Bending of Open & Closed Cross Section Beams

Bending of open and closed section beams

Prof. Ravi kumarAM.AeSI, M.Tech (Aero)

SoME, SASTRA University

Introduction

An aircraft is basically an assembly of stiffened shell structures ranging from single cell closed section fuselage to multi-cellular wings and tail surfaces, each subjected to bending, torsional, shear and axial loads.

Other , smaller portions of the structure consist of thin walled channel, T-, Z-, ‘top hat’ – or I-sections, which are used to stiffen the thin skins of the cellular components and provide supports for internal loads from floors, engine mountings, etc. structural members such as known as open section beams.

Cellular components are termed as closed section beams.

MONOCOQUE

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TYPICAL WING COMPONENTS:• Stressed-skin Wing Construction

BENDING

Symmetrical bending• Symmetrical bending arises in beams which have either

singly or doubly symmetrical cross –sections.

• The neutral axis of the cross-section is found to be coincident with the axis of couple.

Un-symmetrical bending Situations where couples do not act in a plane of symmetry of the

member or the member does not possess any plane of symmetry.

Neutral axis of cross section does not usually coincide with the axis of couple.

Anticlastic bending

as shown above direct stress distribution due to negative B.M.applied in a vertical plane varies from compression in the upper half of the beam to tension in the lower half. However due to the Poisson effect, compressive stress produces a lateral elongation of the upper fibers of the beam section, while the tensile stress produces a lateral contraction of the lower. The section does not therefore remain rectangular but distorts as shown in Fig.c ; the effect is known as anticlastic bending.

Neutral Surface Definition

In the process of bending there is an axial line that do not extend or contract. The surface described by the set of lines that do not extend or contract is called the neutral surface. Lines on one side of the neutral surface extend and on the other contract since the arc length is smaller on one side and larger on the other side of the neutral surface. The figure shows the neutral surface in both the initial and the bent configuration.

Direct stress calculation due to symmetrical bending

Consider a length of beam, subjected to pure sagging moment , M , applied in vertical plane. Bending moment will cause the length of beam to bend in a similar way as shown in Fig.

Direct stress calculation due to symmetrical bendingSince fiber ST has changed in length ( original length was δz), it has

suffered a strain which is given by We know that

Then,

i.e.,

So that (1)

From hookes law (2)

Direct stress calculation due to symmetrical bending

The direct or normal force on the c/s of the fiber ST is However, since the direct stress in the beam section is due to pure bending moment( as there is no axial load), the resultant normal force on the complete c/s of the beam must be zero.

Or

In which both E and R are constants for a beam of a given material subjected to given bending moment. Therefore,

Which states that the first moment of area of the c/s of the beam with respect to N.A. is zero. Thus we see that the N.A. passes through the centroid of area of the c/s.

Direct stress calculation due to symmetrical bendingConsider now the elemental strip δA as shown in Fig. as strip is above

the N.A. so there will be a compressive force acting on its c/s, which is numerically equal to (Ey/R)δA. This force will exert a clockwise moment about the N.A.

So

Or (3)

The term is known as second moment of area of c/s about N.A.So

(4)Combining Equations (2) and (4), we have

, Resultant bending Equation.

Example Problem

Q.1. the c/s of a beam has the dimensions as shown in Fig., If the beam is subjected to a negative bending moment of 100 kNm applied in vertical plane, determine the distribution of the direct stress through the depth of the section.

Solution

As beam is doubly symmetric.

So direct stress, therefore, varies linearly through the depth of thesection from y= +150 mm to y= -150 mm.So at the top of the beam,

And bottom of the beam,

Example problem

Q.2. The beam of Q.1, is subjected to a bending moment of 100 kNm applied in a plane parallel to the longitudinal axis of the beam but inclined at 30̊ to the left of vertical . The sense of bending moment is clockwise when viewed from the left hand edge of the beam section. Determine the distribution of direct stresses.

Solution

Bending moment can be resolved in two planes as : ,

So,

Separate distribution then can be determine and superimposed. A moredirect method is to combine the two equations. As,

(1)Now,

is the positive B.M. producing tension in the upper half of the beam where y is positive. Also produces tension in the left hand half of the beam where x is negative; we shall therefore call as negative.

