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TORSION OF

NON-CIRCULAR SHAFTS

INTRODUCTION

Generally for power transmission, circular shafts are used because there is

uniform stress distribution along any radius of the shaft. Plane sections of the

shaft remain plane after the application of twisting moment, as a result there is

no distortion in the sections of shafts and change in volume of the shaft is zero.

For stationary torque application, non-circular shafts of different sections such

as square, rectangular, triangular, elliptical solid, or hollow are used.

Figure 1 Formation of ridges and valleys

The assumption that transverse sections of the shaft remain plane after the

applications of torque does not hold good for shafts of non-circular section. Only

the lines of symmetry remain straight; all other lines in the section go out of

plane and the section gets severely distorted. Figure 1 shows the undeformed

shaft of square section and deformed shape after the application of a twisting

moment along the axis of the shaft.

Figure 2 shows the formation of ridges and valleys in the square section of a

shaft. Shaft of rectangular section is subjected to axial torque T.

Figure 2 (a) Change of shape of cross-section (b) Warping

A rectangular shaft of section

2b × 2a, fixed at one end, is

subjected to axial torque T as

shown in Figure 3 which

produces shear stresses in the

shaft which are zero at centre

and maximum at the outer

surface, but varies from one

radial line to another radial line.

Figure 3 Shaft of rectangular section subjected to axial torque T

Figure 4 shows that the original plane cross-section has deformed or worked

out of its own plane. In general, torsion of a shaft which does not possess

circular symmetry produces deformation that involves rigid body rotation of

one cross-section with respect to another cross-section accompanied by

warping out of the original plane. Figure 4 shows four dotted lines AC, BD,

ac, bd, lines of symmetry remain straight, but all other lines in the cross-

section deform when the shaft is twisted.

Figure 4 Distortion in rectangular section

All non-circular sections are distorted under torsion to a greater or lesser

degree. For sections close to circle, these effects are less marked as in the

case of elliptical section.

The detailed analysis on torsion of non-circular shafts including the warping of

the sections requires a complicated analysis. However, the results of the

theory developed by St. Venant and Prandtl for the calculation of maximum

shear stress and angular twist in non-circular shafts will be summarized here.

section and 2a is the shorter side of the section

as shown in Figure 5.

RECTANGULAR SECTION Torque T = GJθ [Similar to M=EI(d2y/dx2)]

where GJ = Torsional rigidity of the shaft,

θ = angular twist per unit length, and

J = Polar moment of inertia: Ka3b

Angular twist,

The value of constant K depends upon the ratio of

, where 2b is the longer side of the rectangular

Figure 5 Rectangular section

r

L T

lengthunitLL

GJ

T

L

r

GJ

rT

GJ

rTG

rL

,

,

ab

bGKa

T3

The values of K for various ratios of are given in tables, such as

presented here.

Table Values of constants for rectangular section

Maximum shear stress, [for circular sections ]

where K1 is another constant again depending upon the ratio of .

J

Tr

JTa

K1

ab

ab

or

where

The constants K, K1, and K2 are presented in the Table.

Expressions for and can be approximately given as follows, and

then one does not need to refer to the table of constants.

where

For the sake of comparison let us take =1.5.

From tables K = 3.136, K1 = 1.696, K2 = 0.541

Maximum shear stress,

Angular twist per unit length

From approximate analysis Maximum shear stress,

ab


If we compare the results of maximum shear stress and angular twist, from

two analyses, we can find only negligible difference between the two analysis.

The maximum intensity of shear stress, , occurs at the centre of the longer side

for rectangular cross section as shown in Figure 6. Figure 7 shows the distortion of

the ends of a shaft of square section.

Figure 6 Shear stress distribution in shaft of rectangular section

Figure 7 warping of a square section

Example

A 50 mm × 25 mm rectangular steel shaft is subjected to a torque of 1.0 kNm.

What is the maximum shear stress developed in the shaft and what is the

angular twist per unit length? G = 80 GN/m2.

G = 80 GN/m2 = 80 × 109 N/m2 = 80 × 103 N/mm2

Longer side, 2b = 50 mm, (b = 25 mm)

Shorter side, 2a = 25 mm, (a = 12.5 mm)

Torque, T = 1.0 kNm = 1.0 × 106 Nmm


From tables for ba

TK

22

Angular twist per unit length,

θ

=

= 0.06987 × 10−3 radian / mm length

= 0.06988 radian / meter length

= 4.0 ° / meter length

Example

A rectangular shaft 6 cm × 4 cm made of steel is subjected to a

torque of 3000 Nm. What is the maximum shear stress developed in

the shaft and what is the angular twist per metre length? G = 80

kN/mm2. Use approximate relationship.

