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7.4 The Elementary Beam Theory
In this section, problems involving long and slender beams are addressed. As with pressure vessels, the geometry of the beam, and the specific type of loading which will be considered, allows for approximations to be made to the full three-dimensional linear elastic stress-strain relations.
The beam theory is used in the design and analysis of a wide range of structures, from buildings to bridges to the load-bearing bones of the human body.
7.4.1 The Beam
The term beam has a very specific meaning in engineering mechanics: it is a component that is designed to support transverse loads , that is, loads that act perpendicular to the longitudinal axis of the beam, Fig. 7.4.1. The beam supports the load by bending only. Other mechanisms, for example twisting of the beam, are not allowed for in this theory.
Figure 7.4.1: A supported beam loaded by a force and a distribution of pressure
It is convenient to show a two-dimensional cross-section of the three-dimensional beam together with the beam cross section, as in Fig. 7.4.1. The beam can be supported in various ways, for example by roller supports or pin supports (see section 2.3.3). The cross section of this beam happens to be rectangular but it can be any of many possible shapes.
It will assumed that the beam has a longitudinal plane of symmetry , with the cross section symmetric about this plane, as shown in Fig. 7.4.2. Further, it will be assumed that the loading and supports are also symmetric about this plane. With these conditions, the beam has no tendency to twist and will undergo bending only 1.
Figure 7.4.2: The longitudinal plane of symmetry of a beam
(^1) certain very special cases, where there is not a plane of symmetry for geometry and/or loading, can lead also to bending with no twist, but these are not considered here
longitudinal plane of symmetry
roller support pin support
applied force applied pressure
cross section
Imagine now that the beam consists of many fibres aligned longitudinally, as in Fig. 7.4.3. When the beam is bent by the action of downward transverse loads, the fibres near the top of the beam contract in length whereas the fibres near the bottom of the beam extend. Somewhere in between, there will be a plane where the fibres do not change length. This is called the neutral surface. The intersection of the longitudinal plane of symmetry and the neutral surface is called the axis of the beam , and the deformed axis is called the deflection curve.
Figure 7.4.3: the neutral surface of a beam
A conventional coordinate system is attached to the beam in Fig. 7.4.3. The x axis coincides with the (longitudinal) axis of the beam, the y axis is in the transverse direction and the longitudinal plane of symmetry is in the x y plane, also called the plane of
bending.
7.4.2 Moments and Forces in a Beam
Normal and shear stresses act over any cross section of a beam, as shown in Fig. 7.4.4. The normal and shear stresses acting on each side of the cross section are equal and
opposite for equilibrium, Fig. 7.4.4b. The normal stresses will vary over a section
during bending. Referring again to Fig. 7.4.3, over one part of the section the stress will be tensile, leading to extension of material fibres, whereas over the other part the stresses will be compressive, leading to contraction of material fibres. This distribution of normal stress results in a moment M acting on the section, as illustrated in Fig. 7.4.4c. Similarly, shear stresses act over a section and these result in a shear force V.
The beams of Fig. 7.4.3 and Fig. 7.4.4 show the normal stress and deflection one would expect when a beam bends downward. There are situations when parts of a beam bend upwards, and in these cases the signs of the normal stresses will be opposite to those shown in Fig. 7.4.4. However, the moments (and shear forces) shown in Fig. 7.4.4 will be regarded as positive. This sign convention to be used is shown in Fig. 7.4.5.
