Pure-bending of curved bar
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Transcript of Pure-bending of curved bar
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 …