Shear Center Theory

8/4/2019 Shear Center Theory

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AE3145 Shear Center Lab (S2k) Slide 1

Shear Center in Thin-Walled Beams Lab

Shear flow is developed in beams with thin-walled cross sections shear flow (qsx): shear force per unit length along cross section

qsx=sxt

behaves much like a flow, especially at junctions in cross section

shear flow acts along tangent (s) direction on cross section

there is a normal component, nx, but it is very small

e.g., because it must be zero at t/2

shear force: qsx

ds (acting in s direction)

Shear flow arises from presence of shear loads, Vy or Vz needed to counter unbalanced bending stresses, x to determine, must analyze equilibrium in axial (x) direction

Shear center: resultant of shear flow on section must equal Vyand Vz moment due to qsxmust be equal to moment due to Vyand Vz shear center: point about which moment due to shear flow is zero

not applying transverse loads through shear center will cause atwisting of the beam about the x axis


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Approach for Lab

Apply transverse loading to tip of a cantilever thin-walled beam

use cross-arm at tip to apply both a lateral force and twisting mom.

measure bending deflection

measure twisting vary location of load point along cross-arm

repeat for beam rotated 90 deg. about x axis

Data analysis

record deflections using LVDT

plot twisting versus load position on cross-arm

determine location on cross-arm where load produces no twisting

Compare the measured shear center with theoretical location shear flow calculations used to compute shear center

consider both y axis and z axis loading (rotated 90 deg)


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Review from AE2120 (2751), AE3120

Bending of beams with unsymmetrical cross sections bending stress depends on Iy, Izand Iyz neutral surface is no longer aligned with z or y axes

Shear stresses are computed from axial force equilibrium

shear stress needed to counter changingx analysis strictly correct for rectangular sections only

Thin-walled cross sections

thin walls support bending stress just like a solid section (no change)

thin walls support shear stress in tangential direction

transverse shear component is negligable...

because it must vanish at the free surfaces (edges of cross section)

shear flow: xst (force/unit length along section)

shear flow must be equivalent to Vyand Vzso it must: produce same vertical and horizontal force (Vxand Vy)

produce same mumoment about any point in cross section

point about which no moment is developed: SHEAR CENTER

lateral load must be applied through SC to avoid twisting beam twisting loads will cause section to twist about SC (center of rotation)


4/15AE3145 Shear Center Lab (S2k) Slide 4

LVDT

cross arm

weight

Test Configuration

Lab Apparatus

Cantilever with thin-walled C section

Cantilever with thin-walled C section

LVDT measures tip

deflection on cross-arm

LVDT measures tipdeflection on cross-arm

Small weight used to applyload at point on cross-arm

Small weight used to applyload at point on cross-arm



Lab Procedure

1. Determine the beam material properties from reference material (e.g., referencedtextbooks or MIL Handbook 5 which can be found in the GT Library).

2. Find the centroid of the given beam cross-section.3. Determine Iz, Iy, Iyz for the given section.

4. Determine the shear flow distribution on the cross-section for a Vy shear load.5. Determine the shear flow distribution on the cross-section for a Vz shear load.

6. Determine the shear center for the cross-section.7. Using data from the lab, determine the measured location of the shear center andcompare this with the location determined in step 6 above.



Beam Cross Section

2

2

yy

A

zz

A

yz

A

I z dA

I y dA

I yz dA

=

=

=

0

0

A

A

z dA

ydA

=

=

Centroidal Axes: Area Moments (of Inertia):

1.353in.

1.330in.

0.420in.

0.050in.

Z

Y

Use single line approx forcross section (t



Bending of Beam with Unsymmetrical Cross Section

q

Z

Y

A1

2( ) ( ) yy yz z yz zz y

x

zz yy yz

y I z I M y I z I M I I I

+

=

zx

zz

y MI

=

Symmetric cross section, Mz=0:

General:

But also consider

equilibrium ofsegment A1 (see

next slide!)

But also consider

equilibrium ofsegment A1 (seenext slide!)

