# FLEXURE EM 327: MECHANICS OF MATERIALS LABORATORY

Mechanics

Jul 18, 2012 (5 years and 10 months ago)

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FLEXURE
EM 327: MECHANICS OF MATERIALS LABORATORY
68
EXPERIMENT:FLEXURE TEST
OBJECTIVES:
(1)Develop a load-deflection diagram from a flexure
test of wood.
(2)Predict material properties based on the results
from the experiment.
(3)Explore the effects of the moment of inertia, I
INTRODUCTION:
In this experiment, beams of three types of wood are
tested in flexure to failure. The deflection, bending
stress, shear stress and failure modes associated with
FIGURE 1
L
P
P/2
P/2
P/2
-P/2
PL/4
V
M
L/2
x
y
P
x
x
x
Beams loaded in flexure are among the most
common type of engineering structures in existence.
Examples range from tree branches to diving boards
to airplane wings. Although simple in nature, such
beams develop relatively complex stress
distributions and may fail in a variety of manners.
Figure 1 shows a simply supported beam loaded at
its midpoint. The free body diagram with deflected
shape as well as the corresponding shear and
moment diagrams are also presented. Four aspects of
the behavior of this beam are next reviewed; beam
deflection, bending stresses, shear stresses, and
failure modes.
BEAM DEFLECTION:
There are several methods for determining the
equations for beam deflection in the EM324
textbook. The integration method shown here for
determining the deflection starts with the differential
equation for the elastic curve:
EI (d
2
y/dx
2
) = M(x)

(1)
Using singularity functions, the moment at any
section of the beam may be expressed as:
M(x) = (P/2)x - P<x-L/2> (2)
Combining the above equations results in:
EI (d
2
y/dx
2
) = (P/2)x - P<x-L/2> (3)
The boundary conditions for this beam are:
y = 0, at x = 0 and x = L
Integrating the differential equation twice gives:
EIy
(x)
= (P/12)x
3
-(P/6)<x-L/2>
3
+ C
1
x +C
2

