/mm , at the inner fiber The stress distribution on the critical section is as shown in the figure.
σri
Combined stress
σro
=-42.112 N/mm
=67.926 N/mm2
2
σbi
Bending stress σ bo=-44.362 N/mm
=65.676 N/mm
2
Direct stress (σd)
σd
=2.25 N/mm
A A C N
b = 50 mm
h =80 mm
2
2
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: c2
F
h = 40mm e
c1
m m 6 = o b
Draw the critical section as shown in the figure.
m m 8 1 = i b
co
Inner radius of curved beam r i = 25mm
ci
ri = 25mm
100mm
rc ro A A C N
Outer radius of curved beam ro = 25+40 = 65mm
Distance of centroidal axis from inner fiber c = 1
=
= 16.667mm
rn C L
F
∴
Radius of centroidal axis rc = ri
c1
= 25+16.667 = 41.667 mm
Radius of neutral axis r = = =38.8175mm n
Distance of neutral axis to centroidal axis e = r c rn
= 1.667 38.8175 =2.8495mm
Distance of neutral axis to inner radius c i = rn
ri
= 38.8175 25=13.8175mm
Distance of neutral axis to outer radius c o = 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 =
= = 480mm
2
Bending moment about centroidal axis M b = FI = 5000x141.667 = 708335 N-mm
= 10.417N/mm (tensile) Bending stress at the inner fiber σ = = Direct stress
2
σd = =
bi
2
= 286.232N/mm (tensile) Bending stress at the outer fiber σbo = =
=
2
208.606N/mm (compressive)
Combined stress at the inner fiber σri = σd + σbi = 10.417+286.232 2
= 296.649N/mm (tensile)
Combined stress at the outer fiber σro = σd + σbo = 10.417 286.232 = Maximum shear stress
max =
2
198.189N/mm (compressive)
0.5x σmax = 0.5x296.649 2
= 148.3245 N/mm , at the inner fiber
The figure shows the stress distribution in the critical section.
σri
Combined stress
=296.649 N/mm
2
=-198.189 N/mm
σro
σbi
Bending stress
2
=-208.606 N/mm
σbo
=286.232 N/mm
2
2
=10.417 N/mm
Direct stress (σd)
σd
A A C N
bo = 6 mm
bi = 18 mm mm
h =40 mm
2
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
75
750 mm 300
75
85 kN
2 5 0 0 5 5
Solution: Draw the critical section as shown in the figure. Inner radius of curved beam r i = 250mm
m 5 7 = a
225 mm
i
F
B
a1
b2 =75mm
m m 0 0 3 = i b
a2
Outer radius of curved beam ro = 550mm co
ci X
A ri = 250 250 mm rn
e
rc
A A C N
Radius of neutral axis rn =
ai = 75mm; bi = 300mm; b2 = 75mm; ao = 0; bo = 0 2
A=a1+a2=75x300+75x225 =39375mm
∴
rn
=
= 333.217mm
Let AB be the ref. line
750
ro=550 mm
C L
F
= = 101.785mm Radius of centroidal axis r c = ri +
= 250+101.785=351.785 mm
Distance of neutral axis to centroidal axis e = r c
rn
= 351.785-333.217=18.568mm Distance of neutral axis to inner radius c i = rn
ri
= 333.217 Distance of neutral axis to outer radius c o = ro
= 550
250=83.217mm
rn 333.217=216.783mm
Distance from centroidal axis to force = 750+ rc = 750+351.785 = 1101.785mm Applied force
F = 85kN
Bending moment about centroidal axis M b = FI = 85000x1101.785 = 93651725N-mm Direct stress
σ = = = 2.16N/mm (tensile) 2
d
Bending stress at the inner fiber σbi =
= 2
= 42.64N/mm (tensile)
Bending stress at the outer fiber σbo
= = =
2
50.49N/mm (compressive)
Combined stress at the inner fiber σri = σd + σbi = 2.16+42.64 2
= 44.8N/mm (tensile)
Combined stress at the outer fiber σro = σd + σbo = 2.16 =
Maximum shear stress
max =
50.49
2
48.33N/mm (compressive)
0.5x σmax = 0.5x48.33 2
= 24.165N/mm , at the outer fiber The below figure shows the stress distribution. σri
Combined stress
σro
=44.8 N/mm
