2014 8th wl torsion f

Prof. Younane N. Abousleiman Email: [email protected] Office: SEC, Suite P119

Torsion

Dr. A. Younane N. Abousleiman ‘s lecture notes

iPMI, www.pmi.ou.edu

The moment of a force about a point or an axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis.

oz ⊥ plane xoy in which Fx lies; Fx causes the pipe to turn about the z-axis ⇒Fx causes a moment about the z-axis (Mo)z


Fy passes through O ⇒ Fy does not cause the pipe to turn because the line of action of the force passes through O.


ox ⊥ plane yoz in which Fz lies; Fz causes the pipe to turn about the x-axis ⇒Fz causes a moment about the x-axis (Mo)x

Drilling a well

Torsion resistance to balance TB

Torque transmission

Shallow drilling

T

F

F

Torque is a moment that causes twisting along the length of a bar.

T

Applications

T

Shear Stress Due To Torsion

The twist is the torsional deformation. For a circular shaft, the torque (or torsional moment) rotates each cross-section relative to the nearby ones.

The shear stress varies linearly along each radial line of the cross section

JTρτ =

Shear stress

J is the polar moment of inertia of the cross- sectional area, geometric property

T

JTc

=maxτ

ρ is the radial distance τ is the shear stress at a radial distance ρ τmax is the maximum shear stress in the shaft, which occurs at the outer surface T is the resultant internal torque acting at the cross section. Its value is determined from the method of sections and the equation of moment equilibrium applied about the shaft’s longitudinal axis J is the polar moment of inertia of the cross- sectional area c is the outer radius of the shaft

Shear stress

JTρτ =

JTc

=maxτ

T, reaction torque at fixed head

Φ L

Φ(x)

x

γ⋅∆= xBD dxdφργ =

Note: dΦ/dx is a function of distance x and is a constant for fixed x

In OBD:

O

φρ∆=BD

In ABD: γφρ ⋅∆=∆ x

(4)

τxy

z

x

y

The shear stress τxy (or σxy) acts on the surface normal to x (the first subscript) axis, pointing the y (the second subscript) direction

A A

xyτ

yxτ

yxτ

xyτ

At section A-A

Also a dimensionless quantity θπγ −=2

strainShear xy

γxy xyτ

y

x

xyτ

yxτ

yxτ

Pure shear: τxy = τyx

θ0 = π/2 θ = π/2- γxy

xyτxyτ

yxτ

yxτ

θ0 θ

max)( γρφργcdx

d==

o

(1), (2)

T

τ τ γ

Shear strain by definition θπγ ′−=2

θ'

τyx τ = τxy y

x

z

Element A A

C

o

(1), (2)

The shear strain within the cross section varies linearly along the radial line, from zero to maximum γmax.

γmax

γ

Derivation of shear stress τ = Tρ/J

maxγργc

=

T Element A

Element C

γ = 0 at center

dxdφργ =

dxdc φγ =max


max)( γρφργcdx

d==

o

(1), (2)

T

τ τ γ

τyx

Hooke’s law:

maxmax )()( τργργτcc

GG ===Again the shear stress varies linearly along the radial line

τ = τxy y

x

z Element A

From torque equilibrium equation

∫

∫∫

=

==

A

AA

dAc

dAc

dAT

2max

max ])[()(

ρτ

τρρτρ

J = polar moment inertia

Jc

T maxτ=

JTc

=maxτ


maxτρτc

=

JTρτ =

T

Shear stress

JTρτ =

JTc

=maxτ

ρ is the radial distance τ is the shear stress at a radial distance ρ τmax is the maximum shear stress in the shaft, which occurs at the outer surface T is the resultant internal torque acting at the cross section. Its value is determined from the method of sections and the equation of moment equilibrium applied about the shaft’s longitudinal axis J is the polar moment of inertia of the cross- sectional area c is the outer radius of the shaft

