7laha - "Torsion Pt.2"

"7laha - Torsion 2"

Source: mind of Kites

Ex:

Cylindrical Composite shaft (steel and brass)
 Steel core (G=77Gpa) 50mm diameter and 5mm brass (39Gpa)
Jacket the shaft consist from two parts
Connected by bolts as shown
The right part (L =.5m) is fitted with torsion spring (K=3N.m/ rad) and the left part (L=.4 m) is fitted with fixed thin steel angle (.127*.127*.0095 m) length =.3 m
If a worker made a force=70N by a tool (L=.2m) on the left sec of the square box of bolts, the bolts permit a2 deg rotation of one flange with respect to the other before the flanges start to rotate as single part. Note that the relative rotation angle does not depend on if the ends of parts are fixed or not.

Determine :
1-The max shear stress on the steel core and the brass jacket in the right part.
2-The corresponding angle of twist in the steel angle

Notes 7 " Torsion << Pt.2>> "

Torsion "Pt.2"

* Example :

 
* Shear stress direction :

For Single Piece :

For Closed Section :

Example :

Structural aluminum tubing of 60x100mm rectangular cross section was fabricated by extrusion . Determine the shearing stress in each of the four walls of a portion of such tubing when it is subjected to a torque of 2.7 KN.m , assuming that :

a) a uniform wall thickness .

b)  as a result of defective fabrication , walls AB and AC are 3-mm thick, and walls BD and CD 5-mm thick .

Solution :

(a) Tubing of uniform wall thickness. 

The area bounded by the center line :

Since the thickness of each of the four walls is t=4mm . We find that the shearing stress in each wall is :

 (b) tubing with variable wall thickness

Observing that the area   bounded by the center line is the same as in part a
And substituting t=3 mm and t=5mm and t=5mm

* We note that the stress in a given wall depends only upon its thickness  .

7laha - "Torsion"

"7laha - Torsion"

Smart structure problems for never stop thinking

Source:  Mechanics of material by

Ferdinand p. Beer and E .Russell Johnston

(3.53) The composite shaft shown consist of a 5-mm-thick brass jacket (G brass=39Gpa)  bonded to a 40-mm diameter steel core (G steel =77.2 G pa) . Knowing that the shaft is subjected to a 600(N .m) torque .

Determine:

(1) the maximum shearing stress in the steel core .

(2) The maximum shearing stress in the brass jacket.

(3) The angle of twist of B relative to A .

Solution:

J=π(C^4)/2

J b =π*(.025^(4)  -.02^(4) )/2 =3.6 *(10^-7)

J s =π*(.02^4)/2 =2.5*(10^-7)

T steel /T brass =J b /J s  =1.4     so T s =350 N. m     T b =250 N. m

τ=T*c/J    so   τ b max =250*.025/J b   =17.3 M pa  τ s =350*.02/J s  =28 M pa

φ r = T s *2/(G*J)s  =2.05 deg

(3.58) Two solid steel shafts are fitted with flanges that are  Then connected by bolts as shown .the bolts are slightly Undersized and permit a 1.5 deg rotation of one flange With respect to the other before the flanges begin to rotate As a single unit .knowing that (G=77 Gpa) . 

Determine the maximum shearing stress in each shaft when a torque
Of T magnitude is 570  N .m is applied to the flange C.

Solution:

J=π(c^4)/2

J AB=π*(.015^4)/2   =7.9(10^-8) J CD =π*(.018^4)/2 =1.6*(10^-7) Φ r=.026 Rad Φ=T*L/GJ Φ r = (570*.9)/ (7700*1.6)  - T AB*.6 /(770*8) T AB = 120 N .m TCD total =570-120=450 N .m τAB max =120*0.015/ J AB = 22.5 M pa.    τCD max =450*.018/J CD   =50.6 Mpa