2
2
=-48.33 N/mm
σbi
2
=42.64 N/mm
2
Bending stress σbo =-50.49 N/mm
Direc t stre ss ss (σd)
σd
A A C N
b2 = 75 mm mm
a2
225
a =75 =75mm mm i
a1
m m 0 0 3 = i b
2
=2.16 N/mm
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 1 0 0 m straight beam for the given load? m 90mm 30mm
60mm
25mm
F = 25 kN
A C N
Solution:
h = 90 mm c2
c1
Draw the critical section as shown in the figure. Inner radius of curved beam r i = 100mm Outer radius of curved beam ro = 100+90 = 190mm Distance of centroidal axis from inner fiber c1 =
=
= 40mm
ci
co
ri rn
e
rc ro l
F
C L
Radius of centroidal axis rc = ri + c1 = 100+40 = 140 mm
Radius of neutral axis r = = = 135.42mm n
Distance of neutral axis to centroidal axis e = r c
rn
= 140
135.42=4.58mm
Distance of neutral axis to inner radius c i = rn
ri = 135.42
100
=35.42mm Distance of neutral axis to outer radius c o = 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 =
= = 4050mm
2
Bending moment about centroidal axis M b = FI = 25,000x115 = 2875000 N-mm Direct stress
= 6.173N/mm (tensile)
σd = =
2
=
Bending stress at the inner fiber σbi =
2
= 54.9 N/mm (tensile) Bending stress at the outer fiber σbo =
=
2
=
44.524N/mm (compressive)
Combined stress at the inner fiber σri = σd + σbi = 6.173+54.9 2
= 61.073N/mm (tensile)
Combined stress at the outer fiber σro = σd + σbo = 6.173 = Maximum shear stress
44.524
2
38.351N/mm (compressive)
τmax = 0.5x σmax = 0.5x61.072 2
= 30.5365 N/mm , 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 2
A = 4050 mm
2
Mb = 28750000 N/mm Also C2 =
---------------------- From DDHB
C2 =
= 50mm
c1 = 90-50= 40mm
Moment of inertia I = = 4
= 2632500mm
= 6.173N/mm (tensile) = Bending stress at the inner fiber σ = Direct stress
2
σb = =
bi
2
= 43.685 N/mm (tensile) Bending stress at the outer fiber σbo = -
= 2
= -54.606N/mm (compressive) Combined stress at the inner fiber σri = σd + σbi = 6.173+43.685 2
= 49.858N/mm (tensile) Combined stress at the outer fiber σro = σd + σbo = 6.173-54.606 2
= -48.433N/mm (compressive) The stress distribution on the straight beam is as shown in the figure.
σri
2
= 49.858 N/mm
2
=-48.433 N/mm
σro
σbi
σbo
2
= 43.685 N/mm
2
=-54.606 N/mm
σd
σd
2
= 6.173 N/mm
c =40mm
c =50mm
1
2
b /2 = 15 o
b = 30 mm
A C , A N
m m 0 6
b
b /2 = 15 o
h =90 mm
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 2 tension is 75N/mm
h=60mm b=30mm
100
1 2 5 m m
Solution: F = ?
h = 60mm
Draw the critical section as shown in the figure. Inner radius of curved beam r i = 125mm Outer radius of curved beam ro = 125+60
c2
c1 e n i l d a o L
m m 0 3 = b
= 185mm Radius of centroidal axis rc =100+
e
100 co
ri = 125mm
ci ro
rn rc l
= 130mm
Radius of neutral axis r = =
A A C N
n
= 153.045mm
Distance of neutral axis to centroidal axis e = r c - rn = 155-153.045 = 1.955mm
F
C L
Distance of neutral axis to inner radius c i = rn - ri = 153.045-125 = 28.045mm Distance of neutral axis to outer radius c o = ro - rn = 185-153.045 = 31.955mm Distance from centroidal axis to force l = r c -25 = 155-25 = 130mm 2
Area of cross section A = bh = 30x60 = 1800mm
Bending moment about centroidal axis M b = Fl = Fx130 = 130F Direct stress
σd = =
Bending stress at the inner fiber σbi =
+
Combined stress at the inner fiber σri = σd + σbi i.e., 75 =
+
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.
