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EM 327: MECHANICS OF MATERIALS LABORATORY
VIDEO TITLES: HELICAL SPRING,
BACKGROUND:
OBJECTIVES:
axial and flexural loads, and thus require special
consideration when calculating theoretical
(1)Study the combined effects of bending and
stresses and strains.
for a member subjected to an eccentric tensile
The eccentrically loaded specimen used in this
where the load can be considered as a
(1)Study the combined effects of torsion and
an eccentric load (P) applied at some distance, e,
helical spring in compression.
from the centroidal axis.
(3) Study the combined effects of torsion and
P
circular shaft loaded in torsion and bending.
INTRODUCTION:
t
In this task of the experiment, an aluminum
e
specimen with a series of longitudinal strain
gages is loaded eccentrically. The strain
w
distribution is determined and compared to the
theoretical distribution and to the case of centric
FIGURE 1
will be investigated utilizing a steel helical spring
loaded in compression. The spring will undergo
Equal and opposite loads with a magnitude P can
be added at the centroidal axis as shown in
of the specimen.
In this task of the experiment, a cantilevered
hollow circular shaft with a long bar attached to
the end will be used to investigate the combined
effects of torsion, shear, and flexure loads at
moment tending to bend the member about the
different locations on the shaft.
EM 327: MECHANICS OF MATERIALS LABORATORY
neutral axis and an axial load applied at the σ = P + My (1.1)
centroid as shown in Figure 3. AI
Where: M= Pe
y is measured from the centroidal
axis
P P
3
I = bh /12
Since the stress is uniaxial, the following form of
P
Hooke's Law, equation 1.2, can be used to
convert stress to strain:
ε = σ /E (1.2)
PP
M
M
FIGURE 2
+=
P
M
FIGURE 4
Figure 5 shows the strain distribution over the
Whether or not compressive stresses will be
developed on one edge depends on the amount of
FIGURE 3
ε
The internal stress distribution produced by the
axial load, P, and the bending moment, M, are
w
shown in Figure 4. In both load components the
resulting stress is uniaxial. The stresses can be
combined using the principle of superposition as
FIGURE 5
long as the stresses remain within the elastic
limit.
The resultant stress distribution is shown in
Compression of a helical spring is an example of
Figure 4 and is given by equation 1.1:
EM 327: MECHANICS OF MATERIALS LABORATORY
a loading that can be treated as a combination of a higher stress from the torsional load. When these
are combined with the stresses developed by the
The relationship between the load and deflection
direct shear load (P), the greater shear stress
of the spring is referred to as the spring constant,
would exist on the inside of the coil.
plotted along the Y axis and deformation plotted
along the X axis, the spring constant is
determined as the slope of the straight line.
P
PR
P
R
PR
Combined Stress State
P
FIGURE 7
dl
ds
The strains at both the inside and outside of the
P
helix can be determined experimentally by
locating strain gage rosettes at these points.
PR
Strain transformation equations can be used to
R
determine the maximum shearing strains. If
material properties are known, stresses may be

determined using the stress-strain relationships.
The modulus of rigidity, G, and Poisson's ratio, ν,

are the required material properties for this case.
FIGURE 6
The following equations are used:
τ = =γ G (2.1)
max max
The critical stress component is the maximum
2 2 1/2
ε =(1/2ε +1/2ε )+[(1/2ε -1/2ε ) +(1/2γ ) ] (2.2)
p1p2 x y x y xy
shearing stress developed in the rod making up
the helix. Figure 6 shows a free body diagram of
2 2 1/2 (2.3)
the spring where both vertical forces and =γ = 2[(1/2ε -1/2ε ) +(1/2γ ) ]
max x y xy
torsional forces are applied to maintain
equilibrium. Figure 7 indicates the stresses
ε =(1/2ε +1/2ε )+(1/2ε -1/2ε )(cos2θ+1/2γ sin2θ) (2.4)
n x y x y xy
developed by these forces. The path length along
the inside of the helix is shorter than along the
outside, resulting in a greater angle of twist and
EM 327: MECHANICS OF MATERIALS LABORATORY
For the strain gage rosette shown in Figure 8 shearing load is acts on the cut surface to
0 0
maintain equilibrium. Applying the torsion
where θ=45 (2θ = 90 ), gage 1 is in the x
formula and the relationship for direct shearing
direction and gage 3 is in the y direction.
stresses, the maximum shearing stress becomes
Equation (2.4) thus takes the form of:
ε =ε =1/2ε +1/2ε +0+1/2γ (2.5)
τ = (Tr/J) + (P/A) (2.7)
n 2 1 3 xy
max
Where: T is the torque, = PR
Equation (2.5) can be rearranged to give γ .
