A Shear Flow Visualization Tool for Mechanics of Materials

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Jul 18, 2012 (5 years and 1 month ago)

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A Shear Flow Visualization Tool for Mechanics of Materials
James Mack and Judy Wood
Department of Mechanical Engineering, Clemson University
239 Fluor Daniel Engineering Innovation Building
Clemson, SC 29634-0921
Abstract
As shear flow is one of the more difficult concepts in a Mechanics of Materials class, an I-beam has been created as a
classroom visualization tool. This I-beam is composed of clear polycarbonate layers loosely bolted together. The flanges of
the beam are autonomous units fastened to the web only by wooden skewers. When the beam is loaded with a transverse
shear force, the shear flow along the web-flange interface fractures the skewers, simulating a fastener failure.
Introduction
The purpose of this demonstration tool is to illustrate the concept of shear flow to an introductory Mechanics of Materials class.
Shear flow is often a challenging topic for the student to understand. In an I – shaped beam, it involves a shear stress in the
web resulting in a shear stress at the web-flange interface. This tool provides an accurate visualization of this concept. As the
beam is composed of polycarbonate layers, the slippage between layers provides a visualization of the shear stress present.
Most importantly, the flanges of the beam are a separate unit so that the slippage between the flange and the web can
illustrate the shear stress and shear flow at that location.
Methods and Materials
Using 3/8” thick polycarbonate, the beam is assembled with the cross-section shown in Figure 1. Polycarbonate was selected
for its transparency, its relatively low modulus of elasticity (for large deflections), and its high modulus of toughness (for
repeated use). The length of the beam is 4 ft, and it is sized such that it is clearly visible from the back of a classroom. Pairs
of wooden skewers fasten the flanges to the web at each end and at the middle. For ease of reassembly, the layers of each
flange are joined together with screws at each end. This results in four separate flange units. The screw heads are flush to
prevent interference with the web. The web layers are joined together with bolts at each end. During assembly, chalk dust
may be added between layers. This provides an aural indication of the shear present in the beam.
Figure 1: Cross-Section
Figure 2: Left View
Figure 3: Isometric View
Results
First, an order-of-magnitude verification should be performed to ensure that excessive force will not be required to produce a
visible effect. It is assumed that the beam will be loaded as a cantilever beam. If the beam is simply supported and loaded in
the middle, it will support twice the external load, as the internal shear is reduced by one-half.
The internal shear force which can be supported by one fastener is calculated below. This maximum internal force is based
on the shear strength of structural grade Douglas fir, S
shear
. Then, using the principle of shear flow in a double-fastener
situation, the required external load, V
max
, is calculated.
( )
( )
( )
lbV
in
inV
in
lb
I
QV
q
in
lb
q
in
lb
q
spacing
nV
q
lbV
inin
inV
psi
tI
QV
S
beam
flange
skewer
skewer
skewer
skewerskewer
skewerskewer
shear
90.8
25.65
0625.5
690.0
(3)
:Shear Transverse External Maximum
690.0
24
28.8*2
(2)
:FlowShear Maximum
28.8
125.010*2.1
10*63.1
900
(1)
:Skewerin Shear Internal Maximum
max
4
3
max
max
max
max
max
max
45
34
=
=
=
=
=
=
=
=
=


