Mech sol ch7

7/31/2019 Mech sol ch7

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Chapter 5. Transverse Loading

5.1 Introduction

One of the most common types of loading that an engineer is likely to encounteris the transverse loading (lateral). One reason for this is the fact that stresses

and displacement due to transverse loading are usually greater than that causedby other types of loading such as axial loading. While it is not difficult to

imagine the physical bending of a steel beam and the associated change in theshape of that beam, any axial displacement in a steel bar is unlikely to beobservable without any measuring device. The calculation of stresses and

displacements due to bending is more complex than the correspondingcalculations for an axially loaded bar. For these reasons, understanding the

bending behaviour of beams requires more effort than studying the axial stressesand displacement. Initially we will derive an equation for the normal stress due

to bending of beams whose cross-sections have at least one axis of symmetry.

5.2 Symmetrical Bending of Beams

By definition, normal stress is the intensity of normal force and is given by theequation:

σ = dF/dA

We have seen that, in the absence of anystress concentration, the normal stressdue to centric axial loading may be

taken as uniform and is therefore givenby the formula:

σ = F/A.

In this chapter, we will derive anequation for the normal stress in a

beam due to bending.

Limitations, Definitions andAssumptions:

The derivations presented are forsymmetrical bending of beams. This

means the cross section of the beamsmust have at least one axis of

symmetry.

σσ = F/A F

M σσ = ?

x

y

z

Plane of bending

Neutral PlaneM



2

The loading (forces that cause bending) are assumed to act in the plane of symmetry, thus eliminating the possibility of twisting. This plane will also bereferred to as the plane of bending.

As the beam is transversely loaded it bends; for example the top fibers mayshorten while the bottom fibers stretch. The type of stress on the extreme

surfaces will be opposite to each other (one side in tension the other incompression.) Between the two outer surfaces, there will exist a plane (or

surface) where there will be neither tension nor compression. Thissurface/plane is called the neutral surface or neutral plane. Strictly speaking,

as the loaded beam is not straight, this will not be a flat plane, but since alldisplacements concerned are small, the neutral surface may be referred to as

'neutral plane'. In a pure symmetrical bending, the neutral plane will beperpendicular to the plane of bending.

Assumptions

1. The beam is made of a homogeneous, isotropic material having linear elasticproperties.

2. The stresses are within the proportional and elastic limits.3. Young’s modulus is the same in tension and compression.

4. The beam is initially straight and stress free.5. The beam is symmetrical about the plane of bending.

6. Transverse sections of the beam remain plane during bending.

7. The deflections are small. 8. The sectional dimensions of the beam are small compared to the span andthe deflection due to shearing is negligible.

First let us consider the case of pure bending. That is to say that the only

internal action in a beam is the bending moment.



3

Geometry of Deformation:

Consider an infinitesimal element of the

beam, having length dx at unloaded state. Asthe beam undergoes bending, the eleveation

of the segment would change its shape.From a rectangular shape, it will take the

shape of a curved element, with two straightedges as shown in the diagram. The edges

will be straight, because of the assumptionthat the transverse planes will remain plane

during bending. Let the extended lines alongthe transverse sections meet at point O. The

infinitesimal length of the fibers along the

beam will have changes, except at the neutralplane which is free of stress. In ourillustration, the beam element is shown tohave a positive bending, and all fibres above

the neutral axis will undergo a shortening,while those below the neutral axis are stretched.

Our first task is to obtain an expression for the strain at distance y from the

neutral axis. Since the deflections are assumed to be small, we can take the

curved length as approximatley circular arcs, with which we can obtain thefollowing expression for the length of the element at distance y above the N.A.Length of element at distance y from the N.A. =

Length of element at the N.A. =But since the N.A. is free from stress, the original length at the N.A. = the new

length at the N.A.Prior to bending, the element had a rectangular elevation, and thereforethe original length at distance y = the length at the N.A.

This means the change in length at distance y =

By definition the normal strain = change in length/original length.

Therefore the strain at distance y may be expressed as:

ε =

From the above equation we may conclude that the strain varies linearly with y.

r

y

r - y

y

dy

dθθ

O



4

Hooke’s Law:

Stress σ = ε ×

σ =

Force in area dA is,

dF = σ × =

Net force on the cross section =

F = ∫ ∫ =AA

dF dA

=

−

r

E∫ A

Longitudinal (axial) equilibrium:

However, for a beam undergoing pure bending, the net force on a section =

Therfore, =

−

r

E∫ A

y. dA = 0

i.e. ∫ A

y. dA = 0

This means the f about the N.A. is zero.

