Bending of Curved Beams – Strength of Materials Approach N M V r θ cross-section must be symmetric but does not have to be rectangular assume plane sections remain plane and just rotate about the neutral axis, as for a straight beam, and that the only significant stress is the hoop stress θθ σ θθ σ
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curved beam strength - Rice University · PDF fileBending of Curved Beams ... Comparison of the ratio of the max bending stresses 5.0 0.9994 0.9331 ... Curved beam formula R/h R h
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Bending of Curved Beams – Strength of Materials Approach
N
M
V
r
θ
cross-section must be symmetric but does not have to be rectangular
assume plane sections remain plane and just rotate about the neutral axis, as for a straight beam, and that the only significant stress is the hoop stress
θθσ
θθσ
N
M
centroid neutral axis
nRR
r
R = radius to centroidRn = radius to neutral axisr = radius to general fiber in the beam
N, M = normal force and bending momentcomputed from centroid
θ∆
A
B
A
B
P
P
nR r−
φ∆
( ) 1n nR r Rlel r rθθ
φω
θφωθ
− ∆∆ ⎛ ⎞= = = −⎜ ⎟∆ ⎝ ⎠∆
=∆
Let ∆φ = rotation of thecross-section
Reference: Advanced Mechanics of Materials : Boresi, Schmidt, andSidebottom
From Hooke’s law
1nREe Erθθ θθσ ω ⎛ ⎞= = −⎜ ⎟
⎝ ⎠
Then the normal force is given by
( )
nA A A
n m
dAN dA E R dAr
E R A A
θθσ ω
ω
⎡ ⎤= = −⎢ ⎥
⎣ ⎦= −
∫ ∫ ∫
wherem
A
dAAr
= ∫ has the dimensions of a length
Similarly, for the moment ( )
( )
( )
1
A
n
A
n nA A A A
n m
M R r dA
RE R r dAr
dAE R R R dA R dA rdAr
E R RA A
θθσ
ω
ω
ω
= −
⎛ ⎞= − −⎜ ⎟⎝ ⎠
⎡ ⎤= − − +⎢ ⎥
⎣ ⎦= −
∫
∫
∫ ∫ ∫ ∫
( )( )
n m
n m
M E R RA A
N E R A A
ω
ω
= −
= −
from (1)
(1)
(2)
nm
ME RRA A
ω =−
from (2) ( )n m
m
m
N E R A E AMA E A
RA A
ω ω
ω
= −
= −−
so solving for Eω( )
m
m
MA NEA RA A A
ω = −−
1n
n
REe Er
E R Er
θθ θθσ ω
ω ω
⎛ ⎞= = −⎜ ⎟⎝ ⎠
= −
Recall, the stress is given by
so using expressions for ,nE R Eω ω
we obtain the hoop stress in the form
( )( )
m
m
M A rANA Ar RA Aθθσ
−= +
−
axialstress
bendingstress
nr R=
setting the total stress = 0 gives
0N ≠
( )0m m
AMrA M N A RAθθσ =
=+ −
0N =
setting the bending stress = 0 and gives nm
ARA
=
which in general is not at the centroid
location of theneutral axis
For composite areas
A1
A2
1R2R
i
m mi
i i
i
A A
A A
R AR
A
=
=
=
∑∑∑∑radii to centroids
areas
Example
P
P
100 mm30 mm
For a square 50x50 mm cross-section, find the maximum tensile and compressive stress if P = 9.5 kN and plot the total stress across the cross-section