Unit 48: Structural Behaviour and Detailing for Construction
Deflection of Beams
4.1
Introduction
This topic investigates the deformation of beams as the direct effect of that
bending tendency, which affects their serviceability and stability, and does so in
terms of their deflection.
A beam may be strong enough to resist safely the bending moments due to the
applied loading and yet not be suitable because its deflection is too great.
Excessive deflection might not only impair the strength and stability of the
structure but also give rise to minor trouble such as cracking of plaster ceilings,
partitions and other finishes, as well as adversely affecting the functional needs
and aesthetic requirements or simply being unsightly.
The relevant BS specifications and codes of practice stipulate that the deflection
of a beam shall be restricted within limits appropriate to the type of structure. In
the case of structural steelwork the maximum deflection due to unfactored
imposed loads for beams carrying plaster or other brittle finish must not exceed
1/360 of the span, but for all the other beams it may be span/200 (Clause 2.5.1
of BS 5950: Part 1: 2000). For timber beams, on the other hand, the figure is
0.003 of the span when the supporting member is fully loaded (Clause 2.10.7 of
BS 5268: Part 2: 1996). In reinforced concrete the deflection is generally
governed by the span/depth ratio (Clause 3.4.6.3 of BS 8110: Part 1: 1997).
4.2
Factors affecting deflection
For many beams in most type of buildings, e.g. flats, offices, warehouses, it will
usually be found that, if the beams are made big enough to resist the bending
stresses, the deflections will not exceed the permitted values. In beams of long
spans, however, it may be necessary to calculate the deflections to ensure that
they are not excessive. The derivation of formulae for calculating deflections
usually involves calculus. In this chapter, therefore, only a general treatment will
be attempted and deflection formulae for a few common cases of beam loadings
will be given without proof. General methods of calculation of deflections are
given in standard books on theory of structures or strength of materials.
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Unit 48: Structural Behaviour and Detailing for Construction
4.2.1 Load
AB (figure 1) represents a beam of span L metres supported simply at its ends
and carrying a point load of W kN at mid-span. Let us assume that the deflection
due to the load is 5mm. It is obvious that, if the load is increased, the deflection
will increase. It can be proved that the deflection is directly proportional to the
load, i.e. a load of 2W will cause a deflection of 10mm, 3W will produce a
deflection of 15mm and so on. W must therefore be a term in any formula for
calculating deflection.
W
½l
½l
A
B
5mm
l
Figure 1: Deflection of a beam under loading
4.2.2 Span
In figure 2 (a) and (b) the loads W are equal and the weights of the beams,
which are assumed to be equal in cross-section, are ignored for the purposes of
this discussion. The span of beam (b) is twice that of beam (a). It is obvious that
the deflection of beam (b) will be greater than that of beam (a), but the
interesting fact (which can be demonstrated experimentally or proved by
mathematics) is that instead of the deflection of (b) being twice that of (a), it is
8 times (e.g. 40mm in this example). If the span of beam (b) where 3L, its
deflection would be 27 times that of beam (a). In other words, the deflection of
a beam is proportional to the cube of the span, therefore L3 is a term in the
deflection formula.
W
W
5mm
40mm
l
2l
(a)
(b)
Figure 2: Effect of span upon deflection: the span of (b) is twice that of (a)
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Unit 48: Structural Behaviour and Detailing for Construction
4.2.3 Size and Shape of Beam
Figure 3 (a) and (b) represents two beams (their weights being ignored) of equal
spans and loading but the moment of inertia of beam (b) is twice that of beam
(a). Obviously, the greater the size of the beam, the smaller the deflection
(other conditions being equal). It can be proved that the deflection is inversely
proportional to the moment of inertia, e.g. the deflection of beam (b) will be
one-half that of beam (a). Moment of inertia I is therefore a term in the
denominator of the deflection formula. (It may be noted that, since the moment
of inertia of a rectangular cross section beam is bd3/12, doubling the breadth of
a rectangular beam decreases the deflection by one-half, whereas doubling the
depth of a beam decreases the deflection to 1/8 of the previous value.)
W
W
l
(a)
l
(b)
Figure 3: Effect of size and shape upon deflection: (a) moment of inertia of beam = 1 unit; (b) moment of inertia of
beam = 2 units
4.2.4 ‘Stiffness’ of Material
The stiffer the material of a beam, i.e. the greater its resistance to bending, the
smaller will be the deflection, other conditions such as span, load, etc.,
remaining constant. The measure of the ‘stiffness’ of a material is its modulus of
elasticity E and deflection is inversely proportional to the value of E.
4.3
Derivation of Deflection Formulae
A formula for calculating deflection must therefore contain the load W, the cube
of the span L3, the moment of inertia I, and the modulus of elasticity E. For
standard cases of loading, the deflection formula can be expressed in the form
, where c is a numerical coefficient depending on the disposition of the load
and also on the manner in which the beam is supported, that is, whether the
ends of the beam are simply supported or fixed, etc. For figure 4 the values of c
are respectively 1/48 and 5/384. W and L3 are in the numerator of the formula
because an increase in their values means an increase of deflection, whereas E
and I are in the denominator because an increase in their values means a
decrease of deflection.
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Unit 48: Structural Behaviour and Detailing for Construction
Referring to figure 4 it should be obvious (neglecting the weights of the beams)
that although the beams are equally loaded, the deflection of beam (b) will be
less than that of beam (a). In fact, the maximum deflection of beam (a) is
and the maximum deflection of beam (b) is
100kN
3m
W = 100kN UDL
l = 6m
(b)
l = 6m
(a)
Figure 4: Deflection Formulae
Table 1 gives the values of c for some common types of loading, etc.
