11CHAPTER

Design of Curved Beams

NOTATIONS AND SYMBOLS USED

σ = Stress, MPa.σd = Direct stress, tensile or compressive, MPa.

σi = Stress at inner fibre, MPa.

σo = Stress at outer fibre, MPa.

σbi = Normal stress due to bending at inner fibre, MPa.

σbo = Normal stress due to bending at outer fibre, MPa.

bM = Bending moment for critical section, N-mm.

ic = Distance of neutral axis from inner fibre, mm.

oc = Distance of neutral axis from outer fibre, mm.

e = Eccentricity, mm.

ir = Distance of inner fibre from centre of curvature, mm.ro = Distance of outer fibre from centre of curvature, mm.

cr = Distance of centroidal axis from centre of curvature, mm.

nr = Distance of neutral axis from centre of curvature, mm.d = Diameter of circular rod used in curved beam, mm.h = Depth of curved beam [square, rectangular, trapezoidal or I-section], mm.

τmax = Maximum shear stress, MPa.A = Area of cross-section of member, (curved beam), mm2.P = Load on member, N.

INTRODUCTION

Machine frames having curved portions are frequently subjected to bending or axial loads or toa combination of bending and axial loads. With the reduction in the radius of curved portion,the stress due to curvature become greater and the results of the equations of straight beamswhen used becomes less satisfactory. For relatively small radii of curvature, the actual stresses

2 Design of Machine Elements - II

may be several times greater than the value obtained for straight beams. It has been found fromthe results of Photoelastic experiments that in case of curved beams, the neutral surface doesnot coincide with centroidal axis but instead shifted towards the centre of curvature. It has alsobeen found that the stresses in the fibres of a curved beam are not proportional to the distancesof the fibres from the neutral surfaces, as is assumed for a straight beam.

Differences between Straight and Curved Beams

Straight beam Curved beam

1. The neutral axis of beam coincides The neutral axis is shifted towardswith centroidal axis. the centre of curvature by a distance

called eccentricity i.e. the neutralaxis lies between centroidal axisand centre of curvature.

2. The variation of normal stress due The variation of normal stress due to bending is linear, tensile at the inner to bending across section isfibre and compressive at the outer fibre non-linear and is hyperbolic.with zero value at the centroidal axis.

Derivation of Expression to Determine Stress at any Point on the Fibres ofa Curved Beam

Consider a curved beam with rc, as the radius of centroidal axis, rn, the radius of neutralsurface, ri, the radius of inner fibre, ro, the radius of outer fibre having thickness ‘h’ subjectedto bending moment Mb.

Let AB and CD be the two adjacent cross-sections separated from each other by a smallangle dφ.

Because of Mb the section CD rotates through a small angle dα. The unit deformation ofany fibre at a distance y from neutral surface is

Deformation( )n

ydl r y dδ α

∈= =− φ

… (1.1)

The unit stress on this fibre is,Stress = Strain × Young’s modulus of material of beam

( )n

EydE

r y dα

σ = ∈ =− φ … (1.2)

For equilibrium, the summation of the forces acting on the cross sectional area must bezero.

i.e., σ =∫ 0dA

or ( ) 0–n

E yd dAr y d

α =φ∫

α=

φ ∫ 0–n

d ydAE

d r y … (1.3)

Design of Curved Beams 3

Also the external moment Mb applied is resisted by internal moment. From equation 1.2we have,

( )y dA Mσ =∫

i.e., ( )α =

φ∫2

–n

E y d dAM

r y d

( )α =φ −∫

2

n

d y dAE M

d r y … (1.4)

i.e. , ( )α= − +

φ −∫ ∫( ) nn

Ed ydAM y dA r

d r y … (1.5)

Note: In equation 1.5, the first integral is the moment of cross sectional area with respect toneutral surface and the second integral is zero from equation 1.3.

Therefore, α= φ

dM E Ae

d … (1.6)

Here ‘e’ represents the distance between the centroidal axis and neutral axis.

i.e., c ne r r= −Rearranging terms in equation 1.6, we get

α

=φ

d Md AeE ... (1.7)

Substituting α

=φ

d Md AeE in equation 1.2,

Stress ( )ασ =

− φn

Ey dr y d becomes

Figure: 1.1

B

dφ

C1

dα

dα

O

ro

rc

rnri

e

A

H

C

y

v

D

h

Mb

D1

Centre of curvature

Inner fibre

Mb

Centroidal axis

Outer fibre

Neutral axis

4 Design of Machine Elements - II

( )=−n

Ey Mr y AeE

i.e., ( )σ =–n

Myr y Ae … (1.8)

From figure 1.1,rn = v + y

or y = rn – vTherefore, from equation 1.3

( )( )

( )=∫ ∫–

– –n

n n

r v dAydAr y r y

– 0ndA

r dAv

= =∫ ∫

0ndA

r AV

= − =∫

or nA

rdAv

=

∫…(1.9)

Note: Since e = rc – rn, equation 1.9 can be used to determine ‘e’. Knowing the value of ‘e’, equation 1.8 is used to determine the stress σ.

Straight and Curved Beams

(a) Straight beamConsider a straight beam having moment of inertia ‘I’ subjected to bending moment Mb as

shown in (Fig. 1.2).

Centre of gravity axiscoinciding with neutral axis −σb

−y

+y

σb

B

Normal stressdue to bending (Compressive)

Normal stressdue to bending (Tensile)

MbMb

A

Figure: 1.2

Design of Curved Beams 5

From fundamental equation of bending,

σ= =b bM E

I y R ; σ

=b bMI y

i.e., Bending stress σ = bb

My

I

For a given beam, Mb and ‘I’ are constant and hence σb αy, i.e., the variation of bendingstress is linear and is directly proportional to its distance from centre of gravity axis whichcoincides with neutral axis. The maximum bending (tensile) stress σb tensile at A andcompressive at B are equal in magnitude at ‘A’ and ‘B’ as shown in the Fig. 1.2.

(b) Curved beamFigure 1.3 shows a curved beam subjected to bending moment Mb.

Let ri = Distance of inner fibre from centre of curvature, Cro = Distance of outer fibre from centre of curvaturerc = Distance of centroidal axis (CG axis) from centre of curvaturern = Distance of neutral axis from centre of curvature

CCentre of curvature

Mb

Mb

ri

ro

rc

rn

σbo

co

e

σbi

ci

Figure: 1.3

The neutral axis is shifted towards the centre of curvature by a distance called eccentricity‘e’. The value ‘e’ should be computed very accurately since a small variation in the value of ‘e’causes a large variation in the values of stress.

i.e., e = rc – rn

ci = Distance between neutral axis and inner fibre = rn – ri

co = Distance between outer fibre and neutral axis = ro – rn