Mechanical Engineering Principles

BENDING OF BEAMS 105

But from equation (8.5), σ=E

y

R

Therefore,

∫y 1

−y 2

Ey

R

bdy= 0

Now,EandRare constants, hence

E

R

∫y 1

−y 2

ybdy= 0

However,

∫
ybdy=the first moment of area about
the centroid, and where this is zero, coincides with
the centroidal axis, i.e. theneutral axislies on the
same axis as thecentroidal axis.

Moment of resistance (M)

From Figure 8.4, it can be seen that the system of
tensile and compressive stresses perpendicular to the
beam’s cross-section, cause a couple, which resists
the applied momentM,whereM=

∫y 1

−y 2 σ(bdy)y

But from equation (8.5), σ=E

y

R

Hence, M=

E

R

∫

y^2 bdy

or M=

EI

R

(as required)

8.3 Worked problems on the bending

of beams

Problem 1. A solid circular section bar of

diameter 20 mm, is subjected to a pure

bending moment of 0.3 kN m. If

E= 2 × 1011 N/m^2 , determine the resulting

radius of curvature of the neutral layer of

this beam and the maximum bending stress.

From Table 7.1, page 91,

I=

πd^4

64

=

π× 204

64

=7854 mm^4

Now, M= 0 .3kNm×

N

kN

1000 × 1000

mm

m

= 3 × 105 Nmm

and E= 2 × 1011

N

m^2

× 1

m

1000 mm

× 1

m

1000 mm

= 2 × 105 N/mm^2

From equation (8.8),

M

I

=

E

R

hence, radius of curvature,

R=

EI

M

= 2 × 105

N

mm^2

×

7854 mm^4

3 × 105 Nmm

i.e. R=5236 mm= 5 .24 m

From equation (8.9),

σ

y

=

M

I

and σˆ=

Myˆ

I

where σˆ=maximum stress due to bending

and yˆ=outermost fibre fromNA

=

d

2

=

20

2

=10 mm.

Hence,maximum bending stress,

σˆ=

Myˆ

I

=

3 × 105 Nmm×10 mm

7854 mm^4

=382 N/mm^2 = 382 × 106 N/m^2

=382 MPa

Problem 2. A beam of length 3 m is simply

supported at its ends and has a cross-section,

as shown in Figure 8.5. If the beam is

subjected to a uniformly distributed load of

2 tonnes/m, determine the maximum stress

due to bending and the corresponding value

of the radius of curvature of the neutral layer.

0.1 m

0.2 m Thickness = 0.02 m

Figure 8.5