COMBINED LOADING

EM 327: MECHANICS OF MATERIALS LABORATORY

EXPERIMENT: COMBINED LOADING

VIDEO TITLES: HELICAL SPRING,

BACKGROUND:

ECCENTRIC LOADS

TASK 1: ECCENTRIC LOADS

OBJECTIVES:

Loading conditions can often be a combination of

axial and flexural loads, and thus require special

TASK 1: ECCENTRIC LOADS

consideration when calculating theoretical

(1)Study the combined effects of bending and

stresses and strains.

axial loading by measuring the strain distribution

for a member subjected to an eccentric tensile

load.

The eccentrically loaded specimen used in this

laboratory is an example of combined loading

where the load can be considered as a

TASK 2: HELICAL SPRING

combination of flexural loading and axial

(1)Study the combined effects of torsion and

loading. Figure 1 shows a rectangular bar with

axial loading, by observing the behavior of a

an eccentric load (P) applied at some distance, e,

helical spring in compression.

from the centroidal axis.

TASK 3: CIRCULAR SHAFT

(3) Study the combined effects of torsion and

P

flexural loading, by observing the behavior of a

circular shaft loaded in torsion and bending.

INTRODUCTION:

TASK 1: ECCENTRIC LOADS

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

loading.

TASK 2: HELICAL SPRING

FIGURE 1

In this task of the experiment combined loading

will be investigated utilizing a steel helical spring

loaded in compression. The spring will undergo

Equal and opposite loads with a magnitude P can

both torsional loading and shear loading.

be added at the centroidal axis as shown in

Figure 2 without changing the loading condition

of the specimen.

TASK 3: CIRCULAR SHAFT

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

This loading can then be represented by a

effects of torsion, shear, and flexure loads at

moment tending to bend the member about the

different locations on the shaft.

67COMBINED LOADING

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

width of the bar for some eccentricity of loading.

Whether or not compressive stresses will be

developed on one edge depends on the amount of

eccentricity in the loading.

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.

TASK 2: HELICAL SPRING

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:

68COMBINED LOADING

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

direct shear load and a torsional load. stresses, resulting from the torsional load (PR)

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.

with units of load per deflection. With load

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

Stresses from Torsional Load

PR

Stresses from Direct Shear Load

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

dθ

determined using the stress-strain relationships.

The modulus of rigidity, G, and Poisson's ratio, ν,

dδ

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

69COMBINED LOADING

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

70COMBINED LOADING

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

direct shear loading,

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

TASK 3: CIRCULAR SHAFT

Where: T is the resisting torque at the

There are many examples of machine members

point where the stress is to be

that are subjected to combined loading. In this

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

71COMBINED LOADING

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

72COMBINED LOADING

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

73COMBINED LOADING

EM 327: MECHANICS OF MATERIALS LABORATORY

For uniaxial normal stress ε = -νε .

PROCEDURE:

y x

PRELIMINARY CALCULATIONS:

MATERIAL TO BE TESTED:

TASK 1: ECCENTRIC LOADS

TASK 1: ECCENTRIC LOADS

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

your group and verify your conclusions with the

TASK 2: HELICAL SPRING

lab instructor.

The Hot-Rolled Steel helical spring considered in

this laboratory is from an automobile suspension

system.

The specimen will be loaded to a maximum load

of 3000 lbs. The maximum hydraulic pressure of

the table top tester that will be used must be

TASK 3: CIRCULAR SHAFT

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.)

TASK 1: ECCENTRIC LOADS

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.

TASK 2: HELICAL SPRING

55-K MTS testing machine

TASK 3: CIRCULAR SHAFT

Strain indicator

Preliminary calculations are not required for the

Switch and balance unit

circular shaft.

TASK 3: CIRCULAR SHAFT

TEST PREPARATIONS:

Strain indicator

TASK 1: ECCENTRIC LOADS

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.

74COMBINED LOADING

EM 327: MECHANICS OF MATERIALS LABORATORY

TASK 2: HELICAL SPRING TASK 2: HELICAL SPRING

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

TASK 1 and 2:

between the punch marks with no load applied.

1.) Follow Start- up Procedures

Station Manager spring

TASK 3: CIRCULAR SHAFT

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

5.) 'AUTO OFFSET' Load.

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

Load 100 lbf -400

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*.

TASK 1: ECCENTRIC LOADS

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

load the spring to about 550lb.

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.

TASK 3: CIRCULAR SHAFT

6.) Release the hydraulic pressure.

1.) Zero the strain indicator.

7.) When all testing is complete be sure to turn

2.) Apply loading condition(s) as specified by

off the strain indicator.

your instructor.

75COMBINED LOADING

EM 327: MECHANICS OF MATERIALS LABORATORY

3.) Record the strains for each loading condition

at each strain gage rosette location. The report outline found in Appendix A should

be used.

4.) Remove the load and apply addition loading

condition(s) as instructed.

REPORT REQUIREMENTS:

5.) When all testing is complete be sure to turn

off the strain indicator. TASK 1: ECCENTRIC LOADS

(1) Develop a single plot of the strain versus

specimen width showing the:

REQUIREMENTS:

a. Theoretical strain distribution for centric

TASK 1: ECCENTRIC LOADS

loading

Load the specimen centrically; measure and

b. Experimental strain distribution for

record the resulting strains. Load the specimen

centric loading.

eccentrically; measure and record the resulting

c.Theoretical strain distribution for

strains.

eccentic loading

d. Experimental strain distribution for

TASK 2: HELICAL SPRING

eccentric loading.

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

the spring during loading should also be made.

placed to obtain zero stress on one side of

the specimen? Provide both an explanation

and a numerical value.

TASK 2: HELICAL SPRING

TASK 3: CIRCULAR SHAFT

(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

(assuming that the combined loading

with. The computer output should not replace

consists of a torsional load plus a direct

your calculations.

load), using equation (9)

(4) Calculate the theoretical spring constant per

TASK 2: HELICAL SPRING

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:

76COMBINED LOADING

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

TASK 2: HELICAL SPRING:

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.

TASK 3: CIRCULAR SHAFT

(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

TASK 3: CIRCULAR SHAFT

moment of the cross-section at the neutral

(3.1) Where are the following stresses maximum?

axis, Q.

a. Longitudinal and transverse shear stress

3.) For assigned loading condition(s), use the

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

actual applied load values.

8.) Determine the maximum load that may be

applied to the specimen so that the following

77

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