Torque, T = 3 kNm = 3 × 106 Nmm

Longer side, 2b = 6 cm, b = 3 cm = 30 mm

Shorter side, 2a = 4 cm, a = 2 cm = 20 mm

G = 80 kN/mm2


Angular twist per mm length

where

Exercise 6.1

A rectangular shaft of section 60 mm × 24 mm made of steel is subjected

to a torque such that maximum shear stress developed in shaft is 40

N/mm2. What is the magnitude of the torque and what will be the angular

twist in 1 m length of the shaft? G = 80 kN/mm2.

Take 1) values of constants from the table,

2) approximate values.

Compare the two results.

TORSION OF ELLIPTICAL SECTION SHAFT For the elliptical section shaft, the expressions for maximum shear stress and angular twist per unit length are

and

where b = semi-major axis and

a = semi-minor axis of ellipse.

Maximum shear stress occurs at the

ends of the minor axis as shown in

Figure 8, i.e. at the points B and D. Figure 8 Maximum shear stress occurs

at the ends of the minor axis

Figure 9 shows the contour lines of constant displacement. The convex

portions of the cross-section, where displacements in the direction of axis

of the shaft are positive, are shown by continuous lines. Where the surface

is depressed, depressions are shown by dotted lines.

Figure 9 Contour lines of displacement

Example A shaft of elliptical section with minor axis 2a and major axis 2b is subjected to a torque of 2 kNm. If the maximum shear stress in the shaft is not to exceed 80 N/mm2, determine the major and minor axes, if b = 1.5a. What will be the angular twist in a meter length in this shaft under the given torque? G = 80000 N/mm2. Maximum shear stress,

or

a3

=

Minor axis = 44 mm

Major axis = 66 mm

Angular twist per mm length

θ per meter length = 3.27 × 10–2 = 1.87°

Exercise 6.2

A shaft of elliptical section with major axis 60 mm and minor axis

40 mm is subjected to an axial twisting moment of 0.5 kNm.

What is the maximum shear stress developed in the section and

what is the angular twist per meter length?

G = 40 kN/mm2.

TORSION OF A SHAFT WITH EQUILATERAL TRIANGULAR SECTION

Figure 10 shows an equilateral triangle section of a shaft subjected to the

twisting moment T. Let’s assume a is the side of the equilateral triangle.

Maximum shear stress occurs at the center of the sides.



At the corners of the triangle,

i.e. at A, B, and C, shear

stress is zero. Figure 10 Shear stress distribution in equilateral triangular section

Example

A shaft of equilateral triangular section of side 40 mm is subjected to an axial

twisting moment T. Determine the magnitude of T if the maximum shear stress

is not to exceed 100 N/mm2. What will be the angular twist in 2 metres length

of the shaft?

G = 80000 N/mm2.

= 100 N/mm2

Side, a = 40 mm

Angular twist per meter length,

θ

=

= 0.0625 × 10−3 radians/mm length

= 0.0625 radians/m length

Angular twist in

2 meters length

= 0.0625 × 2

= 0.125 radians

= 7.162°

Exercise 6.3

A shaft of equilateral triangular section of side 60 mm is subjected to a

torque of 1.0 kNm.

Determine (i) maximum shear stress developed in shaft and (ii) angular

twist per meter length of the shaft.

G = 84 kN/mm2.

PRANDTL’S STRESS FUNCTION FOR TORSION

The stress function approach is also useful when dealing with torsion on a

prismatic element with a noncircular cross section. Airy’s stress function cannot

be used here since the general torsion problem does not fall into the category

of a plane elastic problem. Thus, for pure torsion, the equations of equilibrium

and compatibility must be reformulated.

Let’s consider the general cross-

section undergoing torsion in

Figure 11. A surface isolation

perpendicular to the rod axis is

also shown, where it can be seen

that the shear stresses xy and zx

act over a dydz element. Cross

shears xy and yx and xz and zx

are equal.

