Chapter 2 Torsion Stresses in Thin-Walled Multi-Cell Box-Girders

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1 Chapter Torsion Stresses in Thin-Walled Multi-Cell Box-Girders. Torsion of Uniform Thin-Walled Two-Cell Box-Girders The thin-walled box section with uniform thickness t as shown in Fig.., is subjected to a torsion moment T. The shear flow and angle of twist for the thin-walled two cell structure shown in Fig.. could be determined as follows. The flexural warping coefficients are given by d ¼ =G s ¼ =Gt ðac þ CD þ DB þ BAÞ d ¼ =Gt ðdc þ CF þ FH þ HDÞ d ¼ =G CD=t Since the angle of twist is the same for the two cells, then the basic equations are given by d q þ d q A h ¼ 0 d q þ d q A h ¼ 0 From equations (..) and (..) we get ð:þ ð:þ d d q þ d d q ¼ d A h d q þ d d q ¼ d A h Hence q ðd d d Þ¼d A h d A h M. Shama, Torsion and Shear Stresses in Ships, DO: 0.007/ _, Ó Springer-Verlag Berlin Heidelberg 00

2 Torsion Stresses in Thin-Walled Multi-Cell Box-Girders Fig.. Shear flow due to torsion of a thin-walled box girder with two unequal cells The solution of equations (.) and (.) gives q ¼ h ðd A d A Þ d d d ¼ D h q ¼ h ða D d Þ=d ¼ D h D ¼ ðd A d A Þ d d d The equilibrium condition gives Hence D ¼ ða D d Þ=d T ¼ A q þ A q ¼ D 3 h D 3 ¼ðA q þ A q Þ h ¼ =D 3 T q ¼ D =D 3 T q ¼ D =D 3 T q ¼ q q ¼ ðd D Þ=D 3 T: Example. Determine the torsion shear stress and angle of twist for the two uniform thickness thin-walled box-girder shown in Fig... Solution The shear flow and angle of twist for the thin-walled two cell structure shown in Fig.. could be determined as follows.

3 . Torsion of Uniform Thin-Walled Two-Cell Box-Girders 3 Fig.. A uniform thinwalled box-girder with two cells The torsional moment is given by T ¼ q ð A þ q A Þ ð:3þ The angle of twist for cells and are given by 0 h ¼ q GA q 0 h q GA q Z A A Since the angle of twist is the same for the two cells, then we have ð:4þ ð:5þ h ¼ h ¼ h 3 Reformulating equations (.4) and (.5), we get 0 G q q A ¼ A h G q q A ¼ A h Let d ¼ warping flexibility d ¼ G d ¼ G ð:6þ ð:7þ

4 4 Torsion Stresses in Thin-Walled Multi-Cell Box-Girders d ¼ d ¼ Z G Substituting in equations (..4) and (..5), we get d q þ d q A h ¼ 0 d q þ d q A h ¼ 0 ð:8þ ð:9þ Solving equations (..3), (..8) and (..9), q and q could be determined. Example. Determine the torsion shear stresses and the rate of twist for the thinwalled -cell box-girder shown in Fig..3. The girder is subjected to a constant torque T. Fig..3 A thin-walled boxgirder with two unequal cells Solution Area of cell () is given by A ¼ a Area of cell () is given by A ¼ a Let d ¼ warping flexibility d ¼ =G ¼ 6a=Gt d ¼ 4a=Gt d ¼ a=gt The basic equations are d q þ d q A h ¼ 0 d q þ d q A h ¼ 0 ð:0þ ð:þ

5 . Torsion of Uniform Thin-Walled Two-Cell Box-Girders 5 The equilibrium equation gives T ¼ q ð A þ q A Þ ¼ a ðq þ q Þ ð:þ From equations (.0) and (.), we get Hence d q A þ d q A d q A d q A ¼ 0 q ðd A d A Þþq ðd A d A Þ ¼ 0 d A d A d d q ¼ q ¼ q d A d A d d From equation (.), we get T ¼ a q d d þ d d T ¼ a 4d 4d þ d q d d From which q is given by q ¼ T d d a 4d 4d þ d q ¼ T d d a 4d 4d þ d Substituting in equation (.0), we get h ¼ ðd q A þ d q Þ ¼ 4a T a d d d d þ d d d 4d 4d þ d ¼ T 4a 4 d d d 4d 4d þ d Substituting for d, d and d, we get q ¼ T a 8 a=t þ a=t G ½ð Þ= ð 6 a=t þ 4 a=t þ 6 a=t ÞŠ ¼ T a G 9 6 q ¼ T a G ½ð 6 a=t þ a=tþ= ð 6a=t ÞŠ ¼ T a G 8 6 h ¼ T 4a 4 G 4 a t a t T ð6 a=tþ ¼ 4a 4 G 3=6 a=t

