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:haniazizameen@yahoo.com

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)