Engineering science lesson 6 1

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Chapter 1- Static engineering systems

Structural members • Bending of structural member• General bending equation• Position of neutral axis• Second moment of area• Parallel axis theorem

Bending of structural member• When a beam bends, one surface becomes extended and so in

tension and other surface become reduced in length and so in compression. This implies that between upper and lower surfaces, there is a plane which is unchanged in length when the beam is bent. This plane is called the neutral plane.

• Consider the bending scenario

Stress variation across a beam• With a uniform rectangular cross-section beam, the

maximum bending stress will be on the surfaces since these are the furthest distance from the neutral axis.

• The next figure shows how the stress will vary across the section of the beam

Example

A uniform square cross-section steel strip of side 4mm is bent into a circular arc by bending it around a drum of radius 4m. Determine the maximum strain and stress produced in the strip. Take the modulus of elasticity of the steel to be 210GPa.

Solution

The neutral axis of the strip will be central and so the surfaces will be 2mm from it and the radius of the neutral axis will be 4.002m. Thus

Maximum strain=y/R = 2X10-3/(4.002) =0.5X10-3

This will be the value of compressive strain on the inner surface of the strip and that of the tensile strain on the outer surface. Hence:

Maximum stress = E X (maximum strain)

= 210X109X0.5X10-3 =105MPa

General bending equation

• Since the stress on a layer a distance y from the neutral axis is yE/R then M can also be written as

• The above equations are generally combined and written as the general form bending formula

y

IM

R

E

yI

M

Example

An I-section beam is 5.0m long and supported by both ends. It has a second moment of area of 120X10-6m4, a depth of 250mm and a uniformly distributed weight of 30kN/m. Calculate

a) The maximum bending moment

b) The maximum bending stress

c) The radius of curvature of the beam where the bending moment is a maximum. The modulus of elasticity for the beam is 210GPa.

Solution

a) The beam loading is simply supported beam with uniformly distributed load and so the maximum bending moment is at the beam centre and wL2/8, where w is the weight per unit length and L the total length.

Position of the neutral axis• It can be shown that the neutral axis passes though the centroid of the

beam by considering the longitudinal forces acting on the beam

• The total longitudinal force will be the sum of all the forces acting on each segment.

• But the beam is only bent and so there is no longitudinal force stretching the beam. Thus since E and R are non zero, we must have

The above integral is called the first moment of area of the section. The only axis about which we can take such a moment and obtain 0 is an axis through the centre of the area of the cross section, i.e. the centroid of the beam. Thus the neutral axis must pass through the centroid of the section when the beam is subjected to bending

Example• Determine the position of the neutral axis for

the T section beam shown in the figure.

Solution

The neutral axis will pass through the centroid. We can consider the T-section to be composed of two rectangular sections. The centroid of each will be at its centre. Hence, taking moments about the base of T-section

Moment = 250X30X115 + 100X50X50 = 1.11X106 mm4

Hence the distance of the centroid from the base is (total moment)/(total area), and hence the neutral axis

Distance from base = (1.11X106)/(250X30 + 100X50) = 89mm

Second moment of area• The second moment of area I of a section about an axis is given by

• The value of the second monemt of area about an axis depends on the shape of the beam section concerned and the position of the axis.

• The following figure gives the second moment of area about the neutral axis.

Parallel axis theorem• To illustrate the deviation of the second

moment of area, consider a rectangular cross section of breadth b and depth d.

• For a layer of thickness y a distance y from the neutral axis which passes through the centroid, the second moment of area for the layer is

• The total second moment of area for the section is

• We will consider the second moment of area about an axis which is distance h away from the neutral axis. The new second moment of area Ih would be

• This is called the theorem of parallel axes and is used to determine the second moment of area about a parallel axis

Exercise• Calculate the second moment of area about the neutral axis of a

rectangular cross-section beam of depth 40mm and breadth 100mm

Example• Determine the second moment of area about the neutral axis of the I-

section shown in the figure

Solution

Thus for the rectangle containing the entire section, the second moment of area is I = bd3/12 = 50X703/12 – 1.43X106mm4. For the two missing rectangles, each will have a second moment of area of 20X503/12 = 0.21X106mm4. Thus the second moment of area of the I-section is

1.43X106 – 2X0.21X106 = 1.01X106mm4

Example

A horizontal beam with a uniform rectangular cross-section of breadth 100mm and depth 150mm is 4m long and rests on supports at its ends. It has negligible weight itself and supports a concentrated load of 10KN at its midpoint. Determine the maximum tensile and compressive stresses in the beam.

Solution

The second moment of area is I = bd3/12 = 0.100X0.1503/12 = 2.8X10-5 m4. The

reactions at each support will be 5kN and so the maximum bending moment,

which occur at the mid point, is 10kNm. The maximum bending stress will occur

at the cross-section where the bending moment is a maximum and on the outer

surfaces of the beam, i.e. y = 75mm. Thus:

MPaX

XX

I

My8.26

108.2

075.010105

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Example

A rectangular cross-section timber beam of length 4m rests on supports at each end and carries a uniformly distributed load of 10kN/m. If the stress must not exceed 8MPa, what will be a suitable depth for the beam if its width is to be 100mm?

Solution

For simply supported beam with a uniformly distributed load over its full length, the maximum bending moment = wl2/8 = 10X42/8 = 20kNm. Hence: