Mechanics Torsion Presentation

12005 Pearson Education South Asia Pte Ltd

5. Torsion

*5.7 THIN-WALLED TUBES HAVING CLOSED CROSS SECTIONS

• Thin-walled tubes of noncircular shape are used to construct lightweight frameworks such as those in aircraft

• This section will analyze such shafts with a closed x-section

• As walls are thin, we assume stress is uniformly distributed across the thickness of the tube


5. Torsion


Shear flow• Force equilibrium requires the

forces shown to be of equal magnitude but opposite direction, thus AtA = BtB

• This product is called shear flow q, and can be expressed as

q = avgt

• Shear flow measures force per unit length along tube’s x-sectional area


5. Torsion


Average shear stress

avg = average shear stress acting over thickness of tube

T = resultant internal torque at x-section

t = thickness of tube where avg is to be determined

Am = mean area enclosed within boundary of centerline of tube’s thickness

avg = T

2tAm


5. Torsion


Average shear stressSince q = avgt, the shear flow throughout the x-section is

q = T

2Am

Angle of twist

Can be determined using energy methods

= ∫ TL

4Am2G

ds

tO


5. Torsion


IMPORTANT• Shear flow q is a product of tube’s thickness and

average shear stress. This value is constant at all points along tube’s x-section. Thus, largest average shear stress occurs where tube’s thickness is smallest

• Both shear flow and average shear stress act tangent to wall of tube at all points in a direction to contribute to resultant torque


5. Torsion

EXAMPLE 5.16

Square aluminum tube as shown.

Determine average shear stress in the tube at point A if it is subjected to a torque of 85 N·m. Also, compute angle of twist due to this loading.

Take Gal = 26 GPa.


5. Torsion

EXAMPLE 5.16 (SOLN)

Average shear stress

Am = (50 mm)(50 mm) = 2500 mm2

avg = = ... = 1.7 N/mm2T

2tAm

Since t is constant except at corners, average shear stress is same at all points on x-section.

Note that avg acts upward on darker-shaded face, since it contributes to internal resultant torque T at the section


5. Torsion

EXAMPLE 5.16 (SOLN)

Angle of twist

Here, integral represents length around centerline boundary of tube, thus

= ∫ TL

4Am2G

ds

tO = ... = 0.196(10-4) mm-1 ∫ dsO

= 0.196(10-4) mm-1[4(50 mm)] = 3.92 (10-3) rad


5. Torsion

5.8 STRESS CONCENTRATION

• Three common discontinuities of the x-section are:

a) is a coupling, for connecting 2 collinear shafts together

b) is a keyway used to connect gears or pulleys to a shaft

c) is a shoulder fillet used to fabricate a single collinear shaft from 2 shafts with different diameters


5. Torsion


• Dots on x-section indicate where maximum shear stress will occur

• This maximum shear stress can be determined from torsional stress-concentration factor, K


5. Torsion


• K, can be obtained from a graph as shown

• Find geometric ratio D/d for appropriate curve

• Once abscissa r/d calculated, value of K found along ordinate

• Maximum shear stress is then determined from

max = K(Tc/J)


5. Torsion


IMPORTANT• Stress concentrations in shafts occur at points of

sudden x-sectional change. The more severe the change, the larger the stress concentration

• For design/analysis, not necessary to know exact shear-stress distribution on x-section. Instead, obtain maximum shear stress using stress concentration factor K

• If material is brittle, or subjected to fatigue loadings, then stress concentrations need to be considered in design/analysis.


5. Torsion

EXAMPLE 5.18

Stepped shaft shown is supported at bearings at A and B. Determine maximum stress in the shaft due to applied torques. Fillet at junction of each shaft has radius r = 6 mm.


5. Torsion

EXAMPLE 5.18 (SOLN)

Internal torque

By inspection, moment equilibrium about axis of shaft is satisfied. Since maximum shear stress occurs at rooted ends of smaller diameter shafts, internal torque (30 N·m) can be found by applying method of sections


5. Torsion

EXAMPLE 5.18 (SOLN)

Maximum shear stress

From shaft geometry, we have

D

d

r

d

2(40 mm)

2(20 mm)

6 mm)

2(20 mm)

= = 2

= = 0.15

Thus, from the graph, K = 1.3

max = K(Tc/J) = ... = 3.10 MPa


5. Torsion

EXAMPLE 5.18 (SOLN)

Maximum shear stress

From experimental evidence, actual stress distribution along radial line of x-section at critical section looks similar to:


5. Torsion

*5.9 INELASTIC TORSION

• To perform a “plastic analysis” for a material that has yielded, the following conditions must be met:

1.Shear strains in material must vary linearly from zero at center of shaft to its maximum at outer boundary (geometry)

2.Resultant torque at section must be equivalent to torque caused by entire shear-stress distribution over the x-section (loading)


5. Torsion


• Expressing the loading condition mathematically, we get:

T = 2∫A 2 dEquation 5-23


5. Torsion


A. Maximum elastic torque• For maximum elastic shear strain Y,

at outer boundary of the shaft, shear-strain distribution along radial line will look like diagram (b)

• Based on Eqn 5-23,

TY = (/2) Yc3

• From Eqn 5-13,

d = (dx/)


5. Torsion


B. Elastic-plastic torque• Used when material starts yielding, and

the yield boundary moves inward toward the shaft’s centre, producing an elastic core.

