SHEAR DEFLECTIOIN IN 4PB TESTS
A.C. Pronk & M. Huurman
Delft University of Technology, Delft, ZH, The Netherlands
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
Introduction
Problem
Solution
Comparison
Conclusions
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
Introduction of problem
When in a 4PB test a beam is bended not only a deflection due to bending will occur but also a deflection due to the shear force will be present.
If the ratio of height H over the length L is small (slender beam) e.g. < 1/10 the deflection due to shear can be neglected
However, in our practice the ratio is higher
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
Problem
In the analytical approach (beam theory) the actual beam is replaced by a single line (1D) with no real geometrical dimensions.
As a consequence an assumption has to be made for the shear force distribution over a cross section of the beam
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
Possible Solution:
Determine the real 3D solution by Finite Element Calculations (FEM).
Introduced new (minor) problems:
1) Finite Element Calculations are not exact.
2) For bending a beam in FEM you have to touch (grab) the beam.
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
External touching the material introduces constraints e.g. stress
concentrations
The following slides are made by Rien Huurman and presented already at the
1st 4PB Workshop in Delft
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
In case of a 3D model the effects of
grabbing the beam becomes even
more visible.
Due to this touching the beam is ‘hampered’ in its ‘freedom’
EXAMPLES, 4PB3D static elastic ⇒
AA
ABAQUS
Why different behaviour at different clamps?
Rien Huurman 1st 4PB Workshop
EXAMPLES, 4PB3D static elastic ⇒
AA
Bending > longitudinal stress.
Material >contract/expand perpendicularly.
Clamp blocks this >Extra system stiffness
Rien Huurman 1st 4PB Workshop
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
So, the problem to be solved is:
How to bend a 3D beam without touching or grabbing the beam which will result in constraints?
Solution: Apply a shear load distribution over the cross section at the inner supports.
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
But you don’t know on forehand the shear load distribution over a cross section.
Solution: Make use of the St Venant principle
The form of the shear load distribution “far” away from the source loading is “independent” of the form of the source loading.
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
Iterative Solution Procedure
1.Apply a homogenous shear load distribution at the inner supports.
2.Determine the shear load distribution halfway the beam (L/2)
3.Use this distribution as the new input.
4.Repeat step 2 and 3.
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
1st Step 2nd Step
Deviations
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
-25.0
-16.7
-8.3
0.0
8.3
16.7
25.0
0.00 0.05 0.10 0.15Shear stress [MPa]
Hei
ght [
mm
]
mid clamp response
1st response at clamp
2nd response at clamp
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
-0.4
-0.3
-0.2
-0.1
0
0.1
0 50 100 150 200 250Distance to middle of beam [mm]
Def
lect
ion
[mm
]
-0.00005
0.00005
0.00015
0.00025
0.00035
0.00045
Diff
eren
ce [m
m]
Applied shear #2 Applied shear #1 Appliead shear #1 - Applied shear #2
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
Cowper 1966 (10+10μ)/(12+11μ) 0,8517
Timoshenko 1974 5/6 0,8333
Timoshenko 1922 (5+5μ)/(6+5μ) 0,8710
Olson 1935 (20+20μ)/(24+15μ) 0,9231
Pickett 1945 24,612(1+μ)/(29,538+5,942μ+64,077μ2) 0,8419
Tanji 1972 (6+12μ+6μ2)/(7+12μ+4μ2) 0,9354
Pai 1999 5/(6+(μ/(1+μ))2(H/B)4[1-
(90/π4)Σ{Tanh(nπB/H)/(n5(π(B/H))}])
0,8269
Hutchinson 2001 5(1+μ)/(6+5μ-(μ/(1+μ))2(H/B)4[1-
(90/π4)Σ{Tanh(nπB/H)/(n5(π(B/H))}])
0,8763
ABACUSABACUS 20092009 Comparison analytical solution and finite element Comparison analytical solution and finite element calculationscalculations
0,8590,859
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
CONCLUSIONS
Using an internal shear loading over the cross section at the clamp it is possible to bend a beam without touching or grabbing it. In this way unwanted constraints are avoided.
Application of the Saint-Venant principle enables the determination of the unknown shear force distribution at the clamps of a 4PB test.
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
CONCLUSIONSThis is done by replacing the homogenous shear force distribution (per element) at the clamp in the first calculation by the calculated distribution half way the inner and outer clamp in the second calculation.
Equalizing the analytical solution for the deflection profile and the deflection calculated by FEM leads to a value of 0,859 for the shear correction coefficient. This is close to the value of 0,85 used in the 1D calculations in ABACUS.
2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal
CONCLUSIONS
The method described in this paper opens the possibility to obtain in a numerical way the correct expression for the shear correction coefficient.