J, the polar moment of inertia,
xy
4
is equal to (1/2)πr
2
A is the cross-sectional area, πr
r is the radius of the coiled rod
4 2
y
τ = [(PRr)/(1/2πr )] + [P/(πr )] (2.8)
max
o
o
Where R is the radius of the spring helix
θ = 45
θ = 45
2
1
Simplifying (2.8) gives:
3
3
τ = [(2 PR)/ πr ] x [1 + (0.5)/(R/r)]
max
x
(2.9)
The accuracy of this equation depends on how
well the actual behavior of the spring follows the
FIGURE 8 assumptions.
Wahl's formula, equation 2.10, is based on more
The following relationships are those required for
accurate assumptions.
equation (2.3), which is used to determine the
maximum shear stress.
4R

()−1
γ = 2[ε -(1/2ε +1/2ε )] (2.6)
xy 2 1 3 2PR 0.615
r

τ = + (2.10)
max
3
ε = ε
4R R
x 1 πr
()− 4 ()
r r
ε = ε
y 3
To determine the spring constant, an expression
Theoretical relationships for the spring modulus
for the spring deflection, δ, is required. Figure 6
and the maximum shearing stress can be
shows the rotation resulting from a torque applied
developed provided certain assumptions are
to the differential element. The diagram also
made. The assumptions are that the curvature of
shows that by considering similar triangles the
the rod making up the helix is such that the
following relationship for dδ results:
torsion formula is reasonably accurate, and that
the coils are tight enough so that the length of a
dδ/(ABdθ) = R/ AB
coil approaches the circumference of a circle (π
times the mean diameter of the coil, L = 2πR).
dδ = Rdθ
torsion
Referring to the free body diagram shown in
Figure 6, a torsional load as well as a direct
EM 327: MECHANICS OF MATERIALS LABORATORY
The diagram also indicates that the following 9a shows the specimen and the loading
relationship can be written for the differential conditions.
element due to the direct load:
dδ = γdL
shear
The bending stresses are found by using equation
3.1
Mc
(3.1)
σ =
Combining these results for the torsion and the
I
Where: M is the bending moment at the
dδ = Rdθ=+ γdL
location where the stress is to be
determined
The torsional displacement may be written as
c is the distance from the neutral axis to
the point where the stress is to be
dθ = (T/JG)dL
determined
= (PR/JG)dL
I is the moment of intertia with respect to
4 4
the centroidal axis, πr /4-πr /4
o i
And from Hooke's Law,
γ = (τ/G) = P/AG
The cross shear stresses are given in equation 3.2.
Substituting,
VQ
(3.2)
τ =
dδ = ( RT/JG)dL + (τ/G) dL
It
π

2 2
Q = (r − r ) [] y
max o i

2
Integrating,

2
δ = PR L/JG + PL/AG (2.11)
Where: V is the shear force
Therefore, the spring constant, K, may be written
Q is the first moment of the cross
as
sectional area, see Appendix C.
2
K = P/δ = 1/ {(R L/JG) + (L/AG)}
t is the thickness of the section where
(2.12)
the stress is to be determined
Where L is the length of a coil, which may
be approximated as π times the mean
The torsional shear stress is given in equation 3.3.
diameter of the coil, L=2πR.
