chapter 3 torsion - seoul national university

Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]

Chapter 3 Torsion

• Torsion: twisting of a straight bar loaded by moments (or

torques) • Example (): torsion by a ‘couple’ of forces

(moment or torque = 𝑃𝑃 × 𝑑𝑑)

• Vector representation in (b): use r_____-hand rule

• SI Unit of the moment, torque and couple: N ⋅ m

3.2 Torsional Deformations of a Circular Bar

“Pure Torsion”

1. Prismatic bar of circular cross section twisted by torques at the ends can prove

that cross sections do not change in shape as they r_______ about the axis

2. For small deformations neither the length nor the radius changes

3. Angle of twist (𝜙𝜙) or angle of rotation: (See figure below): angle 𝜙𝜙(𝑥𝑥) changes

linearly over the axis as the straight line 𝑝𝑝𝑝𝑝 becomes a helical curve 𝑝𝑝𝑝𝑝′

Shear Strains at the Outer Surface

1. Deformation by torsion: 𝑎𝑎𝑎𝑎𝑎𝑎𝑑𝑑 𝑎𝑎𝑎𝑎′𝑎𝑎′𝑑𝑑

2. This is pure s_______ condition because

there is no n________ s________

3. The change in the angle of the corner:

𝛾𝛾max = 𝑎𝑎𝑎𝑎

=𝑟𝑟 𝑑𝑑𝑥𝑥

~ this equation relates the shear strain to the a_______ of the tw______

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4. Introducing the rate of twist (the angle of twist per unit length) 𝜃𝜃 = 𝑑𝑑𝜙𝜙/𝑑𝑑𝑥𝑥,

𝛾𝛾max = 𝑟𝑟 ⋅

5. For pure torsion, 𝛾𝛾max = 𝑟𝑟𝜃𝜃 = 𝑟𝑟𝑟𝑟𝐿𝐿

Shear Strains within the Bar

1. Considering another bar with the radius ρ, the

shear strain is derived as 𝛾𝛾 = 𝜌𝜌𝜃𝜃

2. Noting 𝛾𝛾max = 𝑟𝑟𝜃𝜃 and γ = ρθ,

𝛾𝛾 =

𝛾𝛾max

Circular Tubes

1. Using the l_______ variation of the shear strain of a

circular bar,

𝛾𝛾max = ⋅ 𝜙𝜙𝐿𝐿

𝛾𝛾min = 𝑟𝑟2𝛾𝛾max =

⋅ 𝜙𝜙𝐿𝐿

Note: All the equations for the strains in a circular bar can be applied to any materials

because the derivations are solely based on g_________ concepts. The equations are

limited to bars having small angles of twist and small strains.

3.3 Circular Bars of Linearly Elastic Materials

Shear Stresses

1. Hooke’s law in shear

𝜏𝜏 = 𝐺𝐺𝛾𝛾

2. For a shaft under torsion:

𝜏𝜏max = 𝐺𝐺𝛾𝛾max =

𝜏𝜏 = 𝐺𝐺𝛾𝛾 = 𝐺𝐺𝜌𝜌𝜃𝜃 =

𝜏𝜏max

Note: unit: 𝜏𝜏 and 𝐺𝐺 ~

force/(length)2 and 𝛾𝛾 ~

radian (unit-less)



3. Shear stresses on longitudinal planes: the same

magnitude as those on the cross-sectional plane (See

Section 1.7)