This V
max
is the maximum external shear force which the beam assembly will sustain before the fasteners fracture. This is
relatively small, compared to the force required to cause a noticeable deflection in the beam. This makes the beam a useful
classroom tool, since a minimum of force is needed to produce the desired result.
It is also of interest to determine the deflection at a load level of 9 lb. This load level gives the centerline shear stress
distribution shown in Figure 4. To estimate the deflection, the deflections of an average web layer and an average flange layer
were computed. Friction between layers was neglected, and the two deflections were averaged. It is roughly estimated that
the deflection under a 9 lb load is 9 inches. The calculations for this deflection are included in the Appendix.
Shear Stress in I-Beam
0
1
2
3
4
5
6
7
0 0.5 1 1.5 2 2.5
Shear Stress (psi)
Vertical Position (in)
Figure 4: Shear Stress Distribution
This shear stress distribution applies to the centerline of the beam, and the resulting stress at the centerline can be seen on
the volume element in Figure 5.
Figure 5: State of Stress at Point B
The state of stress at Point A is shown on the volume elements in Figure 6. The left element’s right face is at the web-flange
interface, as is the right element’s left face.
Figure 6: State of Stress at Point A
Web Element
τ
τ
Point A (Figure 1)
Flange Element
Point B
τ
Between the element in Figure 5 and the elements in Figure 6, the shear gradually changes from a vertical shear to a
horizontal shear. This change is most pronounced in thin-flanged beams, where the flange bears almost wholly horizontal
shear. In thick-flanged beams, like the I-beam under consideration, the change is less dramatic. However, it is still there, and
this change is the cause of shear flow. The effects of the shear flow can be seen in Figure 7.
Figure 7: Effects of Shear Flow
In Figure 7, an exploded view is used to examine the flange-web interface before and after loading. Since the flange has a
greater stiffness due to the vertical alignment of its layers, it will deflect less than the web. Since the web deflects a relatively
large amount, its hole pattern is distorted as shown in Figure 7. The top hole slides horizontally, relative to the corresponding
hole in the flange. Therefore, a force is required to align the two holes. This force must be absorbed by the fasteners. When
this force is normalized to the fastener spacing, the term for shear flow results.
The fasteners, simulated by the skewers, are the heart of the tool. The shear flow, or the distributed shear force along the
web-flange junction, must be resisted by the fasteners. When the shear stress experienced by the fasteners exceeds the
shear strength of the fastener material, the fastener fails and the flange separates at that point. The beam then loses stiffness
and can be deflected further, causing another fastener to fail. It is possible to continue loading the beam until all the fasteners
have broken, simulating catastrophic failure.
After failure of one or more fasteners, it is a simple matter to replace the failed fasteners with new skewers and repeat the
demonstration, if necessary. It is left up to the user’s discretion to partially replace the wooden skewers with permanent bolts,
making reassembly a less complicated task.
Conclusion
The polycarbonate I-beam described above will be an effective classroom tool for illustrating the concept of shear flow. The
benefits of this tool lie in the fact that this is not an abstract or unrealistic model but rather an exact portrayal of a built-up
beam, complete to the fasteners. The topic can be hard to understand with basic lectures and diagrams. However, having a
model in the classroom which clearly describes the phenomenon will bridge a learning gap for many students.
Fastener
Web
Flange
Before Load
After Load
Fastener Hole
Appendix
This Maple 9 worksheet analyzes the deflection of the beam. It is assumed that the beam is loaded as a cantilever beam, and the 9 lb load is
applied vertically at the end. The symmetry of the stress distribution about the neutral axis will be used to simplify computations.
> restart;
Shear Force:
First, the total shear force acting on each layer must be determined.
> V:=9: lb, external applied shear force
> Inertia:=65.25: in^4, moment of inertia of I-beam
For layer 1-2 (Figure 1), Q12 = Q_ + Qtop:
> Qtop:=2.25*6: in^3, Q of top section..this section includes both the flange and the four upper web layers.
Parameters which define Q_
> hweb:=3: in, height of web section
> bweb:=1: in, thickness of web
> Q_:=0.5*(hweb^2/4-y^2)*bweb;
:= Q_ − 1.125000000 0.5 y
2
> Qweb:=Q_+Qtop;
:= Qweb − 14.62500000 0.5 y
2
Now that we have an expression for Q as a function of height above the neutral axis, y, we can integrate the shear stress over each layer to
determine the total shear force acting on that layer.
> V12:=int(V*Qweb/(Inertia*bweb)*bweb,y=0..(1*3/8));
:= V12 0.7552532330
> V23:=int(V*Qweb/(Inertia*bweb)*bweb,y=(1*3/8)..(2*3/8));
:= V23 0.7479795261
> V34:=int(V*Qweb/(Inertia*bweb)*bweb,y=(2*3/8)..(3*3/8));
:= V34 0.7334321123
> V45:=int(V*Qweb/(Inertia*bweb)*bweb,y=(3*3/8)..(4*3/8));
:= V45 0.7116109916
>
Parameters which define Q_top (general form of Qtop)
> htop:=6:
> btop:=4:
>
> Q_top:=0.5*(htop^2/4 - y^2)*btop;
:= Q_top − 18.0 2.0 y
2
> V67:=int(V*Q_top/(Inertia*btop)*btop,y=(4*3/8)..(5*3/8));
:= V67 0.6352370690
> V78:=int(V*Q_top/(Inertia*btop)*btop,y=(5*3/8)..(6*3/8));
:= V78 0.4897629311
> V89:=int(V*Q_top/(Inertia*btop)*btop,y=(6*3/8)..(7*3/8));
:= V89 0.3151939656
> V910:=int(V*Q_top/(Inertia*btop)*btop,y=(7*3/8)..(8*3/8));
:= V910 0.1115301724
> Vtot:=2*(V12+V23+V34+V45+V67+V78+V89+V910);
:= Vtot 9.000000002
This checks out, so the shear distribution calculated above is accurate.
Deflection:
Next, find the deflection of selected layers.
The general formula for deflection of a cantilever beam is shown below:
vmax:=-PL^3/(3EI)
Error, missing operator or `;`
To calculate the deflection for the beam as a unit, one could average the deflections of each layer, when the deflection for each layer was
computed using the above values for V. However, since the deflection equation is linear in P, or V in this case, the V's for the web can be
averaged, and this value used in computing the average value for the deflection.
> Inertia_weblayer:=1/12*bweb*(0.375)^3;
:= Inertia_weblayer 0.004394531250
> Inertia_flangelayer:=1/12*0.375*1^3;
:= Inertia_flangelayer 0.03125000000
> Vweb_avg:=(V12+V23+V34+V45)/4;
:= Vweb_avg 0.7370689657
> vweb_avg:= -Vweb_avg*48^3/(3*350000*Inertia_weblayer);
:= vweb_avg -17.66566464
The same procedure can be followed for the average deflection of the top:
> Vtop_avg:=(V67+V78+V89+V910)/4;
:= Vtop_avg 0.3879310345
> v12:=-V12*48^3/(3*350000*Inertia_weblayer);
:= v12 -18.10149518
> V_C:=int(V*Q_top/(Inertia*btop)*(3/8),y=(4*3/8)..(8*3/8));
:= V_C 0.1454741380
> v_C:=-V_C*48^3/(3*350000*Inertia_flangelayer);
:= v_C -0.4903093599
It can be seen that the deflection of layer C (from Figure 1) is much less than the deflection of the average layer in the web. The deflection
of the combined beam, then, must lie somewhere between the two. To get an idea of this actual deflection, a weighted average can be
taken. One could base this average on the cross-sectional area of each layer. Since there are 16 web layers and 16 flange layers, the
weighted average is equivalent to the simple average of the two deflections.
> v_beam:=(vweb_avg+v_C)/2;
:= v_beam -9.077987000
This is only an estimate It does not account for friction between layers, and it greatly simplifies the relationship between the web stiffness
and the flange stiffness. However, it is reasonable to expect that the computed deflection of 9 inches will be accurate to within an order of
magnitude.