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MSU / Engineering and Tech / Eng 205 / statically indeterminate torsion

statically indeterminate torsion

statically indeterminate torsion

Description

School: Montana State University
Department: Engineering and Tech
Course: Mechanics of Materials
Professor: Joseph fedock
Term: Spring 2016
Tags: EGEN 205 and mechanics of materials
Cost: 25
Name: EGEN 205, Mechanics of Materials, Week 5
Description: Angle of Twist and examples Power transmission Statically Indeterminate Torsion and examples
Uploaded: 02/13/2016
9 Pages 182 Views 0 Unlocks
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Angle of TwistDon't forget about the age old question of What is the definition of by orbital speed?

For design purposes, the amount of rotation / twist of the shaft may be the critical factor

  •  = angle of twist of one end of the shaft relative to the other end (1 cross section to another)
  • Similar to the case of axial deformations, we can look at cross sections of a shaft separated by a distance, dx, and observe that one cross section will rotate an amount  relative to other (figure 6.4b page 149)
  • The shear strain at a location on cross - section is related to the incremental twist d, by d = (p being the distance from center)

If we assume linear elastic behavior,Don't forget about the age old question of What refers to the measure of the effect of gravity on the object?

४ = and t at any location x along shaft

We also discuss several other topics like What is the overview structure of the protein?

t(x) =

४ =  =  substitute for ४

d - =(p’s cancel)If you want to learn more check out What is meant by rule of 72?

 = =

T(x) = internal torque

G = shear modulusWe also discuss several other topics like What is the difference between strain and sprain?

J(x) = polar moment of inertia

Recall                 S analogous relationshipIf you want to learn more check out scantron 882-es

2/8/16 if torque and cross - section are constant over length →  =

If the shaft is subject to different amounts of torque along its length or if the cross - section changes

 = overall cross sections

Sign conversion

Right hand rule

Positive tend  thumb pants away from member

Dimensions

 =

From geometry: if d = , we can integrate and get

 ४p = shear strain at p

If p = c (total radius) →  =

  1. If Ci = 38mm and angle of twist between ends is 0.60, what is shear strain at the inner surface of the tube?

 = 0.60= 0.01047rad

४min = ४at c1 = == 3.18 x 10-4 rad

  1. If max allowable shear strain = .0004rad and angle of twist remains at 0.60 by adjusting torque, What is maximum permissible outer radius?

४max = ४at c1 = = 47.8mm

Log Wrench Problem - removing a tire

Given: driver applies force of P = 100N to each, each arm of wrench = 225mm, G = 78MPa, dia = 12mm

Find:

  1. Maximum shear stress in arm turning lugnut
  2. Angle of twist (in degrees) of the arm

Solution:

  1. Torque T = P(450mm) = 45Nm

Tmax = = for solid shaft

= 133 Mpa

  1.  = = (45Nm)(0.225m) = .0638rad

        (78 x 107N/m2)(.006m)4        = 3.650

Power Transmission

Recall: Power = work / unit time and for a rotating shaft

        Work = torque x angle of rotation of shaft

Recall: work = force x distance

Assumption: of a shaft rotates d during time dt

        P = T

But d/dt = angular velocity(w)

        P = Tw

Dimensions → FL/T

Units → SI → 1 watt =

→ US Customary → 1 HP(horsepower) = 550

Note that w has dimensions of rad / sec but we typically express angular velocity in revolutions / sec or cycles / sec

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