3.5 STRESS AND STRAIN IN PURE SHEAR The next element is in ...

3.5 STRESS AND STRAIN IN PURE SHEAR

■ The next element is in a state of pure shear.

Fig. 3-20 Stresses acting on a stress element cut from a bar in torsion (pure shear)

■ Stresses on inclined planes

Fig. 3-21 Analysis of stresses on inclined planes: (a) element in pure shear, (b) stresses acting on a triangular stress element, and (c) forces acting on the triangular stress element (Free-body diagram)

We are ready to write two equations of equilibrium for the element.

The first equation, obtained by summing forces in the direction of is

The second equation, obtained by summing forces in the direction of is

by the following trigonometric identities.

Fig. 3-22 Graph of normal stresses σθ and shear stresses τθ

versus angle θ of the inclined plane

: ,

, :

:

:

:

Fig. 3-23 Stress elements oriented at θ = 0 and θ = 45° for pure torsion

Fig. 3-24 Torsion failure of a brittle material by tension cracking along a 45°

helical surface

■ Strains in Pure Shear

Shear distortion

: shear strain

: The element changes its shape from a rectangular parallelepiped to an oblique

parallelepiped without changing the lengths under pure shear

Fig. 3-25 Strains in pure torsion: (a) shear distortion of an element oriented at θ = 0, and (b)

distortion of an element oriented at θ = 45°

Hooke's law

at (fig 3-25)

The tensile stress acting at

produces a positive normal strain in that direction equal to .

Since ( ) 1)

Therefore, the normal strain in the direction is

Since ( ) 1) 2)

2)

3.6 RELATIONSHIP BETWEEN MODULI OF ELASTICITY

AND

Fig. 3-28 Geometry of deformed element in pure shear

Consider the stress element abcd shown in Fig.3-28

-----> When this element is subjected to pure shear by stresses,

(fig.3-28(b))

------> To obtain the required geometric relationships, consider triangle

abd(Fig.3.28c)

Now using the law of cosines for triangle abd, we get,

Substituting for

from

we get,

in which ( )

in which( can be disregarded)

= (Because is very small)

∴

↑ substituting

in which( = , , = = )

∴

This equation can be used to calculate the maximum shear strains and

maximum normal strains in pure torsion when the shear stress is known.

￭ Example 3-6

A circular tube with an outside diameter of 80mm and an inside diameter of

60mm is subjected to a torque (Fig. 3-26). The tube is made

of aluminum alloy 7075-T6.

(a) Determine the maximum shear, tensile, and compressive stresses in the

tube and show these stresses on sketches of properly oriented stress elements.

(b) Determine the corresponding maximum strains in the tube and show these

strains on the sketches of the deformed elements.

©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Fig. 3-26 Example 3-6. Circular tube in torsion.

Solution

a) maximum stresses :

(b) maximum strain :

maximum shear stress :

maximum normal strain :

∴


Fig. 3-27 Stress and strain elements for the tube of Example 3-6: (a) maximum shear stresses, (b) maximum tensile and compressive stresses; (c) maximum shear strains, and (d) maximum tensile and compressive strains.


■ Sign Convention for Shear Stresses and Strain as

1) A shear stress acting on a positive face of an element is

positive if it acts in the positive direction of the coordinate axes and

negative if it acts in the negative direction of the coordinate axes


Fig. 3-27 Stress and strain elements for the tube of Example 3-6: (a) maximum shear stresses, (b) maximum tensile and compressive stresses; (c) maximum shear strains, and (d) maximum tensile and compressive strains.

3.7 TRANSMISSION OF POWER BY CIRCULAR SHAFTS

Fig. 3-29 Shaft transmitting a constant torque T at an angular speed

Work is

( Torque × angle of rotation)

where( : angle of rotation in radians)

= angle displacement

Power is the rate at which work is done,

(Watt) (1 W=1(N․m/s) or 1(J/s))

T(N․m),

:

Angle speed (: ) is often expressed as the frequency of rotation, which is the number of revolutions per unit of time.

※ units

One revolution equals radians, we obtain

∴

Another commonly used unit is (the number of revolutions per minute(rpm))

∴

※ In U.S engineering practice, power is expressed horse-power(hp).

(in which hp = 550 ft-lb/s)

∴

1 hp = 746 watt T

￭Example 3-7 A motor driving a solid circular steel shaft transmits 30 kW to a gear at B (Fig 3-30). The allowable shear stress in the steel is 42 MPa.

(a) What is the required diameter of the shaft if it is operated at 500 rpm?

(b) What is the required diameter if it is operated at 4000 rpm?


Fig. 3-30 Example 3-7. Steel shaft in torsion.

