Stress In Bars Of Small Initial Curvature.
Where the radius of curvature is large compared to the dimensions of the cross section, the analysis of stress is similar to that for pure bending.
MISSING IMAGE!
23287/Curved-Beams-007.png cannot be found in /users/23287/Curved-Beams-007.png. Please contact the submission author.
Let

be the initial (unstrained) radius of curvature of the neutral surface and

the radius of curvature under the action of a pure bending moment

.
Then the strain in a

element at a distance

from the neutral axis is given by:
Moment of resistance is a term in structural engineering. It is found from the moment of inertia and the distance from the outside of the object concerned to its major axis.
Strain =
Since
&space;=&space;R_0\theta)
= length along the neutral axis
If

is neglected in comparison with

and noting from
&space;=&space;R_0)
that
Then strain,
Neglecting lateral stress, the normal stress,

strain
Substituting in equation (
1)
Total normal stress = 0, i.e.
which shows that the neutral axis passes through the centroid of the section.
Moment of resistance,
\displaystyle\int&space;f\;y^2\;dA)
from equation (
3)
Combining equations (
2) and (
4),
the strain energy of a short length

(measured along the neutral surface) under the action og bending moment

is:
From equation (
3)
Application To The Design Of A Piston Ring
MISSING IMAGE!
23287/Curved-Beams-008.png cannot be found in /users/23287/Curved-Beams-008.png. Please contact the submission author.
A piston ring is a split ring that fits into a groove on the outer diameter of a piston in a reciprocating engine such as an internal combustion engine or steam engine.
Suppose it is required to design a split ring so that its outside surface will be circular in both
the stressed and unstressed conditions and the radial pressure exerted will be uniform.
If

is the uniform pressure on the outside then the bending moment at

is given by:
R\,sin\phi)
approx
where

is the depth of the ring in the axial direction
integrating
But
)
= a constant for a given condition
i.e.
}{d\,\displaystyle\frac{t^3}{12}})
= constant

when

and
Which is the required variation of thickness.
Using equation (
6). The maximum bending stress at any section
which has it's greatest value when

i.e.
From which,
which determines the initial radius when values for

and

are assumed.
Stresses In Bars Of Large Initial Curvature.
When the radius of curvature is of the same order as the dimensions of the cross section, it is no
longer possible to neglect

in comparison to

and it will be found that the neutral axis does not
pass through the centroid. Further the stress is NOT proportional to the distance from the neutral
axis
MISSING IMAGE!
23287/Curved-Beams-007.png cannot be found in /users/23287/Curved-Beams-007.png. Please contact the submission author.
where

is the strain,

is the distance from the neutral axis as before and

is the initial

radius of the
neutral surface.
For pure bending the Total normal force on the cross section =

.
Moment of resistance,
But
Where

is the distance between the neutral axis and the principle axis which is through the
centroid (

is positive when the neutral axis is on the same side of the centroid as the centre of
curvature)
Substituting in equation (
8)
Rearranging,
In this equation

is positive measured outwards, a positive bending moment being one that tends to
increase the curvature.
Rectangular Cross-section
MISSING IMAGE!
23287/Curved-Beams-003.png cannot be found in /users/23287/Curved-Beams-003.png. Please contact the submission author.
From equation (
7),
Let

= the distance from the centroid. Also the mean radius of curvature

and
Then,
i.e.
Hence,
Giving,
As

is small compared to

and

, it is difficult to calculate with sufficient accuracy from
this equation and the expansion of the log term into a convenient series is of advantage.
Then,
[imperial]
Example - Example 1
Problem
A curved bar, initially unstressed, of square cross section, has

sides and a mean radius of
curvature of
If a bending moment of

is applied to the bar tending to straighten it, find the
stresses at the outer and inner faces.

and

Workings
But

and
At the inside face,
Thus,

Tension
At the outside face,

compression
The actual stress distribution is shown in the diagram.
MISSING IMAGE!
23287/Curved-Beams-004.png cannot be found in /users/23287/Curved-Beams-004.png. Please contact the submission author.
Solution
Tension
Compression
Trapezoidal Cross-section.
MISSING IMAGE!
23287/Curved-Beams-005.png cannot be found in /users/23287/Curved-Beams-005.png. Please contact the submission author.
By Moments,
By putting

and

equation (
7) becomes
i.e.
\displaystyle\int&space;\displaystyle\frac{dA}{(R_m&space;+&space;z)}&space;=&space;0)
or,
From which,
And since,

can be evaluated from equations (
9) and (
10).
[imperial]
Example - Example 2
Problem
A crane hook whose horizontal cross-section is trapezoidal,

wide on the inside and

wide
on the outside by

thick, carries a vertical load of one ton whose line of action is

from the inside edge of this section. The centre of curvature is

from the inside edge.
Calculate the
maximum tensile and
compressive forces set up.
Workings
Referring to the last figure.
MISSING IMAGE!
23287/Curved-Beams-005.png cannot be found in /users/23287/Curved-Beams-005.png. Please contact the submission author.
From equation (9)
Direct stress = load / area =