It is worth recalling that this is also a property of a centroidal axis.

Therefore we can conclude that:

the N.A. passes through the centroid of the cross-section of the beam.

This property may be used to locate the neutral axis of a given section.

r

y

r - y

y

dy

dθθ O



5

Moment-Curvature Relationship:

We already have, σσ = - Ey/r

Taking moments due to dF

(on the positive face) about

the N.A. gives:

dM = −

The minus sign isintroduced to reconcile theconflict between the sense

of the moment due to a positive (tensile) force

acting at a positive y(above the N.A.) and the

definition of a positivebending moment.

But, dF = σ × dA = -(Ey/r)dA

Therfore dM =

To find the net moment on the cross section we may integrate dM with respectto dA.

i.e. M = ∫ ∫ =AA

dM dA

At a given cross section, the only variables are y and dA. Therefore the aboveequation may be simplified to yield:

M = (E/r) ∫ A

y2

dA

The integral may be recognised as the

We may rewrite the above equation as:

M = EI/r

But we lready have, from equation ( ), (E/r) = -σ /yTherefore, in terms of stress, M = -σI/y

Since we want to find stress in terms of M, we may rearrange the above

equation to get the following formula: σ = -My/I

This is referred to as the Euler-Bernoulli beam bending formula.

y

dydF



6

Flexural Stress Calculations

• Beam Bending Formula:

σσ = -My/I ⇐⇐ (Use S. I. Units)

• Sign Convention:

♦ Sagging moment positive

♦ y positive above the neutral axis

♦ Tensile stress (σσ) positive

• Procedure:

♦ Locate the neutral axis

♦Obtain the second moment of area about the neutral axis

♦ Apply the beam bending formula

• Problem:

A 1.2 m long cantilever beam is subject to a 25 kN load at the free end asshown. For each of the cross-sections given, determine the maximum tensile

and compressive stresses induced. All sections have the same cross-sectional area.

First let us consider a square section having an area of 10000 mm2.

Geometric Properties of Areas:

This set of handouts does not contain notes on:Parallel Axis Theorem

Perpendicular Axis TheoremFormulae for the second moment of area of rectangles, triangles and circles

For equilibrium,M =

Magnitude of M ismaximum at the clamp.

Use:

25 kN

1.2 m

x

B.M.D.



7

(a) A Square Section

First step:Locate the neutral axis (N.A.)

This section has an axis of symmetryperpendicular to the plane of

bending. Therefore the axis of symmetry is the N.A.

Second Step:Calculate the second moment of area.For rectangular sections, I =

b = , d =Therefore I =

Final Step:

Apply the flexure formula.

σ = -My/I

The magnitude of moment is maximum at the clamp. Hence, use M =

The stresses are maximumat the top and bottom (i.e. at y = ± 0.05 m).

To determine the stress at the top, put y = which gives,

σ =

Similarly, at the bottom, y = which gives, σ =

Hence Maximum tensile stress = , andMaximum compressive stress =

(b) A Rectangular Section:

25 kN 25 kN

Plane of

bending

N.A.

100 mm

100 mm

N.A.

50 mm

100 mm

100 mm



8

Consider a rectangular section having the same area.i.e. Let the depth be (100a) mm and the breadth be (100/a) mm.

Area = 100a × 100/a = 10000 mm2

I = bd3 /12 = (0.1/a)*( 0.1 a)

3 /12 = 8.33 × 10

-6a

2m

4.

Extreme y values are: ± (0.05) a m

Extreme stresses are given by:

σ = -My/I

=

This means that the extreme stresses decrease with the aspect ratio. Does itmean that a very tall but narrow cross-sectional beam is ideal for carrying

transverse loading?

This is not the case, since there may be problems with instability, anddifficulty in applying loading. Therefore, a practical solution is to provide

flanges to apply the loading, and design a ‘web’ to withstandcompression so that it doesn’t ‘buckle’. Structural beams

such as ‘I’ beams are designed to minimse bendingstress, because the flanges are away from the

centroid, thus giving a very large second

moment of area.



9

Datum

N.A.

y

30 mm 10 mm

200 mm

100 mm

30 mm

(c) A Mono-symmetrical I- sectionFirst step:Locate the neutral axis (N.A.)

From the first moment of area theorem,( )yAyA iii∑ ∑=

A1 = A2 = A3 =

1y = 2y = 3y =

y =

Second Step:Calculate the second moment of area about the N.A.

