Embed Size (px)
Citation preview
Metal Structures
Lecture XIII
Steel trusses
Contents
Definition → #t / 3
Geometry and cross-sections → #t / 7
Types of truss structures → #t / 15
Calculations → #t / 29
Example → #t / 57
Results of calculations - verification → #t / 79
Death weight of truss → #t / 90
Examination issues → #t / 92
Truss – theory (idealization):
• Straight bars only;• Hinge joints;• Forces in joints only;
There are axial forces only
Schwedler-Żurawski formula for straight bar:
d M(x) / dx = Q(x)d Q(x) / dx = q(x)
Forces in joints only → no loads along bar (q(x) = 0):
q(x) = 0 → Q(x) = const = C → M(x) = C x + A
Hinges:
M(0) = 0 → A = 0 ; M(L) = 0 → C = 0
M(x) = 0 ; Q(x) = 0
Definition
Photo: Author
→ Des #2 / 2
Truss – real:
• Bars with imperfections;
• No ideal hinge joints;
• Gravity along bars;
It’s rather framePhoto: Author
Effort (calculation as a truss) ≈ Effort (calculation as a frame)
Time of calculations (truss) << Time of calculations (frame)
Because of these reasons, we calculate real truss as ideal truss.
Chord
Web members = brace members = cross bars
Chord
There is common name for every not-chord bars (web member or brace member or cross bar).
In Polish language exist two words: „słupki” (vertical bars) and „krzyżulce” (inclined bars).
Common mistake is to design different type of cross-section only because of different name or
direction - for web members important are only axial forces and lengths, not names.
Photo: waldenstructures.com
There are many different shapes of trusses.
Geometry and cross-sections
Photo: tridenttruss.com
Photo: steelconstruction.info
Photo: e-plytawarstwowa.pl
Photo: domgaz.com.plPhoto: konar.eu
Photo: i435.photobucket.com Photo: community.fansshare.net
Specific subtype are trusses with paralell chords. For these trusses, forces in top and
bottom chord are similar.
| NEd, top | / | NEd, bottom | ≈ 0,90 – 1,10
Photo: waldenstructures.com
Photo: gillmet.com.pl
max (h1; h2) ≤ 3,40 m
max (L1; L
2) ≤ 18,00 m
max (L3; L
4) ≤ 12,00 m
Initial drawings - geometry
h = L (1/10 ~ 1/15)
H = L (1/5 ~ 1/10)
a ≥ 5o
30o ≥ b ≥ 60o or b ≈ 90o
Photo: Author
→ Des #2 / 9
PN B 03200 EN 1993
Elements Each types of cross-sections are accepted
Joints No additional requirements Additional requirements → many
types of cross-sections are not
accepted
Cross-sections of truss members
Modern types of cross-sections
(EN)
Old types of cross-sections (PN-B)
Chords
Web
members
Photo: Author
Whenever we use hollow sections, we must
hermetically close the ends by welding to
prevent corrosion inside HS. Such type of
corrosion can develop without apparent
external symptoms and can cause damage
of structure without warning signs.
Photo: architectureau.com
Photo: Błędy wykonawcze podczas realizacji konstrukcji stalowych, Litwin M, Górecki M, Budownictwo i Architektura 4 (2009) 63-72
D
C
DD
Number of different cross-sections, designed in truss: 2 - 5:
Chords Brace members
Top and bottom - the same I-beam
or
Two different I-beams, first for top, second for bottom
1-3 different RHS
1-3 different CHS
Top and bottom - the same RHS
or
Two different RHS, first for top, second for bottom
1-3 different RHS
1-3 different CHS
Top and bottom - the same CHS
or
Two different CHS, first for top, second for bottom
1-3 different CHS
Types of truss structures
Truss purlins
„Classical” trusses
Multi-chords trusses
Truss-grates
Truss-frames
Spatial trusses
Laced columns
Span [m]
Photo: Author
Truss purlins
Photo: structural-steelbuilding.com
Photo: CoBouw Polska Sp. z o. o.
"Normal" truss - forces are applied in nodes; there
are axial forces in chords and cross-bars only.
Truss purlin - continuous load from roofing;
there are axial forces in chords and cross-
bars; in addition top chord is bending.
Photo: Author
Truss purlin
Photo: construdare.com
→ #7 / 37
I-beam purlin: bi-axial bending.
