Torsion of circular sections lab report

Solid Mechanics Lab Report (Torsion)

Lab 4. Torsion of Member with Circular Cross-Section: Principal Strains and Stresses

Thus far our only focus has been on axial deformations which cause elongation or compression

of a member along their axis of application. Other types of deformation exist, though, such as

shear deformations which translate or slide the face on which they act. To help visualize this

behavior, in this lab a solid steel shaft with circular cross-section is loaded under pure torsion.

Torsion is caused when a moment is applied in-plane with the cross-section of the member,

twisting it and causing it to deform in shear. A visualization of this is provided below in Figure

4-1.

Figure 4-1: Deformation of a body under shear conditions

Recall that the measurement devices used thus far only measure elongation or contraction

along their primary axis. Meaning that strain gages cannot directly measure these values.

Looking at more detail at Figure 4.1, it is clear that the shear strain, γ , is not a linear quantity

but an angle of distortion. We thus need additional techniques to be able to somehow

determine shear strain from linear-based measurement equipment.

For a bar in pure torsion, each section along the longitudinal axis rotates strictly in a plane; that

is, all points within the member follow a strictly a circular arc in the plane and do not translate

in either direction along the member axis. Note that this is not the case for non-circular cross-

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sections, as torsion develops warping deformations which displace points axially in addition to

rotationally. This behavior is shown below in Figure 4-2, where it can be seen that point B at the

outermost fiber of the cross-section travels along the circumference of the circle to B’ at an

applied torsion of To.

Figure 4-2. Member deformation from applied torsion To

Torsional loading at a section causes shear stresses that are zero at the center of the member

and increase linearly with radius r to a maximum at the surface of the cylinder at r = R. Shear

stress is found from the section torsion T, radius r, and polar moment of inertia J as

J

Tr  (4-1)

For a circular section, the polar moment of inertia is equal to the sum of moments of inertia

about the x and y axes as

A B

To To

L

B’

3

244

444 Rrr IIJ yx

  (4-2)

Shear strain is found by dividing the shear stress by the shear modulus G, where the shear

modulus is found for a given material from material testing as the slope of the shear stress

versus shear strain plot. Thus shear strain is found from

JG

Tr

G 

  (4-3)

Note the similarities in this equation to those seen in the previous labs. To calculate axial

stresses, we had a driving axial load P that was divided by the cross-sectional area – a geometric

property of the member. Then, to calculate strains, we divided this stress by the modulus of

elasticity – a material property of the section – and concluded that the deformation is a

function of the driving load P, the geometrical resistance of the member, and the material

resistance of the member. In the case of torsional strain, we see a similar pattern: a torque T

drives the deformation of the member, a geometric resistance forms via r and J, and a material

resistance forms via the shear modulus G.

Fixing a member on one end and applying a torsion To to the end free to rotate yields the

deformed shape given in Figures 4-2 and 4-3. At the left end of the member all points of the

cross-section are restrained from rotation and remain in the same place before and after

torsion is applied. At the right end of the member, however, torsional loading causes the cross-

section to rotate through an angle indicated by the movement of point B to B’. This point starts

at the top of the section but shifts along the perimeter of the cross-section. All points on the

surface at the right end of the member move through the same angle as point B, though,

generating a rotation of the full cross-section.

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For the example shown in Figure 4-3, the torsion is constant along the member length which

results in a linear variation of deformation from Points A to B. In other words, the amount of

rotation at half the member length is one-half of the maximum rotation at the right end of the

member. To demonstrate this, two lines are given in Figure 4-3 that were originally parallel to

the member geometry and have deformed from load application. A stress element shows that

the original rectangular shape has been deformed due to shear stresses; the length of the sides

remain unchanged, but the amount of deformation of the corners on the left face and the

corners on the right face differ in magnitude. Applied and reacting torsions are indicated with

double arrows: arrows pointing away from the member indicate positive torsion while arrows

facing toward the member indicate negative torsion, with the rotational direction following

that defined by the right hand rule. Positive torsion definition is given for any slice of the beam

in Figure 4-4 and the torsion diagram for the example provided in Figure 4-3 is given in Figure

4-5.

