Basis of Structural Design Arch Truss →

Basis of Structural Design
Course 3
Structural action: trusses and beams
Course notes are available for download at
http://www.ct.upt.ro/users/AurelStratan/
Arch
→
Truss
rafter
tie
Linear arch supporting a
concentrated force: large
spreading reactions at supports
Relieving of support
spreading: adding a tie
between the supports
1
Truss forces
Truss members connected by
pins: axial forces
(direct stresses) only
Supports:
– one pinned, allowing free rotations
due to slight change of truss shape
due to loading
– one roller bearing support ("simple
support") - allowing free rotations
and lateral movement due to
loading and change in temperature
- (C)
- (C)
+ (T)
Forces in the truss:
– tie is in tension (+)
– rafters are in compression (-)
Truss forces
If more forces are present within
the length of the rafter ⇒
bending stresses
To avoid bending stresses,
diagonal members and vertical
posts can be added
-
+
+
-
+
More diagonals and posts can
be added for larger spans in
order to avoid bending stresses
2
Alternative shape of a truss
For a given loading find out the shape of a linear arch
(parabolic shape)
Add a tie to relieve spreading of supports
Highly unstable shape
Alternative shape of a truss
Add web bracing (diagonals
and struts) in order to
provide stability for the
pinned upper chord
members
If the shape of the truss
corresponds to a linear arch
web members are
unstressed, but they are
essential for stability of the
truss
Reverse bowstring arches:
– advantage: longer members
are in tension
– disadvantage: limited
headroom underneath
3
Truss shapes
Curved shape of the arch: difficult to fabricate ⇒ trusses
with parallel chords
Trusses with parallel chords: web members (diagonals
and struts) carry forces whatever the loads
Pratt truss:
– top chord in compression
– bottom chord and diagonals in tension
– economical design as
longer members
(diagonals) are in tension
Truss shapes
Howe truss:
– top chord in compression
– bottom chord in tension
– diagonals in compression
Warren truss:
– top chord in compression
– bottom chord in tension
– diagonals in tension and
compression
– economy of fabrication: all
members are of the same
length and joints have the
same configuration
4
Truss joints
Pinned joints ⇒ statically
determinate structures ⇒ member
forces can be determined from
equilibrium only
Rigid joints ⇒ small bending
stresses will be present, but which
are negligible due to the triangular
shape
Traditionally trusses are designed
with pinned joints, even if members
are connected rigidly between them
Space trusses
The most common plane truss
consists of a series of triangles
The corresponding shape in three
dimensions: tetrahedron (a)
The truss at (b) is a true space
truss
– theoretically economical in material
– joints difficult to realise and
expensive
Two plane trusses braced with
cross members are usually
preferred
5
Statically indeterminate trusses
Indeterminate trusses:
large variety
Example (a): cross
diagonals in the middle
panel, so that one of
the diagonals will
always be in tension
Example (b): Sydney
Harbour Bridge,
Australia - both
supports pinned
Beams
Beam: a structure that supports
loads through its ability to resist
bending stresses
Leonardo da Vinci (1452-1519): the strength of a timber
beam is proportional to the square of its depth
Leonhard Euler and Daniel Bernoulli
were the first to put together a useful
theory around 1750
6
Beams: analogy with trusses
Forces in a
Pratt truss
loaded by a unit
central force
Forces in a
Howe truss
Forces in a
truss with
double diagonals
(reasonable estimate)
Beams: analogy with trusses
Chords:
– The forces in the top and bottom chord members in any panel are
equal, but of opposite signs, and they increase with the distance
from the nearest support
– Chords have to resist the bending moment, proportional to the
distance from the nearest support
Diagonals:
– The forces in the diagonal members are equal, but opposite in
sign, and have the same values in all panels
– Diagonals have to resist the shear forces, the same in all panels
7
Beams: analogy with trusses
Bending and shear deformations in a truss
Steel plate girder
Steel plate girder: heavy flanges and thin web welded
together, and reinforced by transversal stiffeners
Unit vertical force at the midspan
Top flange: in compression
Bottom flange: tension
Web: shear, with principal tension and compression
stresses similar to those in a truss
After web
buckling, only
tensile loads are
resisted by the
web, plate girder
acting as a
Pratt truss
8
Beams: bending action
Top flange in compression
linear variation of
normal stress
Bottom flange in tension
Normal stress proportional to distance from the neutral
plane
Simplifications:
– Thin web, thick flanges ⇒ web has a small contribution to the
bending resistance (ignore it)
– Normal stress can be considered uniform on flanges
Beams: bending action
Moment resistance
– Idealised double T beam: M = σ⋅A⋅⋅d/2
– Rectangular beam of the same area and depth:
M = σ⋅b⋅⋅d2/6 = σ⋅A⋅⋅d/6
The best
arrangement
of material
for bending
resistance:
away from
the neutral axis
A/2
F = σ ·(A/2)
d
d
−σ
+σ
F = σ ·(A/2)
A/2
F = σ ·(0.5d·b/2)
d
2d/3
−σ
A
+σ
M = σ ·A·d/2
M = σ ·A·d/6
F = σ ·(0.5d·b/2)
b
9
Beams: bending action
Examples of efficient location of material for bending
resistance
– light roof beams (trusses)
– hot-rolled and welded girder
Beams: bending action
Examples of efficient location of material for bending
resistance
– panel construction
10
Beams: bending action
Examples of efficient location of material for bending
resistance
– corrugated steel sheet
Beams: bending action
Examples of efficient location of material for bending
resistance
– castellated joist
11
Beams: bending action
Examples of efficient location of material for bending
resistance
– columns requiring bending resistance in any direction: tubular
sections
Beams: shear stresses
Simply supported beam of uniform rectangular crosssection loaded by a concentrated central force W:
– can carry a moment M = σ⋅b⋅⋅d2/6
– has a deflection δ
If the beam is cut in two parts along the neutral plane:
– sliding takes place between the two overlapped beams
– the two overlapped beams can carry a moment
M = 2⋅⋅[σ⋅b⋅⋅(d/2)2/6] = σ⋅b⋅⋅d2/12, half of the uncut beam
– the deflection of the two overlapped beams is 4δ
12
Beams: shear stresses
In the uncut beam stresses should be present along the
neutral plane to prevent sliding of the lower and upper
halves of the beam: shear stresses
Smaller stresses would be required to keep the unity of
action if the beam were cut above the neutral plane
Shear stresses
– parabolic variation in a rectangular cross-section
– carried mainly by the web, on which they can be considered to be
constant for a steel double T beam
Structural shapes
Simply supported beam
subjected to a uniformly
distributed load
The "perfect" use of
material for bending
resistance in a beam with
idealised double T crosssection (M = σ⋅A⋅⋅d/2):
parabolic variation of
height
A/2
A/2
13
Structural shapes
Simply
supported truss
subjected to a
uniformly
distributed load
The "perfect" use
of material for
"bending" action:
parabolic
variation of
height
Structural shapes
Bridge with a simply supported central span and two
cantilevered sides
The shape of the truss must resemble the bending
moment diagram in order to make efficient use of material
in upper and bottom chords
Quebec
railway bridge
14
Structural shapes
Forth
bridge,
Scotland
Angel
Saligny
bridge,
Romania
15