Torsion Formula

1. Resultant of shear stresses 𝜏𝜏 torque 𝑇𝑇 (cf. 𝜎𝜎 𝑁𝑁)

2. Moment force on the infinitesimal element 𝑑𝑑𝑑𝑑 caused by the shear stress 𝜏𝜏:

𝑑𝑑𝑑𝑑 = 𝜏𝜏𝜌𝜌𝑑𝑑𝑑𝑑 = 𝜏𝜏max𝜌𝜌2

𝑟𝑟 𝑑𝑑𝑑𝑑

𝑇𝑇 = �𝑑𝑑𝑑𝑑𝐴𝐴

=𝜏𝜏max𝑟𝑟

�𝜌𝜌2𝑑𝑑𝑑𝑑𝐴𝐴

=𝜏𝜏max𝑟𝑟

𝐼𝐼𝑃𝑃

3. 𝐼𝐼𝑃𝑃: the polar moment of inertia - 𝐼𝐼𝑃𝑃 in Appendix E

- For circular cross-section, 𝐼𝐼𝑃𝑃 = 𝜋𝜋𝑟𝑟4

2= 𝜋𝜋𝑑𝑑4

32

4. “Torsion Formula”: maximum shear stress on the outer surface for torsion 𝑇𝑇,

radius 𝑟𝑟 and the polar moment of inertia 𝐼𝐼𝑃𝑃

𝜏𝜏max =𝑇𝑇 ⋅ 𝑟𝑟𝐼𝐼𝑃𝑃

5. Torsion formula for circular bar: 𝜏𝜏max = 16𝑇𝑇𝜋𝜋𝑑𝑑4

6. Shear stress at distance 𝜌𝜌:

𝜏𝜏 = 𝜏𝜏max𝜌𝜌𝑟𝑟

=𝑇𝑇 ⋅ 𝜌𝜌𝐼𝐼𝑃𝑃

Angle of Twist

1. From 𝜏𝜏max = 𝐺𝐺𝑟𝑟𝜃𝜃 and the torsion formula 𝜏𝜏max = 𝑇𝑇𝑟𝑟𝐼𝐼𝑃𝑃

,

𝜃𝜃 =𝑇𝑇𝐺𝐺𝐼𝐼𝑃𝑃

where 𝐺𝐺𝐼𝐼𝑃𝑃 is termed “torsional rigidity”

2. For a bar in pure torsion (𝜙𝜙 = 𝜃𝜃𝐿𝐿),

𝜙𝜙 =𝑇𝑇𝐿𝐿𝐺𝐺𝐼𝐼𝑃𝑃



3. Torsional stiffness: 𝑘𝑘𝑇𝑇 = 𝐺𝐺𝐼𝐼𝑃𝑃/𝐿𝐿

Torsional flexibility: 𝑓𝑓𝑇𝑇 = 𝐿𝐿/𝐺𝐺𝐼𝐼𝑃𝑃

Circular Tubes

1. Polar moment of inertia:

𝐼𝐼𝑃𝑃 =𝜋𝜋2

(𝑟𝑟24 − 𝑟𝑟14) =𝜋𝜋

32(𝑑𝑑24 − 𝑑𝑑14)

2. Alternatively, using 𝑟𝑟 = (𝑟𝑟1 + 𝑟𝑟2)/2 and 𝑡𝑡 = 𝑟𝑟2 − 𝑟𝑟1

𝐼𝐼𝑃𝑃 =𝜋𝜋𝑟𝑟𝑡𝑡

2(4𝑟𝑟2 + 𝑡𝑡2) =

𝜋𝜋𝑑𝑑𝑡𝑡4

(𝑑𝑑2 + 𝑡𝑡2)

3. For a thin tube,

𝐼𝐼𝑃𝑃 ≅ 2𝜋𝜋𝑟𝑟3𝑡𝑡 =𝜋𝜋𝑑𝑑3𝑡𝑡

4

Example 3-2: A steel shaft as a solid circular bar

(a) or as a circular tube (b) under the torque 𝑇𝑇 =

1,200 N ⋅ m. Allowable shear stress is 40 MPa

and allowable rate of twist is 0.75°/m. The shear

modulus of elasticity 𝐺𝐺 = 78 GPa.

(a) Required diameter 𝑑𝑑0 of the solid shaft

(b) Required outer diameter 𝑑𝑑2 of the hollow

shaft if 𝑡𝑡 = 𝑑𝑑2/10



3.4 Nonuniform Torsion

Bar consisting of prismatic segments

1. Use the same procedure as that in

Section 2.3 (axially loaded members)

2. The angle of twist is

𝜙𝜙 = �𝜙𝜙𝑖𝑖

𝑛𝑛

𝑖𝑖=1

= �𝑇𝑇𝑖𝑖𝐿𝐿𝑖𝑖

𝐺𝐺𝑖𝑖(𝐼𝐼𝑃𝑃)𝑖𝑖

𝑛𝑛

𝑖𝑖=1

Bar with continuously varying cross sections or torque

1. The differential angle of rotation

𝑑𝑑𝜙𝜙 =𝑇𝑇𝑑𝑑𝑥𝑥𝐺𝐺𝐼𝐼𝑃𝑃(𝑥𝑥)