Table 1: Values of coefficient c for deflection formula
Condition of loading
Value of c (max at A)
W
½l
½l
A
B
A
𝑙
W/2
A
Jesmond Agius: Chapter 10
𝑙
A
W/2
𝑙
B
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Unit 48: Structural Behaviour and Detailing for Construction
𝑙
W/2
𝑙
W/2
𝑙
A
A
UDL =W
l
W
𝑙
𝑙
A
Fixed Beam
A
UDL =W
l
Fixed Beam
W
l
Cantilever
A
A
UDL =W
l
Cantilever
When the load system is complicated, e.g. several point loads of different
magnitudes, or various combinations of point loads and uniformly distributed
loads, the deflections must be calculated from first principles.
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Unit 48: Structural Behaviour and Detailing for Construction
In certain simple cases it is possible to derive deflection formulae
mathematically without using the calculus, and the following example is given
for the more mathematically minded student.
Practical Example 1
Work out the value of the coefficient of the deflection formula for the following
scenario and verify your answers with table 1.
W
½l
½l
A
A
B
Answer
See lecturer explanation during the lesson.
Try to work out the coefficient of the deflection formula of the other diagrams
found in table 1.
Practical Example 2
A 406 × 178 UB54 simply supported at the ends of a span of 5m carries a
uniformly distributed load of 60 kN/m. Calculate the maximum deflection. (E =
205 000 N/mm2)
Answer
The formula for the maximum deflection is (from table 1)
Where
W = 60 × 5 = 300 kN = 0.3 × 106 N
l = 5000 mm
E = 205 000 N/mm2
I = 18720 cm4 = 187 200 000 mm4 = 187.2 × 106 mm4 (Taken from table of
universal beams)
Therefore
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Unit 48: Structural Behaviour and Detailing for Construction
Practical Example 3
Calculate the safe inclusive uniformly distributed load for a 457 × 152 UB52
simply supported at its ends if the span is 6m and if the span is 12m. The
maximum permissible bending stress is 165N/mm2 and the maximum
permissible deflection is 1/360 of the span. E is 205 000 N/mm2.
Answer Part 1
Z = 950 cm3 = 950 000 mm3 (Taken from table of universal beams)
From last year notes we have deducted that the elastic modulus can be
expressed as
So
Also, Maximum bending moment can be calculated by using the equation
. This has to be equal to
.
Therefore
Implies
So Maximum Deflection for simply supported uniformly distributed loads is equal
to
Where
W = 209 000 N
l = 6000 mm
E = 205 000 N/mm2
I = 21370 cm4 = 213 700 000 mm4 = 213.7 × 106 mm4 (Taken from table of
universal beams)
Therefore
Maximum permissible deflection =
Therefore safe load = 209.0 kN
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Unit 48: Structural Behaviour and Detailing for Construction
Answer Part 2
For a 12m span
Z = 950 cm3 = 950 000 mm3 (Taken from table of universal beams)
From last year notes we have deducted that the elastic modulus can be
expressed as
So
Also, Maximum bending moment can be calculated by using the equation
. This has to be equal to
.
Therefore
Implies
So Maximum Deflection for simply supported uniformly distributed loads is equal
to
Where
W = 104 500 N
l = 12000 mm
E = 205 000 N/mm2
I = 21370 cm4 = 213 700 000 mm4 = 213.7 × 106 mm4 (Taken from table of
universal beams)
Therefore
Maximum permissible deflection =
This means that, although the beam is quite satisfactory from the strength point
of view, the deflection is too great, therefore the load must be reduced.
Now
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Unit 48: Structural Behaviour and Detailing for Construction
Therefore
Giving W = 64.2 kN and this is the maximum permitted load for the beam.
(Instead of being obtained from the deflection formula, W can also be obtained
from
)
Practical Example 4
Calculate the safe inclusive uniformly distributed load for a 200 mm × 75 mm
timber joist, simply supported at its ends, if the span is 4m and if the span is
8m. The maximum permissible bending stress is 6N/mm2 and the maximum
permissible deflection is 0.003 of the span. E is 9500 N/mm2.
Answer Part 1
The cross section of a timber joist is usually a rectangle.
Thus
(See lesson 9)
From lesson 9 we have deducted that the elastic modulus can be expressed as
So
Also, Maximum bending moment can be calculated by using the equation
(this is the equation for maximum bending moment of
uniformly distributed loads. This has to be equal to
.
Therefore
Implies
So Maximum Deflection for simply supported uniformly distributed loads is equal
to
Where
W = 6 000 N
l = 4000 mm
E = 9500 N/mm2
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Unit 48: Structural Behaviour and Detailing for Construction
(See lesson 9 for second moment of area of
rectangular cross section shapes)
Therefore
Maximum permissible deflection =
Therefore safe UDL for the 4 m span = 6.0 kN
Answer Part 2
The cross section of the second timber joist is same.
Thus
and
Also, Maximum bending moment can be calculated by using the equation
(this is the equation for maximum bending moment of
uniformly distributed loads. This has to be equal to
.
Therefore
Implies
So Maximum Deflection for simply supported uniformly distributed loads is equal
to
Where
W = 3 000 N
l = 8000 mm
E = 9500 N/mm2
(as before)
Therefore
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Unit 48: Structural Behaviour and Detailing for Construction
Maximum permissible deflection =
This means that, although the timber joist is quite satisfactory from the strength
point of view, the deflection is too great, therefore the load must be reduced.
Now
Therefore
Implies
Giving W = 1.7 kN and this is the maximum permitted load for the timber joist.
(Instead of being obtained from the deflection formula, W can also be obtained
from
Jesmond Agius: Chapter 10
)
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