6 6 Torsion Stresses in Thin-Walled Multi-Cell Box-Girders But h ¼ T=GJ Hence J ¼ T=Gh ¼ 04=3 a 3 t:. The General Case of a Uniform Two-Cell Box Girder This is an indeterminate structural problem and its solution is based on the assumption that the rate of twist for each cell is the same as for the whole section, see Fig..4. i.e., h ¼ h ¼ 0 and ðh ¼ du=dzþ The torque T is given by h ¼ T=GJ T ¼ q A þ q A h ¼ q=t ds GA and h ¼ GA q=t ds Fig..4 dealized section and torsion shear flow of a thin-walled two cell structure

7 . The General Case of a Uniform Two-Cell Box Girder 7 i.e., or i.e., 3 4 G q q=t ds q ðþ 5 ¼ A h ð:3þ 4 G q8 Equations (.3) and (.4) are simplified to 3 q ðþ 5 ¼ A h ð:4þ d q þ d q ¼ A h d q þ d q ¼ A h q ¼ A q A d d d d ½dŠfqg ¼ hfag The shear flow in each cell is given by fqg ¼ ½dŠ h fag h fqg ¼ h ½dŠ fag d ¼ G d ¼ G ð:5þ d ¼ d ¼ G ½Š The torque is given by T ¼ ðq A þ q A Þ ð:6þ Solving equations (.5) and (.6), we get q,q and h The torque T is given by T ¼ G J h

8 8 Torsion Stresses in Thin-Walled Multi-Cell Box-Girders Hence J is given by J ¼ T=Gh: Example.3 Determine the torsion shear stresses and angle of twist for the thinwalled box section having uniform thickness t as shown in Fig..5. The section is subjected to a torsion moment T. Fig..5 Shear flow due to torsion of a box-girder with two cells Solution Condition for Compatibility (Consistent Deformation). The warping flexibilities are given by d ¼ =Gt ðab þ BC þ CD þ DAÞ d ¼ =Gt ðch þ HF þ FN þ NCÞ d ¼ =GCD=td ¼ =GCD=t The basic equations of consistent deformation are given by d q þ d q ¼ A h d q þ d q ¼ A h ð:7þ ð:8þ Solving equations (.7) and (.8) we get q ¼ d A d A d d d h ¼ D h q ¼ A d h D h ¼ D h d d Equation for equilibrium condition is given by T ¼ðA q þ A q Þh ¼ D 3 h ð:9þ The solution of equations (.7), (.8) and (.9) gives h ¼ T=D 3 q ¼ T D =D 3

9 . The General Case of a Uniform Two-Cell Box Girder 9 q ¼ T D =D 3 q ¼ q q ¼ T ðd D Þ=D 3 :.3 Torsion Stresses in a Two dentical Cells Box-Girder The two identical thin-walled cells box girder, see Fig..6, behaves exactly as a single cell box girder. Since the two cells are identical we have q ¼ q ¼ q: Fig..6 A thin-walled box girder with two identical cells.3. Shear Flow q The shear flow q is the same for the two cells and is given by q ¼ T=A.3. Shear Stress s A ¼ B D: The shear stress in the sides, top, bottom and the internal partition plating are given by s S ¼ q=t s ; s D ¼ q=t D ; s B ¼ q=t B ; s L ¼ 0:

10 30 Torsion Stresses in Thin-Walled Multi-Cell Box-Girders.3.3 Rate of Twist h The rate of twist is given by h ¼ T=GJ J ¼ 4A ¼ B=t B þ D=t S.4 Torsion of Three-Cell Box-Girder Following the same principle that the angle of twist is the same for the three cell box-girder shown in Fig..7. Then h ¼ h ¼ h 3 ¼ h The equations of consistent deformation are given by The torque is given by d q þ d q A h ¼ 0 d q þ d q A h ¼ 0 d 3 q þ d 33 q A 3 h 3 ¼ 0 ð:0þ ð:þ ð:þ T ¼ðA q þ A q þ A 3 q 3 Þh ð:3þ Solving equations (.0) (.3), we get q,q,q 3,q 4 and h. Fig..7 Torsion of a threecell box-girder