Polar moment of inertia J

4

0

3

0

22

22

)2(

cd

ddAJ

c

c

A

πρρπ

ρπρρρ

==

==

∫

∫∫

Solid shaft

ci

co

ρ

)(2

2

)2(

443

2

io

c

c

c

c

ccd

dJ

o

i

o

i

−==

=

∫

∫

πρρπ

ρπρρ

Tubular shaft

Shaft Design

τmax is maximum shear stress [N/m2 or psi] τallow is τfail/F.S. τallow is allowable shear stress [N/m2 or psi] τfail is failure shear stress [N/m2 or psi] F.S. is factor of safety (F.S. > 1) J is polar moment of inertia [m4 or in4] c is outer radius of the shaft [m or in]

allow

TcJ

τ≥

Shaft design:

Design criteria: the maximum shear stress τmax should be less than the allowable stress τallow

allowJTc ττ ≤=max

τmax τmax

τmax < τallow

τ

γ

τallow

τfail

τallow = τfail/F.S. τpl

T T

Usually, geometric parameters need to be designed!

allow

TcJ

τ≥

Shaft design:

Design criteria: the maximum shear stress τmax should be less than the allowable stress τallow

allowJTc ττ ≤=max

τmax τmax

τmax < τallow

τ

γ

τallow

τfail

τallow = τfail/F.S. τpl

T T

T dt

TdP θ= fTTP πω 2==

Tdθ is work [N·m or lb·ft] ω = dθ/dt is the shaft’s angular velocity [rad/s] f (ω = 2πf) is frequency measuring the cycles the shaft rotates per second [Hz (1 Hz = 1 cycle/s)] P is power [W (1 Watts = 1 N·m/s) or hp (1 horsepower = 550 lb·ft/s)]

or

The power transmitted from a machine through shafts (tubes) is defined as the work performed per unit time:

Motor M provides an output power of P, developing a torque T in the shaft AB

Motor: output power P

f (ω = 2πf) is frequency [Hz (1 Hz = 1 cycle/s)] P is power [W (1 Watts = 1 N·m/s) or hp (1 horsepower = 550 lb·ft/s)] J is πc4/2 for solid shaft and π(co

4-ci4)/2 for

tubular shaft [m4 or in4]

allowfP

cJ

τπ2≥

Shaft design:

If the power P and rotation frequency f, instead of torque T, are given:

allowfJPc τπ

≤2

τmax τmax

τmax < τallow

ci

co c

Solid Tubular

)P/(2T fπ=

T T

Example 5.1

The motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for material is τallow = 25 ksi.

torque T delivered from motor

Motor: output power P

T

Reaction torque from gear G on AB, equal to T

T

Gear G

Example 5.1 Basic parameters: P = 500 hp co = 2 in/2 = 1 in ci = 1.84 in/2 = 0.92 in τallow = 25 ksi

T

T

Torque delivered from motor

Reaction torque

FBD of shaft AB

A

B

ci

co Need to calculate: the smallest angular velocity: ωmin

ffTTP

πωπω

22

===

f is frequency [Hz (1 Hz = 1 cycle/s)] P is power [W (1 Watts = 1 N·m/s) or hp (1 horsepower = 550 lb·ft/s)] J is πc4/2 for solid shaft and π(co

4-ci4)/2 for

tubular shaft [m4 or in4]

Shaft design:

If the power P and rotation frequency f instead of torque T are given:

allowfJPc τπ

≤2

τmax τmax

τmax < τallow

c

Solid Tubular

allowfP

cJ

τπ2≥

ci

co

T T

Solution 5.1

allowo fP

cJ

τπ2≥

The geometric parameters needed to satisfy the design criteria:

)psi100025(2in/ft12lb/s)/hpft(550hp500

in1in445.0 4

××⋅×

≥fπ

444444 in445.0in)92.01)(2/())(2/( =−=−= ππ io ccJci

co

Solution 5.1

rad/s2962 ≥fπ

rpm2830/radrevolution)2/1(s/min60rad/s296

rad/s296)2( minmin

=××=

==π

πω f

So the smallest angular velocity:

)psi100025(2in/ft12lb/s)/hpft(550hp500

in1in445.0 4

××⋅×

≥fπ

Angle of Twist

The angle of twist Φ(x) at location x is proportional to x, at x = L, Φ = TL/JG


JGTL

=φ

Angle of Twist

Φ L

Φ(x)

x


Φ L

Φ(x)

x

ρτ

ργφ dx

Gdxd ==

JTρτ =

JGTLdx

JGTdx

JGTd

L

=== ∫0

φφ

dxdφργ =

Derivation of Φ

T, reaction torqueat fixed head

ΦL

JGTL

=φ

The angle of twist for constant J, G, and T:

Φ(x)

The angle of twist Φ(x) at location x is proportional to x, at x = L, Φ = TL/JG

Angle of Twist

Φ is the angle of twist between two ends [rad] T is the applied torque [N·m or lb·ft] J is the shaft’s polar moment of inertia [m4 or in4] G is the shear modulus [Pa (1 Pa = 1 N/m2) or psi (1 psi = 6895 Pa]

x

∑= JGTLφ

If J, G, and T are constants for each segment: Angle of Twist

T(x) is the internal torque at arbitrary position x [N·m or lb·ft] J(x) is the shaft’s polar moment of inertia at x [m4 or in4]

Constant J, G, T for each segment

J(x), T(x) varying with x

In general case both J and T are functions of x:

∫=L

dxGxJ

xT

0 )()(φ

x

Φ

Φ

x

dx

γ

dΦ

ρτ

ργφ dx

Gdxd ==

)()(xJ

xT ρτ =∫==L

dxGxJ

xTdxGxJ

xTd0 )(

)()()( φφ

ρdxγdφdxγρdφ =⇒⋅=⋅

Derivation of Φ

∫=L

dxGxJ

xT

0 )()(φ

Φ is the angle of twist between two ends T(x) is the internal torque at the arbitrary position x, found from the method of sections and the equation of moment equilibrium applied about the shaft’s axis J(x) is the shaft’s polar moment of inertia, a function of position x G is the shear modulus

The angle of twist

Similarity between rotation Φ under torsion and elongation δ under axial load:

EAFL

=δ

∫=L

dxGxJ

xT

0 )()(φ ∫=

L

dxExA

xF

0 )()(δ

JGTL

=φ

T → F G → E J → A

(Torsion) (Axial load)

OR: ∑= JGTLφ ∑= EA

FLδ

The A-36 steel bar (Est = 200 Gpa, cross-sectional area 300 mm2) is subjected to loads as shown in the figure. Determine the displacement of end D relative to end A.

0.25 m 0.25 m 0.5 m

10 kN 20 kN

50 kN

FBD: FA

FA+10-20-50=0

FA=60 kN

Calculation of FA:

0=→+ ∑ xF

10 kN 20 kN 50 kN

60 kN PAB

PAB= 60 kN

Calculation of PAB:

60-PAB= 0

FBD:

FBD:

60+10-PBC=0

PBC=70 kN

60 kN PBC

Calculation of PBC:

10 kN

0=→+ ∑ xF

0=→+ ∑ xF

PCD-50=0

PCD

Calculation of PCD:

PCD=50 kN

D

50 kN

0=→+ ∑ xF

P (kN)

x A B

-60

-70

C

-50

D

60 kN

Tension: positive; Compression: negative.

10 kN 20 kN 50 kN

The displacement of end D relative to end A.

]m )0GPa][300(1 [200m) kN)(0.5 50

]m )0GPa][300(1 [200m) kN)(0.25 70

]m )0GPa][300(1 [200m) kN)(0.25 60

262626 −−−−

+−

+−

==∑ (((/ EA

FLADδ

mm 9.0/ −=ADδ

Negative sign means that end D moves towards end A.

P (kN)

x A B

-60

-70

C

-50

D

Right-hand rule: The Torque and angle will be positive, if the thumb is directed outward from the shaft when the fingers curl to give the tendency for rotation.

FBDs:

Torque Diagram:

Torque Diagram:

The angle of twist of the end A with respect to the end D:

Example 5.2

The 20-mm-diameter A-36 steel shaft is subjected to the torque shown. Determine the angle of twist of the end B.