Tc
τ = (3.3)
J
Where: T is the resisting torque at the
There are many examples of machine members
point where the stress is to be
determined
task the combined effects of flexure, shear, and
J is the polar moment of inertia with
torsion will be examined and the location of the
respect to the longitudinal axis of the
most severe stress will be determined. It is
4 4
shaft, πr /2 - πr /2
o i
usually necessary to calculate the stresses at more
than one point to determine where the highest
stresses occur. As long as the stresses remain
Figure 9b shows the addition of two equal and
below the proportional limit of the material the
opposite collinear forces of magnitude P. The
superposition method can be used to combine the
result is now a couple of magnitude Pl and a force
stresses at any point of a loaded member. Figure
EM 327: MECHANICS OF MATERIALS LABORATORY
P that produces cross shear and flexural stress in
z
Pl
B
the shaft as shown in figures 9c and 9d. .
y
τ
xt
x
A
.
.
A
L
(a)
B z
.
B
.
r
.
A
.
σ
A
x
l
(a)
P
(b)
P
z
B
P
.
z
y
A
A
x
. .
x
P (c) P
(b)
P
z
y
FIGURE 10
x
Each of the three stresses is maximum at some
P
point(s) on the plane at the wall. Therefore, one
(c)
P only needs to examine the plane section at the
wall to determine where the maximum stress is
P
z
y
located. Figure 11 shows the resulting stress
states at points A and B. The torsional and cross
x
shear stresses are in the same direction on the
right side of the shaft and opposite on the left
(d)
side. This means the maximum stress condition
resulting from the combination of these stresses
will be at point A. The maximum shear stress at
FIGURE 9
A is thus given by the superposition of equations
3.2 and 3.3, shown in equation 3.4.
Tc
VQ
Figure 10 shows a breakdown of the resulting
τ = +
(3.4)
torsion&shear
stresses at points A and B (fig. 9a). The couple, J It
Pl, produces a torsional shearing stress at all
surface points on the shaft (fig 10a). The load, P,
At a point on the opposite side of the shaft from
produces a bending stress that is maximum in
A, the combined shear stress would equal the
tension at the top and maximum in compression
difference of the torsional and transverse shear
at the bottom of the shaft (fig 10b). The load, P,
stress, given in equation 3.5.
also produces a cross-shear stress, equal to the
longitudinal shearing stress, that is maximum on
Tc VQ
(3.5)
τ = − +
plane yz (fig 10c).
torsion&shear
J It
EM 327: MECHANICS OF MATERIALS LABORATORY
The bending (flexural) stress is zero at point A Equations 3.7 and 3.8 give the relationship for the
since it is located on the neutral axis. At point B gages on either side of the center.
the bending stress is maximum and given by
ε +ε γ
x y xy
equation 3.1. Since the bending stress and the ε = ε = + sin(2* 45 ) (3.7)
45
a
2 2
transverse shear stress are never maximum at the
ε +ε γ
same point, the transverse shear stress is zero at
x y xy
ε = ε = + sin(2 * (−45 )) (3.8)
b −45
point B. The same conclusion can be reached by
2 2
considering the magnitude of Q at point B.
The relationship between the shear strain and the
shear stress for point B is given by equation 3.9.
τ + τ τ
torsion shear torsion
Tc
τ
xy
J
σ γ = = (3.9)
A B xy
G G
A 90° strain gage rosette is also installed on the
side at point A. Since there is no normal stress at
point A the gage aligned with the axis of the
FIGURE 11 shaft, shown in equation 3.10 should measure
zero strain. The strains for the gages at 45° from
the center gage will measure normal strains as
given in equations 3.7 and 3.8.
It is necessary to relate the strains to the stresses.
There are 90° strain gage rosettes located on the
− My
test specimen at point A and point B, as well as
σ
(3.10)
I
ε = =
x
on the bottom, opposite of B.
E E
Figure 12 shows the gages mounted on the top of
The relationship between shear strain and shear
the shaft (point B).
stress at point B is given in equation 3.11.
VQ
Tc

τ
y xy
J It
γ = = (3.11)
xy
G G
45° 45°
b
a
Once the shear strain is known, the shear stress
x
can be calculated from Hooke’s law, equation
FIGURE 12
3.12.
τ = γ G
(3.12)
xy xy
Equation 3.6 gives the relationship between the
stress and strain for the center gage, which is
If the normal strains, ε and ε are known,
x y
aligned with the longitudinal axis of the test
Hooke’s Law can be used to determine the
specimen.
normal stress, as given in equation 3.13.