in. tensile
Bending stress =
At the inside edge,
\;=&space;-&space;3.39\,ton.in.)
(tending to decrease the curvature)
Bending stress =
}{3\;\times&space;\;0.11(2.89&space;-&space;0.89\0}\right)=4.02\;tons/sq.in.)
in tension
The combined stress =

tensile.
At the outside edge,
Bending stress =
Combined stress =

in compression
Circular Cross Section
MISSING IMAGE!
23287/Curved-Beams-006.png cannot be found in /users/23287/Curved-Beams-006.png. Please contact the submission author.
The analysis follows the same method as was used in the previous section on Trapezoidal cross
sections.
Hence,
And
To evaluate the above expand:
And
})
Deflection Of Curved Beams (direct Method)
If the length

of an initially curved beam is acted upon by a bending moment

it
follows from equation (
4) that:
Deflection is a term that is used to describe the degree to which a structural element is displaced under a load.
But

is the change of angle subtended by

at the centre of curvature and consequently is the angle through which the tangent at one end
of the element rotates relative to the tangent at the other end.
i.e.
MISSING IMAGE!
23287/Curved-Beams-001.png cannot be found in /users/23287/Curved-Beams-001.png. Please contact the submission author.
The diagram shows a loaded bar which is fixed in direction at

and it is required to find the
deflection at the other end

.
Due to the action of

on

at

only, the length

is rotated through an angle

.

moves to

', where
The vertical deflection of
The horizontal deflection of
Due to the bending of all the elements along
The vertical deflection at
And the horizontal deflection =
[imperial]
Example - Example 3
Problem
A steel tube having an outside diameter of

and a bore of

is bent into a quadrant of

radius. One end is rigidly attached to a horizontal base plate to which the tangent at that end is perpendicular.
If the free end supports a load of

, determine the
vertical and
horizontal deflection of the free end.
MISSING IMAGE!
23287/Curved-Beams-002.png cannot be found in /users/23287/Curved-Beams-002.png. Please contact the submission author.
Workings

and
Vertical deflection =
Horizontal deflection =
Solution
- Vertical deflection is

- Horizontal deflection is
Deflection From Strain Energy ( Castigliano's Theorem)
Castigliano's method is a method for determining the displacements of a linear-elastic system based on the partial derivatives of the strain energy.
Theorem:
If

is the total strain energy of any structure due to the application of external loads,

at

in the direction

and to the couples

then the deflections at

in the directions

are

and

and the angular rotations of the couples are

,

at their applied points.
Proof for concentrated loads:
If the displacements (in the directions of the loads) produced by gradually applied loads

are

then,
Let

alone be increased by
then,

= increase in external work done
Where,
are increases in
But if the loads

were applied gradually from zero,
the total strain energy,
Subtracting equation (
11) and neglecting the products of small quantities,
Subtracting equation (
12),

or
Similarly for

and

and the proof can be extended to incorporate couples.
It is important to stress that

is the total strain energy, expressed in terms of loads and not
including statically determinate reactions and the partial derivative with respect to each load in
turn (treating the others as constant) gives the deflection at the load points in the direction of
the load.
The following principles should be observed in applying the theorem
- 1) In finding the deflection of curved beams and similar problems, only strain energy due to
bending need normally be taken into account (i.e.

)
- 2) Treat all loads as variables initially carry out the partial differentiation and integration
and only putting in numerical values at the final stage.
- 3) If the deflection is to be found at a point where, or in a direction there is no load, a load
may be put in where required and given a value of zero in the final reckoning (i.e.
![\inline x\;=\left[ \displaystyle\frac{\delta U}{\delta W}\ \right]_{W=0} \right](https://latex.codecogs.com/svg.image?\inline&space;x\;=\left[&space;\displaystyle\frac{\delta&space;U}{\delta&space;W}\&space;\right]_{W=0}&space;\right)
)
Generally it will be found that the strain energy method requires less thought in application than
the direct method, it being only necessary to obtain an expression for the bending moment; also
there is no difficulty over the question of sign as the strain energy is bound to be positive and
deflection is positive in the direction of the load. The only disadvantage occurs when a case such
as mentioned in note 3 above has to be dealt with in which case the direct method will probably be
shorter.
[imperial]
Example - Example 4
Problem
MISSING IMAGE!
23287/Curved-Beams-009.png cannot be found in /users/23287/Curved-Beams-009.png. Please contact the submission author.
Obtain an
expression for the vertical displacement of

in the above diagram. If

and

find the
displacement when

. and
Workings
The bending moments in the various sections can be written as follows:-

(at

' from

)

Constant

(at

from

)

(at

from

)
The displacement of the load at

vertically
An allowance could be made for the linear extension of
Which is clearly negligible compared to the deflection due to bending.