Beams, trusses and vierendeel frames
Iain MacLeod
Introduction
This document explains how beams, diagonally braced trusses and vierendeel frames
share behaviour in relation to bending and shear effects. Study of these common
features is beneficial in developing understanding of the features themselves.
Bending and shear action are fundamental issues in the understanding of structural
behaviour of frames.
Vierendeel
frame action
Beam action
Truss action
Shear mode
deformation
Shear force
Shear stress
Behaviour of beams,
trusses and vierendeel
frames
Bending mode
deformation
Axial deformation
Axial force
Axial stress
Bending
Pin-jointed Truss
Truss with moment
connections
Vierendeel frame
Beam
Bending moment
Bending stress
Global
bending


Shear




1


Local
bending


Bending Mode Deformation
Bending of an I Section
Image of a beam here
Figure 1 shows a Universal I beam section and the corresponding internal force
actions . The bending moment at the section is 10.0 kNm. The calculations for the
force actions are here.
In this case the flange takes 90% of the moment at the section and for checking
models it is common to neglect the effect of the web when calculating the I value.
For an I section the second moment of area about the major axis - Iz - is expressed as:
Iz = Iweb + Ay2flange + Iowncg,flange
(1)
More information on calculation of I values
For an approximate I value, the first and third terms of Equation (1) are ignored
giving the simplified expression.
Ig = Ay2flange
(2)
Bending stress
Flange force = 75.1 kN
Flange moment = 4.49 kNm
Web force = 13 kN
Web moment = 0.51 kNm
(a) UB 127 x 76 x 13
(b) Stress block
Figure 1 Stress, internal force actions in an I beam
If the distance between the centrelines of the flanges is hf then
Ig = 2 Af (hf/2)2 = Af hf2 /2
(3)
hf
where Af is the area of one flange
This expression underestimates the I value and therefore gives
conservative results for stress and deflection. It is only used for
‘back of an envelope’ checks (more). It is the basis for the estimation of global
bending in frames.
Global bending of a parallel chord truss
2
Image (or images) of a real truss here
Diagonal
bracing
Post
Chords
10.0 kN
Ac
d
Ac
(a) Analysis model of a parallel chord truss
(b) Equivalent beam model
Figure 2 Parallel chord truss
The truss model of Figure 2(a) supports a nominal 10 kN load at the centre. The top
and bottom chords act like the flanges of an I beam and therefore the truss can be
modelled as an equivalent beam (Figure 2(b)) having an I value based on Equation
(3). The equivalent I value for the truss is:
Ig = Ac d2/2
(4)
Where Ac is the area of a chord and d is the vertical distance between the chords.
This mode of bending is known as global bending as distinct from local bending
(The ‘g’ subscript is for ‘global’). Global bending results from axial deformation in
the chords of the truss and is normally a dominant feature of the behaviour of a truss
of the type shown in Figure 2. Local bending results from bending of the members of
the truss when they are moment connected. The latter effect tends not to be dominant
in trusses but is a major issue for vierendeel frame behaviour (see here),
Bending mode deformation
Figure 3(a) is a plane stress model of a rectangular beam in bending modelled using
plane stress elements (more on plane stress). The deformed shape is dominated by
bending mode behaviour the governing equation for which is:
EId2v/dx2 = M
(5)
where E is Young’s Modulus, I is the second moment of area of the section, v is the
vertical displacement and M is the bending moment. The bending moment diagram is
linear [refer to diagram] and therefore integrating Equation (5) twice result in a cubic
curve (more) for v
3
(a) Deformed shape for a beam modelled using plane stress elements
(b)
(b)Deformed
Deformedshape
shapefor
foraabending
bendingdominant
dominanttruss
truss
Figure 3 Bending mode deformation for point load case
Figure 3(b) is the deformed shape for a model of the truss of Figure 2(a) with the post
and diagonal members having significantly higher axial stiffness to that of the chord
members. That is, the deformation shown in Figure 3(b) is dominantly due to the
global bending action - to the truss acting like a beam in bending.
Note the similarity between the deformed shapes in Figures 3 and 4. Note also how
the vertical lines remain straight in the deformed shape of Figure 4 - a main
assumption in bending theory
Summary
The chords of a truss behave in an analogous way to the flanges of an I beam.
Equation 3 can be used to estimate the second moment or area of an I beam and of a
parallel chord truss.
Shear mode deformation
Shear deformation of an I Section
Figure 4(a) shows the deformed shape for a model of an I beam using plane stress
elements for the web and bar elements for the flanges. The area of the bar elements
was such that the bending deformation is not evident in the shape. (Vertical bar
element stiffeners were also inserted above the supports to remove local
deformations.)
The shear mode shape of the deformation is evident from this diagram.
The governing equation for shear deformation of beam is:
As dv/dx = S
(5)
where As is the shear area (more), v is the vertical displacement and S is the shear
force.
Since the shear force is constant, integrating Equation (5) once will give an linear
expression for v - as is evident from Figure 4(a). (With distributed load the deflected
4
curve will be parabolic and therefore not easily distinguished from deflections with
other curved shapes.)
(a) Deformed shape for an I beam with very stiff flanges
(b) Deformed shape for a truss with very stiff chords
Figure 4 shear mode deformation for point load
case
Figure 4(b) shows the deformed shape for a truss for which the areas of the chord
members were made artificially high so as to eliminate bending mode effects from the
deformed shape. The straight line shape due to shear deformation is evident as in
Figure 4(a)
Summary
Both the beam and the truss have analogous shear mode components of deformation.
For an I beam the shear deformation is in the web; for a truss it is due to axial
deformation of the posts and the diagonals.
5
Image here
Vierendeel frame
Figure 5(a) shows an analysis model of a vierendeel frame. Figure 5(b) is an
equivalent beam model.
Post
Chords
10.0 kN
Ac
d
Ac
(a) Analysis model of a vierendeel frame
(b) Equivalent beam model
Figure 5 Vierendeel frame
Figure 6(a) shows the deformed shape for a vierendeel frame with central vertical
point load. The vertical displacement if quite linear indicating dominant shear mode
deformation (more). This shear type deformation is due to the local bending in the
chords and posts of the frame (more about the effect of local bending).
(a) Deformed shape of vierendeel frame
(b) Deformed shape of vierendeel frame with low axial stiffness of chords
stiffnes
Figure 6 Deformed shapes for a vierendeel frame
6
Figure 6(b) shows the same frame with the areas of the chord members artificially
reduced by a significant amount. Now the vertical displacement is controlled by the
axial stiffness of the chords i.e by global bending (more).
7
Beam models
Deflection formulae for beams and properties for equivalent beam
models for parallel chord trusses and vierendeel frames
Cb and Cs from
Table 3
Deflection due to
shear deformation
Beam
deflection
 = b + b =
Deflection due to
bending
deformation
W - total load
L - span
EI - stiffness parameter
C WL
C bWL3
+ s
Ks
EI
Expressions for Ks
see Table 1
Expressions for I
see Table 1
Table 1 Expressions for I and Ks
Structure
Beam
Parallel chord truss
Vierendeel frame
I
I
Ig
Ig
Ks
AsG - Table 2
Kst - Eq (A)
Ksv - Eq(B)
Ig 
Ac b 2
2
Table 2 Values for shear area As
Section
Rectangle b x d
As
5
bd
6
I section bent about major axis
I section bent about minor axis
G
E
2(1   )
Area of web
Area of flanges
Parallel chord truss
K st 
1
1
1

