Closed-form approximate formulas for torsional analysis of hollow tubes with straight and circular edges
A. Doostfatemeh, M. R. Hematiyan*, S. Arghavan Department of Mechanical Engineering, School of Engineering, 71345, Shiraz University, Shiraz, Iran
*
Corresponding author
Tel: +989173114762, Fax: +987116287294, Email:
[email protected]
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Closed-form approximate formulas for torsional analysis of hollow tubes with straight and circular edges
Abstract Some analytical formulas are presented for torsional analysis of homogeneous hollow tubes. The cross section is supposed to consist of straight and circular segments. Thicknesses of segments of the cross section can be different. The problem is formulated in terms of Prandtl's stress function. The derived approximate formulas are so simple that computations can be carried out by a simple calculator. Several examples are presented to validate the formulation. The accuracy of formulas is verified by accurate finite element method solutions. It is seen that the error of the formulation is small and the formulas can be used for analysis of thin to moderately thick-walled hollow tubes.
Keywords: Torsion; hollow tubes; homogenous; arbitrary shapes.
1- Introduction Torsion is one of the main modes of loading beside bending, extension and shear in structural and mechanical design. Analytical and numerical solutions for torsion of hollow tubes have been the interest of many researchers since hollow members have high strength and rigidity in torsion. A simple analytical formulation for torsional analysis of thin-walled hollow members can be found in elementary text books of mechanics of materials [1]. This formulation which commonly called thinwalled theory can not predict the variation of shear stress along the thickness and just provides an average value of shear stress. Analytical and numerical solutions for torsional analysis of hollow tubes, which can predict the variation of shear stress along the thickness, will be more attractive. In 1903, Prandtl presented a membrane analogy for torsional analysis and proved the accuracy and efficiency of his approximation [2]. Baron investigated torsion of hollow tubes with multiply connected cross sections [3]. He used an iterative method to satisfy the equilibrium and compatibility conditions.
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Nguyen used Prandtl’s stress function and derived finite element formulation for linear elastic torsion [4]. He used hierarchical shapes as interpolation bases for the Prandtl's stress function. Wang presented a numerical method for torsional analysis of a flattened tube consisting of two half annular and two rectangular pieces [5]. He used an eigenfunction expansion and point match method. He generalized the method to treat arbitrary cross sections consisting of circular arcs and straight lines, all with a uniform thickness [6]. The method proposed by Wang can be considered as a semi-analytic method. His method can accurately evaluate the shear stress at re-entrant inner corners at which high local shear stresses exist. Li et al. presented a computational method for calculating torsional stiffness of multi-material bars with arbitrary shape [7]. Their method is based on the theory of elasticity and finite element method. They considered additional compatibility and equilibrium equations in common boundaries of different materials in their formulation and got good results. Sapountzakis and Mokos used boundary element method for torsional analysis of different types of members [8-11]. Hassenpflug analyzed torsion of uniform bars with polygonal cross sections by means of conformal mapping [12]. He presented series solutions for the problem. Kolodziej and Fraska presented an analytical study about torsion of bars possessing a regular polygonal cross-section by means of boundary collocation method [13]. Hematiyan and Doostfatemeh recently derived simple formulas for torsional analysis of hollow tubes with polygonal shapes [14]. They considered a linear variation for shear stress across the thickness and used governing equations and boundary conditions of the problem in terms of Prandtl's stress function to derive the formulas. In this paper, a simple analytical method is presented for torsional analysis of hollow tubes of arbitrary shape. Cross section of the member consists of straight and curved segments. A linear variation of shear stress along the thickness is considered for straight segments and a nonlinear variation is assumed for arcs. The derived formulas for shear stress and angle of twist are simple enough to be carried out by a simple calculator. The proposed method is applicable for hollow members of any shape of the cross section. Here, closed-form formulas for some important cross sections, such as rounded rectangular and flattened tubes are derived. Several examples are presented to show the accuracy and efficiency of the method. As it will be seen, the method can be used for thin to moderately thick-walled hollow members.