E
Mc
σ = () ε +νε
x x y
σ 2
I (3.13)
(3.6)
1 −ν
ε = =
x
E E
E
σ = (ε +νε )
y y x
2
1 −ν
EM 327: MECHANICS OF MATERIALS LABORATORY
For uniaxial normal stress ε = -νε .
PROCEDURE:
y x
PRELIMINARY CALCULATIONS:
MATERIAL TO BE TESTED:
The moment of inertia of the specimen cross-
A 2024-T4 aluminum bar with a series of strain
section will be required. It is critical that the
gages oriented parallel to the longitudinal axis.
proper cross-sectional dimensions be used to
calculate this. Discuss this with the members of
lab instructor.
The Hot-Rolled Steel helical spring considered in
this laboratory is from an automobile suspension
system.
of 3000 lbs. The maximum hydraulic pressure of
the table top tester that will be used must be
determined based on this load limit. The cross-
A hollow aluminum tube is used as a cantilevered
sectional area of the piston in the hydraulic
shaft. A straight bar is fixed to one end.
2
cylinder is 4.1 in . (Do not confuse this cross-
sectional area with the cross-sectional area
EQUIPMENT TO BE USED:
required for the specimen stress calculations.)
Check your maximum hydraulic pressure with the
lab instructor prior to testing.
Table top tester
Strain Indicator
Switch and Balance unit TASK 2: HELICAL SPRING
Preliminary calculations are not required for the
helical spring.
55-K MTS testing machine
Strain indicator
Preliminary calculations are not required for the
Switch and balance unit
circular shaft.
TEST PREPARATIONS:
Strain indicator
Switch and balance unit
The cross-sectional dimensions of the aluminum
specimen need to be measured along with the
SAFETY CONSIDERATIONS:
locations across the specimen width where strain
gages are located as well as the distance of the
Never operate the MTS when someone's hands
eccentricity, e. These dimensions should be
are between the actuator. Be sure all lab
shown on a sketch and the gages should be
participants are clear of machine before operating
properly identified. Extreme care should be taken
the hydraulics.
not to damage the gages. Do not measure gage
locations and cross-section dimensions on the
gages themselves. Lines are drawn on the
specimen to be used for this purpose.
EM 327: MECHANICS OF MATERIALS LABORATORY
Observe the type of strain gage rosettes and their 1.) Determine the radius of the round bar
placement on the spring. Trace the gages to the forming the spring, r; the mean radius of the
strain indicator box. helix, R; and the spacing of the punch marks .
Measure and record the diameter of the spring MTS SET-UP
wire and spring helix. Measure the distance
between the punch marks with no load applied.
Station Manager spring
MPT spring.000
Observe the type of strain gage rosettes and their
2.) Turn hydraulics on.
placement on the shaft. Trace the gages to the
3.) Make sure MANUAL OFFSET = 0 for
strain indicator box. Make a sketch of the gage
Stroke.
configurations.
4.) Adjust 'SET POINT'' to 0.0
Measure and record the inside and outside
6.) Set-up Scope to plot a/b.
diameters of the circular shaft, the length of the
shaft, and the length(s) to the load application
location. (Your instructor will provide you with
Stroke 0.2 in/in -0.8
the nominal location(s) to use during the
Time 15 min
experiment.)
TESTING PROCEDURE:
TESTING PROCEDURE:
1) Create a specimen file spring*.
2) Start the Scope.
1.) Install specimen between grips. Keep the
3) Close SAFETY SHIELD!
specimen as vertical as possible to prevent
unwanted bending. For the first test the pins
4) Lock the MPT and select a Specimen.
should be placed through the center holes in the
5) Press `RUN' and let test proceed. It should
sample. The test will then be repeated with the
pins in the eccentric holes of the specimen.
6) Press `STOP' button.
2.) Trace the hydraulic lines to confirm that the
7) Take readings for strains and spacing
appropriate valves are open and closed so that the
between punch marks.
specimen will be loaded in tension.
8) Unlock the MPT.
3.) Zero the strain indicator.
9) Adjust 'SET POINT' to 0.0.