Solution 5.2

Basic parameters: c = 20 mm/2 = 10 mm G = 75 Gpa LBC = 800 mm LCD = 600 mm LDA = 200 mm

In order to determine the angle of twist of end B, ΦB/A, we first need to determine the internal torque along the shaft BA

80 N·m

B

TBC = -80 N·m

80 N·m

B

TCD = -60 N·m

C

20 N·m

80 N·m

B

TDA = -90 N·m

C

20 N·m

D

30 N·m

C D

A

-80

T (Nm)

-60 -90

B C D A

Torque diagram

The angle of twist of end B:

48

44

4

m1057.1m)01.0)(2/(

)2/(

−×=

=

=

π

π cJc

{ }

°−=−=××

⋅−+⋅−+⋅−=

++=++=

−

74.5rad1.0)N/m1075)(m1057.1(

m)2.0m)(N90(m)6.0m)(N60(m)8.0m)(N80(2948

//// JGLT

JGLT

JGLT DADACDCDBCBC

ADDCCBAB φφφφ

-80

T (N·m)

-60 -90

B C D A

Torque diagram

Solution 5.2 x

Solution 5.2

{ }

°−=−=××

⋅−+⋅−+⋅−=

++=++=

−

74.5rad1.0)N/m1075)(m1057.1(

m)2.0m)(N90(m)6.0m)(N60(m)8.0m)(N80(2948

//// JGLT

JGLT

JGLT DADACDCDBCBC

ADDCCBAB φφφφ

300 Nm 400 Nm

500 Nm

600 Nm

400 mm

600 mm

300 mm

300 Nm 400 Nm

500 Nm

400 mm

600 mm

300 mm E600 Nm

Figure (a) Figure (b)

The solid 20-mm-diameter A-36 steel (G = 75 GPa) shaft is used to transmit the torques applied to the gears, as shown in FIGURE a.

(a) Draw the free body diagrams and determine the three internal torques in sections between A-C, C-D, and D-B;

(b) Draw the torque diagram along the shaft AB;

(c) Calculate the absolute maximum shear stress on the shaft;

(d) Calculate the twist angle of end A relative to C;

(e) Calculate the twist angle of end A to B;

(f) If the external torque of 600 Nm originally applied at the end B is moved to point E (in the middle of BD, i.e., DE = EB = 300 mm, see FIGURE b), redraw the torque diagram along the shaft AB and determine the twist angle of end B relative to D.

600 Nm 300 Nm

500 N

TAC = -300 Nm

600 Nm 300 Nm

TB = ?

500 mm

300 mm

TCD = -300+400 = 100 Nm

TCD

600 Nm 300 Nm

TB = ?

500 mm

600 mm

300 mm TDB

TDB = -300+400+500 = 600 Nm

500 Nm

-300

T (Nm)

100

600

A C D B

Torque diagram

300 Nm 300 Nm 400 Nm

300 Nm 400 Nm

(a)

(b)

Maximum shear stress on the shaft:

48

44

4

m1057.1m)01.0)(2/(

)2/(

−×=

=

=

π

π cJc

MPaJ

cTJ

T

382)m1057.1(m01.0mN600

48maxmaxmax

max

=×

×⋅=== −

ρτ

-300

T (Nm)

100

600

A C D B

Torque diagram

(c)

c

°−=−=××

⋅−== − 4.4rad0764.0

)N/m1075)(m1057.1(m)3.0m)(N300(

2948/ JGLT ACAC

CAφ

48

44

4

m1057.1m)01.0)(2/(

)2/(

−×=

=

=

π

π cJ

-300

T (Nm)

100

600

A C D B

Torque diagram

(d)

{ }°==

××++−

=

++=++=

− 1.15rad263.0)1075)(1057.1(

)6.0)(600()4.0)(100()3.0)(300(98

//// JGLT

JGLT

JGLT DBDBCDCDACAC

BDDCCABA φφφφ(e)

600 Nm 300 Nm

TB = ?

500 mm

600 mm

300 mm TDE

TDE = -300+400+500 = 600 Nm

500 Nm

-300

T (Nm)

100

600

A C D B

Torque diagram

300 Nm 400 Nm

(f)

300 Nm 400 Nm

500 Nm

400 mm

600 mm

300 mm E600 Nm

E 0

TEB

TEB = -300+400+500-600 = 0

c

{ }°==

××⋅+

=

+=+=

− 75.8rad152.0)N/m1075)(m1057.1(

m)3.0m)(N600(02948

/// JGLT

JGLT EDEDBEBE

DEEBDB φφφ

48

44

4

m1057.1m)01.0)(2/(

)2/(

−×=

=

=

π

π cJ

-300

T (Nm)

100

600

A C D B

Torque diagram

E 0

(f)

2014 8th wl torsion f

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