11 .4 Torsion of Three-Cell Box-Girder 3 Hence " d d 0 #( ) q d d d 3 q 0 d 3 d 33 q 3 ½dŠfqg ¼ hfag ¼ h ( ) A A A 3 ð:4þ Hence the shear flow in each cell is given by fqg ¼ d fagh and T ¼ h X A i q i d ¼ d ¼ d 33 ¼ 3 d ¼ d ¼ ½ Š d 3 ¼ d 3 ¼ ½ Š 3 : Example.4 Determine the shear flow, shear stress and rate of twist for the threecell box girder shown in Fig..8. Fig..8 Shear flow due to torsion of a 3-cell thin-walled box girder

12 3 Torsion Stresses in Thin-Walled Multi-Cell Box-Girders Solution Following the same principle that the angle of twist is the same for all cells, see Fig..8, we get d q þ d q ¼ A h d q þ d q þ d 3 q 3 ¼ A h d 3 q þ d 33 q 3 ¼ Ah 3 ð:5þ but h ¼ h ¼ h 3 ¼ h Then " d d 0 #( ) q d d d 3 q 0 d 3 d 33 q 3 ¼ h ( ) A A A 3 ½dŠfqg ¼ hfag fqg ¼ d fagh ð:6þ ð:7þ The torque is given by Solving equations (.6) and (.8) we get T ¼ ðq A þ q A þ q 3 A 3 Þ ð:8þ q ¼ hh A q ¼ hh A q 3 ¼ hh A 3 Substituting in equation (.5), we get h ¼ =A ðd q þ d q Þ d ¼ =G d ¼ =G d 33 ¼ =G 3 d ¼ d ¼ =G½Š d 3 ¼ d 3 ¼ =G½Š 3:

13 .5 Torsion of Uniform Thin-Walled Multi-Cell Box-Girder 33.5 Torsion of Uniform Thin-Walled Multi-Cell Box-Girder The multi-cell thin-walled structure when subjected to pure torsion is a statically indeterminate problem; see Fig..9. The torque T is given by T ¼ Xn i¼ A i q i ¼ GJhJ T = applied uniform torque; A i = enclosed area of the ith cell; J = torsion constant J ¼ 4 Xn The angle of twist per unit length i¼ A i d A i h ¼ du=dz h i ¼ h j ¼ h ij ¼ h jn h i ¼ =GA i q i The angle of twist for cell i is given by Z h i ¼ =GA i q i q i q iþ Z ð:9þ Equation (.9) represents a series of simultaneous equations which gives q ; q ; q 3 ;...; q n. The set of equations of consistent deformation is given by d q þ d q A h ¼ 0 ð:30þ Fig..9 Torsion of a multi-cell thin-walled box-girder

14 34 Torsion Stresses in Thin-Walled Multi-Cell Box-Girders d q þ d q A h ¼ 0 ð:3þ d 3 q þ d 33 q A 3 h ¼ 0 This set of equations could be put in the following form ½dŠfqg ¼ hfag Hence, for a multi cell box girder, the shear flow in each cell is given by fqg ¼ ½dŠ h fag ði ¼ ; ;...; nþ 0 d d d d d d d ¼ 3 d 33 d B 0 0 d 43 d 44 d 45 0 A d 54 d 55 d d 65 d 66 The torsion shear stresses are given by s ¼ q =t ; s ¼ q =t ; s 3 ¼ q 3 =t 3 : ð:3þ.6 Combined Open and Closed Thin-Walled Sections For the combined open and closed section, see Fig. 3., the angle of twist is the same for the whole section whether it is an open or closed section..6. Combined Open Section with One Closed Cell The total torque T for the thin-walled section shown in Fig..0 is given by T= X ¼ T i ¼ G J h Fig..0 Combined open and closed one-cell thinwalled section

