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strength of materials, torsion of non circular cross section, Hani Aziz Ameen
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Strength of Materials Handout No.6
Torsion of Non- Circular Section
Asst. Prof. Dr. Hani Aziz Ameen Technical College- Baghdad Dies and Tools Eng. Dept. E-mail:[email protected]
www.mediafire.com/haniazizameen
Strength of materials- Handout No.6 - Torsion of Non- Circular Section- Dr. Hani Aziz Ameen
6-1 Introduction
One of the important methods for the analysis of non circular shafts
is based on experimental results) . Mathematical analysis for non circular solid sections based on theory of elasticity is rather difficult .So in this chapter we are setting only the formula of shearing stress for several non- circular cross sections . 6-2 Torsion of Nom-circular Solid Sections
1- Rectangular Section From Fig(6-1)
Fig(6-1) where b = larger side of the rectangular shaft h= smaller side of the rectangular shaft L= length of the shaft G= modulus of rigidity
bh8.13
T = applied torque
2bhT
for square section 3bT8.4
angle of twist G
L.Tbh3
Strength of materials- Handout No.6 - Torsion of Non- Circular Section- Dr. Hani Aziz Ameen
where 2
bh15.3
2- Elliptical Section
From Fig(6-2)
Fig(6-2) where
2abT16
22 b1
a1
abGLT16
3- Equilateral Triangle From Fig (6-3)
Fig(6-3)
Strength of materials- Handout No.6 - Torsion of Non- Circular Section- Dr. Hani Aziz Ameen
where section
T ....... applied torque
Ad217.0T
GAd133.0T
2
6-3 Torsion of Thin Walled Tube with Non circular Cross Section The tube with variable thickness is as shown in Fig(6-4)
Fig(6-4)
Strength of materials- Handout No.6 - Torsion of Non- Circular Section- Dr. Hani Aziz Ameen
where
Consider the equilibrium element along the longitudinal axis
q2 dL = q1 dL
q2 = q1 = q
0Mo 0r.dL.qT
T= q . 2 r.2
dL
T q. 2 A = 0 -1)
The shear stress at any point across the wall thickness is
tq -2)
From eq. (6-1) &eq(6-2) it can be obtained
At2T
6-4 Examples Example(6-1) Fig(6-5) shows a torque 600 N.m applied to the rectangular section . Find the wall thickness ( t ) so that shearing stress 2m/MN60
Fig(6-5)
Strength of materials- Handout No.6 - Torsion of Non- Circular Section- Dr. Hani Aziz Ameen
Solution
At2T
)06.0*03.0(*t*260010*60 6
t = 2.78 mm
Example(6-2) Fig(6-6) shows a hollow member having the cross section and subjected to the action of torque of 203.3 N.m . If the section is symmetrical about
shearing stress across its wall thickness The mean radius r = 25.4 mm , L = 76.2 mm , t = 6.35mm and 60
Fig(6-6) Solution A = 3 ( twist the area subtending the angle rL) + area of triangle abc.
22
2
mm56.1032223*76.2*76.2*0.5 )2.76*4.25
64.25**2(3A
60sinL*L*5.0)L*r)360/60(*r**2(3A
36 10*35.6*10*56.10322*23.203
t*A*2T =15.51 MPa
Strength of materials- Handout No.6 - Torsion of Non- Circular Section- Dr. Hani Aziz Ameen
6-5 Problems 6-1) Fig(6-7) shows a rectangular strip , find the maximum shear stress in
terms of the applied torque T for the section .
Fig(6-7) 6-2) Fig(6-8) shows the section which is subjected to a torque of
1739.7N.m , find the minimum thickness of the section , if the shearing stress does not to exceed 41.37 Mpa.
Fig(6-8) 6-3) Fig(6-9) shows the upper and lower portions which constitute a full
ellipse of 153.4mm major axis and 63.5mm minor . Find the maximum safe torque , if the shearing stress in the material does not exceed 41.37MPa .
Strength of materials- Handout No.6 - Torsion of Non- Circular Section- Dr. Hani Aziz Ameen
Fig(6-9)