Figure 4-3. Side view of member under torsion with stress element indicated

A B

To To

L

B’

Stress Element

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Figure 4-4. Definition of positive torsion for a short length of the member (slice)

Figure 4-5. Torsion diagram over member length

When the torsion is constant along a member length as in Figure 4-4 the rotation at its right

end, called the angle of twist, is given as

JG

Tl  (4-4)

TT +

T

X A B

To

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If the torsion varies along the member length, however, rotation needs to integrated over the

length of the member similar to how distributed loads must be integrated for beam

displacements. To do this the twist of the member, which is the rotation per unit length, is

calculated for a given cross-section

JG

T 

(4-5)

The rotation at this cross-section is then calculated by integrating over the length

 l

dx JG

T

0



(4-6)

An example of torsion that varies along the member length is given in Figure 4-6, with evenly

distributed torsion of To/L applied for the entire member length. There is no torsion in the

member at its right end and the torsion increases linearly to a maximum at the left of the

member. From statics, the torsion reaction at the fixed end is To and the torsion diagram can

be developed as shown in Figure 4-6. The torsion in the member at any section along its length

is given as

 

  

 

L

x Tx

L

T TT o

o o 1

(4-7)

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Figures 4-6. Distributed torsion along member length with torsion diagram

Torsion given as a function of x in Eq. (4-7) is substituted into Eq. (4-6) to find member rotation.

Thus

  

  

 

 l ol

dx JG

L

x T

dx JG

T

00

1



(4-8)

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If the material is the same for the entire member and the cross-section is constant then J and G

are constant and can be moved out of the integral, as can the constant To, giving

JG

lT dx

L

x

JG

T o l

o B

2 1

0

  

  

 

(4-9)

This is half the rotation of a member that has constant torque along its length with the same

reacting torsion at its left end of To.

The stress element shown in Figure 4-3 distorts due to an upward shear stress on its left side

and, from statics, an equal and opposite shear stress on its right side (down) similar to how the

block in Figure 4-1(b) deforms. However, while this satisfies equilibrium of forces in the vertical

direction, equilibrium of the stress block as a whole is not satisfied: by themselves these shear

stresses form a couple and thus give a tendency for the block to rotate.

To correct for this, moments can be taken about one of the bottom corners of the stress block

to generate a balancing horizontal stress on the top face that imparts an equal but opposite

rotation to the vertical shear stresses. It is found that this horizontal shear stress is equal in

magnitude to the two vertical shear stresses, and equilibrium in the horizontal direction shows

that an equal but opposite shear stress forms on the bottom face. To help visualize this process,

an element is also shown at the top right of Figure 4-7 with these shear stresses indicated. The

stress element has dimensions of dx and dy, and when it has a square or rectangular shape it

vanishes to a single point.

With shear stresses defined at a stress element from torsion, it is possible to determine

maximum compressive and tensile stresses of the stress element that act on different faces of

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the element (at different angles). These are called principal stresses and can be determined

through the use of Mohr’s circle as shown in Figure 4-7. For a given set of shear and normal

stresses on two planes of a stress element, Mohr’s circle is constructed. The horizontal axis is

for normal stresses on a plane and the vertical axis is for shear stresses on a plane. X and Y

planes are defined by passing the x and y axes through the element and the faces that are

intersected are defined, as indicated at the bottom right of Figure 4-7. There are two X planes

and two Y planes, and either one can be used in the following discussion as they are completely

interchangeable.

Figure 4-7. Mohr’s circle with stress element and plane definitions

t

t

dx

dy

t

s

X (0, t )

Y (0, -t )

s 1s 2

f 1

f 2

x

y

x Planex Plane

y Plane

y Plane

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To construct Mohr’s circle an X coordinate and Y coordinate are plotted, representing the

normal and shear stresses on X and Y faces, respectively, of the stress element being

considered. On the X plane of our stress element there is a positive shear and no normal stress,

resulting in the X point given in Figure 4-7. As discussed before, the X plane on either left or

right sides of the element can be used to determine the shear and normal stresses to plot in

Mohr’s circle.

If the shear stress on the face being considered wants to turn the element in the clockwise

direction we call this positive shear. The stress element in Figure 4-7 shows that the upward

shear stress on the left of the element wants to rotate the element in the clockwise direction,

and so does the downward shear stress on the right of the element, thus they both give the

same positive shear stress and no normal stress. On the Y plane (top or bottom of the element)

the shear stress wants to turn the element in the counterclockwise direction and this is defined

as negative shear. There is no normal stress on the Y plane, yielding an X coordinate of zero

and a Y coordinate equal to the negative shear stress found from the torsion expression given

in Eq. (4-1).