Figure 11 Torsion of a prismatic element

For equilibrium in the x direction, it will be assumed that sx is negligible. We

should note, however, that when torsion is applied to noncircular cross sections

without constraint in the x-direction, plane surfaces perpendicular to the

longitudinal axis will not remain plane and the surfaces warp. If the rod is

constrained from warping in the x direction, sx will develop. We are going to

assume that the rod is free in the x direction. Thus, for equilibrium of forces in

the x direction, neglecting the body force and the normal stress sx,

0

0

xzxxyx

zxxy

Fzyx

zy

s

xF

If a stress function (y, z), called the Prandtl’s stress function, exists such

that

y

z

zx

xy

The equilibrium equation is automatically

satisfied.

When deflections in terms of the angle of twist

are considered, the deflection of a point on the

isolated surface is as shown in Figure 12, from

which it can be seen that for small

Figure 12 Isolated surface

zvyw

022

yzzy

w

y

v

z

0

zy

zxxy or

Assuming that the rod is fixed at the origin of

xyz, we can represent the angle of twist per unit

length ', where =‘ x, and

xzv

xyw

For noncircular cross sections, the surface warps,

and in general the point will also deflect in the x

directon u. However, this is independent of x and

u = u (y, z). The strains xy and zx are

z

uy

z

u

x

w

y

uz

y

u

x

v

zx

xy

Since = [E/2(1+n)],

z

uy

E

y

uz

E

zx

xy

12

12

Writing the expression, using the above expressions, zy

xyzx

1

E

zy

xyzx

yz

uEE

z

zy

uEE

y

xy

zx

2

2

1212

1212

Substitution of results in

This equation is called the Poisson’s equation, it is the governing

equation for the torsional stress function .

At the boundary, the net shear stress must be tangent to the boundary.

Thus,

0 dzdyordz

dyxyzx

zx

xy

yzzxxy

1

2

2

2

2 E

zy

1

E

zzyy(x -1)

dy

dz

xy

zx

net

Substituting in results in

However, since = (y, z), the above equation can be written

yzzxxy

0

dz

zdy

y

0d

Since this requirement applies to the boundary, is constant along the

boundary of the cross section. The value of this constant is arbitrary

and is normally chosen to be zero. If the boundary of the cross section

is a well-behaved function of y and z such as a circle, ellipse, etc., the

equation of the boundary becomes an excellent stress function.

To relate the stress function to the transmitted torque T, as seen in the

figure, the net torque about the x axis due to the stresses on the dydz

element is (yzx – zxy)dydz. Thus, the total torque is

dydzzyT xyzx

Substitution of

gives

yzzxxy

dydzz

zy

yT

z

y

Let’s consider the first term in the integral. Integrating this with respect to

y by parts gives

dzdyydzdy

yydydz

yy

y

y

2

1

where y1 and y2 are boundary points for the dz slice. However, as stated

earlier, is zero at the boundary. Thus is zero at y1 and y2, and the first

term within the integral disappears, resulting in

dzdydzdy

yy

In a similar manner, the second integral gives identical results, so that we

have

dzdyT 2

Example

A solid circular shaft of radius ro is transmitting a torque. Determine the

corresponding shear stress distribution.

The equation for the boundary of a circle of radius ro in the xz plane is

y2 + z2 = ro2. Let’s try the following stress functon:

222

orzyk

where k is a constant. = 0 along the entire

boundary. To establish the value of k we can use

. It can be seen from the above

Prandtl’s stress function that polar coordinates

are more suitable to the problem. Let r2 =y2 + z2,

where r is a variable radial position.

dzdyT 2

y

ro

z

z

y

The infinitesimal area dydz can be replaced by 2prdr since at any given

position r, the stress function is constant. Thus, for this problem, the double

integral of reduces to a single integral, and dzdyT 2

r

dr rdrrrkTor

o 40

22

p

Integrating and solving for k results in

4 or

Tk

p

The polar moment of inertia of a circular cross section is J = (p/2)ro4. Thus,

J

Tk

2

Substituting k into the stress function, we get

222

2orzy

J

T

The shear stresses are determined as

J

Ty

J

Tzzxxy

We should note that at any given point, the net shear stress is given by

22

)( zxxynetx

J

Tryz

J

Tnetx 22

)(

Therefore,

which is identical to that used in

elementary strength of materials.

