Talat Lecture 2301: Design Of Members Example 8.1: Torsion Constants For Open Cross Section

TALAT Lecture 2301

Design of Members Torsion Example 8.1 : Torsion constants for open cross section 6 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm

Date of Issue: 1999  EAA - European Aluminium Association

TALAT 2301 – Example 8.1

1

Example 8.1. Torsion constants for open cross section

Comment: The following expressions are applicable to open cross sections. When there are branches, go back the same way but with thickness = 0. Example: The co-ordinates for node 7 and 8 are the same as for 5 and 4 and t = 0 for element 7 and 8.

Nodes, co-ordinates and thickness

i

0

34

0

6

1

10

0

6

2

8

9

6

3

25

9

6

4

25

3

6

5

59

0

6

6 7

59

y

59

. mm

13

z

0

. mm

15

t

0

8

25

3

0

9

25

34

6

10

0

37

6

11

0

245

6

12

62

252

6

13

62

289

10

. mm 300 0 250

200

150 z

Nodes Area of cross section elements Cross section area

i

1 .. rows ( y )

1

mm 100

dAi

ti .

yi

2

yi 1

zi

zi 1

2 50

rows ( y )

1 A = 3.199 . 10 mm 3

dAi

A

0

0

2

i =1 50

First moment of area, gravity centre

rows ( y )

zi

Sy

dAi zi 1 . 2

z gc

i =1

Second moment of Iy area of effective cross section

rows ( y )

1 zi

2

zi 1

2

dAi zi . zi 1 . 3

i =1

rows ( y ) First moment of area, gravity centre

1

Sz

0

50

Sy A

100

y mm

z gc = 117.14 mm Iy

Iy

A . z gc

2

I y = 3.596 . 10 mm 7

1 yi

dAi yi 1 . 2

y gc

i =1

TALAT 2301 – Example 8.1

50

2

Sz A

y gc = 19.073 mm

4

Second moment of area

Second moment of area

Principal axis

rows ( y )

Iz

1 2 . yi 1 . zi 1

I yz

2 . yi . zi

dAi yi . zi 1 . 6

yi 1 . zi

I yz

i =1

ω0

if I z 1. 2 1. 2

Iz

Iz

Iy

Iy

Iz

Iz

Iy

2

4 . I yz

2

4 . I yz

α.

.z

180

π

ωi

1

6

4

ω 0 i

rows ( y )

1

ω

Iω

i

1

ω

.

i

dAi

ω mean

2

Iω A

i =1

ω mean = 799.492 mm 1 2 . yi 1 . ω i 1

I yω

2 . yi . ω i

yi 1 . ω i

dAi yi . ω i 1 . I yω 6

1 2 . ω i 1 . zi 1

2 . ω i . zi

ω

.z

i

1 i

ω i . zi

dA . Ii 1 6 zω

2

ω

i

ω

dAi ω i .ω i 1 . 3

2 i

1

I yω . I yz

I y .I z I ωω

2

I yz

z sc . I yω

rows ( y )

TALAT 2301 – Example 8.1

1

I yω . I y

z sc

I y .I z

y sc . I zω

to

i

2

3

I zω . I yz 2

I yz

if t i > 0 . mm , t i , 100 . mm

Iω

I ωω

I ωω

I ωω

= 7.377 . 10

i =1

Iw

A 8

1

5

S y .I ω

I zω

I zω = 9.181 . 10 mm

I ωω

I zω . I z

A 8

I zω

y sc

S z.I ω

I yω

I yω = 2.921 . 10 mm

rows ( y )

Warping constant

4

I η = 1.983 . 10 mm

i =1

Shear centre

= 3.414 7

2

4

yi . zi 1

rows ( y )

Sectorial constant

A

I ξ = 3.608 . 10 mm

2

i =1

Sectorial constant

S y .S z

I yz

2

1 i

i

4

I yz = 2.026 . 10 mm

2 . I yz 1 4 0 . mm , 0 , . atan 2 Iz Iy

Iy

ω 0 yi i ω

2

6

Iy

0 . mm

A . y gc 6

rows ( y )

α

Iz

I z = 2.104 . 10 mm

rows ( y ) Sectorial constant

dAi yi . yi 1 . 3

2

yi 1

i =1

Iη

ωmean = mean of sectorial co-ordinates

2

yi

Iz

Iξ

Sectorial co-ordinates

1

5

2

A 10

mm

y sc = 18.728 mm z sc = 120.773 mm I w = 2.129 . 10

10

mm

6

6

rows ( y ) Torsion constant

1

ti

dAi .