15 .6 Combined Open and Closed Thin-Walled Sections 35 Hence h ¼ T=GJ T ¼ GJ h J = torsion constant of the open section; J = Torsion constant of the closed section. For the open part of the structure, the shear flow q is given by q ¼ T t =J T ¼ G J h For the closed section of the structure, the shear flow q is given by q ¼ T= A T ¼ A q ¼ G J h:.6. Combined Open Section with Two Closed Cells The applied torque T for the thin-walled structure shown in Fig.. is given by T ¼ Xn q j A j þ Xm GJ i h j¼ i¼ n the above particular example T ¼ q A þ q A þ GJ 3 h J 3 = the torsion constant of the open section part of the structure and is given by J 3 Fig.. Combined thinwalled open and closed two-cell structure T ¼ =3 Xn i¼ b i t 3 i

16 36 Torsion Stresses in Thin-Walled Multi-Cell Box-Girders The shear flow in the two cells is given by d q þ d q ¼ A h d q þ d q ¼ A h d ¼ =G d ¼ =G d ¼ d ¼ =G½Š n the general case, for combined open and closed sections, the shear flow in each cell is given by q i ¼ d hh i ði ¼ ; ;...; nþ And in each open member the shear flow is given by And the angle of twist is given by T = the torque and is given by T ¼ 4 Xn i¼ q i ¼ T=J t i h ¼ T=GJ A 0 i d A i þ 3 Example.5 Determine the shear flow distribution and rate of twist for the idealized ship section shown in Fig... The ship section is subjected to a torque T. Solution The torque T is distributed among the thin-walled structural members of the ship section as follows T ¼ X4 T i i¼ X m j¼ b j t 3 j : T ¼ A q ; T ¼ A q ; T 3 ¼ A 3 q 3 ; T 4 ¼ GJ 4 h ¼ q 4 J 4 t ; J 4 ¼ =3 St 3 4 4

17 .6 Combined Open and Closed Thin-Walled Sections 37 Fig.. dealized ship section because of symmetry of the ship section, we have T ¼ T ; T 4 ¼ T 5 Hence T ¼ T ¼ A q ¼ G J h Thus T ¼ GJ h þ GJ 4 h þ GJ 3 h ¼ X3 J ¼ 4A G ¼ E ð þ tþ ¼ E 6 J ¼ J þ J4 þ J3 ds t j¼ G J h ¼ 4ða b) ðb=t þ a=tþ J 3 ¼ A(B h) ðb=t 3 þ h=t 3 Þ J 4 ¼ =3 S 4 t 3 4 ¼ =3 ðd a h) t3 4 Hence h ¼ T.X GJ Substituting, we get the torque carried by each structural element.

18 38 Torsion Stresses in Thin-Walled Multi-Cell Box-Girders Hence T ¼ GJ h T 3 ¼ GJ 3 h T 4 ¼ GJ 4 h Substituting, we get the shear flow in each structural element as follows q ¼ T =A ; q 3 ¼ T 3 =A 3 ; and q 4 ¼ T 4 =A 4 : Example.6 Determine the shear flow and rate of twist for the ship section of bulk carrier shown in Fig..3. Solution The torque T is given by T ¼ A ½ q þ q A þ GJ 6 h þ A 3 q 3 þ A 4 q 4 þ A 5 q 5 Š The torsion constant J is given by J ¼ ( ) X5 i¼ The rate of twist h is given by 4A 0 i d A i þ =3 k 6 t 3 6 h ¼ T=GJ The set of equations of consistent deformation for cells () and () is given by d q þ d q ¼ A h ¼ T=J A d q þ d q ¼ A h ¼ T=J A Fig..3 An idealized section of a bulk carrier

19 .6 Combined Open and Closed Thin-Walled Sections 39 This set of equations can be put in the matrix form as follows d d q A ¼ h d d A i.e., q ðdþfqg ¼ T=J fag Hence, the torsion shear flow in cells () and () are given by q ¼ d T=J A q ¼ d T=J A Similarly, the torsion shear flow in cells (3), (4) and (5) are given by q 3 ¼ d T=J A 3 q 4 ¼ d T=J A 4 q 5 ¼ d T=J A 5 3 d 33 d 34 0 d ¼ 4 d 43 d 44 d d 54 d 55 d ii ¼ =G Pm k j tj ; i ¼ ; ;...; n ¼ N of cells J¼ d rj ¼ =G k rj trj i r, and j are cells, having a common boundary; i = cell No. i.

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