Solution (a) Motor operating at 500 rpm

from

from

---- >

∴ required d=41.1mm

(b) Motor operating at 4000 rpm the same procedure as in part (a)

This example illustrates that the higher the speed of rotation, the smaller the required size of the shaft. (for the same power and the same allowable stress)

￭Example 3-8 A solid steel shaft ABC of 50mm diameter (Fig 3-13a) is driven at A by a motor that transmits 50 ㎾ to the shaft at 10 ㎐. The gears at B and C drive machinery requiring power equal to 35 ㎾ and 15 ㎾, respectively.

Compute the maximum shear stress in the shaft and the angle of twist between the motor at A and the gear at C. (Use G = 80 ㎬.)

Solution

Torques acting on the shaft.

: torque applied at end shaft

: torque applied at B

: torque applied at C

The internal torques in the two segments

∴

∴

Shear stresses and angles of twist.

the total angle of twist

3.8 STATICALLY INDETERMINATE TORSION MEMBERS

Fig. 3-32 Statistically indeterminate bar in torsion

1) equation of equilibrium 2) equation of compatibility

When the torque is applied at the end of the bar (in fig. 3-32)

: diameter of internal of the bar

: diameter of outer of the bar

1) equation of equilibrium

3) torque-displacement relation( )

2) equation of compatibility

: angle of the twist of the solid bar : angle of the twist of the tube bar

3) torque-displacement relation( )

from ( ) 1 2

∴ 1 2

from 1) and 3)

￭Example 3-9

The bar ACB shown in Fig. 3-33a is fixed at both ends and loaded by a torque at point C. Segments AC and CB of the bar have diameters and ,, length and and polar moments of inertia and , respectively. The material of the bar is the same throughout both segments. Obtain formulas for (a) the reactive torques and at the ends, (b) the maximum shear stresses and in each segment of the bar, and (c) the angle of rotation at the cross section where the load iis applied.

Fig. 3-33 Example 3-9. Statically indeterminate bar in torsion.

Solution

1) equation of equilibrium

2) equation of compatibility

3) torque-displacement relation

3) ---> 2)

or

∴

∴

maximum shear stresses :

As a special case, if ( )

∴

Angle of twist :

※ in special case if

B

3.9 STRAIN ENERGY IN TORSION AND PURE SHEAR

Consider a prismatic bar AB in pure torsion under the acting of a torque.

Fig. 3-34 Prismatic bar in pure torsion

Fig. 3-35 Torque-rotation diagram for a bar in pure torsion (linearly elastic material)

From

Since W = U

compare

■ Nonuniform Torsion

Fig Bar in nonuniform torsion

: the strain energy of the segment

: the number of segment

-----------------------------------------------------------------------

■ Strain-Energy Density in Pure Shear

: Because the individual elements of a bar in torsion are stressed in pure shear, it is useful to obtain expressions for the strain energy associated with the shear stresses.

In Fig. 3-36

shear force :

Fig. 3-36 Element in pure shear

Total Strain Energy :

(from : volume of element)

Strain-Energy Density :

from

∴

￭Example 3-10 A solid circular bar AB of length L is fixed at one end and free at the other(fig. 3-37). Three different loading conditions are to be considered: (a) torque acting at the free end; (b) torque acting at the midpoint of the bar; and (c) torque and acting simultaneously.

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Fig. 3-37 Example 3-10. Strain energy produced by two loads.

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Fig. 3-37 Example 3-10. Strain energy produced by two loads.

For each case of loading, obtain a formula for the strain energy stored in the bar. Then evaluate the strain energy for the following data: = 100 N․m, = 150 N․m, L = 1.6 m, G = 80 ㎬, and = .

Solution

(b) torque acting at the midpoint of the bar;

(a) torque acting at the free end;

torque of CB :

torque of AC : +

(c) torque and acting simultaneously.

The strain energy of a structure supporting more than one load cannot be obtained by adding the strain energies obtained for the individual loads acting separately

(d) Numerical results :

(1J=1N․m)

(a) Derive a formula for the strain energy of the bar.

A prismatic bar AB, fixed at one end free at the other, is loaded by a distributed torque of constant intensity t per unit distance along the axis of the bar(Fig. 3-38)

(b) Evaluate the strain energy of the shaft if the data are as follows:

￭Example 3-11


Fig. 3-38 Example 3-11. Strain energy produced by a distributed torque.

Solution

(a) Derive a formula for the strain energy of the bar.

(from )

(b) Numerical results :

￭Example 3-12

Determine the angle of rotation at end A of the bar by equating the strain energy to the work done by the load.

A tapered bar AB of solid circular cross section is supported at the right-hand end and loaded by a torque T at the other end(Fig. 3-39). The diameter of the bar varies linearly from at the left-hand end to at the right-hand end.


Fig. 3-39 Example 3-12. Tapered bar in torsion.

Solution

U=W

Substituting

,

∴

Thus, the integration equals

in which

strain energy :

From

∴

3.5 STRESS AND STRAIN IN PURE SHEAR The next element is in ...

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