H i
o
b = M
Z
5
Data: Load capacity F = 100kN = 10 N; Trapezoidal section; FS = 2 Solution: Approximately 1kgf = 10N
∴
5
10 = 10,000 kgf =10t
Selection the standard crane hook dimensions from table 25.3 when safe load =10t and steel (MS)
∴
A A C N
c =11933; Z = 14mm; M = 71mm and h = 111mm
h = 111 mm c2 c1
bi= M = 7133 bo = 2xZ = 2x14 = 28 mm
= 59.5mm
r1 = =
h = 111mm
bo =28
C L
b=71 i ci
co e
ri=59.5 mm rn rc= l ro
Assume the load line passes through the centre of hook. Draw the critical section as shown in the figure. Inner radius of curved beam r i = 59.5mm Outer radius of curved beam ro = 59.5+111 = 170.5mm
Radius of neutral axis r = = n
= 98.095mm
= 47.465mm =
Distance of centroidal axis from inner fiber c 1 =
Radius of centroidal axis rc = ri + c1 = 47.465+59.5= 106.965 mm Distance of neutral axis to centroidal axis e = r c - rn =106.965-98.095 =8.87mm Distance of neutral axis to inner radius c i = rn - ri = 98.095-59.5=38.595mm Distance of neutral axis to outer radius c o = ro - rn = 170.5-98.095=72.0405mm Distance from centroidal axis to force l = r c -106.965
Applied force
5
F = 10 N
= = 5494.5mm
Area of cross section A =
2
5
Bending moment about centroidal axis M b = Fl = 10 x141.667 5
= 106.965x10 N-mm Direct stress
σd = =
2
= 18.2N/mm (tensile)
Bending stress at the inner fiber σ = = bi
2
= 142.365/mm (tensile) Bending stress at the outer fiber σbo
= = 2
= -93.2 N/mm (compressive) Combined stress at the inner fiber σri = σd + σbi = 18.2+142.365 2
= 160.565N/mm (tensile) Combined stress at the outer fiber σro = σd + σbo = 18.2-93.2 2
= -75N/mm (compressive) Maximum shear stress
τmax = 0.5x σmax = 0. 160.565 2
= 80.2825 N/mm , at the inner fiber The figure shows the stress distribution in the critical section.
σri
σro
=160.565 N/mm
2
2
=-75 N/mm
=142,365 N/mm
σbi
=-93.2 N/mm
σbo
2
σd
σd
h = 111 mm
A A C N
b = 28 mm o
b = 71 mm i
2
=18.2 N/mm
2
Problem no.13 The figure shows a loaded offset bar. What is the maximum offset 2 distance ’x’ if the allowable stress in tension is limited to 50N/mm
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 r = ro
n
2
4
= 150
4
93.3mm
ri
50
2
=
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
2
2
2
A = x d = x 100 = 7853.98mm
Mb = F = 5000
Combined maximum stress at the inner fiber (i.e., at B)
σ ri= Direct stress + bending stress
=
∴
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-Q as shown in the figure . Radius of centroidal axi
rc =100mm
Inner radius ri =100
= 87.5mm
Outer radius ro = 100+
= 112.5mm
Radius of neutral axis
rn =
=
= 99.6mm Distance of neutral axis rom centroidal axis e =rc - rn =100 - 99.6 = 0.4mm Distance of neutral axis o inner fiber
ci = rn – ri = 99.6 – 87.5 =12.1 mm
Distance of neutral axis to outer fiber c o = ro -rn =112.5 – 99.6 = 12.9 mm Area of cross-section
A=
Distance from centroidal axis
π 4
2
π
2
2
d = x25 = 490.87 mm 4
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 F
M b Co
A
Aeo
σro= -
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
M b ci
A
Aer i
= +
=
1000
+
490 87
100000 X 12 1
490 87 X 0 4 X 87 5
= 72.466 N/MM2 (tensile) (ii) Consider the section R -S Redraw the critical section at R –S as shown in fig.
