Unsymmetrical Bending
PURE BENDING



1.
2.
3.
4.
Bending is a very severe form of stressing a structure
The simple bending theory applies when bending takes place about an
axis which is perpendicular to a plane of symmetry.
Bending moments acts along the axis of the member.
Assumptions made in pure bending
The normal planes remain normal even after bending.
There is no net internal axial force.
Stress varies linearly over cross section.
Zero stress exists at the centroid and the line of centroid is the
Neutral Axis (N.A)
•Bending stress and strain at any point may be computed as
Symmetrical bending : The plane of loading or the plane of
bending is co-incident with or parallel to, a plane containing
principal centroidal axes of inertia of the cross-section of the
beam.
 Bending stress is given by

σz =

Mx
Ix
y+
Bending stress along N.A is σz = 0
My
Iy
x
UNSYMMETRICAL BENDING
Assumptions
1.
The plane sections of the beam remain plane after bending
2.
The material of the beam is homogeneous and linearly elastic.
3.
There is no net internal axial force.
Sign conventions and notation
 u,
v and w are the
displacement components
of
any point within beam parallel to
x, y, z axes.
 P = axial load and T = torque
 𝑤𝑥 (z) and 𝑤𝑦 (z)are
distributed loads
𝑀𝑥 and 𝑀𝑦 are applied bending
moments
Fig. Representation of positive internal and external force systems
•We assume 𝑀𝑥 and 𝑀𝑦 as positive when they each induce tensile stresses in the
positive xy quadrant of the beam section.
PRODUCT SECOND MOMENT OF AREA

The second moments of area of the surface about the X and Y axes are defined
as Ixx = x 2 dA and Iyy = x 2 dA

Similarly, the product second moment of area of the section is defined as
follows
Ixy = xy dA

Since the cross-section of most structural members used in bending applications
consists of a combination of rectangles the value of the product second moment
of area for such sections is determined by the addition of the Ixx value for each
rectangle.
Ixx = Ahk

Where h and k are the distances of the centroid of each rectangle from the X and
Y axes respectively (taking account of the normal sign convention for x and y)
and A is the area of the rectangle.
DETERMINATION OF PRINCIPAL AXIS OF SECTION
Let,
 U-U, V-V be the Principal Centroidal Axes,
 X-X, Y-Y be the pair of orthogonal axes,
 α be the angle between both the axes system,
 𝜕𝑎 Be the elementary area with co-ordinates (u, v) referred to the principal
axes i.e., U-V axes and (x, y) referred to the X-Y axes, Since, U-V axes are
principal axes the product of inertia = 0
Iuv =
uv 𝜕A = 0
By definition
Ixx = y 2 . 𝜕A ; Iyy =
x 2 . 𝜕A ; Ixy =
xy. 𝜕A ;
v 2 . 𝜕A ; Ivv =
u2 . 𝜕A ; Iuv =
uv. 𝜕A ;
Iuu =
 The relationship between (x, y) and (u, v) co-ordinates are given by
u = AB+DP = 𝑥 cos ∝ + 𝑦 sin ∝
v = GP-HG =GP-EF = 𝑦 cos ∝ − 𝑥 sin ∝
 Now substituting for u and v in the above equations we get,
Iuu =
{y cos ∝ − x sin ∝}2 . 𝜕A
Iuu = cos2 ∝
y 2 . 𝜕A + sin2 ∝
x 2 . 𝜕A - 2 sin ∝ cos ∝
Iuu = cos 2 ∝ Ixx + sin2 ∝ Iyy - sin 2 ∝ Ixy
xy. 𝜕A
Similarly,
Ivv = cos2 ∝ Iyy + sin2 ∝ Ixx + sin 2 ∝ Ixy
Iuv = cos 2 ∝ Ixy - sin2 ∝ Ixy - sin 2 ∝ (Ixx - Iyy )
we have Iuv = 0, therefore we can write
tan 2 ∝ =
cos 2
Iuu =
Ivv =
∝=
2 Ixy
Ixx − Iyy
(1+cos 2∝)
2
Ixx + Iyy
2
Ixx + Iyy
2
+
-
and
Ixx − Iyy
2
Ixx − Iyy
2
sin2
∝=
(1−cos 2∝)
2
cos 2 ∝ - Ixy sin 2 ∝
cos 2 ∝ + Ixy sin 2 ∝
we can write for sin 2 ∝ and cos 2 ∝ as follows
−Ixy
sin 2 ∝ =
(
Ixx − Iyy
2 +I 2
)
xy
2
Ixx −Ixy
cos 2 ∝ =
(
Ixx − Iyy
2 +I 2
)
xy
2
Substituting the values of sin 2 ∝ and cos 2 ∝
Iuu =
Ivv =
Ixx + Iyy
2
Ixx + Iyy
2
Ixx − Iyy 2
) +Ixy 2
2
+ (
-
Ixx − Iyy 2
(
) +Ixy 2
2
Thus knowing the values of Ixx , Iyy and Ixy , the principal moments of inertia
Iuu and Ivv can be calculated from the above analytical expression.
Note: moment of inertia of a section about its principal axes have maximum and
minimum values.
DIRECT STRESS DISTRIBUTION
Fig. bending of an unsymmetrical beam section

