# Shear Center in Thin-Walled Beams Lab

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1 Shear Center in Thin-Walled Beams Lab Shear flow is developed in beams with thin-walled cross sections shear flow (q sx ): shear force per unit length along cross section q sx =τ sx 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) Shear flow arises from presence of shear loads, V y or V z 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 AE3145 Shear Center Lab (S2k) Slide 1

2 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) AE3145 Shear Center Lab (S2k) Slide 2

3 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) AE3145 Shear Center Lab (S2k) Slide 3

4 Test Configuration Cantilever Cantilever with with thinwalled thinwalled C section section LVDT LVDT measures measures tip tip deflection deflection on on cross-arm cross-arm cross arm LVDT weight Lab Apparatus Small Small weight weight used used to to apply apply load load at at point point on on cross-arm cross-arm AE3145 Shear Center Lab (S2k) Slide 4

5 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, I y, I yz for the given section. 4. Determine the shear flow distribution on the cross-section for a V y shear load. 5. Determine the shear flow distribution on the cross-section for a V z 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 and compare this with the location determined in step 6 above. AE3145 Shear Center Lab (S2k) Slide 5

6 Beam Cross Section 1.353in. Use Use single single line line approx approx for for 1.330in. cross cross section section (t<<b,h) (t<<b,h) 0.420in. Centroidal Axes: 0.050in. 0 = 0 = A A zda yda Area Moments (of Inertia): 2 z da A 2 y da A AE3145 Shear Center Lab (S2k) Slide 6 I I I yy zz yz = = = yzda A

7 Bending of Beam with Unsymmetrical Cross Section q General: σ x Acts over cross section Symmetric cross section, M z =0: x ( yi zi ) M + ( yi zi ) M = σ = ym I zz A 1 yy yz z yz zz y 2 Izz Iyy Iyz z But But also also consider consider equilibrium equilibrium of of segment segment A 1 (see 1 (see next next slide!) slide!) AE3145 Shear Center Lab (S2k) Slide 7

8 Shear Stresses and Shear Flow σ x q sx s Complementary Complementary q sx acts sx acts on on A 1 in 1 in opposite opposite direction direction A 1 σ x +dσ x Axial force equilibrium for element: X F da q dx da A1 A1 0 = x = σ x + sx σ x x+ dx x AE3145 Shear Center Lab (S2k) Slide 8

9 Shear Flow Result for q sx : V y V z qsx = 2 Iyy yda Iyz zda + 2 Izz zda Iyz yda Iyy Izz I yz A1 A I 1 yy Izz I yz A1 A1 s Shear flow: q sx (s) AE3145 Shear Center Lab (S2k) Slide 9

10 Shear Center Moment Moment due due to to V y y must must be be equal equal to to M 0 0 V y s e z Shear flow: q sx (s) Therefore: Therefore: Shear Shear center center lies lies distance distance e z from z from origin origin where: where: M 0 =V 0 =V y e y z z Moment, Moment, M 0, 0, at at origin origin due due to to shear shear flow, flow, q sx sx AE3145 Shear Center Lab (S2k) Slide 10

11 Examples of Shear Centers V y V y q sx Shear Center lies on y axis Shear Center q sx Section Symmetric about about y axis: axis: Angle Angle Section: Shear Shear center center must must lie lie on on y y axis axis (similar (similar argument argument for for z z axis axis symmetry) symmetry) Shear Shear center center must must lie lie at at vertex vertex of of legs legs (regardless (regardless of of orientation orientation of of section) section) AE3145 Shear Center Lab (S2k) Slide 11

12 Shear Center Must Lie Outside C B q sx V y e q sx h/2 Shear Center A h/2 q sx Sum moments from q sx about A: =force in each flange x h/2 Must equal moment from V y about A: =V y x e e must must be be positive positive for for q sx as sx as shown shown so so shear shear center center lies lies to to left left of of section section AE3145 Shear Center Lab (S2k) Slide 12

13 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 arm Replace dial gages with LVDT s Loading system AE3145 Shear Center Lab (S2k) Slide 13

14 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) AE3145 Shear Center Lab (S2k) Slide 14

15 AE 3145 Lab - Fall 99 Lab name=lab#7 Shear Center Group name = Monday1 Load Position Channel 1 Channel 2 Excitation Voltage 0.00E E E E E E E E E E E E E+00 Convert -7.04E E E E+00 Convert voltages -6.15E+00 voltages to to displacement -3.75E E+00 displacement using 2.50E E+00 using LVDT LVDT -3.38E E+00 calibration 3.00E+00 calibration data -3.87E+00 data -3.67E E E E E E E E E E+00 Cal: Position LVDT 1 LVDT 2 Deflection Rotation Sample Data Reading (inch or radian) Shear Shear Center Center is is point point where where Rotation Rotation = = 0 0 or or point point where where LVDT1=LVDT2 LVDT1=LVDT2 LVDT 1 LVDT 2 Rotation Position Plot Plot your your data! data! Compute Compute avg avg deflection deflection and and rotation rotation from from geometry geometry AE3145 Shear Center Lab (S2k) Slide 15

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