x
y
z fibres extending
fibres contracting
neutral surface
Example 1
Consider the simply supported beam in Fig. 7.4.7. From the loading, one would expect the beam to deflect something like as indicated by the deflection curve drawn. The reaction at the roller support, end A , and the vertical reaction at the pin support^2 , end B , can be evaluated from the equations of equilibrium, Eqns. 2.3.3:
R (^) Ay P / 3 , RBy 2 P / 3 (7.4.1)
Figure 7.4.7: a simply supported beam
The moments and forces acting within the beam can be evaluated by taking free-body diagrams of sections of the beam. There are clearly two distinct regions in this beam, to the left and right of the load. Fig. 7.4.8a shows an arbitrary portion of beam representing the left-hand side. A coordinate system has been introduced, with x measured from A.^3 An unknown moment M and shear force V act at the end. A positive moment and force have been drawn in Fig. 7.4.8a. From the equilibrium equations, one finds that the shear force is constant but that the moment varies linearly along the beam:
x
P
M
P
V
l x (7.4.2)
Figure 7.4.8: free body diagrams of sections of a beam
(^2) the horizontal reaction at the pin is zero since there are no applied forces in this direction; the beam theory does not consider such types of (axial) load; further, one does not have a pin at each support, since this would prevent movement in the horizontal direction which in turn would give rise to forces in the horizontal direction – hence the pin at one end and the roller support at the other end (^3) the coordinate x can be measured from any point in the beam; in this example it is convenient to measure it from point A
x
A
P / 3 V
M
(a ) (b)
x
A
P / 3 V
M
2 l / 3 P
2 l / 3 P deflection
curve
l
A B
RAy RBy
Cutting the beam to the right of the load, Fig. 7.4.8b, leads to
l x
P
M
P
V
( x l
l (7.4.3)
The shear force is negative, so acts in the direction opposite to that initially assumed in Fig. 7.4.8b.
The results of the analysis can be displayed in what are known as a shear force diagram and a bending moment diagram , Fig. 7.4.9. Note that there is a “jump” in the shear force at x 2 l / 3 equal to the applied force, and in this example the bending moment is everywhere positive.
Figure 7.4.9: results of analysis; (a) shear force diagram, (b) bending moment diagram ■
Example 2
Fig. 7.4.10 shows a cantilever , that is, a beam supported by clamping one end (refer to Fig. 2.3.8). The cantilever is loaded by a force at its mid-point and a (negative) moment at its end.
Figure 7.4.10: a cantilevered beam loaded by a force and moment
Again, positive unknown reactions M (^) A and V (^) A are considered at the support A. From
the equilibrium equations, one finds that
(a ) (b)
V M
2 l / 3
2 Pl
P
2 P
l l
5 kN
3 m 3 m
V A
M A
4 kNm
A
Fig. 7.4.13 shows a simply supported beam subjected to a distributed load (force per unit length). The load is uniformly distributed over half the length of the beam, with a triangular distribution over the remainder.
Figure 7.4.13: a beam subjected to a distributed load
The unknown reactions can be determined by replacing the distributed load with statically equivalent forces as in Fig. 7.4.14 (see §3.1.2). The equilibrium equations then give
RA 220 N, RC 140 N (7.4.6)
Figure 7.4.14: equivalent forces acting on the beam of Fig. 7.4.
Referring again to Fig. 7.4.13, there are two distinct regions in the beam, that under the uniform load and that under the triangular distribution of load. The first case is considered in Fig. 7.4.15.
Figure 7.4.15: free body diagram of a section of a beam
The equilibrium equations give
V 220 40 x , M 220 x 20 x^2 0 x 6 (7.4.7)
40 N/m
x
220 V
M
240 N
3 m 3 m 2 m 4 m
120 N
R A RC
40 N/m
A (^) 6 m B (^) 6 m
C
The region beneath the triangular distribution is shown in Fig. 7.4.16. Two possible approaches are illustrated: in Fig. 7.4.16a, the free body diagram consists of the complete length of beam to the left of the cross-section under consideration; in Fig. 7.4.16b, only the portion to the right is considered, with distance measured from the right hand end, as 12 x. The problem is easier to solve using the second option; from Fig. 7.4.16b then, with the equilibrium equations, one finds that
V 140 10 ( 12 x )^2 / 3 , M 140 ( 12 x ) 10 ( 12 x )^3 / 9 6 x 12 (7.4.8)
Figure 7.4.16: free body diagrams of sections of a beam
The results are summarized in the shear force and bending moment diagrams of Fig. 7.4.17.