Acts overcross section



Shear Stresses and Shear Flow

1 1

0 x x sx xA A

x dx x

F dA q dx dA

+

= = +

Axial force equilibrium for element:

Complementary

qsx acts on A1 inopposite direction

Complementary

qsx acts on A1 inopposite direction

sZ

Y

X

qsx

x+dx

x

A1



Shear Flow

1 1 1 1

2 2

y zsx yy yz zz yz

A A A A yy zz yz yy zz yz

V Vq I y dA I z dA I z dA I y dA

I I I I I I

= +

Result for qsx:

s

Z

Y

Shear flow: qsx(s)



s

Z

Y

Shear flow: qsx(s)

Shear Center

Vy

ez

Therefore:Shear center liesdistance ez from

origin where:

M0=Vyez

Therefore:Shear center lies

distance ez fromorigin where:

M0=Vyez

Moment, M0, atorigin due to

shear flow, qsx

Moment, MMoment, M00, at, at

origin due toorigin due to

shear flow, qshear flow, qsxsx

Moment due to Vy

must be equal to M0

Moment due to VMoment due to Vyymust be equal to Mmust be equal to M

00



Examples of Shear Centers

Section Symmetric about y axis:

Shear center must lie on y axis

(similar argument for z axis symmetry)

Section Symmetric about y axis:

Shear center must lie on y axis

(similar argument for z axis symmetry)

Angle Section:

Shear center must lie atvertex of legs (regardless oforientation of section)

Angle Section:

Shear center must lie atvertex of legs (regardless oforientation of section)

Z

Y

Vy

Shear

Centerlies on

y axis

Z

Y

qsx

Vy

Shear

Center

qsx


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Z

Y

A

B

ShearCenter

Vy

qsx

qsx

qsx

Shear Center Must Lie Outside C

Sum moments from qsx about A:=force in each flange x h/2

h/2

h/2

e

Must equal moment from Vy about A:

=Vyx e

e must be positivefor qsx as shown

so shear centerlies to left of

section

e must be positivefor qsx as shown

so shear center

lies to left of

section


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Data Acquisition

Use PC data acquisition program to acquire deflection andstrain data and test machine load

Use 2LVDTdisplacement gages

Measure vertical displacements at ends of cross arm Use to determine vertical deflection and cross arm rotation

Use single weight but move to different locations on cross arm

Replace dialgages with LVDTs

Replace dialgages with LVDTs

Loading systemLoading system

Cross armCross arm


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Data Reduction

Acquired data is voltage from transducers

convert to inch units

Determine vertical displacement per applied load

Determine rotation per applied load Plot rotation vs cross arm location: 0 point defines shear center

or: plot both displacements: crossing point defines shear center

Example (next slide)


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AE 3145 Lab - Fall 99

Lab name=Lab#7 Shear Center

Group name = Monday1

Load Position Channel 1 Channel 2 Excitation Voltage

0.00E+00 -1.04E+01 -3.57E+00 2.50E+00

5.00E-01 -8.93E+00 -3.23E+00 2.50E+00

1.00E+00 -7.70E+00 -3.10E+00 2.50E+00

1.50E+00 -7.04E+00 -3.54E+00 2.50E+002.00E+00 -6.15E+00 -3.75E+00 2.50E+00

2.50E+00 -4.68E+00 -3.38E+00 2.50E+00

3.00E+00 -3.87E+00 -3.67E+00 2.50E+00

3.50E+00 -2.52E+00 -3.42E+00 2.50E+00

4.00E+00 -1.81E+00 -3.81E+00 2.50E+00

Cal: 10 10

Position LVDT 1 LVDT 2 Deflection Rotation

0.0 -0.104 -0.036 -0.036 -0.068

0.5 -0.089 -0.032 -0.032 -0.057

1.0 -0.077 -0.031 -0.031 -0.046

1.5 -0.070 -0.035 -0.035 -0.035

2.0 -0.062 -0.038 -0.038 -0.024

2.5 -0.047 -0.034 -0.034 -0.013

3.0 -0.039 -0.037 -0.037 -0.002

3.5 -0.025 -0.034 -0.034 0.009

4.0 -0.018 -0.038 -0.038 0.020

-0.120

-0.100

-0.080

-0.060

-0.040

-0.020

0.000

0.020

0.040

0.0 1.0 2.0 3.0 4.0 5.0

Position

Reading(in

ch

orradian)

LVDT 1

LVDT 2

Rotation

Sample Data

Compute avg deflectionand rotation from geometry

Compute avg deflectionand rotation from geometry

Convert voltages todisplacement using LVDT

calibration data

Convert voltages todisplacement using LVDT

calibration data

Plot yourdata!

Plot yourPlot your

data!data!

Shear Center is point where

Rotation = 0 or point whereLVDT1=LVDT2

Shear Center is point whereRotation = 0 or point where

LVDT1=LVDT2

Shear Center Theory

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