(4)
Applying the boundary conditions to obtain the
constants results in the following equation for the
elastic curve:
y
(x)
=(P/EI)[(x
3
/12)-(1/6)<x-L/2>
3
-(1/16)L
2
x]
(5)
FLEXURE
EM 327: MECHANICS OF MATERIALS LABORATORY
69
the center deflection, δ
c
, can be determined form the
above equation by substituting x=L/2.
δ
c
= y
(L/2)
= -PL
3
/48EI (6)
BENDING STRESS:
Bending stresses in the beam are determined from
the flexure formula:
σ = My/I (7)
Where:M = bending moment
y = distance form neutral axis to
point of stress
I = moment of inertia of cross-
section with respect to neutral axis.
If the beam has the rectangular cross section shown
in Figure 2, the neutral axis coincides with the
centroidal axis for symmetrical bending and is
located at the middle of the cross section. For this
section, I = bh
3
/12.
FIGURE 2
N
.A.
b
h
The maximum bending stresses occur at the top and
bottom of the beam where the value of y is largest
and at the mid-span of the beam (x=L/2) where the
bending moment, M, is largest. Substituting these
values,
σ
max
= + [(PL/4)(h/2)] / (bh
3
/12) (8)
Note that the maximum bending stress is tensile on
the bottom of the beam and compressive on the top.
SHEAR STRESS:
Shear stresses in the beam are determined from the
shearing stress formula:
τ = VQ
It (9)
Where: V is the shear force
Q is the first moment of cross
section between the location of
stress and outer location of beam.
￿
=
c
h
tydyQ
(10)
Where:I is the moment of inertia of the
cross-section with respect to the
neutral axis.
t is the thickness of the beam
For a beam with the rectangular cross-section shown
in Figure 2, I = bh
3
/12.
The maximum shear stress occurs at the neutral axis
(y=0) where Q is maximum. For this location
Q = bh
2
/8. The maximum shear force in the beam is
+ P/2. Substituting these values
bh
P
4
3
max
±=τ
(11)
FAILURE MODES:
Three types of failure may occur in a wooden beam
loaded in flexure as shown in Figure 3. First, a
tension failure may occur at the bottom of the beam
where the bending stress is the maximum positive
value. Second, a compression failure may occur at
the top of the beam where the bending stress is the
maximum negative value. Finally, a shear failure
may occur at the neutral axis of the beam where the
shear stress is largest. These failure modes are
depicted in Figure 3.
FLEXURE
EM 327: MECHANICS OF MATERIALS LABORATORY
70
FIGURE 3
Tension Failure
Compression
Failure
Shear
Failure
Shear
Failure
A load versus deflection diagram is to be produced
for each specimen tested. The curve in Figure 4 is
typical of the load versus center deflection of a wood
beam in flexure.
FIGURE 4
Stroke,
δ
, (in.)
This type of plot may be used to determine material
properties. The equation for the maximum bending
stress (equation 8) at the center of the span can be
used to determine the Proportional Limit and
Modulus of Rupture. The Modulus of Elasticity can
be determined by rearranging the center deflection
equation (6).
E = (P/δ
c
)L
3
/48I (12)
Where: δ
c
is the center deflection at y=L/2.
The modulus of rupture is the maximum stress
obtained using
σ = Mc/I (13)
The average work is defined as the average force
times the distance through which it moves:
W
avg
= (F
avg
)(d) (14)
At the proportional limit this is
W
avg
= (1/2)P
pl
δ
c,pl
(15)
Where: P
pl
is the applied force at the
proportional limit
δ
c,pl
is the center deflection of the
beam at the proportional limit
The work per unit volume can be determined by
dividing this equation by the volume of the beam.
MATERIAL TO BE TESTED:
Four wood beams will be tested. The specimens are
birch, oak, and (2)pine and are approximately
14"x 1/2" x 1".
EQUIPMENT TO BE USED:
55-kip MTS testing machine
SAFETY CONSIDERATIONS:
!!USE SAFETY PLEXIGLASS SHIELD ON
MTS MACHINE AT ALL TIMES WHEN
TESTING WOOD!!
Never operate the MTS machine when someone's
hands are between the grips. Make sure all lab
participants are clear of equipment before beginning
or resuming testing.
PROCEDURE:
SPECIMEN PREPARATIONS:
TASK 1: Measure cross-sectional dimensions of the
four specimens. Visually inspect the specimens for
flaws and imperfections. Make note of any flaws and
imperfections in the lab report.
MTS SET-UP
FLEXURE
EM 327: MECHANICS OF MATERIALS LABORATORY
71
Station Manager flexure
MPT flexure.000
2.) Turn hydraulics on.
3.) Make sure MANUAL OFFSET = 0 for Stroke.
4.) Adjust 'SET POINT'' to 0.0
6.) Set-up Scope to plot a/b.
Stroke 0.05 in/in -0.2
Time 15 min
TESTING PROCEDURE:
TASK 1: Three specimens (oak, birch and pine)
1) Create a specimen file flex*.
2) Center specimen in flexure fixture. Be sure to
have the 1/2" side as the base.
3) Measure the length between the supports.
4) Start the Scope.
5) Close SAFETY SHIELD!
6) Lock MPT and select specimen.
7) Press RUN' and let test proceed until rupture.
The load will drop off at this point. The test can
be continued from here until a final rupture
occurs
8) Press STOP' button.
9) Unlock MPT.
10) Adjust 'SET POINT' to 0.0.
11) Remove specimen.
12) Repeat procedure for each specimen.
TASK 2: Investigate the effects of moment of
inertia, I
1.) Create a specimen file flex*.
2.) Center the second pine specimen in the flexure
fixture. The base should be the 1" side.
3.) Enable Manual Control on the Remote Control
Pod.
4.) Move the cross-head up until the wood is just
below the test fixture.
5.) Disable manual control on the Remote control
Pod.
6.) AUTO OFFSET Load and Stroke.
7.) Lock MPT and select specimen.
8.) Start the Scope.
9.) CLOSE SAFETY SHIELD!
10.) Press 'RUN' and let the test proceed to rupture.
11.) Press STOP.
12.) Unlock MPT.
13.) Adjust the SET POINT to 0.0.
14.) Remove specimen.
15.) Shut down hydraulics.
16.) Copy all data files (Task 1 and 2) to diskette.
c:\em327data\flex*\specimen.dat
17.) Delete specimens flex*.
REPORT:
The report outline found in Appendix A should be
used.
REPORT REQUIREMENTS:
(1) Label and affix appropriate scales to graphs
(2) Determine the following properties for each
wood specimen.
a. Proportional Limit at top and bottom of
specimen
b. Maximum shear stress in the specimen at
failure
c.Maximum bending stress in the specimen
at failure
d.Modulus of Elasticity
e.Average work per unit volume at the
proportional limit.
f.Type of failure (description and sketch)
(3) Compare properties with reference values.
FLEXURE
EM 327: MECHANICS OF MATERIALS LABORATORY
72
(4) Scale your Modulus of Elasticity results to
predict the Modulus of Elasticity for the wood
sample in compression (recall E in compression
is approximately 10% greater than E for
flexure). How do your estimations for
compression compare with the values you
(5) Discuss the results, possible sources of error
and other conclusions relevant to TASK 1.
(1) Label and affix appropriate scales to graph(s).
(2) Determine the proportional limit and the
maximum bending stress in each specimen.
(3) Compare the results for the two specimen
orientations.
(4) Discuss your results and conclusions relevant to
QUESTIONS:
(1) After the first failure the load once again
increases until the next fracture. Discuss what is
physically happening during this process.
(2) How does the relative importance of shearing
and bending stress vary with the span of the
beam?
(3) Both tension and compression are present in
bending. Would the wood be stronger in
(4) What are the assumptions made in the
derivation of the flexural formula?
(5) Is there any difference between the stresses
measured at the top of the beam and at the same
location (at the same section) on the bottom of
the beam? If so, explain.
(6) How should the Modulus of Rupture and
Modulus of Elasticity of a small clear specimen
compare with the corresponding properties
determined from testing a full-size structural
member?
(7) Is the Modulus of Rupture a true ultimate fiber
stress? Explain.
(8) Is the strength of a beam under a static load
necessarily related to its stiffness?
(9) Distinguish between primary failure and
secondary failure.
(10) Is it always possible to determine the primary
cause of failure by examining the fractured
specimen?