Figure 7.4.17: results of analysis; (a) shear force diagram, (b) bending moment diagram
■
7.4.3 The Relationship between Loads, Shear Forces and
Bending Moments
Relationships between the applied loads and the internal shear force and bending moment in a beam can be established by considering a small beam element, of width x , and
(a ) (b)
V M
6 m 6 m
6 m 6 m
x (^) 12 x
220 140
V
M M V
(a ) (b)
Consider now moment equilibrium, by taking moments about the point A in Fig. 7.4.18:
x p
x V x px x
M x x M x
x p x
x M x V x x M x x px x
Again, as the size of the element decreases towards zero, the left-hand side becomes a derivative and the second and third terms on the right-hand side tend to zero, so that
V ( x ) dx
dM (^) (7.4.13)
This relation can be seen to hold in Eqns. 7.4.2-3, 7.4.5 and 7.4.7-8. It also follows from Eqn. 7.4.13 that the change in moment along a beam is equal to the area under the shear force curve:
M x M x V xdx
x
x
2
1
7.4.4 Deformation and Flexural Stresses in Beams
The moment at any given cross-section of a beam is due to a distribution of normal stress, or flexural stress (or bending stress ) across the section (see Fig. 7.4.4). As mentioned, the stresses to one side of the neutral axis are tensile whereas on the other side of the neutral axis they are compressive. To determine the distribution of normal stress over the section, one must determine the precise location of the neutral axis, and to do this one must consider the deformation of the beam.
Apart from the assumption of there being a longitudinal plane of symmetry and a neutral axis along which material fibres do not extend, the following two assumptions will be made concerning the deformation of a beam:
- Cross-sections which are plane and are perpendicular to the axis of the undeformed beam remain plane and remain perpendicular to the deflection curve of the deformed beam. In short: “plane sections remain plane”. This is illustrated in Fig. 7.4.19. It will be seen later that this assumption is a valid one provided the beam is sufficiently long and slender.
2. Deformation in the vertical direction, i.e. the transverse strain yy , may be neglected
in deriving an expression for the longitudinal strain xx. This assumption is
summarised in the deformation shown in Fig. 7.4.20, which shows an element of length l and height h undergoing transverse and longitudinal strain.
Figure 7.4.19: plane sections remain plane in the elementary beam theory
Figure 7.4.20: transverse strain is neglected in the elementary beam theory
With these assumptions, consider now the element of beam shown in Fig. 7.4.21. Here, two material fibres ab and pq , of length x in the undeformed beam, deform to a b
and p q . The deflection curve has a radius of curvature R. The above two assumptions
imply that, referring to the figure:
p a b a b q / 2 (assumption 1)
ap a p , bq b q (assumption 2) (7.4.15)
Since the fibre ab is on the neutral axis, by definition a b ab. However the fibre
pq , a distance y from the neutral axis, extends in length from x to length x . The
longitudinal strain for this fibre is
R
y R
R y R x
x x xx (^)
As one would expect, this relation implies that a small R (large curvature) is related to a large strain and a large R (small curvature) is related to a small strain. Further, for y 0
(above the neutral axis), the strain is negative, whereas if y 0 (below the neutral axis),
the strain is positive^4 , and the variation across the cross-section is linear.