2. The angle of twist

𝜙𝜙 = � 𝑑𝑑𝜙𝜙𝐿𝐿

0= �

𝑇𝑇𝑑𝑑𝑥𝑥𝐺𝐺𝐼𝐼𝑃𝑃(𝑥𝑥)

𝐿𝐿

0

𝜙𝜙 = � 𝑑𝑑𝜙𝜙𝐿𝐿

0= �

𝑇𝑇(𝑥𝑥)𝑑𝑑𝑥𝑥𝐺𝐺𝐼𝐼𝑃𝑃(𝑥𝑥)

𝐿𝐿

0



Example 3-4: Solid steel shaft

having diameter 𝑑𝑑 = 30 mm. 𝑇𝑇1 =

275 N ⋅ m, 𝑇𝑇2 = 450 N ⋅ m, and 𝑇𝑇3 =

175 N ⋅ m. 𝐿𝐿𝐵𝐵𝐵𝐵 = 500 mm and

𝐿𝐿𝐵𝐵𝐶𝐶 = 400 mm. Shear modulus 𝐺𝐺 =

80 GPa. Determine the maximum

shear stress in each part of the

shaft and the angle of twist between gears B and D.



3.5 Stresses and Strains in Pure Shear

“Pure Shear” Condition Created by

Torsion

1. Check the sign conventions ()

Stresses on Inclined Planes

1. The same sign convention as described in Section 2.6 (counter-clockwise 𝜏𝜏𝜃𝜃

positive, tension 𝜎𝜎𝜃𝜃 positive)

2. 1st equation of equilibrium (normal to the inclined surface):

σ𝜃𝜃𝐴𝐴0 sec𝜃𝜃 = 𝜏𝜏𝐴𝐴0 sin𝜃𝜃 + 𝜏𝜏𝐴𝐴0 tan𝜃𝜃 cos𝜃𝜃

σ𝜃𝜃 = = 𝜏𝜏 sin 2𝜃𝜃

3. 2nd equation of equilibrium (tangential to the inclined surface)

𝜏𝜏𝜃𝜃𝐴𝐴0 sec𝜃𝜃 = 𝜏𝜏𝐴𝐴0 cos𝜃𝜃 − 𝜏𝜏𝐴𝐴0 tan𝜃𝜃 sin𝜃𝜃

𝜏𝜏𝜃𝜃 = = 𝜏𝜏 cos 2𝜃𝜃

4. Plots of 𝜎𝜎𝜃𝜃 and 𝜏𝜏𝜃𝜃 ()

Pure shear at 𝜃𝜃 = , , .

Normal stress becomes

maximum/minimum at 𝜃𝜃 = ±

5. Stresses at θ = 0° and 45° ()

“pure shear” and “normal stresses

only” ~ all τ

Compare with Fig 2.36 (under axial

load): no “pure shear” at any 𝜃𝜃



6. Torsion failure by maximum tensile

stress along the angle 𝜃𝜃 = °

Strains in Pure Shear

1. Shear strain in pure shear (𝜃𝜃 = 0°):

𝛾𝛾 =𝜏𝜏𝐺𝐺

2. Normal strain in the 𝜃𝜃 = 45° direction:

𝜀𝜀max =𝜏𝜏𝐸𝐸

+𝜈𝜈𝜏𝜏𝐸𝐸

=𝜏𝜏𝐸𝐸

(1 + 𝜈𝜈)

Elongation (tension) at 𝜃𝜃 = and shortening (compression) at 𝜃𝜃 =

3. Can show (Section 3.6) that

𝜀𝜀max =𝛾𝛾2

Example 3-6:

(a) Determine the maximum shear,

tensile, and compressive

stresses, and show these

stresses on sketches of properly

oriented stress elements

(b) Determine the corresponding

maximum strains and sketch

them



(c) Maximum torque 𝑇𝑇max if the allowable normal strain is 𝜀𝜀𝑎𝑎 = 0.9 × 10−3?