Wind
Dead weight
Snow
Imposed load
a
(D + S + I) cos a + W
(D + S + I) sin a
Wind
Dead weight
Snow
Imposed load
D + S + I + W cos a
W sin a
Truss purlin: horizontal force
W sin a
has very small value and can be neglected
(acts on roofing, not purlins). All loads act in
plane of truss. There is need wedge to install
truss purlin in vertical position.
a
Photo: Author
→ #7 / 38
„Classical” trusses
Photo: steelconstruction.infoPhoto: domgaz.com.pl
Photo: waldenstructures.com
2-chords flat trusses, used
as a single roof girders.
Multi-chords trusses
Photo: steelconstruction.info
Photo: multimetalgb.ca
3 or 4 trusses connected each other in triangle or square
cross-section. Often used for temporary structures
(bandstand) or for masts and towers.
Photo: rktruss.com
Photo: conference-truss-hire.co.uk
Photo: stretchtents.com.auPhoto: eioba.pl
Truss-grates
Photo: cdn8.muratorplus.smcloud.net
Photo: qdjinfei.en.made-in-china.com
Photo: Author
Complex of trusses of the same height, perpendicular each other.
Frame-truss
Photo: wikipedia
Photo: bbsc303.arch.school.nz
"Classical" portal frame (two columns and
roof girder), but made of trusses, not I-beams.
Spatial trusses
Photo: miripiri.co.in
Photo: urwishengineers.com
Photo: civiltech.ir Photo: shreeengineering.in/
2 or 3 layers of bars + cross-bars
between layers.
Layers can be flat, cylidrical or
spherical. More information about this
type of structures will be presented on
IInd step of studies.
Photo: cnxzlf.com
Photo: wikipedia
Photo: urwishengineers.com
Laced columns
Photo: zksgrzelak.eu
Structure looks like truss, but its behaviour and calculation model is completely different than for truss.
For laced columns, battened columns and closely spaced build-up compression members
we use specific algorithm of calculations → #t / 43
Calculations
NEd / Nb,Rd ≤ 1,0
Nb,Rd = χ A fy / γM0
NEd / Nt,Rd ≤ 1,0
Nt,Rd = A fy / γM1
χ - lecture #5:
Flexural buckling
Torsional buckling
Flexural-torsional buckling
Photo: Author
Critical length for compressed chord of truss is depend on plane of analysis. For vertical
plane, critical length is equal distance between cross-bars. For horizontal plane,
critical length is equal distance between bracings:
Photo: Author
Flexural buckling of chords:
Top chords in compression; buckling in plane: critical length = distance between nodes
Photo: Author
Critical length for compressed chord of truss is depend on plane of analysis and part of
structure. For web members, critical length = distance between nodes.
→ Des #2 / 34
Flexural buckling of chords:
Top chords in compression; buckling out of plane: critical length = distance between
horizontal bracings
Photo: Author
→ Des #2 / 35
Flexural buckling of chords:
Bottom chords in compression; buckling in plane: critical length = distance between
nodes
Photo: Author
→ Des #2 / 36
Flexural buckling of chords:
Bottom chords in compression; buckling out of plane: critical length = distance between
vertical bracings
Photo: Author
→ Des #2 / 37
Bracings are very important for behaviour of structures, including for trusses. For
calculations, there are used special algorithms and models.
More information about bracings will be presented on lecture #15.
Photo: Author
Buckling
Flexural Flexural, torsional,
flexural-torsional
c = cy = cz
(only if lcr, y = lcr, z )
c = min( cy ; cz) c = min( cy ; cz ; cT ; cz, T)
(hot rolled I)(welded I)
The result of calculations is buckling factor c.
It is calculated in different way for different cross-sections.
Photo: Author
→ #5 / 42
The same stiffness on each direction (Jy = Jz ):
Flexural buckling (y);
Flexural buckling (z) if lcr, y ≠ lcr, z (for example – distance between bracings is not the
same as between web members).
Photo: Author
Different stiffness on each direction (Jy ≠ Jz ):
Flexural buckling (y);
Flexural buckling (z);
Additionally, is possible that lcr, y ≠ lcr, z
Hot-rolled
Photo: Author
Welded cross-section or less than two axes of symmetry:
Flexural buckling (y) (axis u for L section);
Flexural buckling (z) (axis v for L section);
Torsional buckling;
Flexural-torsional buckling;
Additionally, is possible that lcr, y ≠ lcr, z ≠ lcr, T
Welded
Photo: Author
Buckling length ratio μ for truss members:
Element Direction Cross-section
I H pipe closely-spaced
build-up members
other
Chord In plane 0,9 0,9 1,0 1,0
Out of plane 1,0 0,9 1,0 1,0
Brace In plane 0,9 0,9; 1,0; 0,75 1,0; A 1,0; A
Out of plane 1,0 1,0; 0,75 1,0; A 1,0; A
Bolted connections
Chords parallel and dtruss bracing / dchord < 0,6
L sections:
leff, i = 0,5 + 0,7 li i = y, z
leff, n = 0,35 + 0,7 ln n = u, v
_ _
_ _
EN 1993-1-1 BB.1.1
Not recommended type of cross-sections
because of requirements for joints
Photo: EN 1993-1-1 fig 1.1
Main axes for differen cross-sections.