(^4) this is under the assumption that R is positive, which means that the beam is concave up; a negative R implies that the centre of curvature is below the beam
plane in deformed beam remains perpendicular to the deflection curve
deflection curve
x
y
h
l dl
, 0 h
dh l
dl xx yy
dh
just as the tangential stresses are much larger than the radial stresses in the pressure vessel, it is found that the longitudinal stresses in a beam are very much greater than the transverse stresses. With this assumption, the first of Eqn. 7.4.17 reduces to a one- dimensional equation:
xx xx / E (7.4.18)
and, from Eqn. 7.4.16, dropping the subscripts on ,
y R
E
Finally, the resultant force of the normal stress distribution over the cross-section must be zero, and the resultant moment of the distribution is M , leading to the conditions
y dA y
y dA R
E
M ydA
ydA R
E
dA
A A A
A A
2 2
and the integration is over the complete cross-sectional area A. The minus sign in the second of these equations arises because a positive moment and a positive y imply a compressive (negative) stress (see Fig. 7.4.4).
The quantity (^) A ydA is the first moment of area about the neutral axis, and is equal to yA ,
where y is the centroid of the section (see, for example, §3.2.1). Note that the horizontal
component (“in-out of the page”) of the centroid will always be at the centre of the beam due to the symmetry of the beam about the plane of bending. Since the first moment of area is zero, it follows that y 0 : the neutral axis passes through the centroid of the
cross-section.
The quantity (^) A y^2 dA is called the second moment of area or the moment of inertia
about the neutral axis, and is denoted by the symbol I. It follows that the flexural stress is related to the moment through
I
My
Flexural stress in a beam (7.4.21)
This is one of the most famous and useful formulas in mechanics.
The Moment of Inertia
The moment of inertia depends on the shape of a beam’s cross-section. Consider the important case of a rectangular cross section. Before determining the moment of inertia one must locate the centroid (neutral axis). Due to symmetry, the neutral axis runs
through the centre of the cross-section. To evaluate I for a rectangle of height h and width b , consider a small strip of height dy at location y , Fig. 7.4.22. Then
/ 2 3
/ 2
I y^2 dA b y^2 dy bh
h
A h
(^)
This relation shows that the “taller” the cross-section, the larger the moment of inertia, something which holds generally for I. Further, the larger is I , the smaller is the flexural stress, which is always desirable.
Figure 7.4.22: Evaluation of the moment of inertia for a rectangular cross-section
For a circular cross-section with radius R , consider Fig. 7.4.23. The moment of inertia is then
(^2 ) 2 3 2 0 0
sin 4
R
A
R
I y dA r drd
(^) (^) (7.4.23)
Figure 7.2.23: Moment of inertia for a circular cross-section
Example
Consider the beam shown in Fig. 7.4.24. It is loaded symmetrically by two concentrated forces each of magnitude 100N and has a circular cross-section of radius 100mm. The reactions at the two supports are found to be 100N. Sectioning the beam to the left of the forces, and then to the right of the first force, one finds that
V M x l/
V M x x
r y dA^ rdrd^
b
h y 0
centroid
y
dy
From equilibrium of forces in the horizontal direction of the surface section:
^0
dA dA b x A x A x x
The third term on the left here assumes that the shear stress is uniform over the section – this is similar to the calculations of §7.4.3 – for a very small section, the variation in stress is a small term and may be neglected. Using the bending stress formula, Eqn. 7.4.21,
(^) dA b I
y x
M x x M x A
and, with Eqn. 7.4.13, as x 0 ,
Ib
VQ
Shear stress in a beam (7.4.28)
where Q is the first moment of area (^) A ydA of the surface section of the cross-section.
Figure 7.4.26: stresses and forces acting on a small section of material at the surface of a beam
As mentioned, this formula 7.4.28 can be used as an approximation of the shear stress in a beam of arbitrary cross-section, in which case b can be regarded as the depth of the beam at that section. For the rectangular beam, one has
(^) ^2
/ (^22)
2 4
y
b h Q b ydy
h
y
so that
x
V ( x )
x
M ( x )
V ( x x )
M ( x x )
x x
h
b
^ ^
^2
2 (^3 )
y
h bh
V
The maximum shear stress in the cross-section arises at the neutral surface:
A
V
bh
V
max (7.4.31)
and the shear stress dies away towards the upper and lower surfaces. Note that the average shear stress over the cross-section is V / A and the maximum shear stress is 150% of this value.
