Ch3 Torsion Lecture

Topic 3: TORSION 1. 2. 3. 4. 5.

Introduction Stresses in Elastic range Angle of twist Design transmission shafts Thin-wall hollow shafts

1. Introduction - Members subjected to loads along the longitudinal axes (Axial deformation) - Members subjected to torsional loads twisting the members about their longitudinal centroid axes are considered in this topic. - Examples: a lug-wrench, power transmission shaft

- A lug-wrench: + A lug-wrench shaft AB, arm CD + Apply equal and opposite forces (P) to ends of CD  AB is a torsion member

Sign convention: + Torque T(x): right-hand-rule sense

+ Angle of rotation, (x)

- Power transmission shaft: + Turbine exerts torque T on the shaft + The shaft transmits T to the generator + Generator creates an equal & opposite torque T’ Generator Rotation Turbine

2. Stresses in Elastic range Shear stress in shaft: How is torque transferred through shaft? An element dA distance  from the center of the shaft has shear force dF. For equilibrium:

𝑀= 0

Since dF = dA Shear stress distribution?

𝑑𝐹 = 𝑇

𝑑𝐹 = (𝑑𝐴) = T

Shaft deformations: Square shaft: wrapped under tension, cross section do not remain plane.

Circular shaft: Every cross section remains plane and undistorted. - Circumferential lines remain in a plane after deformation - Longitudinal lines: Parallel to the axis  become helical - Right angle 𝐴𝐵𝐶  A∗B∗C∗: Shear deformation  - : Torsional deformation

Consider a circular shaft attached to a fixed support. - Apply torque T: + Shaft will twist + Free end will rotate through angle  - By observation: +   T +L Relationship of , T, L? Shear stress distribution?

Shearing strain in circular shaft: Consider a shaft of length L, radius c, angle of twist .  AA’ = L =    = 𝐿

,  [rad]  is proportional to  and    varies linearly with the distance from the axis of the shaft. 𝑐 max = 𝐿

Stresses in the elastic range:

We have:  =

 𝐿

and max =

𝑐 𝐿



  = 𝑚𝑎𝑥 𝑐

From Topic 2:  = G



 G = G 𝑚𝑎𝑥 𝑐



  = 𝑚𝑎𝑥 𝑐

min =

𝑐1  𝑐2 𝑚𝑎𝑥

As long as the yield strength is not exceeded,  varies linearly to the distance  from the axis of the shaft.

Recall that (𝑑𝐴) = T 𝑚𝑎𝑥 2 𝑚𝑎𝑥 .𝐽 T =  𝑑𝐴  T = 𝑐

𝑐

J = 2 𝑑𝐴: the polar moment of inertia of the cross section with respect to its center O. 𝑇 𝑇𝑐  max = and  = 𝐽

𝐽

+ A solid cross section of radius r: dJ = 𝑢2 𝑑𝐴 dA = 2udu J =

𝑟 2 𝑢 2𝑢𝑑𝑢 0



𝑑 4

2

32

 J = 𝑟4 =

= 2

𝑟 3 𝑢 𝑑𝑢 0

+ A tubular shaft of outer radius r0 and inner radius ri:  4 𝐽 = 𝑟0 − 𝑟𝑖 4 2

Example 1: a/ Tmax =? If max = 120 Mpa b/ min = ? 𝐽𝑚𝑎𝑥 𝑐1 Solution: T = min = 𝑚𝑎𝑥 𝑐

𝑐2

Example 2: Determine max in shaft AB, BC. Solution:

Axial shear components: A torque applied to a shaft  shearing stresses on the face perpendicular to the axis. Equilibrium  equal stresses on planes containing the axis.

Normal stresses due to Torsion: - Element a: pure shear - Element b: Normal stress, shear stress or combination of both. - Consider an element at 45 to the shaft axis: F = 2(maxA0) cos45 = maxA0 2 𝐹 𝑚𝑎𝑥 𝐴0 2 45 = = = max 𝐴

𝐴0 2

Torsion failure modes: When subjected to torsion: + Ductile specimen breaks along a plane of maximum shear. + Brittle specimen breaks along planes perpendicular to the direction in which tension is a maximum.

+ Shear deformation  + Torsional deformation  c Shearing strain max =

- Shaft deformation:

- Strain:

- Stress:

L

max =

Tc J

and

=

T J

3. Angle of twist 𝑐 𝑇𝑐 max = max = 𝐿

𝐽

In the elastic range:  = G  𝑐 𝑇𝑐 𝑇𝐿  =G = 𝐽

𝐿

𝐽𝐺

- For shaft with cross-section changes: =

𝑇𝑖 𝐿𝑖 𝑖 𝐽𝐺 𝑖 𝑖

Each segment has constant cross section an torque. - If cross section changes continuously: d =

𝑇𝑑𝑥 𝐽 𝑥 𝐺

 =

𝑇 𝐿 𝑑𝑥 𝐺 0 𝐽(𝑥)

Gear assembly:

E/B = E - B =

𝑇𝐿 𝐽𝐺

Example 3: rA = 2rB Apply T at E Determine E?

