AE 3145 Shear Center in Thin-Walled Section Lab (S2k) Page 1

G

EORGIA

I

NSTITUTE OF

T

ECHNOLOGY

School of Aerospace Engineering

AE 3145 Laboratory

Shear Center in Thin-walled Beams

Objective:

This lab is designed to demonstrate the behavior of thin-walled beams of unsymmetrical cross-

section when subjected to bending and torsion loads. In particular, the lab will study the concept

of shear center, its calculation using the analysis of shear flow in the section, and its actual

measurement using a simple C- section cantilevered beam.

Procedure:

The following experimental procedures are to be performed on a thin-walled aluminum beam

(6063-T5 aluminum) with an open “C” section as shown in Figure 1. Figure 2 shows the actual

apparatus in the lab.

LVDT

cross arm

weight

Figure 1. Experimental Configuration

The beam is to be clamped to the top of the lab table (using a vise) and cantilevered over the

table surface. It will be loaded at its tip using a small weight attached to a rigid cross arm as

shwon in Figure 3. In this way different combinations of bending and torsion can be applied to

the beam. Two LVDT's (or dial gages) will be attached to a steel base plate located on an

adjacent table and arranged to measure the downward deflection of the beam at opposite ends of

the tip cross arm. In this way it will be possible to determine from these two measurements both

the tip rotation and tip deflection separately.

As you will see, the beam will bend and twist under load, but if the load point is relocated at

different points on the cross arm, it is possible to determine a location for which only bending is

AE 3145 Shear Center in Thin-Walled Section Lab (S2k) Page 2

produced and no twisting is developed. This is referred to as the “shear center” and when loads

are applied through this point, only bending is produced.

Figure 2. Experimental Apparatus

You should record the tip deflections as a function of location of the vertical load on the cross-

arm. The load will be applied using a small weight (- 2 lbs.,) and a special bracket. The bracket

allows the load to be applied vertically at an easily measured point shown by the scale mounted

on the cross arm.

Figure 3. Cross Arm for Loading and Recording Deflection & Rotation

The tip deflection and rotation can be computed from the two displacement measurements (and

knowledge of the distance apart). By plotting rotation versus load position, you should be able to

determine the point through which the load can be applied without causing any section rotation,

e.g., the shear center.

Carry out the above steps for the beam oriented with the C section opening facing horizontally.

If the beam you are using has an adjustable cross arm, re-orient it and carry out the above steps

with the C section opening facing vertically.

AE 3145 Shear Center in Thin-Walled Section Lab (S2k) Page 3

Analysis:

The shear flow distribution across the cross-section will have to be calculated for an applied

vertical shear load with the cross section assumed to be oriented either vertically or horizontally

(this involves two independent calculations). From these results, you will then have to determine

the location of the shear center for each section orientation. This will locate the proper 2D

location in the cross section. Make maximum use of symmetry in your calculations. You will

have to calculate the section properties for the beam that is tested. Since the cross section is

somewhat difficult to measure, the actual dimensions are provided in Figure 2. Specific analysis

steps are:

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 I

z,

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.

NOTE: The experimental data from each LVDT (dial gage) must be plotted versus load

position. Two different methods can be used. In the first method, the data for each dial gage is

plotted separately on a single figure. This will yield two straight lines which cross at the location

of the shear center (i.e., the two displacement gage readings will be the same at this paint). In

the second method, the twist of the beam can be computed directly from the LVDT or dial gage

readings for each loading position. If these values are plotted versus loading position, the

intersection with the horizontal (load position) axis will determine the shear center location. You

may use either way, but you should fully understand what you are doing.

AE 3145 Shear Center in Thin-Walled Section Lab (S2k) Page 4

1.353in.

1.330in.

0.420in.

0.050in.

Z

Y

Figure 2. Beam Cross Section Dimensions

AE 3145 Shear Center in Thin-Walled Section Lab (S2k) Page 5

APPENDIX

Shear Flow in Thin-Walled Sections

Background

The treatment of bending of beams with unsymmetrical open cross sections and the study of

shear flow in such sections is covered in AE3120 (AE3105). The textbooks used for these

courses over the years provide a good source of background information

1

. The development

shown below is taken from Crandall, Dahl & Lardner and differs from the other treatments only

in the particular sign convention that is used.