Finally, since the shear stress on a vertical cross-section has been evaluated, the shear stress on a longitudinal section has been evaluated, since the shear stresses on all four sides of an element are the same, as in Fig.7.4.6.
Example
Consider the simply supported beam loaded by a concentrated force shown in Fig. 7.4.27. The cross-section is rectangular with height 100 mmand width 50 mm. The reactions at
the supports are 5 kNand 15 kN. To the left of the load, one has V 5 kNand
M 5 x kNm. To the right of the load, one has V 15 kNand M 30 15 x kNm.
The maximum shear stress will occur along the neutral axis and will clearly occur where V is largest, so anywhere to the right of the load:
- 5 MPa 2
(^3) max max ^ A
V
Figure 7.4.27: a simply supported beam
As an example of general shear stress evaluation, the shear stress at a point 25 mm below the top surface and 1 m in from the left-hand end is, from Eqn 7.4.30, 1. 125 MPa.
The shear stresses acting on an element at this location are shown in Fig. 7.4.28.
- 5 m^20 kN 0. 5 m
The solution for shear stress, Eqn 7.4.33, turns out to be exact; however, the exact solution corresponding to Eqn 7.4.34 is^6
2 1 1 3 5
l p b h
^
^
It can be seen that the beam theory is a good approximation for the case when l / h is large, in which case the term 1/5 is negligible.
Following this type of analysis, a general rule of thumb is this: for most configurations, the elementary beam theory formulae for flexural stress and transverse shear stress are accurate to within about 3% for beams whose length-to-height ratio is greater than about
7.4.7 Beam Deflection
Consider the deflection curve of a beam. The displacement of the neutral axis is denoted by v , positive upwards, as in Fig. 7.4.30. The slope at any point is then given by the first derivative, dv / dx.
For any type of material, provided the slope of the deflection curve is small, it can be
shown that the radius of curvature R is related to the second derivative d^2 v / dx^2 through (see the Appendix to this section, §7.4.10)
2
dx
dv R
and for this reason d^2 v / dx^2 is called the curvature of the beam. Using Eqn. 7.4.19,
Ey / R , and the flexural stress expression, Eqn. 7.4.21, My / I , one has the
moment-curvature equation
2
2 ( ) dx
d v M x EI moment-curvature equation (7.4.37)
Figure 7.4.30: the deflection of a beam
With the moment known, this differential equation can be integrated twice to obtain the deflection. Boundary conditions must be supplied to obtain constants of integration.
(^6) this can be derived using the Stress Function method discussed in Book 2, section 3.
v
Example
Consider the cantilevered beam of length L shown in Fig. 7.4.31, subjected to an end- force F and end-moment M (^) 0. The moment is found to be M ( x ) F ( L x ) M 0 , with x
measured from the clamped end. The moment-curvature equation is then
1 2
2 3 0
1
2 0
2 0
2
EIv FL M x Fx Cx C
FL M x Fx C dx
dv EI
FL M Fx dx
d v EI
The boundary conditions are that the displacement and slope are both zero at the clamped end, from which the two constant of integration can be obtained:
1
2
v C
v C (7.4.39)
Figure 7.4.31: a cantilevered beam loaded by an end-force and moment
The slope and deflection are therefore
^
^ ^2
0
2 3 (^0 )
FL M x Fx dx EI
dv FL M x Fx EI
v (7.4.40)
The maximum deflection occurs at the end, where
^2 ^3
( ) M L FL
EI
v L (7.4.41)
The term EI in Eqns. 7.4.40-41 is called the flexural rigidity , since it is a measure of the resistance of the beam to deflection.
Example
Consider the simply supported beam of length L shown in Fig. 7.4.32, subjected to a uniformly distributed load p over half its length. In this case, the moment is given by
L^ F
M 0
