Curved Beam
Short Description
Download Curved Beam...
Description
CURVED BEAMS CONTENT: WHAT’S A CURVED BEAM BEAM? DIFFERENCE BETWEEN A STRAIGHT BEAM AND A CURVED BEAM WHY STRESS CONCENTRA CONCENTRATION TION OCCUR AT INNER SIDE OR CONCAVE SIDE OF CURV CURVED BEAM? DERIVATION FOR STRES STRESSES IN CURVED BEAM PROBLEMS.
Theory of Simple Bending Due to bending moment, tensile stress develops in one portion of section and compressive stress in the other portion across the depth. In between these two portions, there is a layer where stresses are zero. Such a layer is called neutral layer. Its trace on the cross section is called neutral axis.
Assumption The material of the beam is perfectly homogeneous and isotropic. The cross section has an axis of symmetry in a plane along the length of the beam. The material of the beam obeys Hooke’s law. The transverse sections which are plane before bending remain plane after bending also. Each layer of the beam is free to expand or contract, independent of the layer above or below it. Young’s modulus is same in tension & compression. Consider a portion of beam between sections AB and CD as shown in the figure. Let e1f1 be the neutral axis and g1h1 an element at a distance y from neutral axis. Figure shows the same portion after bending. Let r be the radius of curvature and ѳ is the angle subtended by a1b1 and c1d1at centre of radius of curvature. Since it is a neutral axis, there is no change in its length (at neutral axis stresses are zero.) EF = E1F1 = RѲ
G1H1 = (R+Y) Ѳ GH = R RѲ
Also Stress
OR dF = 0 ∴ there is no direct force acting on the element considered.
Since Σyδaa is first moment of area about neutral axis, Σyδa/a Σ is the distance of centroid from neutral axis. Thus neutral axis coincides with centroid of the cross section. Cross sec sectional tional area coincides with neutral axis.
From (1) and (2)
CURVED BEAM Curved beams are the parts of machine members found in C clamps, crane hooks, frames of presses, riveters, punches, shears, boring machines, planers etc. In straight beams the neutral axis of the section coincides with its centroidal axis and the stress distribution in the beam is linear. But in the case of curved beams the neutral axis of the section is shifted towards the centre of curvature of the beam causing a nonnon linear [hyperbolic] distribution of stress. The neutral axis lies between the centroidal axis and the centre of curvature and will always be present within the curved beams.
Stresses in Curved Beam Consider a curved beam subjected to bending moment Mb as shown in the figure. The distribution of stress in curved flexural member is determined by using the following assumptions: i) The material of the beam is perfectly homogeneous [i.e., same material throughout] and isotropic [i.e., equal elastic properties in all directions] ii) The cross section has an axis of symmetry in a plane along the length of the beam. iii) The material of the beam obeys Hooke's law. iv) The transverse sections which are plane before bending remain plane after bending also. v) Each layer of the beam is free to expand or contract, independent of the layer above or below it. vi) The Young's modulus is same both in tension and compression. Derivation for stresses in curved beam Nomenclature used in curved beam Ci =Distance from neutral axis to inner radius of curved beam Co=Distance from neutral axis to outer radius of curved beam C1=Distance from centroidal axis to inner radius of curved beam C2= Distance from centroidal axis to outer radius of curved beam F = Applied load or Force A = Area of cross section L = Distance from force to centroidal axis at critical section σd= Direct stress σbi = Bending stress at the inner fiber σbo = Bending stress at the outer fiber σri = Combined stress at the inner fiber σro = Combined stress at the outer fiber
co
c2
ci
c1
e
CA NA
F
F
F F
Mb
Mb ri
C L
rn rc ro
Stresses in curved beam Mb = Applied Bending Moment ri = Inner radius of curved beam ro = Outer radius of curved beam rc = Radius of centroidal axis rn = Radius of neutral axis CL = Center of curvature
In the above figure the lines 'ab' and 'cd' represent two such planes before bending. i.e., when there are no stresses induced. When a bending moment 'Mb' is applied to the beam the plane cd rotates with respect to 'ab' through an angle 'd 'dθ'' to the position 'fg' and the outer fibers are shortened while the inner fiberss are elongated. The original length of a strip at a distance 'y' from the neutral axis is (y + rn)θ. It is shortened by the amount ydθ and the stress in this fiber is, σ = E.e Where σ = stress, e = strain and E = Young's Modulus
We know, stress σ = E.e Ѳ We know, stress e Ѳ
i.e., σ
=–E
θ
θ
..... (i)
Since the fiber is shortened, the stress induced in this fiber is compressive stress and hence negative sign. The load on the strip having thickness dy and cross sectional area dA is 'dF' θ i.e., dF = σdA = – dA 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 dF 0
i.e.