Shear Flow

The bending stress, σ

x

, in a beam with a cross section that is not symmetric with respect to either

the y or the z axis is given by a lengthy formula:

2

( ) ( )

yy yz z yz zz y

x

zz yy yz

y I z I M y I z I M

I I I

σ

− + −

= −

−

(1)

where I

y

and I

z

are the area 2

nd

moments (moments of inertia) with respect to the y and z

(centroidal) axes and I

yz

is the area cross product (cross product of inertia). Figure A1 below

shows the assumed axis system and the bending moments are assumed to be positive on a

positive face in the right-handed sense. The shear forces are also assumed to act in a positive

coordinate direction on positive faces. It should also be noted that the origin for the coordinate

system is located at the centroid of the cross section (a centroidal axis system) and the beam is

assumed to be prismatic (doesn’t change shape in the x direction). When only M

z

is acting and

the cross section is symmetric with respect to either y or z so that I

yz

=0, then:

z

x

zz

y M

I

σ = −

(2)

which is the familiar bending stress formula for simple beam bending.

Equation 1 is equally applicable to beam with a thin-walled cross section such as the one shown

in Figure A1 below. It is this kind of cross section that is the subject of study in the present lab

project. While Eq. (1) presents the bending stress, in this case we are interested in the shear

stresses acting in the beam, and particularly the shear flow acting in the cross section. Recall

that the presence of a shear force (V

y

or V

z

) arises whenever the bending moments vary with x,

and these shear forces must be equivalent to the shear stresses distributed across the cross

section. While the computation of shear stresses on the cross section of a beam with arbitrary

cross section is quite involved, the computation for thin-walled cross sections reduces to a

particularly simple formulation as we will see below. The key simplification is that the only

significant shear stress in a thin-walled section is the component that is tangent to the centerline

1

Megson, T.H.G., Aircraft Structures, 2

nd

ed., Halstead Press, John Wiley & Sons, 1990.

Gere, J. and Timoshenko, S., Mechanics of Materials, 3

rd

ed., PWS Publishing Co., 1990.

Crandall, S., Dahl, N. and Lardner, T., An Introduction to the Mechanics of Solids, 2

nd

ed., McGraw Hill, 1978.

AE 3145 Shear Center in Thin-Walled Section Lab (S2k) Page 6

of the cross section (along the s coordinate direction in Figure A1 below). While there will be a

perpendicular component, it is quite small since it must be zero at sides of the cross section.

s

Z

Y

A

1

Figure A1. Cross Section of a Thin-Walled Beam

The presence of bending moments (M

y

and M

z

) that change with x gives rise to values of

bending stress, σ

x

, that also change with x (e.g., Eq. (1). As for beams with solid sections, this

changing σ

x

, in turn, results in the presence of shear flow, q

sx

, as shown in Figure A2 below.

s

Z

Y

X

q

sx

σ

x

+d

σ

x

σ

x

A

1

Figure A2. Axial Equilibrium of a Small Element from a Thin-Walled Beam

Figure A2 shows a small section of length, dx, in the axial direction. The perimetric coordinate,

s, measures the distance along the cross section from one of the free edges and the area in the

cross section is designated as A

1

as shown in Figures A2 and A1.. For this element, the presence

of q

sx

is needed in order to maintain force equilibrium in the axial (x) direction due to the

changing values of σ

x

resulting from the changing bending moment (due in turn to the presence

of V

y

or V

z

). It should be noted that the “shear flow,” q

sx

, is defined simply as the local shear

stress, τ

sx

, times the wall thickness, t. As a result, the shear flow is expressed in units of force

per unit length and when multiplied times the incremental distance, ds, along the section, it

yields a shear force in the same direction. It should also be noted that due to the complementary

nature of τ

sx

on orthogonal planes, there will be a complementary shear flow, q

sx

, acting on the

cross section. This is the shear flow that we are normally concerned with, but as you can see, we

AE 3145 Shear Center in Thin-Walled Section Lab (S2k) Page 7

must compute it by considering its complementary form acting in the x direction. Expressing

axial equilibrium of the element in Fig. A1 yields:

1 1

0

x x sx x

A A

x dx x

F dA q dx dAσ σ

+

= = + −

∑

∫ ∫

(3)

And substituting the expression for σ

x

from Eq. 1 yields the familiar expression for shear flow on

the cross section of a thin walled section.