It

i

i

2

It

Wt

3

I t = 5.856 . 10 mm

4

W t = 9.76 . 10 mm

3

4

min t o

i =1

3

Sectorial i 0 .. rows ( y ) 1 co-ordinate with respect ω s ω i ω mean i to shear centre

ω max

i

z sc . yi

y gc

y sc . zi

if ω mi > ω ma , ω mi , ω ma

A = 3.199 . 10 mm 3

z gc

ω mi min ω s

ω max= 7.804 . 10 mm 3

2

ω ma Ww i

2

y gc = 19.073 mm z gc = 117.14 mm

250

z sc = 120.773 mm I y = 3.596 . 10 mm 7

I z = 2.104 . 10 mm 6

4

6

Torsion constants

I t = 5.856 . 10 mm

4

W t = 9.76 . 10 mm

3

4 3

Warping constants

I w = 2.129 . 10

10

200

4

I yz = 2.026 . 10 mm

4

mm

150

100

6 50

6 4 W w = 2.729 . 10 mm

ω max= 7.804 . 10 mm 3

2 0

α = 3.414 deg I ξ = 3.608 . 10 mm 7

4

I η = 1.983 . 10 mm 6

50 4

Comment: If the load is acting below the shear centre (the point) there is no torsional moment acting on the beam.

TALAT 2301 – Example 8.1

4

50

0

50

ω max

1 .. rows ( y )

300

y sc = 18.728 mm

max ω s Iw

100

1

Sectorial co-ordinate with respect to shear centre

300

250

i

0 .. rows ( y )

1

ω s i

200

2 mm . 1000

i= 0

-7.804

1

-4.906

2

-4.743

3

-0.46

4

0.065

5

3.938

6

2.927

7

3.938

8

0.065

9

1.42

10

-0.618

11

3.278

12

-4.293

13

-1.307

= 150

100

50

0

50

100

100

50

0

50

100

150

sectorial co-ordinate cross section kN 1000 . N

Biaxial bending Moment My

Principal axis bending

R

Torsion

mt

Bi-moment

B3

St Venants torsion

Tw

Warping stress

σ w i

q q.

L

5 . kN . m

2

Mz

8 sin( α )

Mξ

sin( α ) cos ( α )

Mη

cos ( α )

0 . kN

m t .L

L

R.

My

Mξ

Mz

Mη

B3

2

T w = 0 kN . m

2 .ω

TALAT 2301 – Example 8.1

si

5

6 MPa 10 . Pa

Mz

2

B 3 = 0 kN . m

4.8 . m

My

0 . kN . m

m t = 0 kN

8 m t .L

Iw

1

14.4

=

0

=

kN . m

14.374 0.857

kN . m

Rotation of co-ordinate system

i

0 .. rows ( y ) M ξ . yi

σ ξ i

Bending stresses

yi

1

zi y gc

σ η

Iξ

Sum of stresses

σ

σ ξ

Max stresses

σ max

R.

σ η

σ w

yi

y gc

zi

z gc

M η . zi

y gc

R.

z gc

z gc

Iη

i

σ mi

min ( σ ) σ ma

max ( σ )

if σ mi > σ ma , σ mi , σ ma

σ max= 74.1 MPa

Principal axis 300

250

σ w i i=

200

150

100

50

0

0

50

0

50

MPa

=

σ η i MPa

=

σi MPa

=

ti mm

0

-8.78

51.3

42.52

6

2

0

-7.78

55.133

47.36

6

3

0

5.35

54.283

59.63

6

4

0

5.07

49.105

54.17

6

5

0

18.66

49.524

68.18

6

6

0

18.97

55.134

74.1

15

7

0

18.66

49.524

0

0

8

0

5.07

49.105

0

0

9

0

4.33

35.727

40.06

6

10

0

-5.68

35.076

29.39

6

11

0

-10.62

-54.687

-65.31

6

12

0

13.87

-59.304

-45.43

6

13

0

13

-75.271

-62.27

10

100

Cross section Axial stress

TALAT 2301 – Example 8.1

σ ξ i

1

50

100

MPa

=

6

=