rc = 75mm ri = 75 -
25 2
=62.5 mm
25
ro = 75 + = 87.5 mm 2
A=
π 4
d
rn =
2
=
π 4
X 25
2
= 490.87 mm2
r o
ri
2
4
= = 87 5
62 5
2
4
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 F
M b co
A
Aer o
σro = –
=
1000
490 87
-
75000 X13 0245
490 87 X 0 5245 X 62 5
= - 41.324 N / mm2 (compressive)
Combined stress at the inner fiber (at S) = Direct stress + Bending stress F
M b co
A
Aer o
σri = +
=
1000
490 87
+
75000 X 11 9755
490 87 X 0 5245 X 62 5
= 55.816 N / mm2 (tensile)
∴
2
Maximum tensile stress = 72.466 N / mm at P 2
Maximum compressive stress = 56.36 N / mm 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 symmetri al 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 for es would be F/2. No horizontal forces w uld be there at A and B. this argument can be proved by understandin that since the ring and the external forces are symmetrical, the reactio s too must be symmetrical. Assume that two horizo tal 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 i violation of symmetry and hence H must be zero. esides the forces, moments of equal magnitude M0 ct at A and B. It should be noted that these mo ents do not violate the condition of symmetry. hus loads on the section can be treated as that shown in the figure. The unknown quantity is M0. Again Considering symmetry, We conclude that the tangents at A and B must be vertical and must rema n so after deflection or M0 does not rotate. By Castigliano’s theorem, the partial derivative of the strai energy with respect to the lo d gives the displacement of the load. In this case, this would be zero.
……………………….(1)
The bending moment at any point C, located at angle θ, as s own in the figure. Will be
………..(2)
As per Castigliano’s the rem,
From equation (2)
And,
ds = Rdθ
As this quantity is positi e the direction assumed for Mo is correct and it produces tension in the i ner fibers and compression on the outer. It should be noted that t ese equations are valid in the regio , 0
θ = 0 to θ = 90 .
The bending moment M at any angle θ from equation (2) will be:
It is seen that numerically, Mb-max is greater than Mo. The stress at any angle can be found by considering the forces as shown in the fi gure. 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 = - .318FR Mbo = 0.318FR The vertical force F/ can be resolved in two components (creates n rmal direct stresses) and S (creates shear stresses).
The combined norm l stress across any section will be:
The stress at inner (σ1Ai) and outer points (σ1Ao) at A-A will be (at = 0)
On similar lines, the stress at the point of application of loa at inner and 0 outer points will be (at = 90 )
It should be noted that i calculating the bending stresses, t is assumed that the radius is large ompared to the depth, or the bea is almost a straight beam.
A Thin Extended C osed 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 w uld be there at C and D, as disc ssed earlier ring. The unknown quantity is M0. Again considering sy metry, we conclude that the tange ts at C and D must be vertical and must remain so after deflection or M0 does not rotate. There are two regions to be considered in this case: The straight porti n, (0 < y < L) where Mb = MO The curved porti n, where
\
As per Castigliano’s theorem
It can be observed that t L = 0 equation reduces to the sa e expression as obtained for a circul r ring. Mo produces tension in th inner fibers and Compression on the outer. The bending moment M at any angle will be
Noting that the equation are valid in the region,
At = 0 At section B-B bending
= 0 to
p/2,
oment at inner and outer side of t e fiber is
At section A-A the loa point, i.e., at = p/2, the maxi um value of bending moment occurs (numerically), as it is observed th t the second part of the equation is m ch greater than the first part.
It can be observed tha at L = 0, equation (v) reduces to the same expression as obtained f r a circular ring. It is seen that numerically, Mb-max is greater than Mo.
The stress at any angl shown in the figure.
can be found by considering the force as
The vertical force F/2 can be resolved in two compon nts (creates normal direct stresses) a d S (creates shear stresses).