2
fE d Ad sin  cos  E p A p cot an 
Ap, Ep
Ac, Ec
(A)
Ad, Ed
Ld
f = 1.0 for singly braced truss

= 2.0 with compressive cross
bracing
= 0.5 for K bracing
L
Parameters for parallel chord truss
With tensile only cross bracing treat
as singly braced
With compressive cross bracing
ignore flexibility of posts
8
b
K sv 
 
24 EI c
a [1  2 ]
2
Ip
Ac Ic
Vierendeel frame
(B)
b
Ic / a
Ip / b
a
L
Parameters for vierendeel frame
Table 3 Beam deflection coefficients
Structure
Load
Cantilever
Point tip
Cb bending
1/3
Cs shear
1.0
UD
1/8
1/2
Point central
1/48
1/4
UD
5/384
1/8
E,I
L
Simply supported
E,I
L
For more values of this type see Input document
Derivation of Kst
From the Bar Element Document, Equation (21) is:
 = 1 + 2 =
WLd
(EA) d cos 
2
+
WLp
(EA) p
Governing differential equation for shear
deformation:
S = Ks dv/dx
i.e. K s 
y, v
S
dv / dx
v is the displacement in the y direction
From the diagram: dv/dx = /a
Substituting this and S = W into (21):

S Ld
Sb
dv


2
dx
a E p Ap
a E d Ad sin 
a
Note that sin2 is used because the  for the frame is
(90 - ) for Equation (21)
Using a/Ld = cos  and a/b = cot :
Ks 
S

dv / dx
1
1
E d Ad sin  cos 
2

S

1
E p Ap cot 
9
Derivation of Ksv
Figure 1(a) shows a vierendeel frame with points of contraflexure at mid-length of all
members. Such positions for the points of contraflexure is the fundamental
assumption in developing the shear mode deformation of a vierendeel frame as an
equivalent beam.
a/2

a/2
S/2
b
(a) Frame showing panel width section

S/2
b/2
(b) Panel width sub-frame
S/2
(c) Symmetrical half of sub-frame
Figure 1 Vierendeel frame
Also shown on Figure 1(a) is a section of the frame bounded by points of
contraflexure. This is extracted to Figure 1(b) where the shear at the points of
contraflexure S/2. S is the total shear at the section and half is taken by each chord
(assuming them to have the same I value). The final trick is to work on a
symmetrical half of this sub-frame as in Figure 1(c). The deflection under the S/2
load of the frame of Figure 1(c) is calculated (using the principle of virtual work) to
be:

S a3
1  2 
24EI c
where  
Ic / a
Ip / b
hence dv/dx = /a and Ksv = S/(dv/dx)
hence K sv 
24 EI c
a [1  2 ]
2
Note G = E/(2(1+)) where  is Poisson’s Ratio
Constitutive relationships for bending and shear:
Bending
M = EI d2v/dx2
i.e d2v/dx2 = M/EI
Shear
S = Ks i.e. dv/dx = S/ Ks
where Ks is the shear stiffness.
For a beam Ks = AsG where As is the shear area and G is the shear modulus.
Table 3 shows shapes of shear force and bending moment diagrams and
corresponding displacement diagrams.
For bending, the basic relationship needs to be integrated twice to get the
displacement. therefore the function for the displaced shape is two orders higher than
10
that for the bending moment e.g. from Table 3 with UD load, the bending moment is
parabolic whereas the displacement is quartic (fourth order).
For shear, the basic relationship needs to be integrated once to get the displacement
and therefore, for example, with a point load, the shear is constant and the
displacement is linear.
Table 3 Diagram shapes
Loading
Point
y,v
Shear force
Constant
Shear
Displacement
linear
Bending
Bending M
Displacement
linear
cubic
W
x
UD
y,v
linear
w
parabolic
x
11
parabolic
quartic