We know that a beam bends about the neutral axis of its cross section so
that the radius of curvature, R, of the beam is perpendicular to the neutral
axis.
σZ = E
p
R

The beam section is subjected to a pure bending moment so that the resultant direct
load on the section is zero. Hence
𝜎
𝐴 𝑧
𝑑𝐴 = 0
𝑃
𝐸 𝑑𝐴 = 0
𝐴 𝑅

For a beam of a given material subjected to a given bending moment
P dA = 0
A

Above equation states that the first moment of area of the beam section about the
neutral axis is zero. It follows pure bending of beams in which the neutral axis
always passes through the centroid of the beam section.
p = x sin ∝ + y cos ∝
σz =

E
R
(x sin ∝ + y cos ∝)
The moment resultants of the direct stress distribution are
Mx =
A
σz y dA , My =
A
σz x dA

Substituting for σz
E sin∝
E cos∝
2
Mx =
xy
dA
+
y
dA
A
A
R
R
E sin∝
E cos∝
2
My =
x dA +
xy dA
A
A
R
R
Mx =

E sin∝
R
Ixy +
E cos∝
R
Ix ; My =
E sin∝
R
Iy +
E cos∝
R
Ixy
Bending stress is written as
My
Mx
σz =
y+
Ix
Iy
x
Where,
Mx =
My −Mx Ixy /Iy
My −Mx Ixy /Ix
; My =
1− (Ixy )2 /Ix Iy
In the case where the beam section has either Ox or Oy (or both) as an axis of
symmetry, then Ixy is zero and Ox, Oy are principal axes, Mx = Mx , My =
My
My
Mx
σz =
y+
x
Ix
Iy
1− (Ixy )2 /Ix Iy

Position of the neutral axis
The direct stress at all points on the neutral axis of the beam section is zero.
Thus,
O=
Mx
Ix
yN.A +
My
Iy
xN.A
WherexN.A and yN.A are the coordinates of any point on the neutral
axis. Thus
yN.A
xN.A
= −
tan ∝ =
My Ix
Mx Iy
My Ix
Mx Iy
Since ∝ is positive when yN.A is negative and xN.A is positive
DEFLECTION OF BEAM UNDER UNSYMMETRICAL BENDING
Let the bending moment “M” inclined at an angle “θ” with one of principal
planes (Say VV-axis)
Along UU-axis M component will be Mvv = M cos θ
Along VV-axis M component will be Muu = M sin θ

From the application of principal of virtual work (unit load method) the
deflection (δ) of the beam in any direction, due to a bending moment M is
given by
δ=
lMm
0 EI
dx
Where,
M = moment due to applied moment (say M)
m = moment due to unit load applied at the point in the direction of the desired
deflections,
dx = elementary length of beam, measured along the span of the beam
Hence the deflection of the beam in the direction of VV- axis is given by
δv =
l M cos θ
mv
0 E Iuu
dx
Hence the deflection of the beam in direction of UU-axis is given by
δu =
l M sin θ
mu
0 E Ivv
The resultant deflection δ is given by
δ = δ2u + δ2v
dx