TAD = 2T A =

𝑇𝐴𝐷 𝐿 𝐽𝐺

=

2𝑇𝐿 𝐽𝐺

CC’ = CC’’  rAA = rBB  B = (rA/rB) A = 2A = E = B + E/B =

4𝑇𝐿 𝐽𝐺

+

𝑇𝐿 𝐽𝐺

=

4𝑇𝐿 𝐽𝐺 5𝑇𝐿 𝐽𝐺

Example 4: Shaft ABC with d = 60 mm is supported by 2 journal bearings. Shaft EH with d = 80 mm, is fixed at E and supported by a bearing at H. If A/C= 0.04 rad, determine T1 and T2? Solution:

Example 4: d = 40 mm. Determine B/A (rad) =? Solution:

Comparison of axial and torsion:

4. Design of Transmission shaft: - Select the shaft material (G) and cross section (J) to meet performance specifications (power - P and speed - ) without exceeding the allowable  and allowable . Step 1: Determine T? From dynamics: P = T = T(2f)  T = SI units:

US units:

T: torque [N.m] P: power [N.m/s]  [W] : angular velocity (rad/s) f: frequency of rotation in Hz f [rpm], 1 rpm =

1 60

Hz

P [hp – horsepower] 1 hp = 550 ft.lb/s (6600 in.lb/s) = 746 W

𝑃

2𝑓

Step 2: Select a material (set all)

Design requirement: max =

𝑇𝑐 𝐽

 all

Step 3: Find shaft cross section? 𝐽 𝑐



𝑇

𝑎𝑙𝑙

If considering the allowable angle of twist:   all

Example 5: P = 16 hp from motor A  machine tool D Motor A: f = 1260 rpm all = 8 ksi Determine dAB and dCD? Solution: all = 8 ksi = 8 x 103 (lb/in2) P = 16 hp = 105600 in.lb/s 1260 f = 1260 rpm = = 21 Hz 60 𝑃 105600 T= = = 800.72 (lb.in) 2𝑓 2..21 𝑑 𝑇𝐴𝐵 . 𝐴𝐵 2

max = 𝑑 4  all  dAB  0.798 [in.] 𝐴𝐵 32

𝑇𝐴𝐵 3

=

𝑇𝐶𝐷 5

 TCD = 1334.53 (lb.in)

𝑑 𝑇𝐶𝐷 . 𝐶𝐷

max = 𝑑 24  all  dCD  0.947 [in.] 𝐶𝐷 32

Example 6: dAB = dCD; max  80 Mpa, D  1.5 ; G = 70 Gpa Determine d AB = dCD = ? Solution: Design based on allowable stress:

=

𝑇.𝑐 𝐽

𝑑 𝑇𝐴𝐵 𝐴𝐵 2

= 𝑑 4  max 𝐴𝐵

=

32

𝑇.𝑐 𝐽

𝑑 𝑇𝐶𝐷 𝐶𝐷 2

= 𝑑 4  max 𝐶𝐷 32

TAB = rB.F TCD = rC.F

 TAB =

 dAB = 59.6 mm

𝑇𝐶𝐷 .𝑟𝐵 𝑟𝐶

= 2500 (N.m)

Design based on allowable angle of twist:

D = D/C + C =

𝑇𝐶𝐷.𝐿𝐶𝐷 𝐺𝐽

+ C  1.5

C rC = B rB  C dAB = 62.9 mm Design must use the lager value for d: d = 62.9 mm

Statically determinate shaft: Example 7: a/ Determine T1, T2: 𝑀𝑥 = 0

TC – T2 = 0 TC + TB –T1 = 0 b/ Determine C, B - Torque-twist behavior: 1 = 𝑓𝑡1 𝑇1 𝑓𝑡1 = - Geometry of deformation:

𝐿1 𝐺𝐽1 1

2 = 𝑓𝑡2 𝑇2 𝑓𝑡2 =

𝐶 = 2 + 𝐵 ; 𝐵 = 1 + 𝐴 = 1

𝐿2 𝐺𝐽2 2

Statically indeterminate shaft: The unknown torques exceeds the number of applicable equilibrium equations. - Step 1: Write equil. eqs. - Step 2: Formulate compatibility eqs. - Step 3: Use torque-displacement eqs. to relate  and T. Example 8:

- Equilibrium eqs: TA = -T1 T1 – T2 = T B TC = T2 - Element torque-twist behavior: 1 = 𝑓𝑡1 𝑇1 =

𝐿 𝑇 𝐺𝐼𝑝1 1

2 = 𝑓𝑡2 𝑇2 =

- Geometry of deformation: 𝑡𝑜𝑡𝑎𝑙 = 1 + 2 = 0 Results: T1, T2 are determined 𝑇1 =

𝑇𝐵 3

- Determine rotation angle B 𝐵 = 1

𝑇2 =

2𝑇𝐵 − 3

𝐵 = −2

𝐿 T 𝐺𝐼𝑝2 2

1/ a/ Determine expressions of max1, max2, max3 b/ B = ? 2/ Problem 3.53 in “Mechanics of Materials”, 6th edition, Beer and Johnston.

5. Thin-walled hollow shafts Consider a thin-walled circular hollow shaft with inner radius r1, outer radius r2, thickness t = r2 – r1, and median 1 radius rm = (r1 + r2) 2

The stress distribution for a circular shaft increases linearly as we move from the center axis to the surface of the member. 𝑚𝑖𝑛 =

𝑇𝑟1 𝐽



𝑚𝑎𝑥 =

𝑇𝑟2 𝐽

where J = (𝑟2 4 − 𝑟1 4 ) 2

Write r1 and r2 in terms of rm: r2 = rm + t/2 r1 = rm – t/2

J =

 2

𝑟𝑚 +

𝑡 4 2

−

𝑡 4 𝑟𝑚 − 2

t2