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2- Formulation 2-1- Formulation for a general hollow cross section Cross section of a typical hollow tube of arbitrary shape is shown in Figure 1-a. Shear flow is defined as follows:
q avg t
(1)
where:
avg
1 dt t
(2)
In the above equation, τ and t are shear stress and thickness of the cross section respectively. One can show that shear flow is constant in all segments of the cross section (Appendix A). Shear flow in each segment is replaced by its equivalent forces and torque (couple) as illustrated in Figure 1-b. The resultant torque is then evaluated using these forces and torques. For each straight segment of the cross section, a local coordinate system x′y′ is defined as what is illustrated in Figure 2. Length of each straight segment is supposed to be li. For each curved segment, a polar coordinate system is defined. Center of the polar coordinate system is coincided with the center of the arc (Figure 3). Governing equations and boundary conditions of the problem in terms of Prandtl's stress function can be expressed as follow [15]:
2 2G
(3)
T 2 d 21 A1
(4)
(5)
1
d 2GA1
0 at 0
(6)
1
(7)
at 1
where 0 and 1 represent respectively outer and inner boundaries of the domain . A1 is the area bounded by 1 . The shear stress τ can be expressed as:
d d
(8)
4
where μ is the outward normal of the contour line. The constant value 1 must be chosen so that Eq. (5) is satisfied. One can expand Eq. (3) in Cartesian and polar coordinate systems as follow:
2 2 2G x 2 y 2
(9)
1 1 1 r 2G r r r r r
(10)
For a hollow cross-section, contour lines (contours of constant ) are parallel to boundaries with very good approximation and derivatives of along contour lines are zero. As a result the derivatives of in
y -direction in the straight segments and in -direction in the curved segments can be ignored with very little loss of accuracy. The presented examples will show that this approximation is sufficiently accurate. After deleting the second term in the left side, of Eqs. (9) and (10), and by a simple integration, one can find
d d and , which represent shear stress in straight and curved segments respectively. These dx dr
shear stresses are expressed as follow:
( x) 2Gx C1 (r ) Gr
(11)
C2 r
(12)
where C1 and C2 are constants of integration. One can evaluate the average shear stress in straight and curved segments respectively as follow:
straight
curved
q 1 ti ( x ) dx ti ti 0
q 1 tj tj
(13)
ro
r (r) dr
(14)
i
where ti and tj are thicknesses of straight and curved segments respectively. i and j represent number of a straight and a curved segment and ro and ri are outer and inner radii of a curved segment. After substituting Eqs. (11) and (12) respectively in Eqs. (13) and (14) and implicating the fact that shear flow q is constant in all segments, one can find constants C1 and C2 as follow:
C1
q Gt i ti
(15)
5
C2
GRt j q ln
(16)
ro ri
R is the mean radius of the curved segment. By substituting Eqs. (15) and (16) respectively in Eqs. (11) and (12), shear stress will be found. From governing equations, one can expand Eq. (5) as follows:
d i
1i
d 2GA1
(17)
1j
j
If one substitutes shear stresses in Eq. (17), an explicit relation will be found between shear flow and angle of twist per unit length as follows:
2 A1
l j (ri j
q
j
Rt ) r ri ln( o ) ri
lj ri ln(
ro ) ri
i
li t i i
li ti
G
(18)
where li and lj are lengths of inner boundaries of straight and curved segments respectively. Now that the expression for shear flow has been found, one can find the resultant of the shear flow in each segment of the cross section. By replacing the resultant (force and couple) instead of the shear flow in each segment, next steps will be carried out simpler. The effect of the shear flow in a straight segment is replaced by a force at its center and a couple. It must be mentioned that couple is a free vector. The resultant force in a circular segment is expressed by two components acting through the center of the segment and the corresponding resultant couple is computed by taking moment of shear flow about the center of the segment. The resultant force at center of a straight segment and the corresponding couple for shear flow in the segment can be expressed as follow:
Fi avg t i li qli
Ti li
ti
(19)
ti
0 ( x 2 ) ( x)dx
(20)
By substituting the shear stress from Eq. (11) in Eq. (20), the equivalent couple for shear flow in a straight segment will be found as follows:
6
Ti
Gt 3 li 6
(21)
According to Figure 4, the shear flow in a curved segment is replaced with two equivalent forces Fj and an equivalent torque Tj. The forces and torque can be evaluated by the following equations.
