Carpenter Center for the Visual Arts - Harvard University - Le Corbusier architectureFull description
Full description
nice!! my work.Full description
Command CenterFull description
call centerFull description
Shear Center in Thin-Walled Beams Lab •
Shear Shear flow flow is de deve velop loped ed in beam beams s wit with h thin thin-w -wall alled ed cross cross sectio sections ns – shear flow (q sx ): shear force per unit length along cross section • q sx = τ τsx t • 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, τ nx , but it is very small • e.g., because it must be zero at ±t/2
– shear force: q sx ds (acting in s direction)
•
Shea Shearr flo flow w aris arises es from from pres presen ence ce of shea shearr loa loads ds,, Vy or Vz – needed to counter unbalanced bending stresses, σ x – to determine, must analyze equilibrium in axial (x) direction
•
Shear center: – resultant of shear flow on section must equal V y and V z – moment due to q sx must be equal to moment due to V y and V z – shear center: point about which moment due to shear flow is zero – not applying transverse loads through shear center will cause a twisting of the beam about the x axis
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)
Review from AE2120 (2751), AE3120 •
Bending of beams with unsymmetrical cross sections – bending stress depends on I y , I z and I yz – neutral surface is no longer aligned with z or y axes
•
Shear stresses are computed from axial force equilibrium – shear stress needed to counter changing σ x – 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: τ xs t (force/unit length along section) – shear flow must be equivalent to V y and V z so it must: • produce same vertical and horizontal force (V x and V y ) • 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)
Test Configuration Cantilever Cantileverwith withthinthinwalled C section walled C section LVDT LVDTmeasures measurestip tip deflection on cross-arm deflection on cross-arm
cross arm
LVDT
weight
Lab Apparatus
Small Smallweight weightused usedtotoapply apply load at point on cross-arm load at point on cross-arm
Lab Procedure
1. Determine the beam material properties from reference material (e.g., referenced textbooks 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. 5. 6. 7.
Determine the shear flow distribution on the cross-section for a V y shear load. Determine the shear flow distribution on the cross-section for a V z shear load. Determine the shear center for the cross-section. Using data from the lab, determine the measured location of the shear center and compare this with the location determined in step 6 above.
Beam Cross Section Y 1.353in.
Use Usesingle singleline lineapprox approxfor for cross section (t<
Z
0.420in.
Centroidal Axes:
0 = z dA A
0 = ydA A
0.050in.
Area Moments (of Inertia):
I yy
2
z dA
= A
y2 dA
I
zz = A
I
yz =
yz dA
Bending of Beam with Unsymmetrical Cross Section Y
q
A1
Z
But Butalso alsoconsider consider equilibrium equilibriumofof segment segmentAA11(see (see next nextslide!) slide!)
Acts over cross section
General: σ x = −
(yI
yy−
I zz I
Symmetric cross section, Mz=0: σ x = −
z I yz) M
y M z I zz
z+
(yI
yy−
2
I yz
yz−
z I zz) M y
Shear Stresses and Shear Flow Y σx
Complementary Complementary qqsx acts on A1 in sx acts on A1 in opposite oppositedirection direction
qsx
s
Z
σx
A1 X Axial force equilibrium for element:
é 0 = å F =x ê ò σ dA x êë A 1
+
é + q dx sx − ê ò σ dA x êë A x dx 1
x
+dσx
Shear Flow Result for qsx:
qsx
V
− y
=
I
I
yy
− zz
æ ç I yy y dA − I yz z dA 2 I yzçè A A 1
V
− z
+
I
1
I
yy
− zz
æ ç I zz z dA − I yz y dA 2 I yzçè A A 1
Y
s
Z Shear flow: qsx(s)
1
Shear Center Y
Moment Momentdue dueto toVVyy must mustbe beequal equalto toMM0
0
s
Vy
ez
Z
Shear flow: qsx(s)
Therefore: Therefore: Shear Shearcenter centerlies lies distance e from distance ezz from origin originwhere: where: MM0=V =Vyeez 0
y z
Moment, Moment,MM00, ,at at origin due to origin due to shear shearflow, flow,qqsx sx
Sum moments from qsx about A: =force in each flange x h/2 Must equal moment from Vy about A: =V y x e
h/2
qsx
eemust mustbe bepositive positive for q as shown for qsxsx as shown so soshear shearcenter center lies to left lies to leftofof section section
Data Acquisition •
Use PC data acquisition program to acquire deflection and strain data and test machine load – – – –
Use 2 LVDT displacement 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
Cross Cross arm arm
Loading Loading system system
Replace Replace dial dial gages gages with with LVDT’s LVDT’s
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)
Sample Data 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+00to -3.54E+00 2.50E+00 Convert Convertvoltages voltages to-3.75E+00 2.50E+00 2.00E+00 -6.15E+00 displacement using displacement usingLVDT LVDT 2.50E+00 -4.68E+00 -3.38E+00 2.50E+00 calibration data calibration data -3.67E+00 2.50E+00 3.00E+00 -3.87E+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: Position 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