1 1 1 1

2 2

y

z

sx yy yz zz yz

A A A A

yy zz yz yy zz yz

V

V

q I y dA I z dA I z dA I y dA

I I I I I I

−

−

= − + −

− −

∫ ∫ ∫ ∫

(4)

Finally, it should be noted that we are making no distinction between q

sx

and q

xs

which act on

complementary areas just like we make no distinction between τ

xs

and τ

sx

.

Shear Center

When the shear flow acting on a cross section is evaluated in the y direction, the resultant must

be equal to the shear force, V

y

, and similarly, the resultant in the z direction must be equal to the

shear force, V

z

.. (It should be pointed out that the shear flow itself must be integrated along the

cross section as defined by the variable, s, in order to yield a force, and we must then consider

the y and z components of this force.) At the same time, the moment of the shear flow about any

point on the cross section must be equal to the moment about the same point due to V

y

and V

z

.

Since there will always be a point on the cross section where the moment of the shear flow is

zero, this is the point through which the shear forces, V

y

and V

z

, must then act in order to be

equivalent. This point is called the Shear Center, and as we will see it is a function of the cross

section shape alone. More importantly, if the shear force is not applied through this point, then it

will result in a moment about the x axis (in other words, a twisting moment). The practical

results of this is that if a load is applied to a beam, it must be applied through the shear center.

Otherwise, the load will also induce a twisting response in the beam due to the presence of a

twisting moment, M

x

, arising from the moment of the load about the shear center.

The shear center can be computed for a given cross section by considering the shear flow first for

V

y

and then for V

z

. As shown in Figure A3 below, if V

y

is acting alone, then the moment due to

V

y

will depend only on the its location in the z direction (e.g., e

z

which is perpendicular to the

line of action of the force). This means that we can analyze the effect of V

y

to determine the z

coordinate, e

z

, of the shear center location. In a similar manner, we can analyze the effect of V

z

in order to determine the y coordinate, e

y

, of the shear center location. Together, these two

calculations will yield the (y,z) location of the shear center on the cross section.

AE 3145 Shear Center in Thin-Walled Section Lab (S2k) Page 8

s

Z

Y

A

1

V

y

Moment, M

0

, due

to shear flow, q

sx

e

z

Shear center lies

distance e

z

from

origin where:

M

0

=V

y

e

z

Figure A3. Determining the z Coordinate of the Shear Center

Four particular cases are worthy of special note and are shown in Figure A4 below. First, if the

cross section is symmetric with respect to both the y and z axes, then the shear center is at the

origin of the centroid axes (this case is not shown in the figure). Second, if the cross section is

symmetric with respect to the y axis as shown in Fig. A4(a) then the shear center must lie on this

axis. A similar argument can be made for the shear center lying on the z axis if the cross section

is symmetric with respect to the z axis. The third case concerns the angle section such as shown

in Fig. A4(b). It should be clear from the figure that no matter how the angle cross section is

oriented with respect to the y and z axes, the moment of the shear flow in the two legs will

always be zero with respect to the vertex of the legs (e.g., the lines of action both pass through

this point). This means that this is the shear center for this section!

Finally, the case of a C-shaped cross section is shown also (Figure A5). Since it is symmetric

with respect to the z axis, the shear center must lie on this axis somewhere to the right or left of

the shear center (which must lie somewhere within the C itself). It is easy to show that the shear

center must actually lie outside the C as shown. To show this consider summing the moments

due to the shear flow about the point, A. In this case, only the shear flows in the upper and lower

legs will contribute and if the shear force, V

y

, is to create an equivalent moment, it must be

applied outside the C as shown.

As a final note, see if you can reason why it is easiest to compute the moment due to the shear

flow if we take it about point B instead of A. The amount of computation needed to compute the

location of the shear center can often depend on how carefully one chooses the point about which

to compute the moments!

AE 3145 Shear Center in Thin-Walled Section Lab (S2k) Page 9

Z

Y

V

y

Shear

Center

(a) Symmetric about y axis

Z

Y

q

sx

V

y

Shear

Center

q

sx

(b) Angle Section

Figure A4. Shear Center Locations for Special Cases

Z

Y

A

B

Shear

Center

V

y

q

sx

q

sx

q

sx

Figure A5. Shear Center for a “C” Section Must Lie Outside the “C”

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