There is important, that axes are not
horizontal and vertical for L-sections.
Special way of calculations;
Resistance depends on distance between batten plates (a) and number of their
planes.
There were used special type of bars for old-type truss: closely spaced build-up
compression members.
Photo: img.drewno.pl
Photo: Author
Photo: EN 1993-1-1 fig 6.13
laced compression members,
battened compression members,
closely spaced build-up compression members
We must analysed initial deformations for members with two or more chords
(according to EN 1993-1-1 p.6.4). This means, these elements are always bent, even
if axial force acts only.
MEdII = eimperf NEd
More information will be presented on lecture #13
and # 19
→ #6 / 76
Photo: EN 1993-1-1 fig 6.7
Photo: EN 1993-1-1 fig 6.13
n = 29 n = 41
For these three types of members (laced
compression members, battened
compression members, closely spaced
build-up compression members) we
should use special way of calculations.
Of course, we can put full geometry
(chords and lacing system or batten
plates) into computer programm, but
each membes consist of many sub-
members
Photo: Author
Photo: Author
There is possible, that in structure we have more than 100 000 sub-members; it means very long term of calculations.
Photo: s9.flog.pl
laced compression members → lecture #19;
battened compression members → lecture #19;
closely spaced build-up compression members → #t;
Special way of calculations – we no analysed sub-members, but one equivalent uniform
element of effective geometry. Next, cross-sectional forces, calculated for this member,
are recalculated to new values of cross-sectional forces applied to chord and lacing
system or batten plates.
Photo: EN 1993-1-1 fig 6.13
Photo: EN 1993-1-1 fig 6.7
2 or 1 batten
planes:
Small distance: a ≤ 70 imin a ≤ 15 imin
Big distance: a > 70 imin a > 15 imin
imin = iv (for one L section)
EN 1993-1-1 tab. 6.9
There are four possibilities for closely spaced build-up compression members:
Photo: Author
For small distance: calculations as for uniform cross-sections, i.e.:
Ju1, Ju1, JW, JT – according to real cross-section;
Flexural buckling (u1);
Flexural buckling (v1);
Torsional buckling;
Flexural-torsional buckling;
Additionally, is possible that lcr, u1 ≠ lcr, v1 ≠ lcr, T
Photo: Author
No batten plates: calculation as for two independent L section
Axial force = 0,5 NEd
Flexural buckling (u);
Flexural buckling (v);
Torsional buckling;
Flexural-torsional buckling;
Additionally, is possible that lcr, u ≠ lcr, v ≠ lcr, T
Photo: Author
Batten plates exist, but on big distance:
Photo: Author
Buckling about y - y axis (material axis; axid goes through chords):
• No imperfections;
• Global axial force NEd only;
• Critical length = length of member;
• Moment of inertia for cross-section = 2 · moment of inertia for one
chord;
Photo: Author
Buckling about z - z axis (immaterial axis; axis goes out of chords):
• No imperfections;
• Global axial force NEd only;
• Critical length = length of member;
• Moment of inertia for cross-section = effective moment of inertia for
cross-section;
Photo: Author
Buckling about y1 - y1 axis (own axis of chord):
• Bow imperfections;
• Axial force Nch, Ed in one chord;
• Bending moment from imperfections Mch, Ed in one chord;
