torsional shaft function: transmit torques from one plane to the other. design of torsional shafts:...

15
Torsional Shaft tion: transmit torques from one plane to the other. gn of torsional shafts: stress and deformation T T T

Upload: eugene-montgomery

Post on 18-Dec-2015

216 views

Category:

Documents


0 download

TRANSCRIPT

Torsional Shaft

Function: transmit torques from one plane to the other.

Design of torsional shafts: stress and deformation

T TT

Deformation of a Torsional Shaft with Circular Cross Section

Questions: After applied a torque T at section C, 1. What happens to the longitudinal member ab? 2. Does the cross section C change its shape? 3. Is there any out-of-plane deformation at C? 4. What about sections A and B, where B is halfway away from A and C? 5. What happens to the angle between ab and cd?

Deformation of a Torsional Shaft with Circular Cross Section

Answers:

Shear Strain - on the Surface of the Shaft

', is the angle of twist, is the radius of the cross section

2'

2

', is the rate of change of the angle of twist

bb rr

ab Lree r

ae L L

ee d dr

ae dx dxd

rdx

Shear Strain - inside the Shaft

'dd

cd L

is the radial distance of element cd from the center.

Shear strain is linearly proportional to the radial distance.It reaches its maximum value at the surface of the shaft.

Torsional Shaft - Shear Stress G G

0 0

= sr Gr

Shear strain: Shear Stress:

sr

Elastic torsion formula:

2s ss

s s

T ds ds Jr r r

J is the polar second moments of inertial

e.g. r

r1

r2

2 2 32 2s s s

J ds d d

Torsional Shaft - Torque Diagram Example 1: Draw the torque diagram for the following shaft. Find the maximumstresses inside the shaft.

Torsional Displacement

L L

L G

T

J

T L

GJ

Angle of twist:

Assumptions:

With respect to a reference cross section locatedL away from the targeted cross section.

• uniform cross section (J is constant)• material properties are constant• carrying an internal torque of T

Comparison with axially loaded member

PL

EA

Torsional Displacement

i ii

i i i i

T L

G J

Multiple shafts:

Continuously changing shafts:

0

L Tdxx

GJ

x

Torsional Displacement Example 2: Find the twist angle at cross section D if the cross section A is fixed.

Torsional Shaft - Statically Indeterminate Problems

Problem solving procedure:

1. Static equilibrium conditions: Equation of equilibrium

2. Deformation relations: Equation of compatibility

3. Load and deformation relations: Torque - displacement equation

Torsional Shaft - Statically Indeterminate Problems

Example 3: A shaft shown below is fixed between two walls. The outer diameterof the shaft is 20 mm and the diameter of the cavity of 16 mm. Find the reactions.

Torsional Shaft - Statically Indeterminate Problems

Example 4: A hollow circular aluminum alloy (G = 4000 ksi) cylinder has a steel (G = 12,000 ksi) core as shown in the figure. The steel and aluminum parts are securelyconnected at the ends. (1) If T = 80 , what are the torques applied at aluminum and steel respectively?(2) If the allowable stresses in the steel and aluminum must be limited to 12 ksi and 10ksi,determine the maximum torque that can be applied to the right end of the composite shaft.(3) Find the rotation of the right end of the composite shaft when the torque of (2) is applied.

kip in

Torsional Shaft - Statically Indeterminate Problems

Example 5: A circular bar AB with ends fixed against rotation has a hole extending for half of its length (see figure). The outer diameter of the bar is d2 = 100 mm and the diameter of the hole is d1 = 80 mm. The total length of the bar is L = 1250 mm. At what distance x from the left-hand end of the bar should a torque T0 be applied so that the reactive torques at the supports will be equal?

Conclusion: If a shear stress exists at a point on any plane, a shear stress of the same magnitude must also exist at this point on an orthogonal plane.

Torsional Shaft - Stresses on Inclined Planes

0

o xy yx

xy yx

M dydz dx dxdz dy

0 cos sin sin cos 0

0 cos cos sin sin 0

n n xy yx

t nt xy yx

F dA dA dA

F dA dA dA

xyxy

yx

yx

Take an element of dx, dy and dz out from the shaft:

xy

yx

nnt

Vy

Vx

FnFt

sin 2

cos2

n xy

nt xy