θ
∴ As
θ θ
∴
θ
=0
is not equal to zero,
=0
..... (ii)
The neutral axis radius 'rn' can be determined from the above equation. If the moments are taken about the neutral axis, Mb = – ydF Substituting the value of dF, we get Mb =
θ θ
dA
=
θ
y ! " dA
=
θ
ydA
θ
θ
#$
0%
Since ydA represents the statical moment of area, it may be replaced by A.e., the product of total area A and the distance 'e' from the centroidal axis to the neutral axis.
∴
θ
Mb =
From equation (i)
θ
θ θ
A.e
=–
σ
Substituting in equation (iii) Mb = – ∴
σ =
..... (iii)
σ
. A. e.
&'
..... (iv)
This is the general equation for the stress in a fiber at a distance 'y' from neutral axis. At the outer fiber, y = co ∴ Bending stress at the outer fiber σbo i.e.,
σbo= !
&' () )
($ rn + co = ro)
..... (v)
Where co = Distance from neutral axis to outer fiber. It is compressive stress and hence negative sign. At the inner fiber, y = – ci ∴ Bending stress at the inner fiber σbi= i.e.,
σbi =
&' ( *
+(* &' ( * *
($ rn – ci = ri)
..... (vi)
Where ci = Distance from neutral axis to inner fiber. It is tensile stress and hence positive sign.
Difference between a straight beam and a curved beam Sl.no 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 axis and centroidal Neutral axis is shifted axis coincides towards the least centre of curvature
Location of the neutral axis By considering a rectangular cross section
Centroidal and Neutral Axis of Typical Section of Curved Beams
Why stress concentration occur at inner side or concave side of curved beam Consider the elements of the curved beam lying between two axial planes ‘ab’ and ‘cd’ separated by angle θ.. Let fg is the final position of the plane cd having rotated through an angle ddθ about neutral axis. Consider two fibers symmetrically located on either side of the neutral axis. Deformation in both the fibers is same and equal to yd ydθθ.
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 – 0.910h = 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 F = 22X103N b = 20mm h = 150-50 = 100mm A = bh = 20X100 = 2000mm2
e = rc - rn = 100 - 91.024 = 8.976mm Section A-B B will be subjected to a combination of direct load and bending, due to the eccentricity of the force. Stress due to direct load will be,
y = rn – r = 91.024 – r Mb = 22X103X100 = 2.2X106 N-mm
Problem no.2 Determine the value of “t” in the cross section of a curved beam as shown in fig such that the normal stress due to bending at the extreme fibers are numerically equal. Given data; Inner radius ri=150mm Outer radius ro=150+40+100 =290mm Solution; From Figure Ci + CO = 40 + 100 = 140mm…………… (1) Since the normal stresses due to bending at the extreme fiber are numerically equal we have,
i.e
Ci= = 0.51724Co…………… (2)
Radius of neutral axis rn=
rn =197.727 mm ai = 40mm; bi = 100mm; b2 =t; ao = 0; bo = 0; ri = 150mm; ro = 290mm;
=
i.e.,
+83.61t = 4000+100t; 4674.069+83.61t
∴ t = 41.126mm Problem no.3 Determine the stresses at point A and B of the split ring shown in the figure. Solution: The figure shows the critical section of the split ring. Radius of centroidal axis rc = 80mm Inner radius of curved beam
ri = 80 = 50mm
Outer radius of curved beam
ro = 80 + = 110mm
Radius of neutral axis
rn = =
Applied force
= 77.081mm F = 20kN = 20,000N (compressive)
π
A = d2 =
Area of cross section
,
π
,
x602 = 2827.433mm2
Distance from centroidal axis to force l = rc = 80mm Bending moment about centroidal axis Mb = Fl = 20,000x80 =16x105N-mm Distance of neutral axis to centroidal axis e = rc ! rn = 80! 77.081=2.919mm Distance of neutral axis to inner radius ci = rn ! ri = 77.081! 50=27.081mm Distance of neutral axis to outer radius co = ro ! rn = 110 ! 77.081=32.919mm Direct stress
σd = !
.
=
/0000
/1/2.,44
=! 7.0736N/mm2 (comp.) Bending stress at the inner fiber σbi = ! =
&' * *
+ 567508 7/2.015
/1/2.,447/.9597:0
= ! 105N/mm2 (compressive) Bending stress at the outer fiber σbo =
&' ) )
=
567508 74/.959
/1/2.,447/.9597550
= 58.016N/mm2 (tensile) Combined stress at the inner fiber
σri = σd + σbi = ! 7.0736! 105.00 = - 112.0736N/mm2 (compressive) Combined stress at the outer fiber σro = σd + σbo = ! 7.0736+58.016 = 50.9424N/mm2 (tensile) Maximum shear stress τmax = 0.5x σmax = 0.5x112.0736 = 56.0368N/mm2, at B The figure.
Problem No. 4 Curved bar of rectangular section 40x60mm and a mean radius of 100mm is subjected to a bending moment of 2KN 2KN-m m tending to straighten the bar. Find the position of the Neutral axis and draw a diagram to show the variation of stress across the section. Solution Given data: b= 40mm h= 60mm rc=100mm Mb= 2x106 N-mm C1=C2= 30mm rn= =130 =(ri+c1+c2) ro= rc+h/2=100+30=130 ri= rc- h/2 = 100 - 30= 70mm (rc-c1) rn= 96.924mm Distance of neutral axis to centroidal axis e = rc - rn= 100-96.924 =3.075mm Distance of neutral axis to inner radius ci= rn- ri = (c1-e) e) = 26.925mm Distance of neutral axis to outer radius co=c2+e= (ro-rn) = 33.075mm Area A= bxh = 40x60 = 2400 mm2 Bending stress at the inner fiber σbi =
=
= 104.239 N/mm2 (compressive)
Bending stress at the outer fiber σbo =
&' ) )
=
+/;50< ;44.02:
/,00;4.02:;540
= -68.94 N/mm2 (tensile) Bending stress at the centroidal axis = =
+&' =
+/;50<
/,00;500
= -8.33 N/mm2 (Compressive) The stress distribution at the inner and outer fiber is as shown in the figure.
Problem No. 5 The section of a crane hook is a trapezium; the inner face is b and is at a distance of 120mm from the centre line of curvature. The outer face is 25mm and depth of trapezium =120mm.Find the proper value of b, if the extreme fiber stresses due to pure bending are numerically equal, if the section is subjected to a couple which develop a maximum fiber stress of 60Mpa.Determine the magnitude of the couple. Solution ri = 120mm; bi = b; bo= 25mm; h = 120mm σbi = σbo = 60MPa Since the extreme fibers stresses due to pure bending are numerically equal we have, & ' * & ' ) = * ) We have, Ci/ri =co/ro =ci/co =120/240 2ci=co But
h= ci + co 120 = ci+2ci
Ci=40mm; co=80mm rn= ri + ci = 120+40 =160 mm
b=150.34mm To find the centroidal axis, (C2) bo= 125.84mm; b=25mm; h=120mm
= 74.313mm. But C1=C2 rc= ro-c2 =240 - 74.313 =165.687mm e=rc- rn = 165.687 - 160 = 5.6869 mm Bending stress in the outer fiber,
σ >?
M> c Aer 5:0.1,/:5/0
A= / = 1050.4mm 60 =
&' ;10
50::0.,;:.612;/,0
Mb=10.8x106 N-mm
Problem no.6 Determine the stresses at point A and B of the split ring shown in fig.1.9a Solution: Redraw the critical section as shown in the figure. Radius of centroidal axis rc = 80mm Inner radius of curved beam ri = 80!
60 /
= 50mm
Outer radius of curved beam ro = 80 +
60 /
= 110mm
Radius of neutral axis rn =
,
= Applied force
CD) D* E
C√550√:0E ,
=77.081mm
F = 20kN = 20,000N (compressive) π
Area of cross section A = d2 = ,
π
,
x602 = 2827.433mm2
Distance from centroidal axis to force l = rc = 80mm Bending moment about centroidal axis Mb = FI = 20,000x80 =16x105N-mm Distance of neutral axis to centroidal axis e = rc ! rn = 80! 77.081 =2.919mm
Distance of neutral axis to inner radius ci = rn ! ri = 77.081! 50 = 27.081mm Distance of neutral axis to outer radius co = ro ! rn = 110! 77.081 = 32.919mm Direct stress
σ d =!
.
=
/0000
/1/2.,44
=! 7.0736N/mm2 (comp.) Bending stress at the inner fiber σbi = !
&' * *
=
+ 567508 7/2.015
/1/2.,447/.9597:0
= ! 105N/mm2 (compressive) Bending stress at the outer fiber σbo =
&' ) )
=
567508 74/.959
/1/2.,447/.9597550
= 58.016N/mm2 (tensile) Combined stress at the inner fiber σri = σd + σbi =! 7.0736! 105.00 =! 112.0736N/mm2 (compressive) Combined stress at the outer fiber σro = σd + σb = ! 7.0736+58.016 = 50.9424N/mm2 (tensile) Maximum shear stress Gmax = 0.5x σmax = 0.5x112.0736
= 56.0368N/mm2, at B
The figure shows the stress distribution in the critical section.
Problem no.7 Determine the maximum tensile, compressive and shear stress induced in a ‘c’ frame of a hydraulic portable riveter shown in fig.1.6a
Solution:
50
Draw the critical section as shown the figure.
R1 00
9000N
in
80
Inner radius of curved beam ri = 100mm
175 mm
Outer radius of curved beam ro = 100+80 = 180mm 10 /
c2
Radius of neutral axis rn = ln = ln
I) # I* %
10
JK? #J ?%
= 136.1038mm Distance of neutral axis to centroidal axis e = rc - rn = 140-136.1038 = 3.8962mm
Critical Section
co
ri = 100mm
ci
175mm
ro CA NA
= 140mm
c1
b = 50 mm
Radius of centroidal axis rc = 100+
F h = 80mm e
rn rc CL
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 l = 175+ rc = 175+140 = 315mm Applied force
F = 9000N
Area of cross section A = 50x80 = 4000mm2 Bending moment about centroidal axis Mb = FI = 9000x315 = 2835000 N-mm Direct stress
σd =
.
=
9000
,000
= 2.25N/mm2 (tensile)
Bending stress at the inner fiber σbi =
&' * *
=
/14:000746.5041 ,00074.196/7500
= 65.676N/mm2 (tensile) Bending stress at the outer fiber σbo = !
&' ) )
=!
/14:0007,4.196/ ,00074.196/7510
= ! 44.326N/mm2 (compressive)
Combined stress at the inner fiber σri = σd + σbi = 2.25+65.676 = 67.926N/mm2 (tensile) Combined stress at the outer fiber σro = σd + σbo = 2.25! 44.362 = !42.112 N/mm2 (compressive)
Gmax = 0.5x σmax = 0.5x67.926
Maximum shear stress
= 33.963 N/mm2, at the inner fiber The stress distribution on the critical section is as shown in the figure.