For the angle of twist, substitution of the stress function

into yields

EJ

T

E

J

T

J

T

12

1

If the total length of the bar is L, the angle of twist across the entire

length is , and thus,

1

2

2

2

2 E

zy

222

2orzy

J

T

L

12 12

GE

EJ

TL

which again agrees with the elementary strength of materials solution.

MEMBRANE ANALOGY

A German Scientist Prandtl, analyzed that the solution of a partial differential

equation that must be solved in the elastic torsion problem is mathematically

identical to that for a thin membrane, such as a thin rubber sheet stretched

over a hole, and the hole is geometrically similar to the cross-section of the

shaft under study. On one side of the thin sheet, or membrane, there is a light

air pressure. Following rules are followed for the analogous solution:

1.Shear stress at any point of the section (non-circular section of shaft) is

proportional to the slope of the stretched membrane at that point.

2.The direction of a particular shear stress at a point is at right angles to the

slope of the membrane at the same point.

3.Torque on the section is proportional to twice the volume enclosed by the

stretched membrane.

Figure 13 (a) shows that the area

bound by the edges of a thin rubber

sheet- membrane is of the shape of a

shaft of non-circular section. The

stretched sheet or membrane is

subjected to an internal pressure and

the membrane is deflected as shown in

Figure 13 (b). It should be noted that

the initial tension in membrane should

be large enough, so that when

membrane is blown up due to internal

air pressure, changes in tension can be

ignored. With the help of a travelling

microscope, deflection at grid points on

membrane can be noted down.

Figure 13 Deflection contour lines of membrane

(a)

(a)

And deflection contours (lines of

constant deflection) are plotted as

shown in Figure 13 (c). From deflection

contours, slope at any point of the

section can be determined, as shown by

a tangent at any point P of non-circular

section. Knowing the results of a

circular section shaft for a given torque,

the membrane is calibrated for a

circular section under a given

pretension σ, internal pressure p, and

membrane thickness.

Figure 13 Deflection contour lines of membrane

Membrane covering a curved cross-section

x

zy

yzx

yz

yz

xz

xzyz

yz

(from page 57)

yz

yz

yz

yz

T

T

T

T T

T

T

T T

T T

T T

T

TORSION OF

THIN WALLED

SECTIONS

Consider a shaft with thin walled

tubular section subjected to a

twisting moment T. The

thickness of the tube can vary,

and we have considered the

section of variable thickness as

shown in Figure 14 (a). At any

point along the periphery say t is

the thickness and is the shear

stress. Take a small element

abcd on the periphery of tubular

section, with thickness t1, at cd

and t2 at ab. Since there is

variation in thickness, there will

be variable shear stress. Figure 14 (a) Variable thin walled section (b) Sections of variable thickness

At cd, thickness is t1, and shear stress is 1.

At ab, thickness is t2, and shear stress is 2.

Axial length of element is dl. An enlarged view of the element is shown

in Figure 14 (b).

Complementary shear stress on face dd′c′c —1

Complementary shear strain on face a′b′ba —2

For equilibrium of forces

t1 dl 1 − τ2dl 2 = 0

or t1 1 = t2 2 = t q (for any thickness t)

Figure 15 Element of variable thin walled section

Let us consider a small element of length ds along the periphery, say thickness is t

Shear force acting on the small element, dQ = tds =qds

= shear flow × length. (Figure 15)

Moment of the force dQ at the centre O of the shaft

dT = hdQ = hq ds where q is the shear flow q = . t.

where dT = q 2 dA (where dA is the area of small triangle)

Total torque,

ds − base of triangle (shaded area)

h = altitude of the triangle.

dT = q (hds)

where A is the mean area enclosed by the centre line of the thin tubular

section as shown for a rectangular section in Figure 16.

Figure 16 Thin rectangular section

A = B × D, area bounded by outer

line of section. This equation T = 2q A is generally known as Bredt-Botha equation.

qAdAqT 22

Let us determine the angular twist in the shaft. Consider a small element abcd of

axial thickness dL. Due to the twisting moment point d is displaced to d″ and a is

displaced to a″.