The combined norm l stress across any section will be
The stress at inner fiber σ1Bi and outer fiber σ1Bo and at section B-B will be (at = 0):
The stress at inner fiber σ1Ai and outer fiber σ1Ao and at section A-A 0 will be (at the loading p int = 90 ):
Problem 15 Determine the stress ind ced in a circular ring of circular cr ss section of 25 mm diameter subj cted to a tensile load 6500N. The i ner diameter of the ring is 6 mm. Solution: the circular ring and its critical section are as sho n in fig. 1.29a and 1.29b respecti ely. Inner radius ri =
= 30 m
Outer radius = 30+25 = 5mm Radius of centroidal axi rc = 30 +
= 42.5mm
Radius of neutral axi rn = =
=42.5mm
Distance of neutral axis o centroidal axis e = rc - rn = 42.5 – 41.56 = 0.94mm Distance of neutral axis o inner radius ci = rn - ri = 41.56 – 30 = 11.56mm
Distance of neutral axis to outer radius c o = 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 0
At section A – A, θ = 90 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-mm
This couple produces compressive stress at the inner fiber and tensile stress at the at outer fiber
=Direct stress + Bending stress = 0 - =
Maximum stress at the inner fiber σ
2
= - 73.36N/mm (compressive)
= Direct stress + Bending stress =0+ =
Maximum stress at outer fiber σ
2
= 46.52N/mm (tensile) Consider the cross section B – B
0
At section B – B, θ = 0 with respect to horizontal Direct stress
σd =
= = 6.621 N/mm
2
Bending moment Mb = 0.182Fr Where r = rc, positive sign refers to tensile load
= 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
=Direct stress + Bending stress
= σ = 6.621 + d
2
= 48.6 N/mm (tensile) Maximum stress at outer fiber σ
= Direct stress + Bending stress = σ + =6.621+ d
2
= -20 N/mm (compressive)
Problem 16 Determine the stress ind ced in a circular ring of circular cr ss section of 50 mm diameter rod subjected to a compressive load of 2 kNN. The mean diameter of the ring is 100 mm. Solution: the circular ring and its critical section are as sho n in fig. 1.30a and 1.30b respecti ely. Inner radius ri = Outer radius =
+
25mm 75mm
Radius of centroidal axi rc =
= 50mm
Radius of neutral axi rn = =
= 46.65mm
Distance of neutral axis o centroidal axis e = rc - rn = 50 - 6.65 = 3.35mm Distance of neutral axis o inner radius ci = rn - ri = 46.65-25 = 21.65 mm Distance of neutral axis o outer radius co = ro - rn = 75 - 46.65 = 28.35mm Area of cross section A =
2
2
x55 = 1963.5mm
Direct stress at any cross section at an angle θ with horizontal σd =
Consider the cross section A – A 0
At section A – A, θ = 90 with respect to horizontal Direct stress
σd =
= 0
Bending moment Mb = + 0.318Fr Where r = rc, positive sign refers to tensile load
= + 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 =0+
=Direct stress + Bending stress
= 2
= 41.86 N/mm (tensile) Maximum stress at outer fiber =0 -
= Direct stress + Bending stress
= - 2
= - 18.27 N/mm (compressive)
Consider the cross section B – B 0
At section B – B, θ = 0 with respect to horizontal Direct stress
σd =
= 2
= 5.093 N/mm (compressive) Bending moment Mb = -0.1828Fr Where r = rc, negative sign refers to tensile load
= - 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
=Direct stress + Bending stress
= σ = -5.093 + d
2
= - 29.05 N/mm (compressive)
= Direct stress + Bending stress = σ + = -5.093 +
Maximum stress at outer fiber σ
d
2
= 5.366 N/mm (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 sid s of the link are also 80mm.If the ink carries a load of 90kN; estimate the tensile and compressive stres in the link 0 along the section of loa line. Also find the stress at a section 90 away from the load line Solution: refer figure = 80mm; dc = 80mm;
rc = 40mm;
F = 90kN = 90000N Draw the critical cross s ction as shown in fig.1.32 Inner radius ri = 40 Outer radius =
+
20mm = 60mm
Radius of centroidal axi rc = 40mm Radius of neutral axi rn = =
= 37.32mm
Distance of neutral axis o centroidal axis e = rc - rn =40-37.32 = 2.68mm Distance of neutral axis o inner radius ci = rn - ri = 37.32-20 = 17.32 mm