Solution( Contd)

So,

Or,

Thus on upper edge ofSo that direct stress varies linearly with x. At the top left-hand corner of the

top flange ,

AT the top right-hand corner

Neutral axis does not in this case coincide either with x or y axis, although it still passes through the centroid of the section. Its inclination to the x axis can be found by

So,

Which gives, , so distribution of stress is as:-

Solution( Contd)

• Direct stress at a point in the c/s depends on:– Its location in the c/s– The loading– The geometry of the c/s

• Assumption – plane sections remain plane after deformation (No Warping), or cross-section does not deform in plane (i.e. σxx, σyy = 0)

• Sign Conventions!

Direct stress calculation due to Unsymmetrical bending

M – bending momentS – shear forceP – axial loadT – torqueW – distributed load

Direct stress calculation due to bending (cont’d)

• Beam subject to bending moments Mx and My and bends about its neutral axis (N.A.)

• N.A. – stresses are zero at N.A.

• C – centroid of c/s (origin of axes assumed to be at C).

The axial strain in a line element a distance ᶓ above the neutral surface is given by:

• Consider element A at a distance ξ from the N.A.

• Direct Stress:

• Because ρ (bending radius of curvature) relates the strain to the distance to the neutral surface:

Direct stress calculation due to bending (cont’d)

0

0

lll

z

EE zzz

First Moment of Inertia Definition

• Given an area of any shape, and division of that area into very small, equal-sized, elemental areas (dA)

• and given an Cx-Cy axis, from where each elemental area is located (yiand xi)

• The first moment of area in the "X" and "Y" directions are respectively:

dAxAxI

ydAAyI

y

x

• IF the beam is in pure bending, axial load resultant on the c/s is zero:

• 1st moment of inertia of the c/s about the N.A. is zero N.A. passes through the centroid, C

• Assume the inclination of the N.A. to Cx is α

Direct stress calculation due to bending (cont’d)

• Then

The direct stress becomes:

AA

z dAdA 00

cossin yx

cossin yxEEz

Direct stress calculation due to bending (cont’d)

• Moment Resultants:

• Substituting for σz in the above expressions for Mx and My, and using definitions for Ixx, Iyy, Ixy

dAxM

ydAM

zy

zx

dAxyI

dAxI

dAyI

xy

yy

xx

2

2

cossin

cossin

cossin

xyyy

xxxy

y

x

xyyyy

xxxyx

IIIIE

MM

IEIEM

IEIEM

Direct stress calculation due to bending (cont’d)

• Using the above equation in:

• Gives:

From Matrix Form

y

x

xyyy

xxxy

MM

IIIIE

1

cossin

y

x

xyyy

xxxy

xyyyxx MM

IIII

IIIE

2

1cossin

cossin yxEz

yIII

IMIMx

IIIIMIM

xyyyxx

xyyyyx

xyyyxx

xyxxxyz

22

Direct stress calculation due to bending (cont’d)

• Or, rearranging terms:

• If My= 0, Mx produces a stress that varies with both x and y. Similarly for My, if Mx=0.

• If the beam c/s has either Cx or Cy (or both) as an axis of symmetry, then Ixy = 0.

• Then:

22xyyyxx

xyxxy

xyyyxx

xyyyxz III

yIxIMIII

xIyIM

yy

y

xx

xz I

xMI

yM

• Further, if either My or Mx is zero, then:

• We saw that the N.A. passes through the centroid of the c/s. But what about its orientation α?

At any point on the N.A. σz = 0

Direct stress calculation due to bending (cont’d)

xx

xz I

yM

yy

yz I

xMor

tan

022

xyyyyx

xyxxxy

xyyyxx

xyyyyx

xyyyxx

xyxxxyz

IMIMIMIM

xy

yIII

IMIMx

IIIIMIM

Example Problem

The beam shown is subjected to a 1500 Nm bending moment in the vertical plane.

Calculate the magnitude and location of max σz.

0

8 mm40 mm 80 mm

80 mm

8 mm

1st: Calculate location of Centroid

mm 528808120

)880(40)8120(60

xxxxxx

AAx

xc

mm 4.668808120

)880(40)8120(84

xxxxxx

AAy

yc

x

y

Example Problem (cont’d)Calculate Ixx, Iyy, Ixy, with respect to Cxy:

4623

23

23

mm 1009.1)404.66(80812808

)4.6684(812012

812012

cxx dAbtI

4623

23

23

mm 1031.1)4052(80812

808

)5260(812012

812012

cyy dAtbI

Example Problem (cont’d)

46 mm 1034.0)4.6640()5240(808

)4.6684()5260(8120

A

xy xydAI

Mx = 1500 Nm, My = 0

22xyyyxx

xyxxy

xyyyxx

xyyyxz III

yIxIMIII

xIyIM

mm]in y x,N/mmin [

39.05.12

z

xyz By inspection, MAX at y = -66.4 mm and x = -8 mm(Max stress always further awayFrom centroid)

0x

y

Simplifications for thin-walled sections

• Thin-walled t << c/s dimensions. – Stresses constant through thickness– Terms in t2, t3, etc… neglected

• In that case Ixx reduces to:• What about Ixy for this c/s• What about Iyy for this c/s

horizontal members

vertical members

Example:

You should do this on your own

Simplifications for thin-walled sections

• Thin-walled sections frequently have inclined or curved walls which complicate the calculation of section properties.