Once two points are given on a circle then the entire circle is completely defined as shown in

Figure 4-7. In Mohr’s circle we are working in stress space where all values are stresses. Thus,

if a ruler is put on a graph of Mohr’s circle to scale a result, the values will be in stress units such

as psi or ksi. Only results that lie on the circle have meaning, providing normal and shear

stresses at all different angles within the stress element. One can imagine a plane cut through

the stress element at some arbitrary angle and asking what the shear and normal stresses are

on this exposed new plane. The construction of Mohr’s circles provides the answer to this

question in an elegant and efficient manner. It can be shown from solid mechanics that the

angle swept in Mohr’s circle is twice the angle within the element. Or, equivalently, the angle

within the stress element is one-half of the angle swept in Mohr’s circle. The direction of

rotation, clockwise or counterclockwise, is the same within the element and in Mohr’s circle.

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Maximum tensile and compressive stresses 1 and 2, called principal stresses, occur where the

shear stresses are zero, as indicated in Figure 4-7. The angle rotated to the principal stresses

from the X and Y points on Mohr’s circle is 90 degrees in the clockwise direction. Therefore the

angle rotated within the physical element from the X and Y planes to the principal planes is 45

degrees. Also from Mohr’s circle it is clear that for this case the magnitude of the principal

stresses is equal to the magnitude of the shear stresses found from torsion (radius of Mohr’s

circle).

A stress element showing the principal planes is given in Figure 4-8, occurring on a surface at 45

degrees from the element in pure shear that has normal stresses but no shear stresses. Figure

4-9 breaks this down mechanically, showing how tensile and compressive stresses develop

internally in an element to enforce equilibrium in the element. Note that rotating 45 degrees

clockwise from either X plane gives the same values on the principal tensile stress plane, and

rotating 45 degrees clockwise from either Y plane gives the same values on the principal

compressive plane. It is interesting to note that if the whole element is rotated 45 degrees to

the principal planes then only normal stresses are seen on the surfaces, with no shear stresses.

Thus, even though the element deforms in pure shear, pure tensile and compressive stresses

develop at an angle 45 degrees from this axis.

Figure 4-8. Element rotated to principal planes

45

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Figure 4-9. Stresses on principal planes from Mohr’s circle

The reverse is true when considering a wire in tension. From Mohr’s circle for a stress element

within the wire a normal stress is found in the vertical direction and no normal stress in the

transverse direction. A rotation of 90 degrees in Mohr’s circle shows a maximum shear stress

at this angle which is an element rotation of 45 degrees. Yielding of the wire in tension is

caused by shear of the crystals at 45 degrees resulting in Luder’s lines. This is clearly shown on

a polished coupon tensile test.

Experiments

A steel bar with ¾-inch diameter is tested in torsion. The steel material has modulus of

elasticity E and Poisson’s ratio  of 30,000 ksi and 0.25, respectively. Two strain gages are

applied to the surface of the specimen one positioned in the principal tensile stress direction

and the other in the principal compressive stress direction. Thus they are both positioned 45

degrees from the longitudinal member axis and 90 degrees from reach other. The strain gages

utilize a half bridge.

t

t

dx

dy

x

y

x Planex Plane

y Plane

y Plane

t

t

45 x Planet

t

45 x Plane

t

t

y Plane

45

y Plane

t

t

45

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Apply a torsion of 1,000 lb-in. at intervals of 250 lb-in. and record the principal strains at each

load increment. Compare theoretical and measured principal strains. Principal strains are

related to principal stresses in Eq. (4-10) and Eq. (4-11). With measured principal strains, these

equations can be solved for the principal stresses.

EE 21

1

 

  

(4-10)

EE 12

2

 

  

(4-11)

Provide the following plots of theory versus measured (or calculated from measured):

Torsion diagram (torsion along member length)

Torsion versus principal tensile stress

Principal tensile stress versus principal compressive stress

Principal tensile strain versus principal compressive strain

Torsion versus shear stress

REQUIRED DISCUSSION:

How do the measured results in the above plots compare to theoretical results? Provide

detailed explanation for any variations.

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Also, if an axial load was applied to the rod would the orientation of the strain gages need to

change or would the orientation of 45 degrees still be valid? Justify your answer.

Definitions:

Shear Modulus of elasticity G (ksi). This is the initial, linear slope of the stress-strain diagram in

shear. In this range the material (and any structure that is made of this material) will be linear-

elastic, meaning that if the load is doubled the deformations will double (linear), and upon

removal of the load it will return to its original position (elastic) with no permanent

deformation.

Strain gage. An electrical strain gage is a resistor that changes its resistance as a function of

strain and temperature. By calibrating the strain gage (gage factor), this change in resistance

that is associated with strain in the gage is accurately measured as a voltage change and

converted to a strain value. The change in resistance of the strain gage due to temperature

effects is removed from the measured results by use of a Wheatstone bridge.

Polar moment of inertia. For circular sections the polar moment of inertia is equal to the sum

of the section moment of inertias about the x and y axes that run through the section centroid.