(d) 𝑇𝑇 = 4.0 kN ⋅ m and 𝜀𝜀𝑎𝑎 = 0.9 × 10−3 New outer diameter 𝑑𝑑2 to carry the torque?



3.8 Statically Indeterminate Torsional Members

Statically Indeterminate Torsional Members

1. Use the same procedure as that in Section

2.4 (axially loaded members)

2. Example: Statically indeterminate composite

member (bar + tube) in torsion ()

Equation of e_____________

𝑇𝑇1 + 𝑇𝑇2 = 𝑇𝑇 (1)

3. How to solve? Introduce equation of c______________

𝜙𝜙1 = 𝜙𝜙2 (2)

4. Equations (1) and (2) are defined in terms of

different terms (force and displacement) Use f_______-d________ relationship

𝜙𝜙1 =𝑇𝑇1𝐿𝐿𝐺𝐺1𝐼𝐼𝑃𝑃1

𝜙𝜙2 =𝑇𝑇2𝐿𝐿𝐺𝐺2𝐼𝐼𝑃𝑃2

5. Now, Eq. (2) becomes Eq. (3)

𝑇𝑇1𝐿𝐿𝐺𝐺1𝐼𝐼𝑃𝑃1

=𝑇𝑇1𝐿𝐿𝐺𝐺1𝐼𝐼𝑃𝑃1

6. Solving Eq. (1) and (3) together, one can get

𝑇𝑇1 = 𝑇𝑇 �𝐺𝐺1𝐼𝐼𝑃𝑃1

𝐺𝐺1𝐼𝐼𝑃𝑃1 + 𝐺𝐺2𝐼𝐼𝑃𝑃2�

𝑇𝑇2 = 𝑇𝑇 �𝐺𝐺2𝐼𝐼𝑃𝑃2

𝐺𝐺1𝐼𝐼𝑃𝑃1 + 𝐺𝐺2𝐼𝐼𝑃𝑃2�

7. With these internal torques, one can find the angle of twist (using force-displacement

relationship), stresses (torsion formula), and strains (material properties)



Example 3-9: Bar is fixed at both ends and

loaded by a torque 𝑇𝑇0 at point C. Obtain

formulas for

(a) Reactive forces 𝑇𝑇𝐴𝐴 and 𝑇𝑇𝐵𝐵

(b) Maximum shear stresses 𝜏𝜏𝐴𝐴𝐴𝐴 and 𝜏𝜏𝐴𝐴𝐵𝐵

(c) Angle of rotation 𝜙𝜙𝐴𝐴



3.9 Strain Energy in Torsion and Pure Shear

Strain Energy of a Prismatic (Linear Elastic) Bar

under Pure Torsion

1. Recall the linear relationship between the

torque and the angle of twist

𝜙𝜙 =

2. From the linear relationship

𝑈𝑈 = 𝑊𝑊 =𝑇𝑇𝜙𝜙2

=𝑇𝑇2𝐿𝐿2𝐺𝐺𝐼𝐼𝑝𝑝

=𝐺𝐺𝐼𝐼𝑃𝑃𝜙𝜙2

2𝐿𝐿

Nonuniform Torsion

1. Bar consisting of prismatic segments and

constant torque:

𝑈𝑈 = �𝑈𝑈𝑖𝑖

𝑛𝑛

𝑖𝑖=1

= �

𝑛𝑛

𝑖𝑖=1

2. Cross section and/or internal torque varies along the axis:

𝑑𝑑𝑈𝑈 =[𝑇𝑇(𝑥𝑥)]2𝑑𝑑𝑥𝑥2𝐺𝐺𝐼𝐼𝑃𝑃(𝑥𝑥)

→ 𝑈𝑈 =

3. Note: 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 (See Example 2-13)

4. Strain-energy Density in Pure Shear ()

𝑈𝑈 = 𝑊𝑊 =𝑉𝑉𝑉𝑉2

Substituting 𝑉𝑉 = 𝜏𝜏ℎ𝑡𝑡 and 𝑉𝑉 = 𝛾𝛾ℎ,

𝑈𝑈 =𝜏𝜏𝛾𝛾ℎ2𝑡𝑡

2

Dividing by the volume ℎ2𝑡𝑡, the density is

𝑢𝑢 =𝜏𝜏𝛾𝛾2

=𝜏𝜏2

2𝐺𝐺 =𝐺𝐺𝛾𝛾2

2

Note: axially loaded member 𝑢𝑢 = 𝜎𝜎2

2𝐸𝐸= 𝐸𝐸𝜀𝜀2

2 (Section 2.7)