Since mv ,mu are the moments developed due to unit load applied, it can be
taken both equal to m i.e. ( mv = mu = m)
If β is the inclination of neutral axis (NN-axis) with respect to UU-axis we
can write as
tan β =
Iuu
Ivv
tan θ
Let γ be the inclination of resultant deflection in the direction N ! N! -axis
makes with UU-axis
tan γ = tan γ = -
lM sin θ
mu dx
0 E Ivv
lM cos θ
0 E Iuu mv dx
tan γ = tan γ = - tan β = tan(90 + β)
Therefore we can write as
γ = 90 + β
δu
δv
Iuu
tan θ
Ivv
Hence the resultant deflection occurs in the direction exactly perpendicular
to the neutral axis (N ! N ! - axis perpendicular to NN-axis)
Let us consider the case of simply supported beam (SSB) subjected to
UDL, then,
δv =
δ=
5 w cos θ l4
384
E Iuu
; δu =
5 w sin θ l4
384
E Ivv
δ2u + δ2v
δ=
5 wl4 cos θ sec β
384
EIuu
Multiplying and dividing by cos(β − θ)
δ=

5 l4 w cos(β−θ)
384 E
Inn
Thus from the above expression for a simply supported beam (SSB) we
can conclude that the term w cos(β−θ) is the resultant udl acting along
N ! N ! - axis which is perpendicular to neutral axis.
PROBLEMS ON UNSYMMETRICAL BENDING
A Cantilever Problem
1.A horizontal cantilever 2 m long is constructed from the Z-section shown
below. A load of 10 KN is applied to the end of the cantilever at an angle
of 60°to the horizontal as shown. Assuming that no twisting moment is
applied to the section, determine the stresses at points A and B.

(Ixx = 48.3 x 10−6 m4 , Iyy = 4.4 x 10−6 m4 )
2.Determine the principal second moments of area of the section and hence,
by applying the simple bending theory about each principal axis, check
the answers obtained in part1.
3. What will be the deflection of the end of the cantilever? E = 200 GPa.
In the given section Ixy for the web is zero since its centroid lies on both axes
and hence h and k are both zero. The contributions to Ixy of the other two
portions will be negative since in both cases either h or k is negative.
Therefore, Ixy = -2(80 x 18) (40 - 9) (120 - 9) 10−12
= -9.91 x 10−6 m4
Mx = +10000 sin60° x 2 = +17320 Nm
My = -10000 cos60° x 2 = -10000 Nm
But we have,
My
Mx
σz =
y+
x
Ix
Iy
Let
My
Iy
= P,
Mx
Ix
=Q
Mx = P Ixy + Q Ix ; My = -P Iy - Q Ixy
17320 = P (-9.91) x10−6 + Q 48.3x10−6
-10000 = P (-4.4x10−6 ) + Q 9.91x10−6
Solving the above two equations for P and Q,
P = 5725x106 ; Q = 1533x106
The inclination of the N.A relative to the X axis is then given by
tan ∝ =
=
My Ix
Mx Iy
P
Q
° ′
=-
5725 x 106
1533 x 106
= - 3.735
∝ = −75 1
Now
σz =
Mx
Ix
y+
My
Iy
x=Px+Qy
Stress at A = 5725x106 x 9x10−3 + 1533x106 x 120x10−3
= 235 N/mm2
Similarly stress at B = 235 MN/mm2
The points A and B are on both side of neutral axis and equidistant from it.
Stresses at A and B are therefore of equal magnitude but with opposite sign.
2. The principal second moments of area may be found from the following
equations
Iuu =
=
Ixx + Iyy
2
Ixx − Iyy 2
) +Ixy 2
2
+ (
48.3x10−6 +4.4x10−6
2
+
48.3x10−6 − 4.4x10−6 2
(
) +(−9.91
2
x 10−6 )2
= 50.43x10−6
Ivv =
=
Ixx + Iyy
2
Ixx − Iyy 2
) +Ixy 2
2
- (
48.3x10−6 +4.4x10−6
2
-
48.3x10−6 − 4.4x10−6 2
(
) +(−9.91
2
= 2.27x10−6
tan 2 ∝ =
2 Ixy
Iyy − Ixx
=
−2 x 9.91 x 10−6
(4.4− 48.3)x 10−6
2 ∝ = 24° 18′ , ∝ = 12° 9′
= 0.451
x 10−6 )2
The required stresses can now obtained from the above equation
Mv
Mu
σ =
u+
v
Iv
Iu
Mu = 10000 sin(60° -12° 9′ ) x 2
= 10000sin(47° 51′ )
= 14828 Nm and
Mv = 10000cos(47° 51′ ) x 2
= 13422 Nm
And, for A,
u = xcos ∝ + ysin ∝
u = (9 x 0.9776) + (120 x 0.2105) = 34.05 mm
v = ycos ∝ - xsin ∝
v = (120 x 0.9776) - (9 x 0.2105) = 115.5 mm
14828 x 115.5 x 10−3
13422 x 34.05 x 10−3
σ =
+
50.43x10−6
2.27x10−6
= 235 MN/m2 as obtained before
3. The deflection at free end of a cantilever is given by
WL3
δ=
3EI
Therefore the component of deflection perpendicular to the V axis
δv =
Wv L3
3EIv
=
10000 cos(47° 51′ ) x 23
= 39.4 mm
3 x 200 x 109 x 2.27 x 10−6
And component of deflection perpendicular to the U axis
δu =
Wu L3
3EIu
=
10000 sin(47° 51′ ) x 23
= 1.96 mm
3 x 200 x 109 x 50.43 x 10−6
The total deflection is then given by
δ = δ2u + δ2v = 39.42 + 1.962 = 39.45 mm

Its direction is normal to the N.A.
UNSYMMETRICAL CANTILEVER UNIT [11]
 To demonstrate unsymmetrical bending of
beams
 Determines deflections along u and v
directions
 Consist of
1. Main column (cantilever specimen
clamped at its bottom)
2. Loading head at upper end – can rotate
180° with 15° intervals about vertical
axis
3. Set of pulley, located at the loading head,
to apply a horizontal load.
4. 2 Dial gauges of 0-25 mm and 0.01
mm accuracy, to measure and deflections.
 Poligona industrial san jose de valderas,
spain demonstrated this model.
 Limitations
-Dimensions: 400 x 300 x 400 mm approx.
-Weight: 14 Kg. approx.
IMPORTANCE OF UNSYMMETRICAL BENDING





If the plane of bending or plane of loading does not lie in or parallel to the
plane that contains the principal centroidal axes of cross-section, the
bending is called as unsymmetrical bending.
Members that are not symmetrical about the vertical axes and which are
typically composed of thin unsymmetrical sections (e.g. ISA, Channel)
undergo phenomenon of twisting under the transverse loads.
A channel section carrying the transverse load would twist because the line
of action of the load does not pass through shear centre of the member.
whereas rectangular beam would not twist because the loading would pass
through the centre of gravity of the section and for such two axis
symmetrical section the shear centre would coincide with the cg of the
section.
If one is desired to use unsymmetrical sections to carry transverse loads
without twisting, it is possible to do so by locating the load so that it passes
through the shear centre of the beam.
CONCLUSION



The axis of about which the product moment of inertia is zero is called as
principal axis. Hence we can conclude by saying that the simple bending
theory is applicable for bending about principal axes only.
It should be noted that moment of inertia of a section about its principal axes
have maximum and minimum values respectively.
The resultant deflection for simply supported beam subjected unsymmetrical
bending is
5 l4 w cos(β−∝)
δ=
384 E
Inn
The resultant deflection for cantilever beam is
δ=

L3 w cos(β−∝)
3E
Inn
Taking Inn = Iuu cos 2 β + Ivv sin2 β
In order to overcome the effect of twisting when the beam subjected to
unsymmetrical loading, the study of unsymmetrical bending is useful.