F j csc( ) 2
Tj 2
ro
02 r
ro
r (r )r 2 0
r (r ) cos drd
(22)
drd
(23)
i
2
i
By substituting the shear stress from Eq. (12) in Eqs. (22) and (23), equivalent forces and couple for a curved segment will be found as follow:
Fj
(q GRt j )t j G 3 (ro ri3 ) r 3 ln o ri
2 r ri2 G q GRt j T j Rt j o r 2 ln o ri
(24)
(25)
Now, one can write an expression for total torque about an arbitrary point P (Figure 1). After that, the second relation between applied torque and angle of twist per unit length will be found which can be solved simultaneously with Eq. (18) to find the unknowns. It must be mentioned that the total resultant force on the cross-section is zero; therefore, the resultant moment of forces about any point is the same. In other words, the forces on the cross section produce a couple and the point about which moments are to be taken is arbitrary.
2-2- Closed-form formulas for a rounded rectangular hollow cross section A rounded rectangular cross section consisting of four quarter annular and four rectangular pieces is shown in Figure 5. Present formulation predicts shear flow in this cross section as follows:
r t Rt ab 2 Ra b ln o ri q r G t a b ln o ri
(26)
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Resultant torque can be expressed as follows:
t a b t 2 2ri ro T 2a b t ro2 ri2 R G 3 3
r 2t ab Ra b t a b 2R ab 2 Ra b ln o ri r t a b ln o ri
(27)
The torque in the above equation has been evaluated by taking moment about point P (Figure 5). The result is the same if any other point is used instead of point P. Shear stresses in straight and curved segments are expressed as:
( x ) t 2 x G
Rt ab 2 Ra b ln t a b ln
ro ri
(28)
ro ri
(r ) t ab Ra b 1 r r r G t a b ln o
(29)
ri
Thin-walled theory predicts shear stress and angle of twist as follow:
ab 2 Ra b R 2 G a b R G T
a b R
2t ab 2 Ra b R 2
(30)
(31)
2
When lengths of straight segments are equal (a=b), the rectangle is converted to a square. Present formulation predicts shear flow in this rounded square cross section as follows:
r t Rt a 2 4 Ra ln o ri q r G t 2a ln o ri
(32)
Resultant torque can be written as follows:
2ta t 2ri ro T G 3 2
r 2t a 2 2 Ra 2t a R a 2 4 Ra ln o ri 4a t ro2 ri2 R r 3 t 2a ln o ri
Shear stresses in straight and curved segments are expressed as:
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(33)
( x ) t 2 x G
Rt a 2 4 Ra ln
ro ri
(34)
r t 2a ln o ri
(r ) t a 2 2 Ra 1 r r r G t 2a ln o
(35)
ri
2-3- Closed-form formulas for a flattened tube A flattened tube with uniform thickness is shown in Figure 6. Following the present formulation, one can find shear flow as follows:
r Rt t 2a ln o ri q ro G t a ln ri
(36)
The resultant torque is found as follows:
r 2aRt at 2Rt 2aR ln o ri T at 2 2a t 2ri ro t ro2 ri2 R r G 3 3 t a ln o ri
(37)
Shear stresses in straight and curved segments can be expressed as follow:
ro ri ( x ) R t 2 x r G t a ln o ri
(38)
(r ) r G
(39)
aR ln
Rta
1 ro r t a ln ri
Thin-walled theory predicts shear stress and angle of twist as follow:
2aR R 2 G a R G a R T 2t 2aR R 2
(40)
(41)
2
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3- Results Several examples are considered to show the accuracy and efficiency of the formulas. In each example, results are verified by accurate finite elements method (FEM) solutions. Moreover, thin-walled theory solutions are also considered. In the following examples, normalized shear stress and normalized angle of twist are respectively and G . T G For FEM analysis of the problem, the governing equation and boundary conditions are decoupled into two simple problems. Before presenting examples, decoupling of the problem for FEM analysis is described.
3-1- Decoupling of the problem for FEM analysis Governing equation of the problem is decoupled into two simpler problems. The first problem is governed by Poisson’s equation and the second one by Laplace’s equation, both with simple boundary conditions. The first problem is defined as follows: 2 2G , 0 at 0 and 1
(42)
while the second problem is: 2 0, 0 at 0 and 1 at 1
(43)
As a result, one can write the Prandtl’s stress function ( ) as follows:
1
(44)
Substituting Eq. (44) into Eq. (5) one can write:
n 1 n d 2GA
(45)
1
1
The constant 1 can be found as follows:
1
d 1 n d 1 n
2GA1
(46)
where n is the normal direction of the inner boundary. Note that
( ) ( ) . n 1 1
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A MATLAB code has been developed for FEM analysis of Poisson's equation and used for solving the two problems described in Eqs. (42) and (43). For this purpose the method described in [16] has been used. Quadrilateral 8-node finite elements (quadratic elements) [16] are used in the analyses.