Example 3-11: Prismatic bar under

distributed torque of constant intensity 𝑡𝑡

(per unit distance along the axis)

(a) Formula for the strain energy

(b) Evaluate the energy for 𝑡𝑡 = 480 lb-

in./in., 𝐿𝐿 = 12 ft, 𝐺𝐺 = 11.5 × 106 psi,

and 𝐼𝐼𝑃𝑃 = 17.18 in.4

Example 3-12: Tapered bar of solid

circular cross section, loaded by a torque

𝑇𝑇. The diameter varies linearly from 𝑑𝑑𝐴𝐴 to

𝑑𝑑𝐵𝐵. Determine the angle of rotation 𝜙𝜙𝐴𝐴

using the strain energy



3.10 Torsion of Noncircular Prismatic Shafts (Skipped)

Unlike bars with circular cross-sections, a

more advanced theory, developed by Saint-

Venant, is required to develop expressions

for torsion … because of warping

phenomenon.

3.11 Thin-Walled Tubes

Shear Stresses and Shear Flow

1. Assumption: the intensity of the stresses varies so slightly across the thickness that

it is assumed that the stress is constant in the direction (because it’s thin).

2. Forces on faces 𝑎𝑎𝑎𝑎 and 𝑐𝑐𝑑𝑑:

𝐹𝐹𝑏𝑏 = 𝜏𝜏𝑏𝑏𝑡𝑡𝑏𝑏𝑑𝑑𝑥𝑥 and 𝐹𝐹𝑐𝑐 = 𝜏𝜏𝑐𝑐𝑡𝑡𝑐𝑐𝑑𝑑𝑥𝑥

3. From equilibrium of forces, 𝐹𝐹𝑏𝑏 = 𝐹𝐹𝑐𝑐

𝜏𝜏𝑏𝑏𝑡𝑡𝑏𝑏 = 𝜏𝜏𝑐𝑐𝑡𝑡𝑐𝑐

4. This shows that the product of the shear stress and thickness, i.e. shear flow, is

constant

𝑓𝑓 = 𝜏𝜏𝑡𝑡 = constant



Torsion Formula for Thin-Walled Tubes

1. The moment of the shear force on the element

with the length 𝑑𝑑𝑑𝑑

𝑑𝑑𝑇𝑇 = 𝑟𝑟𝑓𝑓𝑑𝑑𝑑𝑑

where 𝑟𝑟 is the perpendicular distance from point

O to the line of action of the force 𝑓𝑓𝑑𝑑𝑑𝑑

2. Total torque

𝑇𝑇 = � 𝑟𝑟𝑓𝑓𝑑𝑑𝑑𝑑𝐿𝐿𝑚𝑚

0=

� 𝑟𝑟𝑑𝑑𝑑𝑑𝐿𝐿𝑚𝑚

0= 2𝐴𝐴𝑚𝑚

where 𝐴𝐴𝑚𝑚 is the area enclosed by the median line of the cross section (NOT the

cross-sectional area)