• Also, outer portion of shaft forms a plastic annulus or ring

• General formula for elastic-plastic material behavior,

T = (Y /6) (4c3 Y3)


5. Torsion


B. Elastic-plastic torquePlastic torque• Further increases in T will shrink the radius of

elastic core till all the material has yielded• Thus, largest possible plastic torque is

TP = (2/3)Y c3

• Comparing with maximum elastic torque,

TP = 4TY / 3

• Angle of twist cannot be uniquely defined.


5. Torsion


C. Ultimate torque• Magnitude of Tu can be determined “graphically”

by integrating Eqn 5-23


5. Torsion


C. Ultimate torque• Segment shaft into finite

number of rings• Area of ring is multiplied

by shear stress to obtain force

• Determine torque with the product of the force and

• Addition of all torques for entire x-section results in the ultimate torque,

Tu ≈ 2 2


5. Torsion


IMPORTANT• Shear-strain distribution over radial line on shaft

based on geometric considerations and is always remain linear

• Shear-stress distribution must be determined from material behavior or shear stress-strain diagram

• Once shear-stress distribution established, the torque about the axis is equivalent to resultant torque acting on x-section

• Perfectly plastic behavior assumes shear-stress distribution is constant and the torque is called plastic torque


5. Torsion

EXAMPLE 5.19

Tubular shaft made of aluminum alloy with elastic - diagram as shown. Determine (a) maximum torque that can be applied without causing material to yield, (b) maximum torque or plastic torque that can be applied to the shaft. What should the minimum shear strain at outer radius be in order to develop a plastic torque?


5. Torsion

EXAMPLE 5.19 (SOLN)

Maximum elastic torqueShear stress at outer fiber to be 20 MPa. Using torsion formula

Y = (TY c/J); TY = 3.42 kN·m

Values at tube’s inner wall are obtained by proportion.


5. Torsion

EXAMPLE 5.19 (SOLN)

Plastic torqueShear-stress distribution shown below. Applying = Y into Eqn 5-23:

TP = ... = 4.10 kN·m

For this tube, TP represents a 20% increase in torque capacity compared to elastic torque TY.


5. Torsion

EXAMPLE 5.19 (SOLN)

Outer radius shear strainTube becomes fully plastic when shear strain at inner wall becomes 0.286(10-3) rad. Since shear strain remains linear over x-section, plastic strain at outer fibers determined by proportion:

o = ... = 0.477(10-3) rad


5. Torsion

*5.10 RESIDUAL STRESS

• Residual stress distribution is calculated using principles of superposition and elastic recovery


5. Torsion

EXAMPLE 5.21

Tube made from brass alloy with length of 1.5 m and x-sectional area shown. Material has elastic-plastic - diagram shown. G = 42 GPa.

Determine plastic torque TP. What are the residual-shear-stress distribution and permanent twist of the tube that remain if TP is removed just after tube becomes fully plastic?


5. Torsion

EXAMPLE 5.21 (SOLN)

Plastic torqueApplying Eqn 5-23,

When tube is fully plastic, yielding started at inner radius, ci = 25 mm and Y = 0.002 rad, thus angle of twist for entire tube is

TP = ... = 19.24(106) N·mm

P = Y (L/ci) = ... = 0.120 rad


5. Torsion

EXAMPLE 5.21 (SOLN)

r = (TPco)/J = ... = 104.52 MPa

Plastic torque

Then TP is removed, then “fictitious” linear shear-stress distribution in figure (c) must be superimposed on figure (b). Thus, maximum shear stress or modulus of rupture computed from torsion formula,

i = (104.52 MPa)(25 mm/50 mm) = 52.26 MPa


5. Torsion

EXAMPLE 5.21 (SOLN)

’P = (TP L)/(JG) = ... = 0.0747 rad

Plastic torque

Angle of twist ’P upon removal of TP is

+ = 0.120 0.0747 = 0.0453 rad

Residual-shear-stress distribution is shown.

Permanent rotation of tube after TP is removed,


5. Torsion

CHAPTER REVIEW

• Torque causes a shaft with circular x-section to twist, such that shear strain in shaft is proportional to its radial distance from its centre

• Provided that material is homogeneous and Hooke’s law applies, shear stress determined from torsion formula, = (Tc)/J

• Design of shaft requires finding the geometric parameter, (J/C) = (T/allow)

• Power generated by rotating shaft is reported, from which torque is derived; P = T


5. Torsion

CHAPTER REVIEW

• Angle of twist of circular shaft determined from

• If torque and JG are constant, then

• For application, use a sign convention for internal torque and be sure material does not yield, but remains linear elastic

=TL

JG

=T(x) dx

JG∫0

L


5. Torsion

• If shaft is statically indeterminate, reactive torques determined from equilibrium, compatibility of twist, and torque-twist relationships, such as = TL/JG

• Solid noncircular shafts tend to warp out of plane when subjected to torque. Formulas are available to determine elastic shear stress and twist for these cases

• Shear stress in tubes determined by considering shear flow. Assumes that shear stress across each thickness of tube is constant

CHAPTER REVIEW


5. Torsion

CHAPTER REVIEW

• Shear stress in tubes determined from = T/2tAm

• Stress concentrations occur in shafts when x-section suddenly changes. Maximum shear stress determined using stress concentration factor, K (found by experiment and represented in graphical form). max = K(Tc/J)

• If applied torque causes material to exceed elastic limit, then stress distribution is not proportional to radial distance from centerline of shaft


5. Torsion

CHAPTER REVIEW

• Instead, such applied torque is related to stress distribution using the shear-stress-shear-strain diagram and equilibrium

• If a shaft is subjected to plastic torque, and then released, it will cause material to respond elastically, causing residual shear stress to be developed in the shaft

Mechanics Torsion Presentation

Documents

Transcript of Mechanics Torsion Presentation