3-2- Example 1: Flattened tube In this example, some torsional analyses are carried out for a flattened tube (Figure 6) with t=5, 10, 20, 30 and 40 mm. Length of straight segments is considered L=100 mm and centerline radius of the curved segments are assumed R=50 mm. Results of the present formulation for normalized shear stress, in comparison with thin-walled theory, and accurate FEM solutions are shown in Figures 7 and 8. It must be mentioned that the thin-walled theory estimates only an average value of shear stress through the thickness and it cannot estimate maximum and minimum values. In the present formulation, shear stress at a connecting point of straight and curved segments (points C and D) is supposed to be the average value of shear stresses of the two segments. Maximum error of the present formulation at the inner and outer boundaries occurs at points C and D respectively. For t=40 mm error values at points C and D are respectively 1.8% and 0.6% in comparison to FEM solutions. As shown in the Figures 7 and 8, the average shear stress estimated by the thin-walled theory is very far from maximum and minimum values of shear stress. Results of the normalized angle of twist are presented in Figure 9. Maximum values of error are 9.4% and 1.2% for thin-walled theory and present study respectively in comparison to FEM solutions. As it can be seen, the present formulation gives very good solutions not only for thin-walled members but also for moderately thick-walled flattened tubes. In the present formulation, constant
contours are supposed to be parallel to boundaries
everywhere in the cross section. This is the only source of error in the present method. Since the exact contour lines in the cross section of a flattened tube are parallel to boundaries with a very good approximation, the results obtained by the present method are also very good.
3-3- Example 2: Rounded square tube In this example, analyses are carried out for a square tube (Figure 5) with a=b=100 mm and t=5, 10, 20, 30 and 40 mm. Centerline radius of the curved segments are assumed R=50 mm. Results of the present
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formulation for normalized shear stresses, in comparison with thin-walled theory, and accurate FEM solutions are shown in Figures 10 and 11. Maximum error occurs at point E for t=40 mm. Error value is 3.2% in comparison to FEM solution. Results of the normalized angle of twist are presented in Figure 12. Maximum values of error are 4.5% and 1% for thin-walled theory and present study respectively in comparison to FEM solutions. The results obtained by the present formulation are satisfactory for both thin-walled and moderately thick-walled rounded square tubes.
3-4- Example 3: A cross section with different edge thicknesses In this example, a hollow tube with different edge thicknesses is considered (Figure 13). Three sets of dimensions are considered. Outer length of straight segments is a=100 mm. Normalized shear stresses for thin-walled theory, present study and FEM are presented in Table 1 for the points that have been shown in Figure 13. Normalized angle of twists are compared in Table 2. Since the thicknesses in the first model are small, the results obtained by the three methods are close to each other. The results obtained by the present formulation for second and third cases are much better than those of thin-walled theory. It must be mentioned that there are stress concentrations at the two internal sharp corners of the considered cross-section which cannot be evaluated by the present method and thin-walled theory.
3-5- Example 4: A multicell cross section Present formulation is useful in torsional analysis of members with multicell cross section. In contrast to previous examples, a cross section with three cells is considered in this example (Figure 14). Distribution of shear flow in different segments has been shown in Figure 14. Similar to thin-walled theory, a system of n equations must be solved in analysis of a cross section with n1 cells. These equations are expressed as follow: 2GAmidi
midi
qi ds i 1, 2, ...,n 1 ti
(47-a)
12
n1
Amid qi
T 2
i 1
(47-b)
i
Using the symmetry of the cross section, one can write q1 q3 . By applying Eq. (18), one can write following relations between shear flows and angle of twist per unit length.
2 b c 9 t q 4t b q 2 r 1 2t t ln o ri G Rt 2 A1 ri2 t b 2c 6t r ln o ri
(48-a)
24t b 2a b 3c 3t q1 q2 t t G 2 A2 2t a 2t
(48-b)
where:
A1
c t b 4t b t 2 2
22
(49-a)
A2 (c t )(b 4t ) (b 2t )( a 2c)
(49-b)
Replacing shear flow in straight and curved segments by their equivalent forces and torques, one can find the third relation between angle of twist per unit length and applied torque. Solving these three equations simultaneously, one can find the unknowns. In this example, analysis is carried out for t=5 mm while a=200 mm, b=100 mm, c=60 mm and d=80 mm. Normalized shear stresses for thin-walled theory, present study and FEM are presented in Table 3 for the points that have been shown in Figure 14. Results for normalized angle of twist are compared in Table 4. Thin-walled theory surprisingly gives a good solution for angle of twist, but its results for shear stress have a considerable error. The results obtained by the present formulation are satisfactory. Values of shear stress at sharp corners at connecting points of segments with different thicknesses cannot be evaluated by the present formulation and thin-walled theory. There are stress concentrations at these sharp corners.