• Its second moment of area abt horizontal axis,

From which

similarly

Product second moment of area,

Simplifications for thin-walled sectionsSo,

For Thin walled- curved section:-

Example Problem

Que: Determine the direct stress distribution in the thin-walled Z-section as shown in figure, produced by a positive B.M.

SolutionThe section is anti-symmetrical with its centroid at the mid-point of the

vertical web. ( My = 0)

Therefore, (1)

Section properties are calculated as,

Substituting these values in (1), we have

Solution ( contd).

Solution ( contd)And

The complete distribution is then,

Additional Problems :Q.1. Figure below shows the section of an angle purlin. A bending moment

of 3000Nm is applied to the purlin in a plane at an angle of 30̊ to the vertical y axis. If the sense of the bending moment is such that its components Mx and My both produce tension in the positive xy quadrant, calculate the maximum direct stress in the purlin stating clearly the point at which it acts.

SolutionSol:

The position of the Centroid of area ,C may be found as,Taking moment about BC,

Now taking Moment about AB,

Solution (Contd)Second Moment of Inertia’s can be found as,

Substituting second moment of inertia’s and moments in stress eqn, we have

Since the coefficients x and y have the same sign, the maximum value of the direct stress will occur in either first and 3rd quadrants.

Solution (Contd)Then,

Ans.

Additional Problems (Contd)Q.2. A thin-walled, cantilever beam of unsymmetrical cross section

supports shear loads at its free end as shown in Fig. Calculate the value of direct stress at the extremity of the lower flange (point A) at the section halfway along the beam if the position of the shear loads in such that no twisting of the beam occurs.

SolutionBending moments half-way along the beam are ;

By inspection the centroid of area is midway between the flanges. Its distance from the vertical web is given as;

Solution (Contd)

Substituting values of moments and Second Moment pf inertia’s in stress eqn. we have;

So,

At point A ( x= 66.67 mm, y= -50 mm )

Ans.

Additional Problems (Contd)Q.3. A Thin walled beam has the cross section as shown in Fig. if the

beam is subjected to a bending moment Mx in the plane of the web 23. Calculate and draw the direct stress distribution in the beam cross section.

SolutionBy inspection centroid of the section will be at the mid point of the web.

Since My is zero so,

Or,

Solution (Contd)Between 1 and 2 , y = -h and stress is linear,

Between 2 and 3 , x = 0 and stress is linear. Then

Ans.

Additional Problems (Contd)Q.4. Thin walled beam section as shown in Fig., is subjected to a

bending moment Mx applied in negative sense. Find the position of neutral axis and the maximum direct stress in the section.

SolutionThe centroid of the section is at the centre of the inclined web. Then,

Since My is zero, so

Solution (Contd)Substituting the values of Moments and I’s in the stress eqn., we have

On the neutral axis, direct stress is zero, then,So,

The greatest direct stress will occur at the points furthest from the neutral axis, i.e., at points 1 and 2. then at the point 1.,

i.e., Maximum direct stress isAns.

Additional Problems (Contd)Q.5. A uniform thin-walled beam has the open cross section as shown in

Fig. The wall thickness t is constant. Calculate the position of neutral axis and the maximum direct stress for a bending moment Mx = 3.5 Nm applied about the horizontal axis Cx. Take r = 5 mm, t = 0.64 mm.

SolutionSection properties are :

Since My =0, so direct stress equation is,

Solution (Contd)The maximum direct stress will occur at a point a perpendicular distance

from the N.A., i.e. by inspection at Point B or D. ThusAt B ( x =0, y = 2r )

Or,

Alternately direct stress equation may be re-written as,

Or

The expression as its greatest value at θ⇒π, i.e. at B or D. Ans.

Solution (Contd)

Position of the neutral axis :-At N.A. direct stress will be zero, i.e.,

Which gives,

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