Shear force dQ on small element, = τtds

Say displacement of the edge ab or cd = δ

Shear strain,

Strain energy for the small element =

dL

But shear strain,

Shear strain energy

but

So, shear strain energy,

Moreover,

Let us take dL = 1, so that we can find out strain energy per unit length

Total strain energy per unit length

modulusShear

stressShear

G

t

dsdL

GA

Tdu

2

2

8

tdsdL

2

1

Using the Castigliano’s theorem, angular twist θ per unit length

Again,

where integral

is the summation of (length/thickness) along the periphery of thin tubular section.

t

ds

GA

Tu

2

2

8

Example

A thin walled box section 3a × 2a × t is subjected to a twisting moment T. A

solid circular section of diameter d is also subjected to the same twisting

moment. Determine the thickness of the box section (a) if the maximum shear

stress developed in box section is the same as that in solid circular section, and

d = 2a, and (b) if the stiffness for both is the same under the same torque.

The figure shows the thin walled section 3a × 2a × t and a solid circular section of diameter d.

Maximum shear stress in circular section,

Shear flow in box section,

Taking a ≫ t (side ≫ thickness)

Maximum shear stress in box section,

But

or

Angular twist for solid circular shaft

or

where

but d = 2a

So,

Angular twist for the thin box section,

where area A = 3a × 2a = 6a2

and

But θ = θ′

So,

Exercise 6.4

A shaft of hollow square section of outer side 48 mm and inner side 40

mm is subjected to a twisting moment such that the maximum shear

stress developed is 50 N/mm2. What is the torque acting on the shaft and

what is the angular twist if the shaft is 1.6 m long?

G = 80000 N/mm2.

TORSION OF THIN RECTANGULAR SECTIONS

Figure 17 shows a thin rectangular section subjected to the torque T.

Thickness t of the section is small in comparison to its width b (b>>t).

This section consists of only one boundary. In this case maximum shear

stress occurs at .

Figure 17 Shear stress flow as in thin rectangular sectors

If

θ = angular twist per unit length

T = torque on the section

T =

2t

y

Gbt 3

31

Angle of twist per unit length,

= maximum shear stress

These results can be applied to sections built up of rectangular strips and

having only one boundary such as angle section, channel section, T section

and I section as shown in Figure 18.

Figure 18 Various cross sections

3

31

bt

TG

2

3

bt

T

In the case of channel section and I section, Torque,

Angle of twist per unit length,

In the case of Angle section and T section,

G = 8 × 104 N/mm2 = 8 × 106 N/cm2

Torque T = 2 kNm = 2 × 105 N cm

2 Flanges, b1 = b3 = 10 cm, t3 = 2 cm

1 Web b2 = 28 cm, t2 = 1 cm

Example An I section with flanges 10 cm × 2 cm (b1 = b3 x t3) and web

28 cm × 1 cm (b2 x t2) is subjected to a torque T= 2 kNm. Find the maximum

shear stress and angle of twist per unit length. G = 80,000 N/mm2.


Exercise 6.5

A T-section with flange 10 cm × 1 cm and web

19 cm × 0.8 cm is subjected to a torque of 200 Nm. Find the

maximum shear stress and angle of twist per meter length.

G = 82 kN/mm2

TORSION OF THIN WALLED MULTI-CELL SECTIONS

The analysis of thin walled closed sections can be extended to multi-cell

sections. Let’s consider a two cell section as shown in Figure 19. Say the

shear flow in cell 1 is q1, in cell 2 it is q2 and in the web, shear flow is q3.

Now consider the equilibrium of shear forces at the junction of the two

cells, taking a small length δl, along the axis of the multi-cell section. The

complementary shear stresses 1, 2, and 3 are shown in the longitudinal

sections of length δl each with thicknesses t1, t2, and t3, respectively.

Figure 19 Multi cell sections

δl

For the equilibrium along the direction of the axis of multi-cell tubular section:

1t1 δl − 2t2 δl − 3t3 δl = 0

or 1t1 = 2t2 + 3t3

or q1 = q2 + q3

shear flow, q1 = shear flow q2 + shear flow, q3.

This is equivalent to fluid flow dividing itself into two streams:

Shear flow in web, q3 = q1 – q2

Twisting moment T1 about O due to q1, flowing in cell 1.

T1 = 2 q1 A1

where A1 = area enclosed by the centre line of cell 1.

δl

Twisting moment T2 about O due to q2 in cell 2: T2 = 2 q2 (A2 + A′1) − 2 q2 A′1 where 2 q2 A′1 is the twisting moment due to shear flow q2 in the middle web.

