UNSYMMETRICAL BENDING OF

BEAMS

Under the guidance

of

Dr. M. V. RENUKA DEVI

Associate Professor

Department of Civil Engineering,

RVCE, Bangalore

By

VENKATESHA A

(1RV13CSE15)

PURE BENDING

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

1. The normal planes remain normal even after bending.

2. There is no net internal axial force.

3. Stress varies linearly over cross section.

4. Zero stress exists at the centroid and the line of centroid is theNeutral 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

Ixy +

My

Iyx

Bending stress along N.A is σz= 0

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

𝑀𝑥and 𝑀𝑦are applied bending

moments

Fig. Representation of positive internal and external force systems

•We assume𝑀𝑥and 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 definedas Ixx = x

2dA and Iyy= x2dA

Similarly, the product second moment of area of the section is defined asfollows

Ixy= xy dA

Since the cross-section of most structural members used in bending applicationsconsists of a combination of rectangles the value of the product second momentof area for such sections is determined by the addition of the Ixx value for eachrectangle.

Ixx = Ahk

Where h and k are the distances of the centroid of each rectangle from the X andY 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

The relationship between (x, y) and (u, v) co-ordinates are given by

Now substituting for u and v in the above equations we get,

Iuv = uv 𝜕A = 0

By definition

Ixx = y2. 𝜕A ; Iyy = x

2. 𝜕A ; Ixy = xy. 𝜕A ;

Iuu = v2. 𝜕A ; Ivv = u

2. 𝜕A ; Iuv = uv. 𝜕A ;

u = AB+DP = 𝑥 cos ∝ + 𝑦 sin ∝v = GP-HG =GP-EF = 𝑦 cos ∝ − 𝑥 sin ∝

Iuu = {y cos ∝ − x sin ∝}2. 𝜕A

Iuu = cos2 ∝ y2. 𝜕A + sin2 ∝ x2. 𝜕A - 2 sin ∝ cos ∝ xy. 𝜕A

Iuu = cos2 ∝ Ixx + sin2 ∝ Iyy - sin 2 ∝ Ixy

Similarly,

Ivv = cos2 ∝ Iyy + sin2 ∝ Ixx + sin 2 ∝ Ixy

Iuv = cos2 ∝ Ixy - sin2 ∝ Ixy - sin 2 ∝ (Ixx - Iyy)

we have Iuv = 0, therefore we can write

tan 2 ∝ = 2 Ixy

Ixx− Iyy

cos2 ∝ =(1+cos 2∝)

2and sin2 ∝ =

(1−cos 2∝)

2

Iuu = Ixx+ Iyy

2+Ixx− Iyy

2cos 2 ∝ - Ixy sin 2 ∝

Ivv = Ixx+ Iyy

2-Ixx− Iyy

2cos 2 ∝ + Ixy sin 2 ∝

we can write for sin 2 ∝ and cos 2 ∝ as follows

Thus knowing the values of Ixx, Iyy and Ixy , the principal moments of inertia

Iuuand 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.

sin 2 ∝ = −Ixy

(Ixx− Iyy

2)2+Ixy

2

cos 2 ∝ = Ixx−Ixy

(Ixx− Iyy

2)2+Ixy

2

Substituting the values of sin 2 ∝ and cos 2 ∝

Iuu = Ixx+ Iyy

2+ (

Ixx− Iyy

2)2+Ixy

2

Ivv = Ixx+ Iyy

2- (Ixx− Iyy

2)2+Ixy

2

DIRECT STRESS DISTRIBUTION

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.

Fig. bending of an unsymmetrical beam section

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

𝐴𝜎𝑧 𝑑𝐴 = 0

𝐴

𝐸𝑃

𝑅𝑑𝐴 = 0

For a beam of a given material subjected to a given bending moment

A

P dA = 0

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 = Aσz y dA , My = Aσz x dA

Substituting for σz

Mx = E sin∝

R A xy dA +

E cos∝

R A y

2dA

My = E sin∝

R A x2 dA +

E cos∝

R A xy dA

Mx = E sin∝

RIxy +

E cos∝

RIx ; My =

E sin∝

RIy +

E cos∝

RIxy

Bending stress is written as

σz = MxIxy + My

Iyx

Where,

Mx =My −Mx Ixy/Iy

1− (Ixy)2/Ix Iy

; My =My −Mx Ixy/Ix

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

σz =MxIxy +My

Iyx

Position of the neutral axis

The direct stress at all points on the neutral axis of the beam section is zero.

Thus,

O = Mx

IxyN.A +

My

IyxN.A

WherexN.A and yN.A are the coordinates of any point on the neutral

axis. Thus

yN.AxN.A

= − My Ix Mx Iy

tan ∝ = My Ix Mx Iy

Since ∝ is positive when yN.Ais 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 θAlong 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 = 0l Mm

EIdx

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 = 0l M cos θ

E Iuumv dx

Hence the deflection of the beam in direction of UU-axis is given by

δu = 0l M sin θ

E Ivvmu dx

The resultant deflection δ is given by

δ = δu2 + δv

2

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

tanβ = Iuu

Ivvtan θ

Let γ be the inclination of resultant deflection in the direction N!N!-axis

makes with UU-axis

tan γ = -δu

δv

tan γ = - 0lM sin θ

E Ivvmu dx

0lM cos θ

E Iuumv dx

tan γ = -Iuu

Ivvtan θ

tan γ = - tanβ = tan(90 + β)

Therefore we can write as

γ = 90 + β

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 = 5

384

w cos θ l4

E Iuu; δu =

5

384

w sin θ l4

E Ivv

δ = δu2 + δv

2

δ =5

384

wl4 cos θ sec β

EIuu

Multiplying and dividing by cos(β − θ)

δ = 5

384

l4

E

w cos(β−θ)

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.

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.

(Ixx = 48.3 x 10−6m4, Iyy = 4.4 x 10−6m4)

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 10−6m4

Mx = +10000 sin60° x 2 = +17320 Nm

My= -10000 cos60° x 2 = -10000 Nm

But we have,

σz = MxIxy + My

Iyx

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

Ixy +

My

Iyx = P x + Q y

Stress at A = 5725x106x 9x10−3+ 1533x106x 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)2+Ixy

2

= 48.3x10−6+4.4x10−6

2+ (

48.3x10−6− 4.4x10−6

2)2+(−9.91 x 10−6)2

= 50.43x10−6

Ivv = Ixx+ Iyy

2- (Ixx− Iyy

2)2+Ixy

2

= 48.3x10−6+4.4x10−6

2- (48.3x10−6− 4.4x10−6

2)2+(−9.91 x 10−6)2

= 2.27x10−6

tan2 ∝ = 2 Ixy

Iyy− Ixx= −2 x 9.91 x 10−6

(4.4− 48.3)x 10−6= 0.451

2 ∝ = 24°18′ , ∝ = 12°9′

The required stresses can now obtained from the above equation

σ =MvIvu +MuIuv

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

σ =14828 x 115.5 x 10−3

50.43x10−6+13422 x 34.05 x 10−3

2.27x10−6

= 235 MN/m2 as obtained before

3. The deflection at free end of a cantilever is given by

δ =WL3

3EI

Therefore the component of deflection perpendicular to the V axis

δv =WvL3

3EIv= 10000 cos(47°51′) x 23

3 x 200 x 109 x 2.27 x 10−6= 39.4 mm

And component of deflection perpendicular to the U axis

δu =WuL

3

3EIu= 10000 sin(47°51′) x 23

3 x 200 x 109 x 50.43 x 10−6= 1.96 mm

The total deflection is then given by

δ = δu2 + δv

2 = 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

δ = 5

384

l4

E

w cos(β−∝)

Inn

The resultant deflection for cantilever beam is

δ =L3

3E

w cos(β−∝)

Inn

Taking Inn = Iuu cos2 β + Ivv sin

2 β

In order to overcome the effect of twisting when the beam subjected to

unsymmetrical loading, the study of unsymmetrical bending is useful.

REFERENCES

1. Arthur P. Boresi, Richard J.Schindt, “Advanced Mechanics of materials”, Sixth editionJohn Wiley &Sons. Inc., New Delhi, 2005

2. Thimoshenko & J N Goodier, “Mechanics Of Solids”, Tata McGraw-Hill publishingCo.Ltd, New Delhi, 1997

3. Seely Fred B. and Smith James O., “ Advanced Mechanics of Materials”, 2nd edition,John Wiley & Sons Inc, New York, 1952

4. Srinath L.S., “Advanced Mechanics of Solids”, Tata McGraw-Hill publishing Co.Ltd,New Delhi, 1980

5. Thimoshenko S., “Strength of Materials”, Part-1, Elementary Theory and Problems, 3rd

Edition, D. Van Nostrand company Inc., New York, 1955

6. Boresi A.P and Chong K.P(2000), “Elasticity In Engineering Mechanics” 2nd editionNew York ; Wiley – Interscience.

7. N Krishna Raju & D R Gururaja, “Advance Mechanics Of Solids & Structures”, 1997

8. B C Punmia & A K Jain. “Strength of Materials and Theory of Structures”, Vol.2Lakshmi publications (P) Ltd.2002

9. Jiang Furu, “Unsymmetrical Bending Of Cantilever Beams”, Appl. Math. & Mech.(English Ed.), 1982

10. G. D. Williams, D. R. Bohnhoff, R. C. Moody, “Bending Properties of Four-Layer Nail-Laminated Posts”, United States Department of Agriculture, Forest Service (ForestProducts Laboratory), Research Paper FPL–RP–528, 1994

11. Poligona industrial san jose de valderas, spain, “Unsymmetrical Cantilever Unit”,ED01/12, 2012