Engineering science

1. 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
1

2. 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

3. 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

4. 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

5. General bending equation

6. • Since the stress σ on a layer a distance y from
the neutral axis is yE/R then M can also be
written as
σI
M=
y
• The above equations are generally combined
and written as the general form bending
formula
M σ E
= =
I y R

7. 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.

8. 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

9. 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

10. 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.

11. 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 δy 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

12. • 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

13. 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

14. 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:
My 10 X 103 X 0.075
σ= =± = ± 26.8MPa
I 2.8 X 10 −5

15. 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: