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Optimisation of a beam in bending subjected to severe inertia impact loading
Engineering Failure Analysis 18 (2011) 117–123
Contents lists available at ScienceDirect
Engineering Failure Analysis journal homepage: www.elsevier.com/locate/engfailanal
Optimisation of a beam in bending subjected to severe inertia impact loading José María Mínguez a,⇑, Jeffrey Vogwell b a b
Dpto. de Física Aplicada II, Facultad de Ciencia y Tecnología, Universidad de Bilbao, Aptdo. 644, 48080 Bilbao, Spain Department of Mechanical Engineering, Faculty of Engineering and Design, University of Bath, UK
a r t i c l e
i n f o
Article history: Received 19 August 2010 Accepted 19 August 2010 Available online 27 August 2010 Keywords: Impact failure Braking system Beam bending energy
a b s t r a c t This paper summarises an optimal analytical study of a bar when it is subjected to shock inertia loading such that the severe bending causes permanent yielding. The findings are related to a practical case study application and help explain the necessary actions that have to be taken to minimise the impact damage. The analysis relates the sudden strain energy which the bar gains following an impact (when either the stationary bar is collected by a body moving with substantial momentum or alternatively when the moving bar is suddenly brought to a halt) and establishes the bending strength as a function of mass, stiffness and other kinematic conditions. The optimal analysis results in a surprising paradox which is borne out in the actual detail design solution to a critical component which had caused major maintenance problems in the braking system of a railed vehicle. The study reveals some interesting findings relating to the best bar configuration in terms of geometrical design, support locations and material choice. Ó 2010 Elsevier Ltd. All rights reserved.
1. Introduction In this work we study a bar which is subjected to a severe impact causing bending and hence determine the subsequent shock stresses that occur due to the inertia beam force created. The objective of the analysis is to determine the optimal arrangement as although maximising bending strength and stiffness is necessary, the adding of material increases mass and thus the magnitude of the inertia force acting. Parameters which have been considered are the bar material properties, its geometry and the spacing position of the stops or, alternatively, the grabbing supports. Two cases are considered, firstly when the bar is initially lying in a stationary state and then collected by a pair of striking grips which are moving at a speed V and with considerable momentum thus producing action forces F and causing the bar to move suddenly to that speed as shown in Fig. 1a. Also, the reverse situation, shown in Fig. 1b, which is where the beam moves with speed V and is then suddenly stopped by a pair of equally spaced rigid stops producing reaction forces F. Such situations occur in practise, in the braking system of the bobsleigh start track of the University of Bath (UK) [1], which is what stimulated the need for the study. 2. Strain energy It is assumed that the bar of mass m is collected from stationary by a heavy vehicle of mass M (M m), which moves at speed V1 and forces the bar to move at speed V by forces F (see Fig. 1a). ⇑ Corresponding author. E-mail address: [email protected] (J.M. Mínguez). 1350-6307/$ - see front matter Ó 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.engfailanal.2010.08.013
118
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
(a)
F
F V
(2)
(1)
(b) V
F
(1)
F (2)
Fig. 1. (a) Stationary bar suddenly taken by forces F and communicated a speed V, (b) moving bar suddenly struck and stopped by forces F.
From the momentum balance
MV 1 ¼ ðM þ mÞV
ð1Þ
together with the conservation of energy
1 1 MV 21 ¼ ðM þ mÞV 2 þ U 2 2
ð2Þ
the strain energy due to the loss of kinetic energy is
U¼
1 mV 2 2
ð3Þ
Similarly, in the second case, assuming that the bar is struck at speed V by a much heavier object and that at a certain instant the bar is suddenly stopped by forces F (see Fig. 1b), the conservation of energy requires that the loss of kinetic energy be transformed into strain energy and so
U¼
1 mV 2 2
ð4Þ
In the two cases considered the strain energy is represented by similar Eqs. (3) and (4), although in Eq. (3) V is the velocity at which the bar, initially stationary, suddenly starts moving, whereas in Eq. (4) V is the velocity from which the bar is suddenly stopped. In both cases the kinetic energy lost in the collision is transferred to all the colliding elements as strain energy. So the strain energy taken by the bar will be a fraction of U, which may be expressed as
U b ¼ bmV 2
ð5Þ
where 0 < b < 1/2. This is going to be the energy employed in bending the bar. 3. Bending stress theory In the two cases the bar is subjected to its own inertia force against the forces F and experiences bending according to the diagram described in Fig. 2. The inertia force (which is m acceleration) is a uniformly distributed load w acting along the length of the bar, since the bar mass is uniformly distributed. Then, as the two symmetrical supporting forces F equilibrate the total inertia force,
F ¼ wL=2
ð6Þ
Now the relationship between the strain energy and the bending stresses depends on the position of the supporting reaction forces, which is given by the distance A = aL in Fig. 2. The bending deflection mode and moment distribution along the bar depends very much upon where it is supported as shown in Fig. 2. The two extreme cases are shown, that is, when the forces F are simultaneously applied at both ends (i.e. when a = 0) and when the forces are exerted at the central position (i.e. when a = L/2) in Figs. 3 and 4.
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
wL/2
119
F = wL/2 w
A= α L
A L
MB
MA Fig. 2. The bar under the inertia force, with supporting forces at intermediate positions, and subsequent bending moment.
wL/2
wL/2 w
L
MB
MA = 0
Fig. 3. The bar under the inertia force, with supporting forces at both ends, and subsequent bending moment.
With uniformly distributed inertia loading, as shown in Fig. 2, the bending moment in a section of the bar is
M1 ¼
wx2 2
ð7Þ
from x = 0 to x = A, whereas from x = A to x = L/2,
M2 ¼
wx2 wL ðx aLÞ þ 2 2
ð8Þ
Now the total strain energy for the tube in bending, as given in any Solid Mechanics text book such as Ref. [2] and considering the symmetry of the tube, is obtained as follows:
120
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
wL
w
L
MA =M B Fig. 4. The bar under the inertia force, with supporting forces at central position, and subsequent bending moment.
U¼
Z
M2 dx ¼ 2EI
L
0
Z
A
0
M 21 dx þ EI
Z
L=2
A
M 22 dx EI
ð9Þ
where E is the modulus of elasticity of the tube material and I the moment of inertia of the cross section. By substituting Eqs. (7) and (8) into Eq. (9) the following relationship is reached
U¼
w2 L5 PðaÞ 240EI
ð10Þ
where
PðaÞ ¼ 1 10a þ 30a2 20a3 10a4
ð11Þ
Then the load acting upon the beam per length unit is
w¼
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 240 UEI PðaÞ L5
ð12Þ
It is noted that the load w per unit of length, for a given strain energy absorbed by the bar, is a function of a, which means it depends on the position of the supporting forces. Consequently, the supporting forces also depend themselves on where they are applied. It is obvious that the critical sections of the bar are those where the supporting forces are applied, x = A, and the middle section at x = L/2. The bending moments at these sections are respectively named MA and MB and, according to Eqs. (7) and (8), are:
MA ¼ M1 ðx ¼ AÞ ¼ M 2 ðx ¼ AÞ ¼
wL2 2 a 2
ð13Þ
and
MB ¼ M2 ðx ¼ L=2Þ ¼
wL2 ð1 4aÞ 8
ð14Þ
or, substituting w from Eq. (12),
sffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffi 60 2 UEI a PðaÞ L
MA ¼
ð15Þ
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
121
and
sffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffi 15 UEI MB ¼ ð1 4aÞ 4PðaÞ L
ð16Þ
Then, accounting for the bending stress formula (r = My/I) and d being the outside diameter of the bar the absolute maximum stresses caused by these bending moments in the sections corresponding to x = A and x = L/2 are respectively
rA
sffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffi 2 15 2 UEd ¼ a PðaÞ IL
ð17Þ
and
rB
sffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi UEd2 15 ¼ 1 4a 16PðaÞ IL
ð18Þ
4. Size and geometry of the bar Apparently rA and rB depend on the size of the bar as well as on its geometry through the square root that appears in Eqs. (17) and (18). Nevertheless the fact that the bending strain energy U is also related to the mass and the size and geometry of the bar makes it necessary to analyze in detail this dependence. Effectively, based on Fig. 6 it may be written that 2
U ¼ bmV ¼ bqp
! 2 2 d di LV 2 4 4
ð19Þ
where q is the density of the material and di the inner diameter of the bar. It is noted that the analysis includes the bar when it has a solid section, which corresponds to di = 0 . The moment of inertia of the section of the bar is
I¼p
! 4 4 d d i 64 64
ð20Þ
Then Eqs. (17) and (18) yield
rA ¼
sffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 15 2 bqE 4Vd a 2 2 PðaÞ d þ di
ð21Þ
and
5
2
σA/B /(βρEV )1/2
4
σB
3
2
1
σA 0
0
0.1
0.2
0.3
0.4
0.5
α = A/L Fig. 5. Variation of stress parameter with distance ratio.
122
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
di d Fig. 6. Section of the bar.
rB
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 15 bqE 4Vd ¼ 1 4a d2 þ d2 16PðaÞ i
ð22Þ
These results make evident that the maximum stresses experienced by the bar at the two critical sections are independent of the length of the bar. Moreover and surprisingly, the greater di the smaller the two stresses are. Consequently the most convenient geometry for the bar is an inside emptied and hollow bar, as long as the section remains stable. In this case di d and Eqs. (21) and (22) are transformed into
rA
sffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi 120 2 ¼ a bqEV 2 PðaÞ
ð23Þ
and
rB
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi 120 2 ¼ 1 4a bqEV 16PðaÞ
ð24Þ
which means that the maximum critical stresses are totally independent of the size of the hollow bar for they do not depend neither on its length nor on its diameter. In contrast had the bar had a solid section (that is with di = 0) the two stresses would be
rA
sffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi 240 2 ¼ a bqEV 2 PðaÞ
ð25Þ
and
rB
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi 240 2 ¼ 1 4a bqEV 16PðaÞ
ð26Þ
exactly double that of the corresponding to the thin-walled tubular bar. 5. Position of the supporting forces Eqs. (23) and (24) effectively demonstrate that, although the critical stresses rA and rB do not depend on the size of the bar, they do depend on the position of the acting forces relative to the bar. To analyze this dependence, the stresses rA and rB, as given by Eqs. (23) and (24), are represented in Fig. 5 versus parameter a, which defines where the supporting forces are exerted. For every position of the forces, determined by one value of a, rA and rB reach certain values. Fig. 5 shows that the most convenient value of a is a = 0, which represents the forces being applied at both ends of the bar. Then the greatest of the two stresses (rA and rB) is:
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi
rmax ¼ rB ða ¼ 0Þ ¼ 2:74 bqEV 2
ð27Þ
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
123
With any other value of a, that is, with the supporting forces at any other position, rA, rB or both would reach a higher qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi rB ¼ 3:63 bqEV 2 , if a = 0.235, rA ¼ 4:68 bqEV 2 and if a = 0.500, rA ¼ rB ¼ 4:47 bqEV 2 ,
value. For instance, if a = 0.185, as can be seen from Fig. 5 6. The material
The stress given by Eq. (27) is the maximum stress that will take place in the bar when loaded with the shocking inertia force, if it is supported at both ends, and will happen at the middle section. This critical value depends very much on the material of the bar – the density and the elastic modulus. Therefore, the design of the bar, besides considering the strength of the material, to assure that it can resist such stress, needs to account for the material characteristics, for the stress itself depends on them and can be notably lowered by choosing adequately the material. Effectively, by using steel (with density q = 7850 kg/m3 and modulus of elasticity E = 210 GPa) rather than using aluminium (with density q = 2800 kg/m3 and elastic modulus E = 70 GPa), according to Eq. (28), the stress is reduced to a ratio
rmax ðsteelÞ ¼ rmax ðalum:Þ
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 7850 2:1 105 2800 7 104
¼ 2:90 times
ð28Þ
7. Discussion The analysis carried out shows how a bar subjected to shocking inertia loading bends due to the strain energy resulting from the loss of kinetic energy. The fact that the loss of kinetic energy, for a given velocity, depends very much on the mass of the bar gives way to certain paradoxes. Effectively, a solid bar is stronger than a tubular bar. Nevertheless the solid section implies an increase in mass and so in the strain energy, thus making the bar more vulnerable. In particular it was proved that the maximum stress would be doubled if the tube was substituted by a solid bar. Also, it could be thought that the maximum bending moment in the bar is minimised when the supporting forces act in such positions, that MA and MB (see Fig. 2) become equal. Thus, it could be convenient to enlarge the bar beyond the position of the supporting forces. Nevertheless, this would increase the mass of the bar and so the strain energy. Besides, it was shown in Eq. (12) how the resulting load per length unit, for a given strain energy, depends on the relative position of the forces on the bar through a and, according to this relationship, it was proved that exerting the supporting forces at both ends of the bar was the best choice to diminish the maximum stress experienced by the bar. Finally, the theory shows how the inertia force is also a function of the elastic behaviour of the material through the modulus E (see Eq. (12)) and shows how the maximum stress is influenced by the elastic modulus value and the material density. Replacing steel by lighter aluminium reduces by 2.9 times the maximum stress reached. This will be advantageous when, say, using a steel grade having a yield strength 400 MPa as this is only 1.33 times the ultimate strength of a typical aluminium grade. 8. Conclusions A theory has been developed to describe the bending of a bar subjected to shock inertia loading, either when it is moving with a given velocity and suddenly stopped, or in the reverse case, when it is communicated the same velocity from a stationary state. The analysis shows the interrelation between the bar stress and other variables such as the material used, its size and geometry, and the velocity and the position of the supporting forces. Based on this some paradoxical conclusions are reached. Effectively, a bar with a solid section is more vulnerable than a tubular bar. With respect to the density and elastic characteristics of the material, the strongest material is not necessarily the most suitable, but can be less convenient. Finally, the supporting forces are best positioned at both ends of the bar, even when it may seem that this makes the bending moment greater. References [1] Vogwell J, Mínguez JM. Damage in a bobsleigh start track braking system. Eng Fail Anal 2009;16:1109–17. [2] Case J et al. Strength of materials and structures. 4th ed. Butterworth-Heinemann; 1999. p. 415.
Contents lists available at ScienceDirect
Engineering Failure Analysis journal homepage: www.elsevier.com/locate/engfailanal
Optimisation of a beam in bending subjected to severe inertia impact loading José María Mínguez a,⇑, Jeffrey Vogwell b a b
Dpto. de Física Aplicada II, Facultad de Ciencia y Tecnología, Universidad de Bilbao, Aptdo. 644, 48080 Bilbao, Spain Department of Mechanical Engineering, Faculty of Engineering and Design, University of Bath, UK
a r t i c l e
i n f o
Article history: Received 19 August 2010 Accepted 19 August 2010 Available online 27 August 2010 Keywords: Impact failure Braking system Beam bending energy
a b s t r a c t This paper summarises an optimal analytical study of a bar when it is subjected to shock inertia loading such that the severe bending causes permanent yielding. The findings are related to a practical case study application and help explain the necessary actions that have to be taken to minimise the impact damage. The analysis relates the sudden strain energy which the bar gains following an impact (when either the stationary bar is collected by a body moving with substantial momentum or alternatively when the moving bar is suddenly brought to a halt) and establishes the bending strength as a function of mass, stiffness and other kinematic conditions. The optimal analysis results in a surprising paradox which is borne out in the actual detail design solution to a critical component which had caused major maintenance problems in the braking system of a railed vehicle. The study reveals some interesting findings relating to the best bar configuration in terms of geometrical design, support locations and material choice. Ó 2010 Elsevier Ltd. All rights reserved.
1. Introduction In this work we study a bar which is subjected to a severe impact causing bending and hence determine the subsequent shock stresses that occur due to the inertia beam force created. The objective of the analysis is to determine the optimal arrangement as although maximising bending strength and stiffness is necessary, the adding of material increases mass and thus the magnitude of the inertia force acting. Parameters which have been considered are the bar material properties, its geometry and the spacing position of the stops or, alternatively, the grabbing supports. Two cases are considered, firstly when the bar is initially lying in a stationary state and then collected by a pair of striking grips which are moving at a speed V and with considerable momentum thus producing action forces F and causing the bar to move suddenly to that speed as shown in Fig. 1a. Also, the reverse situation, shown in Fig. 1b, which is where the beam moves with speed V and is then suddenly stopped by a pair of equally spaced rigid stops producing reaction forces F. Such situations occur in practise, in the braking system of the bobsleigh start track of the University of Bath (UK) [1], which is what stimulated the need for the study. 2. Strain energy It is assumed that the bar of mass m is collected from stationary by a heavy vehicle of mass M (M m), which moves at speed V1 and forces the bar to move at speed V by forces F (see Fig. 1a). ⇑ Corresponding author. E-mail address: [email protected] (J.M. Mínguez). 1350-6307/$ - see front matter Ó 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.engfailanal.2010.08.013
118
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
(a)
F
F V
(2)
(1)
(b) V
F
(1)
F (2)
Fig. 1. (a) Stationary bar suddenly taken by forces F and communicated a speed V, (b) moving bar suddenly struck and stopped by forces F.
From the momentum balance
MV 1 ¼ ðM þ mÞV
ð1Þ
together with the conservation of energy
1 1 MV 21 ¼ ðM þ mÞV 2 þ U 2 2
ð2Þ
the strain energy due to the loss of kinetic energy is
U¼
1 mV 2 2
ð3Þ
Similarly, in the second case, assuming that the bar is struck at speed V by a much heavier object and that at a certain instant the bar is suddenly stopped by forces F (see Fig. 1b), the conservation of energy requires that the loss of kinetic energy be transformed into strain energy and so
U¼
1 mV 2 2
ð4Þ
In the two cases considered the strain energy is represented by similar Eqs. (3) and (4), although in Eq. (3) V is the velocity at which the bar, initially stationary, suddenly starts moving, whereas in Eq. (4) V is the velocity from which the bar is suddenly stopped. In both cases the kinetic energy lost in the collision is transferred to all the colliding elements as strain energy. So the strain energy taken by the bar will be a fraction of U, which may be expressed as
U b ¼ bmV 2
ð5Þ
where 0 < b < 1/2. This is going to be the energy employed in bending the bar. 3. Bending stress theory In the two cases the bar is subjected to its own inertia force against the forces F and experiences bending according to the diagram described in Fig. 2. The inertia force (which is m acceleration) is a uniformly distributed load w acting along the length of the bar, since the bar mass is uniformly distributed. Then, as the two symmetrical supporting forces F equilibrate the total inertia force,
F ¼ wL=2
ð6Þ
Now the relationship between the strain energy and the bending stresses depends on the position of the supporting reaction forces, which is given by the distance A = aL in Fig. 2. The bending deflection mode and moment distribution along the bar depends very much upon where it is supported as shown in Fig. 2. The two extreme cases are shown, that is, when the forces F are simultaneously applied at both ends (i.e. when a = 0) and when the forces are exerted at the central position (i.e. when a = L/2) in Figs. 3 and 4.
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
wL/2
119
F = wL/2 w
A= α L
A L
MB
MA Fig. 2. The bar under the inertia force, with supporting forces at intermediate positions, and subsequent bending moment.
wL/2
wL/2 w
L
MB
MA = 0
Fig. 3. The bar under the inertia force, with supporting forces at both ends, and subsequent bending moment.
With uniformly distributed inertia loading, as shown in Fig. 2, the bending moment in a section of the bar is
M1 ¼
wx2 2
ð7Þ
from x = 0 to x = A, whereas from x = A to x = L/2,
M2 ¼
wx2 wL ðx aLÞ þ 2 2
ð8Þ
Now the total strain energy for the tube in bending, as given in any Solid Mechanics text book such as Ref. [2] and considering the symmetry of the tube, is obtained as follows:
120
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
wL
w
L
MA =M B Fig. 4. The bar under the inertia force, with supporting forces at central position, and subsequent bending moment.
U¼
Z
M2 dx ¼ 2EI
L
0
Z
A
0
M 21 dx þ EI
Z
L=2
A
M 22 dx EI
ð9Þ
where E is the modulus of elasticity of the tube material and I the moment of inertia of the cross section. By substituting Eqs. (7) and (8) into Eq. (9) the following relationship is reached
U¼
w2 L5 PðaÞ 240EI
ð10Þ
where
PðaÞ ¼ 1 10a þ 30a2 20a3 10a4
ð11Þ
Then the load acting upon the beam per length unit is
w¼
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 240 UEI PðaÞ L5
ð12Þ
It is noted that the load w per unit of length, for a given strain energy absorbed by the bar, is a function of a, which means it depends on the position of the supporting forces. Consequently, the supporting forces also depend themselves on where they are applied. It is obvious that the critical sections of the bar are those where the supporting forces are applied, x = A, and the middle section at x = L/2. The bending moments at these sections are respectively named MA and MB and, according to Eqs. (7) and (8), are:
MA ¼ M1 ðx ¼ AÞ ¼ M 2 ðx ¼ AÞ ¼
wL2 2 a 2
ð13Þ
and
MB ¼ M2 ðx ¼ L=2Þ ¼
wL2 ð1 4aÞ 8
ð14Þ
or, substituting w from Eq. (12),
sffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffi 60 2 UEI a PðaÞ L
MA ¼
ð15Þ
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
121
and
sffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffi 15 UEI MB ¼ ð1 4aÞ 4PðaÞ L
ð16Þ
Then, accounting for the bending stress formula (r = My/I) and d being the outside diameter of the bar the absolute maximum stresses caused by these bending moments in the sections corresponding to x = A and x = L/2 are respectively
rA
sffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffi 2 15 2 UEd ¼ a PðaÞ IL
ð17Þ
and
rB
sffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi UEd2 15 ¼ 1 4a 16PðaÞ IL
ð18Þ
4. Size and geometry of the bar Apparently rA and rB depend on the size of the bar as well as on its geometry through the square root that appears in Eqs. (17) and (18). Nevertheless the fact that the bending strain energy U is also related to the mass and the size and geometry of the bar makes it necessary to analyze in detail this dependence. Effectively, based on Fig. 6 it may be written that 2
U ¼ bmV ¼ bqp
! 2 2 d di LV 2 4 4
ð19Þ
where q is the density of the material and di the inner diameter of the bar. It is noted that the analysis includes the bar when it has a solid section, which corresponds to di = 0 . The moment of inertia of the section of the bar is
I¼p
! 4 4 d d i 64 64
ð20Þ
Then Eqs. (17) and (18) yield
rA ¼
sffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 15 2 bqE 4Vd a 2 2 PðaÞ d þ di
ð21Þ
and
5
2
σA/B /(βρEV )1/2
4
σB
3
2
1
σA 0
0
0.1
0.2
0.3
0.4
0.5
α = A/L Fig. 5. Variation of stress parameter with distance ratio.
122
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
di d Fig. 6. Section of the bar.
rB
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 15 bqE 4Vd ¼ 1 4a d2 þ d2 16PðaÞ i
ð22Þ
These results make evident that the maximum stresses experienced by the bar at the two critical sections are independent of the length of the bar. Moreover and surprisingly, the greater di the smaller the two stresses are. Consequently the most convenient geometry for the bar is an inside emptied and hollow bar, as long as the section remains stable. In this case di d and Eqs. (21) and (22) are transformed into
rA
sffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi 120 2 ¼ a bqEV 2 PðaÞ
ð23Þ
and
rB
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi 120 2 ¼ 1 4a bqEV 16PðaÞ
ð24Þ
which means that the maximum critical stresses are totally independent of the size of the hollow bar for they do not depend neither on its length nor on its diameter. In contrast had the bar had a solid section (that is with di = 0) the two stresses would be
rA
sffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi 240 2 ¼ a bqEV 2 PðaÞ
ð25Þ
and
rB
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi 240 2 ¼ 1 4a bqEV 16PðaÞ
ð26Þ
exactly double that of the corresponding to the thin-walled tubular bar. 5. Position of the supporting forces Eqs. (23) and (24) effectively demonstrate that, although the critical stresses rA and rB do not depend on the size of the bar, they do depend on the position of the acting forces relative to the bar. To analyze this dependence, the stresses rA and rB, as given by Eqs. (23) and (24), are represented in Fig. 5 versus parameter a, which defines where the supporting forces are exerted. For every position of the forces, determined by one value of a, rA and rB reach certain values. Fig. 5 shows that the most convenient value of a is a = 0, which represents the forces being applied at both ends of the bar. Then the greatest of the two stresses (rA and rB) is:
