Mechanical Engineering Principles

104 MECHANICAL ENGINEERING PRINCIPLES

Hence, extension ofAB

=δ=(R+y)θ−Rθ=yθ

Now, strainε=extension/original length,

i.e. ε=

yθ

Rθ

=

y

R

( 8. 3 )

However,

stress(σ)

strain(ε)

=E

or σ=Eε ( 8. 4 )

Substituting equation (8.3) into equation (8.4) gives:

σ=E

y

R

( 8. 5 )

or

σ

y

=

E

R

( 8. 6 )

Consider now the stresses in the beam’s cross-
section, as shown in Figure 8.3.

Tensile

Compressive

(a) Beam's cross-section (b) Stress distribution

A

y

N

b

s

dy

Figure 8.3

From Figure 8.3, it can be seen that the stressσ
causes an elemental coupleδMaboutNA,where:

δM=σ×(b×dy)×y

and the total value of the couple caused by all such
stresses

=M=

∑

δM=

∫

σbydy( 8. 7 )

but from equation (8.5), σ=

Ey

R

Therefore , M=

∫

Ey

R

bydy

=

∫

E

R

y^2 bdy

Now,EandRare constants, that is, they do not

vary withy, hence they can be removed from under

the integral sign. Therefore,

M=

E

R

∫

y^2 bdy

However,

∫

y^2 bdy=I=the second moment of

area of the beam’s cross-section aboutNA(from

Table 7.1, page 91).

Therefore, M=

E

R

I

or

M

I

=

E

R

( 8. 8 )

Combining equations (8.6) and (8.8) gives:

σ

y

=

M

I

=

E

R

( 8. 9 )

Position ofNA

From equilibrium considerations, the horizontal

force perpendicular to the beam’s cross-section, due

to the tensile stresses, must equal the horizontal

force perpendicular to the beam’s cross-section, due

to the compressive stresses, as shown in Figure 8.4.

Hence ,

∫y 1

0

σbdy=

∫y 2

0

σbdy

or

∫y 1

0

σbdy−

∫y 2

0

σbdy= 0

or

∫y 1

−y 2

σbdy= 0

y 1

y 2

Tensile stress

Compressive stress

Figure 8.4