4.) Apply the pressure to the hydraulic cylinder
corresponding to a load of 3000 lbs. (Do not
10) Check your results using the program,
apply 3000 psi pressure!)
'SPRING'. Include a printout in your report.
5.) Record the strain at the maximum load at
each strain gage location.
6.) Release the hydraulic pressure.
1.) Zero the strain indicator.
7.) When all testing is complete be sure to turn
off the strain indicator.
EM 327: MECHANICS OF MATERIALS LABORATORY
at each strain gage rosette location. The report outline found in Appendix A should
be used.
condition(s) as instructed.
REPORT REQUIREMENTS:
5.) When all testing is complete be sure to turn
(1) Develop a single plot of the strain versus
specimen width showing the:
REQUIREMENTS:
a. Theoretical strain distribution for centric
Load the specimen centrically; measure and
b. Experimental strain distribution for
record the resulting strains. Load the specimen
eccentrically; measure and record the resulting
c.Theoretical strain distribution for
strains.
d. Experimental strain distribution for
A load-deformation curve will be developed
This plot may be constructed using any suitable
using the MTS testing machine. Note that this test
computer plotting software.
is conducted entirely in the elastic range of the
material. Make copies of the plot from the MTS 2) Comment on the correlation between the
recorder and give the original to your instructor. theoretical and experimental distributions.
Provide possible explanations for
discrepancies.
A record of the measured strains and deflection of
3) Where should an eccentric uniaxial load be
placed to obtain zero stress on one side of
the specimen? Provide both an explanation
and a numerical value.
(1) Determine the spring constant for the entire
spring using the force and total deflection as
Measure and record the strains resulting from the
recorded on the MTS plot.
load configuration(s) you are assigned to
investigate.
(2) Determine the spring constant per turn
using the maximum force as recorded on the
MTS plot and the punch mark
SOFTWARE:
measurements.
The following software may be used to provide
(3) (3)Calculate the maximum shear stress
results that you may compare your calculations
with. The computer output should not replace
consists of a torsional load plus a direct
(4) Calculate the theoretical spring constant per
turn, using equation (12)
Use the 'SPRING' program.
(5) Determine the maximum shear stress on the
inside and on the outside of the coil using
the strain readings and rosette analysis
REPORT:
EM 327: MECHANICS OF MATERIALS LABORATORY
(equations 1, 3, and 6). Would you expect a allowable stresses are not exceeded. (16 ksi
difference in these values? Explain why. T, and 9 ksi shear)
(6) (6)Determine the maximum shear stress 9.) Discuss and discrepancies you found.
using Wahl's formula. (Equation (10) or p.
463, EM324 text.
QUESTIONS:
(7) Determine the percent difference in
maximum shear stress values between the
(2.1) Did the diameter of the bar and the diameter
experimentally determined values and:
of the helix need to be measured with the
a. the stress determined form the derived
same precision? Why?
equation (9)
(2.2) If the bar had a non-circular cross section
b. the stress determined from Wahl's
would the formulas used in the
equation (10).
determination of stress be applicable?
Explain.
(2.3) Why doesn't the torsion formula supply
1.) Sketch the experimental set-up indicating correct values for the properties of a solid
where the strain gages are mounted as well as circular shaft above the proportional
the important physical dimensions. limit?
2.) Calculate the second moment of inertia, I, the
polar moment of inertia, J, and the first
moment of the cross-section at the neutral
(3.1) Where are the following stresses maximum?
axis, Q.
a. Longitudinal and transverse shear stress
b. Flexural stress
strain readings from each of the three rosettes
to calculate the applied load and compare
c. Torsional stress
with the actual loads applied by calculating
the percent error.
4.) Calculate, based on the measured strains, the
shear and normal stress magnitudes at the top
and side strain gage locations for both load
conditions assigned. Show the stresses on
stress blocks for the top and side location.
5.) Determine the theoretical stresses at the top
and side gage location using the actual
applied load values. Compare to the stresses
obtained in the previous step.
6.) Determine the principal stresses for the top
and side strain gage locations for both load
conditions.
7.) Determine the most severely stressed point on
the specimen for load condition 2 using the