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4- Conclusions Relatively simple analytical formulas were presented for torsional analysis of hollow tubes with straight and circular edges. Calculations are so simple that can be carried out by a pocket calculator and there is no need to computer for calculations. Several examples were considered and results were compared with accurate FEM and thin-walled theory solutions. The results obtained by the present formulation are near to accurate FEM solutions and much better than solutions of thin-walled theory. The maximum error of the present formulation for thin-walled (thickness-to-side ratio less than 0.1) flattened and rounded square tubes is less than 1.5%. The proposed formulas also give acceptable solutions for moderately thick-walled hollow members. The maximum error of the derived formulas for flattened and rounded square tubes with thickness-to-side ratio of 0.2 is less than 3.2%. Similar to the thin-walled theory, the proposed formulation cannot estimate the value of shear stress near sharp corners. A sharp corner may be formed at connecting point of two segments with different thicknesses.
Appendix A A differential element of a hollow member under torsion is shown in the Figure A1. As shown in the figure, an arbitrary variation is considered for shear stress through the thickness. Since the element is in equilibrium, one can write (for forces in direction z):
FABCD FEFGH
(A1)
or
(zdt) (zdt) BC
(A2)
EF
which results in:
dt dt BC
(A3)
EF
or
dt cons.
(A4)
which means:
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avgt cons.
(A5)
where:
avg
1 dt t
(A6)
As seen, the identity (A5) is correct for all hollow tubes under torsion, regardless the magnitude of thickness.
References 1. Beer, F. P., Johnston, E. R., Mechanics of Materials, 2nd Edition, McGraw-Hill, New York (1992). 2. Timoshenko, S. P., Goodier, J. N., Theory of Elasticity, McGraw-Hill, New York (1970). 3. Baron, F. M., “Torsion of multi-connected thin-walled cylinders,”J Appl Mech, 9, pp.72-74 (1942). 4. Nguyen, S. H., “An accurate finite element formulation for linear elastic torsion calculations,”Comput Struct, 42, pp.707-711 (1992). 5. Wang, C. Y., “Torsion of a flattened tube,”Meccanica, 30, pp.221-227 (1995). 6. Wang, C. Y., “Torsion of tubes of arbitrary shape,”Int J Solids Struct, 35, pp.719-731 (1998). 7. Li, Z., Ko, J. M., Ni, Y. Q., “Torsional rigidity of reinforced concrete bars with arbitrary sectional shape,”Finite Elem Anal Des, 35, pp.349-361 (2000). 8. Sapountzakis, E. J., “Nonuniform torsion of multi-material composite bars by the boundary element method,”Comput Struct, 79, pp.2805-2816 (2001). 9. Sapountzakis, E. J., Mokos, V. G., “Warping shear stresses in nonuniform torsion of composite bars by BEM,”Comput Method Appl M, 192, pp.4337-4353 (2003). 10. Sapountzakis, E. J., Mokos, V. G., “Nonuniform torsion of bars of variable cross section,”Comput Struct, 82, pp.703-715 (2004). 11. Sapountzakis, E. J., Mokos, V. G., “Nonuniform torsion of composite bars of variable thickness by BEM,”Int J Solids Struct, 41, pp.1753-1771 (2004). 12. Hassenpflug, W. C., “Torsion of uniform bars with polygon cross-section,”Comput Math Appl, 46, pp.313-392 (2003). 13. Kołodziej, J. A., Fraska, A., “Elastic torsion of bars possessing regular polygon in cross-section using BCM,”Comput Struct, 84, pp.78–91 (2005).
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14. Hematiyan, M. R., Doostfatemeh, A., “Torsion of moderately thick hollow tubes with polygonal shapes,”Mech Res Commun, 34, pp.528-537 (2007). 15. Sadd, M. H., Elasticity, Elsevier, Burlington (2005). 16. Reddy, J. N., Finite Element Method, 2nd edition, McGraw-Hill, New York (1993).
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Tables
Table 1: Normalized shear stress at different locations (m), example 3 Point A First model t1 5 mm ,
Point B
Point C
Point D
Point E
Point F
Point G
Point H
Thin-walled theory
0.0614 0.0614 0.0614 0.0614 0.0614 0.0614 0.0491 0.0491
Present
0.0612 0.0640 0.0599 0.0653 0.0586 0.0666 0.0451 0.0551
FEM
0.0607 0.0635 0.0594 0.0648 0.0581 0.0661 0.0447 0.0547
Thin-walled theory
0.0475 0.0475 0.0475 0.0475 0.0475 0.0475 0.0633 0.0633
Present
0.0458 0.0594 0.0389 0.0658 0.0321 0.0721 0.0544 0.0844
FEM
0.0425 0.0575 0.0359 0.0635 0.0296 0.0695 0.0512 0.0811
Thin-walled theory
0.0522 0.0522 0.0522 0.0522 0.0522 0.0522 0.0417 0.0417
Present
0.0527 0.0636 0.0450 0.0705 0.0374 0.0774 0.0209 0.0709
FEM
0.0490 0.0614 0.0416 0.0679 0.0341 0.0742 0.0195 0.0678
t2 4 mm
Second mode t1 15 mm ,
t2 20 mm
Third model t1 25mm ,
t2 20 mm
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Table 2: Normalized angle of twist (m-4), example 3 Thin-walled theory
Present
FEM
First model, t1 5 mm ; t 2 4 mm
15.679×104
15.371×104
15.488×104
Second model, t1 15 mm ; t2 20 mm
5.308×104
4.725×104
4.912×104
Third model, t1 25 mm ; t 2 20 mm
5.038×104
4.446×104
4.619×104
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Table 3: Normalized shear stress at different locations (m), example 4
Location
Thin-walled theory
Present
FEM
Point A
0.0802
0.0808
0.0796
Point A'
0.0802
0.0792
0.0779
Point B
0.0401
0.0500
0.0499
Point B'
0.0401
0.0300
0.0288
Point C
0.0505
0.0605
0.0607
Point C'
0.0505
0.0405
0.0395
Point D
0.1011
0.1060
0.1051
Point D'
0.1011
0.0960
0.0951
Point E
0.0209
0.0261
0.0263
Point E'
0.0209
0.0161
0.0163
19
Table 4: Normalized angle of twist (m-4), example 4 Thin-walled theory 1.319×10
4
Present 1.278×10
20
FEM 4
1.350×104
Figure Captions Figure 1: a) Hollow tube cross section, b) Equivalent forces and torque in each segment Figure 2: A straight segment Figure 3: A curved segment Figure 4: Equivalent forces and torque in a curved segment Figure 5: Rounded rectangular tube Figure 6: Flattened tube Figure 7: Normalized shear stress at some points on the inner boundary, example 1 Figure 8: Normalized shear stress at some points on the outer boundary, example 1 Figure 9: Normalized angle of twist, example 1 Figure 10: Normalized shear stress at some points on the inner boundary, example 2 Figure 11: Normalized shear stress at some points on the outer boundary, example 2 Figure 12: Normalized angle of twist, example 2 Figure 13: A hollow tube with different edge thicknesses Figure 14: A multicell cross section Figure A1: A differential element of a hollow tube under torsion
21
Figures:
Figure 1: a) Hollow tube cross section, b) Equivalent forces and torque in each segment
Figure 2: A straight segment
Figure 3: A curved segment
22
dA
r
Fj
Fj
Tj
Figure 4: Equivalent forces and torque in a curved segment
Figure 5: Rounded rectangular tube
Figure 6: Flattened tube
23
Figure 7: Normalized shear stress at some points on the inner boundary, example 1
Figure 8: Normalized shear stress at some points on the outer boundary, example 1
Figure 9: Normalized angle of twist, example 1
24
Figure 10: Normalized shear stress at some points on the inner boundary, example 2
Figure 11: Normalized shear stress at some points on the outer boundary, example 2
Figure 12: Normalized angle of twist, example 2
25
Figure 13: A hollow tube with different edge thicknesses
Figure 14: A multicell cross section
26
Figure A1: A differential element of a hollow tube under torsion
27