σri=67.926 N/mm2 2
Combined stress σro=-42.112 N/mm
σbi=65.676 N/mm2 Bending stress σbo=-44.362 N/mm2
2
σd=2.25 N/mm NA
CA
Direct stress (σd)
b = 50 mm h =80 mm
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:
bi = 18 mm
Draw the critical section as shown in the figure.
c1
bo = 6 mm
c2
F
h = 40mm e
co
Inner radius of curved beam ri = 25mm
ci
ri = 25mm
100mm
rc ro rn
CA NA
Outer radius of curved beam ro = 25+40
C L
= 65mm Distance of centroidal axis from inner fiber c1 = =
,0 51/76 4
516
>* />) 4
>* >)
" = 16.667mm
"
F
∴ Radius of centroidal axis rc = ri ! c1 = 25+16.667 = 41.667 mm Radius of neutral axis rn =
J >* >) '* I) L ') I* I) + >* +>) M I*
J 7,0516 = JKN) I)XY *
I) " I) LY)
ai = 75mm; bi = 300mm; b2 = 75mm; ao = 0; bo = 0 A=a1+a2=75x300+75x225 =39375mm2 ∴ rn
=
400
4942:
8?X]8 88?L? "2: "0 8? 8?X]8
Let AB be the ref. line
750
C L
I*
NA
CA
rc ro=550 mm
= 333.217mm
F
x^
J 7J 7 J
=
]8
2:7400 2:7//: 2: 4942:
8 "
= 101.785mm
Radius of centroidal axis rc = ri +x^ = 250+101.785=351.785 mm
Distance of neutral axis to centroidal axis e = rc! rn = 351.785-333.217=18.568mm Distance of neutral axis to inner radius ci = rn! ri = 333.217! 250=83.217mm Distance of neutral axis to outer radius co = ro! rn = 550! 333.217=216.783mm Distance from centroidal axis to force l = 750+ rc = 750+351.785 = 1101.785mm Applied force
F = 85kN
Bending moment about centroidal axis Mb = FI = 85000x1101.785 = 93651725N-mm Direct stress
σd =
.
=
1:000 4942:
= 2.16N/mm2 (tensile)
Bending stress at the inner fiber σbi =
&' * *
=
946:52/:714./52
4942:751.:617/:0
= 42.64N/mm2 (tensile)
Bending stress at the outer fiber σbo =!
&' ) )
=!
946:52/:7/56.214 4942:751.:617::0
= ! 50.49N/mm2 (compressive) Combined stress at the inner fiber σri = σd + σbi = 2.16+42.64 = 44.8N/mm2 (tensile) Combined stress at the outer fiber σro = σd + σbo = 2.16! 50.49
= ! 48.33N/mm2 (compressive)
Maximum shear stress
Gmax = 0.5x σmax = 0.5x48.33
= 24.165N/mm2, at the outer fiber The below figure shows the stress distribution. 2
σri=44.8 N/mm Combined stress σro=-48.33 N/mm2
σbi=42.64 N/mm2 Bending stress σbo=-50.49 N/mm2
σd=2.16 N/mm2
NA
CA
Direct stress (σ d)
ai=75mm
b2 = 75 mm
a2
225
bi =300mm
a1
Problem no.10
0 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? 90mm m m
30mm
60mm 25mm
Solution:
NA
CA
F = 25 kN
h = 90 mm c2 c1
Draw the critical section as shown in the figure. Inner radius of curved beam ri = 100mm Outer radius of curved beam ro = 100+90 = 190mm Distance of centroidal axis from inner fiber c1 =
>* />) 4
>* >)
= 40mm
" =
90 4
x
60/740 6040
ci
co
"
ri rn rc ro
e
l F
C L
Radius of centroidal axis rc = ri + c1 = 100+40 = 140 mm Radius of neutral axis rn =
=
J >* >) '* I)L ') I* I) + >* +>) M I*
J 79076040
View more...
Comments