‘I’ for each rectangular area about itsown centroid is given by:

I =

Due to the top flange,I1 =

Due to the bottom flange,I3 =

Due to the web,I2 =

I = I1 + I2 + I3 =

Final Step:

Apply the flexure formula: σ = -My/I

M= I =

Extreme stresses occor at the top and bottom surfaces.At the top, y = , and the stress is given by:

σ = - =

Similarly at the bottom, σ = - =

Datum

N.A.

y

30 mm10 mm

200 mm

100 mm

30 mm

=134 mm



10

(d) A Fully-symmetrical I- sectionConsider this I beam which also hasthe same cross-sectional area

(10000 mm2). Since it is a

symmetrical section, the neutral axis

is the horizontal axis of symmetry.

The second moment of area aboutthe N.A. can be calculated

conveniently as follows:

First calculate the second momentof area of the 100 mm x 240 mm

block:

I1 =

Now calculate the second moment of

area of the ‘hollow’ part (70 mm x 200mm block):

I2 =

Net second moment of area: I = I1 – I2 =

Final Step:

Apply the flexure formula σ = -My/I

I =

We also have M =

Using the formula for stress,

At the top, y = + 0.120 m,

σ =

At the bottom,y = - 0.120 m,

σ =

Datum

N.A.

y

30 mm20 mm

200 mm

35 mm20 mm

35 mm

= 120 mm

Datum

N.A.

y

30 mm20 mm

200 mm

35 mm20 mm

35 mm

= 120 mm



11

bf = tf =

h = tw=

=y

Due to the flange, I1 in =

Due to the web, Iw in =

The second moment of area of the section I = I1 + Iw =

M =

Stress at the top =

Stress at the bottom =

T-Beam

Datum

N.A.

tw

tf

bf



12

t1

Datum

N.A.

tw

t2b2

Datum

h

b1 = t1 =

h = tw=b2 = t2 =

=y

Due to the top flange, I1 in =

Due to the web, Iw in =

Due to the bottom flange, I2 in =

The second moment of area of the section I = I1 + Iw +I2 =

M =

Stress at the top =


I-Beam



13

b = d =bi = di=

e = 0

For the block,IS =

For the hollow part,Ih =

The second moment of area of the section I = IS – Ih =

M =

Stress at the top =


Fully Symmetrical Rectangular hollow Cross section

b

ddi

bi

N.A.Axis of

Symmetry



14

b = d =r =

e =

=y

Now we can calculate the second moment of area of the rectangle about the

N.A.IS =

The second moment of area of the circular hole about the N.A.

Ih =

The second moment of area of the section I =

M=

Stress at the top =


Rectangular Section With a circular Hole

Datum.

N.A..

d

b

e r



15

12.0 kN

1.5 m 0.8 m

23.5 kN

Figure 1a

100 mm

6 mm

80 mm 12 mm12 mm

Figure 1b

12 kN

0.8 m 0.5m

22 kN

Figure 2a

0.8 m

150 mm

Figure 2b

100 mm

60 mm40 mm

Tutorial 5.1

1. A 2.3 m long channel beam is clamped at one end, and is subject to two point loads as

shown in Figure 1a. A transverse section of the beam is shown in Figure 1b.Determine:

(a) The extreme values (maximum and minimum) of the bending moment in thebeam;

(b) The maximum compressive stress in the beam;

2. A 2.1 m long simply supported beam is subject to two point loads as shown in Figure2a. The beam is of rectangular section with an 80 mm diameter hole. The centre of

the hole is located 60 mm above the bottom of the beam. A transverse section of thebeam is shown in Figure 2b. Determine:

(c) The extreme values (maximum and minimum) of the bending moment in thebeam;

(d) The maximum compressive stress in the beam;

(e) The maximum tensile stress in the beam.



16

100 mm

150 mm

20 mm

Figure 3b

80 mm

70 mm

10

Datum

N.A.

12

10

70

Datum

90

3. A 2.1 m long simply supported beam is subject to two point loads as shown in Figure1a. The beam is of rectangular section with an 80mm x 70 mm rectangular hole asshown in Figure 1b. Determine:

(a) The extreme values (maximum and minimum) of the bending moment in thebeam;

(b) The maximum compressive stress in the beam;

(c) The maximum tensile stress in the beam.

4. Repeat Question 1 with the following sections. All dimensions are in mm.

25 kN

0.8 m 0.5m

45 kN

Figure 3a

0.8 m

Datum

N.A.

12

8

100

160

Figure 4

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