Box Shear Flow

Shear Flow & Shear Center in Box Section (1)

Remove@"Global`∗"D H∗ remove all symbols ∗L

ü Bending stiffness:

H22 = EE1

12b3 t +

1

12b3 5 t + 2 H2 bL t

b

2

2

3

2b3 EE t

ü Shear flow fo in open section:

The stiffness static moments are defined as: Qmn where m=axis and n=coordinate subscript.

Q21 = EE s1 t H−b ê 2L;Q22 = EE s2 t H−b ê 2 + s2 ê 2L;Q23 = EE s3 t b ê 2;Q24 = EE s4 5 t H−b ê 2 + s4 ê 2L;

fo1 = −Q21 V3 ê H22

s1 V3

3 b2

fo2 = −Q22 V3 ê H22 + fo1 ê. s1 → 2 b

2 V3

3 b−

2 I− b

2+

s2

2M s2 V3

3 b3

fo3 = −Q23 V3 ê H22 + fo2 ê. s2 → b

2 V3

3 b−

s3 V3

3 b2

fo4 = −Q24 V3 ê H22

−10 I− b

2+

s4

2M s4 V3

3 b3

ü Continuity in cell:

warp = ‡0

2 b 1

G tHfo1 + fcL s1 + ‡

0

b 1

G tHfo2 + fcL s2 + ‡

0

2 b 1

G tHfo3 + fcL s3 − ‡

0

b 1

G 5 tHfo4 − fcL s4

26 b fc

5 G t+

2 V3

G t

8fc< = 8fc< ê. Solve@warp 0, 8fc<D@@1DD

:−5 V3

13 b>

ü Compute final shear flows:

f1 = fo1 + fc êê Simplify

H−15 b + 13 s1L V3

39 b2


I11 b2 + 13 b s2 − 13 s22M V3

39 b3


H11 b − 13 s3L V3

39 b2

f4 = fo4 − fc êê Simplify

5 I3 b2 + 13 b s4 − 13 s42M V3

39 b3

Plot shear flows:

f1n = f1 ê. s1 → b η

H−15 b + 13 b ηL V3

39 b2

2 Box_shear_flow.nb

PlotBf1n b

V3

, 8η, 0, 2<, AxesLabel → 8"s1êb", "f bêV3"<F

0.5 1.0 1.5 2.0s1êb

-0.3

-0.2

-0.1

0.1

0.2

f bêV3

f2n = f2 ê. s2 → b η

I11 b2 + 13 b2 η − 13 b2 η2M V3

39 b3

PlotBf2n b

V3


0.2 0.4 0.6 0.8 1.0s2êb

0.32

0.34

0.36

f bêV3

f3n = f3 ê. s3 → b η

H11 b − 13 b ηL V3

39 b2

Box_shear_flow.nb 3

PlotBf3n b

V3


0.5 1.0 1.5 2.0s3êb

-0.3

-0.2

-0.1

0.1

0.2

f bêV3

f4n = f4 ê. s4 → b η

5 I3 b2 + 13 b2 η − 13 b2 η2M V3

39 b3

PlotBf4n b

V3


0.2 0.4 0.6 0.8 1.0s4êb

0.5

0.6

0.7

0.8

f bêV3

ü Shear Center Calculation:

Compute moment equivalence at point B (see drawing at beginning):

R1 = ‡0

2 bf1 s1

−4 V3

39

R2 = ‡0

bf2 s2

79 V3

234

4 Box_shear_flow.nb

Solve@−e V3 −b R1 − 2 b R2, eD

::e →67 b

117>>

% êê N

88e → 0.57265 b<<

Shear center is located this distance (e) to the left of the right side of the box and on the i2 (symmetry) axis.

Shear Flow & Shear Center in Box Section (2)

Remove@"Global`∗"D H∗ remove all symbols ∗L

ü Bending stiffness:

H22 = EE1

12b3 t +

1

12b3 5 t + 2 H2 bL t

b

2

2

3

2b3 EE t

ü Shear flow fo in open section:

This notebook uses functions instead of explicit statements to calculate the static stiffness moments. Note that subscripts cannotbe used in function names or arguments. The stiffness static moments are defined as: Qmn where m=axis and n=coordinatesubscript.

Q21@s_D := EE s t H−b ê 2L

Q22@s_D := EE s t H−b ê 2 + s ê 2L

Q23@s_D := EE s t b ê 2

Q24@s_D := EE s 5 t H−b ê 2 + s ê 2L

fo1@s_D := −Q21@sD V3 ê H22

fo2@s_D := −Q22@sD V3 ê H22 + fo1@2 bD

fo3@s_D := −Q23@sD V3 ê H22 + fo2@bD

Box_shear_flow.nb 5

fo4@s_D := −Q24@sD V3 ê H22

ü Continuity in cell:

warp = ‡0

2 b 1

G tHfo1@sD + fcL s +

‡0

b 1

G tHfo2@sD + fcL s + ‡

0

2 b 1

G tHfo3@sD + fcL s − ‡

0

b 1

G 5 tHfo4@sD − fcL s

26 b fc

5 G t+

2 V3

G t

fc =. H∗ may be needed if statement is re−executed ∗L

8fc< = 8fc< ê. Solve@warp 0, 8fc<D@@1DD

:−5 V3

13 b>

ü Compute final shear flows:

f1 = fo1@sD + fc êê Simplify

H−15 b + 13 sL V3

39 b2


I11 b2 + 13 b s − 13 s2M V3

39 b3


H11 b − 13 sL V3

39 b2

f4 = fo4@sD − fc êê Simplify

5 I3 b2 + 13 b s − 13 s2M V3

39 b3

ü On your own:

Redesign the code so that the top and bottom thickness are equal but the side thicknesses are different (and not equal). Thismaintains a horizontal symmetry axis but creates a more general problem.

6 Box_shear_flow.nb

Box Shear Flow

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Transcript of Box Shear Flow