Total twisting moment,

T = T1 + T2

= 2 q1 × A1 + 2 q2 × A2

For continuity, the angular twist per unit length in each cell will be the

same. For closed thin sections

But in this case shear flow is changing, therefore

Say,

for cell 1 including web

for cell 2 including web

for the web

For cell 1,

For cell 2,

Shear flow q1, q2, and angular twist θ can be worked out using the equations derived.

Example The figure shows the dimensions of a double walled cross-

section in the form of a rectangle and a triangle. A torque of 4 kNm is

applied. Calculate the shear stress in each part and the angle of twist per

meter length. G = 82 kN/mm2.

Say,

shear flow in rectangular cell= q1 and shear flow in triangular cell = q2

Area, A1 = 150 × 100

= 15 × 103 mm2

Area,

Line integrals,

Now torque,

T = 2q1 × A1 + 2q2 × A2

4000 × 103 Nmm = 2 × 15 × 103 × q1 + 2 × 7.5 × 103 × q2 (i)

2000 = 15q1 + 7.5q2

But θ in cell 1 = θ in cell 2

So,

But

Substituting in Eq. (ii), we get 130q1 − 60q2 = 320q2 − 120q1 or 250q1 = 380q2 or q1 = 1.52q2 (iii)

Substituting in Eq. (i), we get 15 × 1.52q2 + 7.5q2 = 2000 or shear flow

1, shear stress in rectangular part

2, shear stress in triangular part

3, shear stress in web

Exercise 6.6

A steel girder of the section is shown in the figure. It has a uniform thickness

of 12 mm throughout. What is the allowable torque if maximum shear stress

is not to exceed 30 MPa? What is the angular twist per meter length of the

girder? What is the shear stress in middle web of the section?

G = 82 kN/mm2.

Example

A shaft of rectangular cross section is subjected to a torque of 0.8 kNm

and the maximum permissible shear stress in the shaft is 40 N/mm2. If

the ratio of breadth to depth is 1.5, determine the size of the shaft and

the angle of twist in a length of 4 m.

G = 78.4 kN/mm2.

Torque T = 0.8 × 106 Nmm = 800 × 103 Nmm

Larger side = 1.5 × shorter side (a)

b = 1.5 a

Maximum shear stress = 40 N/mm2

(From page 10)

or

14.2857 a3 = 800 × 103 Nmm

a3 = 56 × 103 mm3

Shorter side, a = 38.2 mm

Longer side, b = 1.5 × 38.2 = 57.3 mm

Angle of twist

where

(From page 10)

Example

A closed tubular section of mean radius R and radial thickness t and a

tube of the same radius and thickness but with a longitudinal slit are

subjected to the same twisting moment T. Compare the maximum shear

stress developed in both and also compare the angular twist in these

tubes.

Mean radius = R

Thickness = t

For Closed Tubular Section

Maximum shear stress


tA

T

2

tR

T

2 21p

t

R

RG

TL

t

ds

AG

TL p

p

2

4 m 1

4 422

Tubular section with a small slit. This can be treated as a thin rectangular section

of width 2pR and thickness t.



It is seen that the closed tubular section is much more stronger and stiffer than

the open tubular section with a slit.

Example

A shaft made of plastic is of elliptical cross-section as shown in the figure. If

it is subjected to torsional loading as shown, determine the shear stress at

point A. Also determine the angle of twist at the end B. Gplastic = 15 GPa.

T, Torque at the point A = 60 + 30 = 90 Nm 90 × 103 Nmm

Semi-major axis of ellipse, b = 50 mm

Semi-minor axis of ellipse, a = 20 mm.

Shear stress at point A (end of the minor axis)

2

2

3

2/86.2

5020

109022mmN

ba

T

pp

Torque for 2 m length, T1 = 60 Nm

Torque for another 2 m length, T2 = 60 + 30 = 90 Nm

Angular twist,

So,

Exercise 6.7

A shaft section consists of a hollow

rectangular section and a solid

rectangular section as shown in the

figure. Composite shaft is

subjected to a twisting moment of

100 Nm. Determine (i) torque

shared by hollow and solid section,

(ii) maximum shear stress

developed in both the sections, (iii)

angular twist per meter length in

shaft if G = 25 × 103 N/mm2.

torsion of non-circular shafts - kişisel sayfalarkisi.deu.edu.tr/emine.cinar/asm16-torsion of...

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