• Shear force from imperfections Vch, Ed in one chord;
• Critical length = distance of battens;
• Moment of inertia for cross-section = moment of inertia
for one chord about axis of one chord;
Photo: Author
Nch, Ed = NEd / 2 + 2 MEdII zs Ach / (2 Jeff)
MEd II = NEd e0 / [1 - (NEd / Ncr) - (NEd / SV)]
e0 = L / 500
Ncr = p2 E Jeff, / (m L)2
Numbers of modules between battens or lacings ≥ 3
Length of modules should be equal
Recommended is odd number of modules
EN 1993-1-1 6.4.1
l = m L / i0
i0 = √ [J / ( 2 Ach ) ]
l meff
0
2 - l / 75
1,0
≥ 150
≤ 75
75 ~ 150
SV = min { 24 X / [1 + 4 Jch, v zs / (n Jb a )] ; 2 p X }
Jeff = 2 zs2 Ach + 2 meff Jch
X = E Jch,v / a2
EN 1993-1-1 (6.73), (6.74)
EN 1993-1-1 tab. 6.8
n - number of battens planes
zs - distance between centre of
gravity for total cross-section and
centre of gravity for one chord
Xch - geometical characteristic of
one chord
Jb - moment of inertia for cross-
section of batten
J = 2 zs2 Ach + 2 Jch
VEd = p MEdII / (n L)
h0 = 2 zs
For chord:
Vch, Ed = VEd / 2
Mch, Ed = a VEd / 4
For batten:
Vb, Ed = VEd a / (2 h0)
Mb, Ed = a VEd / 2
Photo: EN 1993-1-1 fig 6.11
S 235
L 60x40x5
Ach = A (L 60x40x5) = 4,79 cm2
Ju1 (L 60x40x5) = 19,75 cm4 ; iu1 = 2,03 cm
Jv1 (L 60x40x5) = 3,50 cm4 ; iv1 = 0,85 cm
Jy1 (L 60x40x5) = 17,2 cm4 ; iy = 1,89 cm
Jz1 (L 60x40x5) = 6,11 cm4 ; iz = 1,13 cm
Ju (2L 60x40x5) = 93,29 cm4 ; iu = 3,14 cm
Jv (2L 60x40x5) = 24,44 cm4 ; iv = 1,60 cm
L = 4,000 m
d = 8 mm
a = 400 mm
NEd = 20,000 kN
L 60x40x5 → Ist class of cross-section
Example
Photo: Author
No batten plates → #t / 58 - 62
Batten plates in one plane → #t / 63 - 72
Batten plates in two planes → #t / 73 - 77
No batten plates: calculation for single L-section chord (according to example, #5 / 45):
NEd, 1 = NEd / 2 = 10,000 kN
NRd = A fy = 112,565 kN
mu = mv = mT = 1,00
L0u = L0v = L0T = 4,000 m
Ncr, u = p2 EJu / (mu L0u)2 = 25,584 kN
Ncr, v = p2 EJv / (mv L0v)2 = 4,534 kN
JW , JT – approximation according to #5 / 41
JW = 0,727 cm6
JT = 0,375 cm4
zs = 2,73 cm
i0 = √ (iu2 + iv
2) = 2,20 cm
is = √ (i02 + zs
2) = 3,50 cm
Ncr, T = [p2 EJw / (mT l0T)2 + GJT] / is2 = 248,036 kN
m = min[√ (mv / mT) ; √ (mT / mv)] = 1,000
x = 1 - (m zs2 / is
2) = 0,392
Ncr, zT = {Ncr, v + Ncr, T - √ [(Ncr, v + Ncr, T)2 - 4 Ncr, v Ncr, T x] } / (2 x) = 4,484 kN
A fy = NRd = 112,565 kN
lu = √(A fy / Ncr, u) = 2,098
lv = √(A fy / Ncr, u) = 4,983
lT = √(A fy / Ncr, T) = 0,674
lvT = √(A fy / Ncr, vT) = 5,010
L 60x40x5 → tab. 6.1, 6.2, 1993-1-1 → au = av = aT = avT = 0,49
Fu = [1 + au (lu - 0,2) + lu2] / 2 = 3,165
Fv = [1 + av (lv - 0,2) + lv2] / 2 = 14,085
FT = [1 + aT (lT - 0,2) + lT2] / 2 = 0,843
FvT = [1 + avT (lvT - 0,2) + lvT2] / 2 = 14,229
cu = min{1/[Fu + √ (Fu2 - lu
2)] ; 1,0} = 0,181
cv = min{1/[Fv + √ (Fv2 - lv
2)] ; 1,0} = 0,037
cT = min{1/[FT + √ (FT2 - lT
2)] ; 1,0} = 0,741
cvT = min{1/[FvT + √ (FvT2 - lvT
2)] ; 1,0} = 0,036
c = min(cu ; cz ; cT ; cvT) = 0,036
A fy = 112,565 kN
c A fy = 4,052 kN
NEd / A fy = 0,089
OK.
NEd / c A fy = 2,212 > 1,000
Wrong, buckling, destruction!
Batten plates in one plane;
Accordin to #t / 47, limit between small distance and long distance is equal
15 imin = 15 iv = 128 mm;
Distance betwen batten plates
a = 400 mm
a > 15 imin → big distance
There is even number of modules, but odd number is only recommended, not obligatory.
