1. Pure Bending of
curved bars
By
Pratish Bhaskar Sardar
(122090025)

2. CONTENTS
Pure Bending of Curved Bars.
Boundary conditions of the problem.
Numerical Examples

3. Pure Bending of curved bars

4. 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.

5. Basic concept

6. Where…
 b = Radius of outer fiber
 a = Radius of inner fiber
 l = Width of section
 ro= Radius of centroidal axis
 M=Bending moment applied

7.  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)

8. 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

9. 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

10.  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)

11.  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)

12. NUMERICAL EXAMPLES

15. Thank You …