3. As a result, the relationship between the torque and the constant shear flow is

𝑓𝑓 =𝑇𝑇

2𝐴𝐴𝑚𝑚

4. From the shear flow definition 𝑓𝑓 = 𝜏𝜏𝑡𝑡, a torsion formula for thin-walled tubes is

𝜏𝜏 =𝑇𝑇

2𝑡𝑡𝐴𝐴𝑚𝑚

5. Example 1: a thin-walled circular tube with 𝐴𝐴𝑚𝑚 =

𝜏𝜏 =𝑇𝑇

2𝜋𝜋𝑟𝑟2𝑡𝑡

Note: the same result by the standard torsion formula with 𝐼𝐼𝑃𝑃 ≈ 2𝜋𝜋𝑟𝑟3𝑡𝑡

6. Example 2: a thin-walled rectangular tube with 𝐴𝐴𝑚𝑚 =

𝜏𝜏vert =𝑇𝑇

2𝑡𝑡1𝑎𝑎ℎ, 𝜏𝜏horiz = 𝑇𝑇/2𝑡𝑡2𝑎𝑎ℎ

Note: The location of maximum shear stress determined

by the thickness



Strain Energy and Torsion Constant

1. Strain energy in the infinitesimal element

𝑑𝑑𝑈𝑈 =𝜏𝜏2

2𝐺𝐺𝑡𝑡𝑑𝑑𝑑𝑑𝑑𝑑𝑥𝑥 =

𝜏𝜏2𝑡𝑡2

2𝐺𝐺𝑑𝑑𝑑𝑑𝑡𝑡𝑑𝑑𝑥𝑥 =

𝑓𝑓2

2𝐺𝐺𝑑𝑑𝑑𝑑𝑡𝑡𝑑𝑑𝑥𝑥

2. Total strain energy

𝑈𝑈 = ∫ 𝑑𝑑𝑈𝑈 =𝑓𝑓2

2𝐺𝐺�

𝑑𝑑𝑑𝑑𝑡𝑡� 𝑑𝑑𝑥𝑥𝐿𝐿

0

𝐿𝐿𝑚𝑚

0=𝑓𝑓2𝐿𝐿2𝐺𝐺

�𝑑𝑑𝑑𝑑𝑡𝑡

𝐿𝐿𝑚𝑚

0

3. Substituting 𝑓𝑓 = 𝑇𝑇/2𝐴𝐴𝑚𝑚,

𝑈𝑈 =𝑇𝑇2𝐿𝐿

8𝐺𝐺𝐴𝐴𝑚𝑚2�

𝑑𝑑𝑑𝑑𝑡𝑡

𝐿𝐿𝑚𝑚

0

4. Introducing torsion constant 𝐽𝐽

𝐽𝐽 =4𝐴𝐴𝑚𝑚2

∫ 𝑑𝑑𝑑𝑑𝑡𝑡

𝐿𝐿𝑚𝑚0

5. Then, the strain energy of a thin-walled tube is

𝑈𝑈 =𝑇𝑇2𝐿𝐿2𝐺𝐺𝐽𝐽

Note: For a circular bar, 𝑈𝑈 = 𝑇𝑇2𝐿𝐿2𝐺𝐺𝐼𝐼𝑃𝑃

6. If the thickness is constant, 𝐽𝐽 = 4𝑡𝑡𝐴𝐴𝑚𝑚2 /𝐿𝐿𝑚𝑚

7. Example 1: thin-walled circular tube 𝐽𝐽 = 4𝑡𝑡(𝜋𝜋𝑟𝑟2)2/2𝜋𝜋𝑟𝑟 = 2𝜋𝜋𝑟𝑟3𝑡𝑡

8. Example 2: thin-walled rectangular tube ∫ 𝑑𝑑𝑑𝑑𝑡𝑡

𝐿𝐿𝑚𝑚0 = 2∫ 𝑑𝑑𝑑𝑑

𝑡𝑡1

ℎ0 + 2∫ 𝑑𝑑𝑑𝑑

𝑡𝑡2

𝑏𝑏0 = 2 �ℎ

𝑡𝑡1+ 𝑏𝑏

𝑡𝑡2�

𝐽𝐽 =2𝑎𝑎2ℎ2𝑡𝑡1𝑡𝑡2𝑎𝑎𝑡𝑡1 + ℎ𝑡𝑡2

Angle of Twist for a Thin-walled Tube

1. Work = strain energy 𝑊𝑊 = 𝑈𝑈 𝑜𝑜𝑟𝑟 𝑇𝑇𝑇𝑇2

= 𝑇𝑇2𝐿𝐿2𝐺𝐺𝐺𝐺

2. Thus, the angle of twist can be determined by

𝜙𝜙 =𝑇𝑇𝐿𝐿𝐺𝐺𝐽𝐽



Example 3-16: Thin-walled, circular and

square tubes constructed of the same

material are subjected to the same torque.

Both tubes have the same length, same wall

thickness, and same cross-sectional area.

The ratios of their shear stresses and angles

of twist?


chapter 3 torsion - seoul national university

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