Pure Bending of curved bars

ByPratish Bhaskar Sardar

(122090025)

CONTENTS

Pure Bending of Curved Bars.

Boundary conditions of the problem.

Numerical Examples

Pure Bending of curved bars

CURVED MEMBERS IN BENDING

The distribution of stress in a curved flexural member is determined by using the following assumptions.

The cross section has an axis of symmetry in a plane along the length of the beam.

Plane cross sections remain plane after bending.

The modulus of elasticity is the same in tension as in compression.

Basic concept

Where…

b = Radius of outer fiber

a = Radius of inner fiber

l = Width of section

ro= Radius of centroidal axis

M=Bending moment applied

In the absence of body forces equilibrium

equations are satisfied by stress function υ(r,θ)

for which stress components in radial and

tangential directions are

σr = (1/r) (∂υ/∂r) + (1/r2) (∂2υ/∂θ2)

σθ = (∂2υ/∂r2)

τrθ = (1/r2) (∂υ/∂θ) - (1/r) (∂2υ/∂r∂θ)

boundary conditions

1 at r = a , σr = 0

2 at r = b , σr = 0

3 τrθ = 0 for all boundaries

at either end of beam circumferential normal stresses

must have a zero resultant force and equivalent to

bending moment M on each unit width of beam

4 ∫ σθ dr = 0 ∫ σθ r dr = M

Standard Relations…

from BC's 1 and 2

(B/a2) + 2C + D(1+ 2ln a) = 0 and

(B/b2) + 2C + D(1+ 2ln b) = 0

from BC 4

υab = B ln (b/a) + C (b2 -a2) + D (b2 ln b - a2 ln a) = -M

B = (4M/Q)a2 b2 ln(b/a)

C = M/Q 2(b2lnb-a2lna) +b2 –a2

D= 2 M/Q(b2-a2)

Where,

Q =(b2-a2)2 -4a2b2ln(b2/a2)

Radial Stress

• σr = (1/r) (∂υ/∂r) + (1/r2) (∂2υ/∂θ2)

=B/r2 +2C+D(1+lnr)

• circumferential stress

σθ = (∂2υ/∂r2)

= -(B/r2)+2C+D(3+2lnr)

NUMERICAL EXAMPLES

Thank You …