Shear Flow Example

Shear Flow Calculation Process

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

ü Initial load & geometry specification and centroid calculation

We will consider only one nonzero shear resultant at a time but the equations will be set up for both V2 and V3. The case shownbelow is for only V3 acting (and V2 = 0). To compute the shear flow due to V2 it is only necessary to set V3 = 0 instead of V2 = 0in the line below.

V2 = 0 ;

h = 2 b; H∗ this simplifies the resulting stiffnesses ∗Ltw = t;tf = t;

ü Centroid and area:

A = 2 b tf + h tw

4 b t

c = 2 b tf

b

2ì A

b

4

ü Bending stiffnesses:

H22 = EE1

12h3 tw + 2 ∗ b tf

h

2

2

8

3b3 EE t

H33 = EE h tw c2 + 21

12b3 tf + b tf −

b

2+ c

2

5

12b3 EE t

H23 = 0

0

∆H = H22 H33 − HH23L2

10

9b6 EE2 t2

ü Shear flow due to V3

ü (1) Shear flow along top flange Hs1L:

In this example we will use Mathematica functions to define shear flow as a function of s. Even though the integral should be anindefininte integral, we must use a definite integral within a function if the function argument is also the integration variable. Wealso cannot use subscripts in function names. Key: first number is axis and second is si index.

Q21@s_D := EEh

2tf s

Q31@s_D := EE ‡0

sH−c + b − ssL tw ss

f1@s_D :=HQ31@sD H23 − Q21@sD H33L

∆H

V3 −HQ31@sD H22 − Q21@sD H23L

∆H

V2

The following statement makes f1s a simple symbolic result to be used in the plotting function. This avoids having to repeatedlyrecompute the symbolic integral in the above function definitions when plotting the result. The result is a somewhat quickerexecution.

f1s = f1@s1D êê Simplify

−3 s1 V3

8 b2

f1n = f1s ê. s1 → b η H∗ need to use a dimensionless variable for plotting ∗L

−3 η V3

8 b

2 Shear_Flow_example.nb

Plot@f1n b ê V3, 8η, 0, 1<, Frame → True,FrameLabel → 8"s1", "f bêV3", "Shear Flow on Top Flange", ""<, PlotRange → AllD

0.0 0.2 0.4 0.6 0.8 1.0

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s1

fbêV

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Shear Flow on Top Flange

ü (2) Shear flow along web Hs2):

Q22@s_D := EE tf ‡0

s h

2− ss ss

Q32@s_D := EE H−cL tw s

f2@s2_D :=HQ32@s2D H23 − Q22@s2D H33L

∆H

V3 −HQ32@s2D H22 − Q22@s2D H23L

∆H

V2 + f1s ê. s1 → b

f2s = Simplify@f2@s2DD

3 I−2 b2 − 2 b s2 + s22M V3

16 b3

f2n = f2s ê. s2 → h η êê Simplify

3 I−1 − 2 η + 2 η2M V3

8 b

Plot@f2n b ê V3, 8η, 0, 1<, Frame → True,FrameLabel → 8"s2", "f bêV3", "Shear Flow on Web", ""<, PlotRange → AllD

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s2

fbêV

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Shear Flow on Web

Shear_Flow_example.nb 3

ü (3) Shear flow along lower flange Hs3):

This shear flow f Hs3L is the mirror image of f Hs1) and need not be recomputed. However, we will illustrate the power ofMathematica to continue with s4 defined from the lower left vertex instead (i.e., s4 = b - s3).

Q24@s_D := EE tf −h

2s

Q34@s_D := EE t ‡0

sH−c + ssL ss

f4@s4_D :=HQ34@s4D H23 − Q24@s4D H33L

∆H

V3 −HQ34@s4D H22 − Q24@s4D H23L

∆H

V2 + f2s ê. s2 → h

f4s = f4@s4D êê Simplify

3 H−b + s4L V3

8 b2

f4n = f4s ê. s4 → b η êê Simplify

3 H−1 + ηL V3

8 b

Plot@f4n b ê V3, 8η, 0, 1<, Frame → True,FrameLabel → 8"s4", "f bêV3", "Shear flow on Lower Flange", ""<, PlotRange → All D

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s4

fbêV

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Shear flow on Lower Flange

ü Shear flow resultants in flanges and web:

R1 = ‡0

1f1n b η

−3 V3

16

R2 = ‡0

1f2n h η

−V3


R4 = ‡0

1f4n b η

−3 V3

16

NOTE: it is clear that the horizontal and vertical resultants are consistent with and equal to the applied shears V2 and V3.

ü Shear center calculation

Moment equivalence is enforced at a point on the line of action of the shear resultant. Only the x2 location will be computedbecause the shear center must lie on the symmetry axis Hi2L.

ü Method A: ⁄M1 at point A on line of action of V3

8e< = 8e< ê. Solve@0 R1 h − R2 e, eD@@1DD

:3 b

8>

x3 k = 0; x2 k = −e − c

−5 b

8

ü Method B: ⁄M1 at point B on cross section

Moment equivalence is enforced at a convenient point on the cross section (A). Again, only the x2 location will be computedbecause the shear center must lie on the symmetry axis Hi2L.

e =.

8e< = 8e< ê. Solve@−V3 e R1 h, eD@@1DD

:3 b

8>

x3 k = 0; x2 k = −e − c

−5 b

8

Both methods yield exactly the same result for the shear center location. Choose the easiest way for a given section.


Shear flow due to V2

Because of symmetry of both the section and load with respect to axis i2, the shear flow must be zero at points on the section thatlie on the symmetry axis (i.e., shear is an antisymmetric quantity).For this reason, perimetric coordinates could be started at thispoint on the bottom of the section, and due to symmetry only one-half of the section needs to be analyzed (the other is a mirrorimage).

Instead, we will continue with the same perimetric coordinates used previously. In the present formulation which includes bothV2 and V3, it is only necessary to set V2 = 0 and V3 ∫ 0.

V3 = 0; V2 =. H∗ re−define shears ∗L

ü (1) Shear flow along side flange Hs1L:

Using the same previously defined functions:


−3 H3 b − 2 s1L s1 V2

5 b3

f1n = f1s ê. s1 → b η H∗ need to use a dimensionless variable for plotting ∗L

−3 η H3 b − 2 b ηL V2

5 b2

Plot@f1n b ê V2, 8η, 0, 1<, Frame → True,FrameLabel → 8"s1", "f bêV2", "Shear Flow on Side Flange", ""<, PlotRange → AllD

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s1

fbêV

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Shear Flow on Side Flange


ü (2) Shear flow along horizontal web Hs2):

Again, using the same previously defined functions:


3 H−b + s2L V2

5 b2

f2n = f2s ê. s2 → h η êê Simplify

3 H−1 + 2 ηL V2

5 b

Plot@f2n b ê V2, 8η, 0, 1<, Frame → True,FrameLabel → 8"s2", "f bêV2", "Shear Flow on Web", ""<, PlotRange → AllD

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fbêV

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Shear Flow on Web

This confirms that f Hs2 = h ê2L = 0 at the point where the section crosses the symmetry axis. The shear flow on the right flange isthe mirror image of f Hs1L so that f Hs3L = f Hs1L and is not recomputed. Note that once again, the shear flow is linear on axesperpendicular to the shear and parabolic elsewhere.

Due to the symmetry, the shear center must lie on the symmetry axis, as noted previously.

For a general combination of V2 and V3, it is necessary to superpose the two shear flow components to find the total shear flow.


Shear Flow Example

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