Unit 1 Engineering Physics

22 CHAPTER 1. PROPERTIES OF MATTER

P x

Q

A

B

C

φ

θ

dx

r

l

Fixed End

Free End

(twisted)

Figure 1.21: Couple per unit twist

The Rigidity Modulus of the wire (n) = (shear stress) / (shear strain).

Hence, shear stress,=n◊(shear strain) (1.6)

We first consider an imaginary cylindrical shell of outer radiusxand thicknessdxas
shown in Figure5.65and calculate the shear stress at this shell. A vertical line AB drawn
on the shell get shifted to the position AC due to the torsional deformation, resulting in
an angle of twist\BQC=◊and an angle of shear\BAC=„.Itcanbeseenfrom
Figure5.65thatBC=l„=x◊. Therefore,

shear strain,„=

x◊

l

(1.7)

Therefore from equations (5.281) and (5.282), at the imaginary shell,

shear stress,=n◊

x◊

l

(1.8)

The area of cross-section of the imaginary shell = circumference◊width =(2fix)◊(dx).
Therefore, force acting in the shell cross-section is

Force = (shear stress)◊(area) =

A

n

x◊

l

B

◊(2fixdx)=

2 fin◊x^2

l

dx (1.9)

The torque about the axisPQacting on the shell is

(Force)◊(Distance) =

A

2 fin◊x^2

l

dx

B

◊x=

2 fin◊x^3

l

dx (1.10)

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