Distance of neutral axis to outer radius c o = 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] 0
At section A – A, θ = 90 with respect to horizontal
= 0 where r = r , Bending moment = - = = 1.4x10 N-mm Direct stress
σd =
c
π
6
π
This couple produces compressive stress at the inner fiber and tensile stress at the at outer fiber
=Direct stress + Bending stress =0+ =
Maximum stress at the inner fiber σ π
2
= - 360 N/mm (tensile)
= Direct stress + Bending stress =0 = -
Maximum stress at outer fiber σ
π
2
= 157.14 N/mm (compressive)
0
Consider the cross section B – B [i.e., 90 away from the load line] 0
At section B – B, θ = 0 with respect to horizontal Direct stress (compressive)
σ = = = 35.81 N/mm θ
2
π
d
where r = r , = - 399655.7N-mm =
Bending moment
π
= -
c
π
π
π
This couple produces compressive stress at the inner fiber and tensile stress at the at outer fiber
=Direct stress + Bending stress = σ - = 35.81 +
Maximum stress at the inner fiber σ
π
d
2
= 138.578 N/mm (tensile) Maximum stress at outer fiber σ
= Direct stress + Bending stress = σ + = 35.81 π
d
2
= - 9.047 N/mm (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 the moment of resist nce M of a curved beam of initial radius R1 when bent to a radius R2 by uniform bending mom nt is M = EAeR1
Consider a curved bea of uniform cross section as sho n in Figure below. Its transverse se tion is symmetric with respect to t e y axis and in its unstressed state; its upper and lower surfaces intersect the vertical xy plane along the arcs of circle AB and EF centered at [Fig. 1(a)]. Now apply two equal and opposite couples M and M' as shown in Fig. 1. (c). The length of neut al surface remains the same. θ a d θ' are the central angles before an after applying the moment M. Since the length of neutral surface remai s the same R1θ =R2θ' ..... (i)
Figure Consider the arc of circle JK located at a distance y above the neutral surface. Let r1 and r2 b the radius of this arc before and fter bending couples have been appli d. Now, the deformation of JK, .... (ii) From Fig. 1.2 a and , r1 = R1 – y; r2 = R2 – y
..... (iii)
∴ δ = (R2 – y) θ' – (R1 – y) θ =R2θ' – θ'y – R1θ + θy = – y θ' – θ)
[ R1θ = R2θ' from equ ( )]
∴ δ =– y∆θ ..... (iv) [ θ − θ = θ + ∆ − θ = ∆θ] The normal strain ∈x in the element of JK is obtaine by dividing '
the deformation δ by the original length r1θ of arc JK. ∴ ∈x
=
..... (v)
The normal stress σx may be obtained from Hooke's law σx = E∈x ∴ σx =–E i.e. σx
= –E
..... (vi) ( r1 = R1 – y) .(vii)
Equation (vi) shows that the normal stress σx does not vary linearly with the distance y fro the neutral surface. Plotting σx ersus y, we obtain an arc of hyperbola as shown in Fig. 1.3.
From the condition f equilibrium the summation of fo ces over the entire area is zero and the summation of the moments due t these forces is equal to the applied b nding moment. ∴ ∫δF =0 i.e., ∫σxdA =0 ..... (viii) and ∫(– yσxdA)=M
..... (ix)
Substituting the value of the σx from equation (vii) into equation (viii)
dA=0 Since is not equal to zero dA = 0 i.e. R1 =0 i.e., . R1 =0 ∴ R1 = ∴ It follows the distance R1 from the centre of curvature O to the neutral surface is obtained by the relation R 1 = ..... (x) –
1
1
1
1
The value of R1 is not equal to the
of the cross-section, since =
1
1
1
distance from O to the centroid
is obtained by the relation, ..... (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) dA = M i.e., = M i.e., i.e., = M [using equations (x) and (xi)]
i.e.,
i.e.,
=M = i.e., 1
i.e.,
=
Substituting
..... (xii)
1
( e=
1–
R1 from Fig. 1.2a)
………xiii)
into equation (VI)
σx = ∴ σx=
..... (xiv)
1
M
1
1
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 M From equation (i),
=
{ From equation (xiii)
i.e.
=
M
EAeR 1
∴ M =EAe R1
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/mm 2 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 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 cr oss 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.
A
5 . 7 8
5 2
B m m 5 2. 6
112.5
F = 100kN
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.