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi
rmax ¼ rB ða ¼ 0Þ ¼ 2:74 bqEV 2
ð27Þ
J.M. Mínguez, J. Vogwell / Engineering Failure Analysis 18 (2011) 117–123
123
With any other value of a, that is, with the supporting forces at any other position, rA, rB or both would reach a higher qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi rB ¼ 3:63 bqEV 2 , if a = 0.235, rA ¼ 4:68 bqEV 2 and if a = 0.500, rA ¼ rB ¼ 4:47 bqEV 2 ,
value. For instance, if a = 0.185, as can be seen from Fig. 5 6. The material
The stress given by Eq. (27) is the maximum stress that will take place in the bar when loaded with the shocking inertia force, if it is supported at both ends, and will happen at the middle section. This critical value depends very much on the material of the bar – the density and the elastic modulus. Therefore, the design of the bar, besides considering the strength of the material, to assure that it can resist such stress, needs to account for the material characteristics, for the stress itself depends on them and can be notably lowered by choosing adequately the material. Effectively, by using steel (with density q = 7850 kg/m3 and modulus of elasticity E = 210 GPa) rather than using aluminium (with density q = 2800 kg/m3 and elastic modulus E = 70 GPa), according to Eq. (28), the stress is reduced to a ratio
rmax ðsteelÞ ¼ rmax ðalum:Þ
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 7850 2:1 105 2800 7 104
¼ 2:90 times
ð28Þ
7. Discussion The analysis carried out shows how a bar subjected to shocking inertia loading bends due to the strain energy resulting from the loss of kinetic energy. The fact that the loss of kinetic energy, for a given velocity, depends very much on the mass of the bar gives way to certain paradoxes. Effectively, a solid bar is stronger than a tubular bar. Nevertheless the solid section implies an increase in mass and so in the strain energy, thus making the bar more vulnerable. In particular it was proved that the maximum stress would be doubled if the tube was substituted by a solid bar. Also, it could be thought that the maximum bending moment in the bar is minimised when the supporting forces act in such positions, that MA and MB (see Fig. 2) become equal. Thus, it could be convenient to enlarge the bar beyond the position of the supporting forces. Nevertheless, this would increase the mass of the bar and so the strain energy. Besides, it was shown in Eq. (12) how the resulting load per length unit, for a given strain energy, depends on the relative position of the forces on the bar through a and, according to this relationship, it was proved that exerting the supporting forces at both ends of the bar was the best choice to diminish the maximum stress experienced by the bar. Finally, the theory shows how the inertia force is also a function of the elastic behaviour of the material through the modulus E (see Eq. (12)) and shows how the maximum stress is influenced by the elastic modulus value and the material density. Replacing steel by lighter aluminium reduces by 2.9 times the maximum stress reached. This will be advantageous when, say, using a steel grade having a yield strength 400 MPa as this is only 1.33 times the ultimate strength of a typical aluminium grade. 8. Conclusions A theory has been developed to describe the bending of a bar subjected to shock inertia loading, either when it is moving with a given velocity and suddenly stopped, or in the reverse case, when it is communicated the same velocity from a stationary state. The analysis shows the interrelation between the bar stress and other variables such as the material used, its size and geometry, and the velocity and the position of the supporting forces. Based on this some paradoxical conclusions are reached. Effectively, a bar with a solid section is more vulnerable than a tubular bar. With respect to the density and elastic characteristics of the material, the strongest material is not necessarily the most suitable, but can be less convenient. Finally, the supporting forces are best positioned at both ends of the bar, even when it may seem that this makes the bending moment greater. References [1] Vogwell J, Mínguez JM. Damage in a bobsleigh start track braking system. Eng Fail Anal 2009;16:1109–17. [2] Case J et al. Strength of materials and structures. 4th ed. Butterworth-Heinemann; 1999. p. 415.