Ach = (L 60x40x5) = 4,79 cm2
Jch, u = Ju1 (L 60x40x5) = 19,75 cm4
Jch, v = Jv1 (L 60x40x5) = 3,50 cm4
Ju = Ju (2L 60x40x5) = 93,29 cm4
Jv = Jv (2L 60x40x5) = 24,44 cm4
e0 = L / 500 = 8 mm
zs = 2,73 cm
m = 1
i0 = √ [Jv / ( 2 Ach ) ] = 1,60 cm
l = m L / i0 = 250 → #t / 55 → meff = 0
Jeff = 2 zs2 Ach + 2 meff Jch, v = 71,40 cm4
X = E Jch,v / a2 = 45,938 kN
Jbatten = 103 ∙ 1 / 12 = 83,33 cm4
SV = min {24 X / [1 + 4 Jch, v zs / (n Jb a )] ; 2 p X } = 288,637 kN
Ncr = p2 E Jeff / (m L)2 = 92,491 kN
MEdII = NEd e0 / [1 - (NEd / Ncr) - (NEd / SV)] = 0,224 kNm
Nch, Ed = NEd / 2 + 2 MEdII zs Ach / Jeff = 18,203 kN
Vch, Ed = p MEdII / n L = 0,088 kN
Mch, v, Ed = Vch, Ed a / 4 = 0,022 kNm
Axis v goes through chords → material axis
Axis u goes out of chords → immaterial axis
Photo: Author
v - v flexural buckling → only axial force; moment of inertia Jv
NEd = 20 kN
NRd = A fy = 2 Ach fy = 225,130 kN
Lcr,u = L = 4,000 m
iv = √ (Jv / A) = 1,60 cm
lu1 = (Lcr,v / iv) (1 / l1) = 2,662
Buckling curve c → a = 0,49
Fv = 4,646
cv = 0,118
NEd / (cv NRd) = 0,753 < 1,000
OK
u - u flexural buckling → only axial force; moment of inertia = Jeff
NEd = 20 kN
NRd = A fy = 2 Ach fy = 225,130 kN
Lcr,u = L = 4,000 m
iu = √ (Jeff / A) = 2,73 cm
l = 1,560
Buckling curve c → a = 0,49
F = 2,050
c = 0,296
NEd / (c NRd) = 0,431 < 1,000
OK
v1 - v1 flexural buckling → interaction axial force - bending moment → interaction flexural-torsional buckling - lateral buckling
Nch, Ed = 18,203 kN
Mch, v, Ed = 0,022 kNm
Vch, Ed = 0,088 kN
Lcr, v1 = a = 0,800 m
Lcr, LT = a = 0,800 m
mz1 = mLT = 1,0
iv1 = √ (Jv1 / Ach) = 0,85 cm
Nch, Rd = Ach fy = 112,565 kN
Mch, Rd, v1 = 0,415 kN
Vch, Rd ≈ Ach fy / √3 = 64,989 kN
Ncr, v = p2 EJv / (mv a)2 = 36,079 kN
Ncr, T = [p2 EJw / (mT a)2 + GJT] / is2 = 249,881 kN
m = min[√ (mv / mT) ; √ (mT / mv)] = 1,000
x = 1 - (m zs2 / is
2) = 0,392
Ncr, vT = {Ncr, v + Ncr, T - √ [(Ncr, v + Ncr, T)2 - 4 Ncr, v Ncr, T x] } / (2 x) = 25,889 kN
lvT = √(A fy / Ncr, vT) = 2,085
avT = 0,49
FvT = 3,135
cvT = 0,183
Mcr = is √ (Ncr, z Ncr, T) = 2,815 kNm
lLT = √ (Wy fy / Mcr) = 0,384
FLT = [1 + aLT (lLT -0,2) + lLT2] / 2 = 0,605
cLT = 0,932
Interaction between flexural-torsional buckling and lateral buckling → lecture #18
Rough approximation (EN 1993-1-1 NA.20):
Nch, Ed / (c NRd) + Mch, Ed / (cLT MRd) ≤ 0,8
Nch, Ed / (c NRd) + Mch, Ed / (cLT MRd) = 0,884 + 0,057 = 0,941 > 0,800
Wrong, buckling, destruction!
Interaction between V and M → lecture #16
Battens resistance, welds between battens and chords → lecture #19
Batten plates in two planes;
Accordin to #t / 47, limit between small distance and long distance is equal
70 imin = 70 iv = 597 mm;
Distance betwen batten plates
a = 400 mm
a < 70 imin → small distance
Jb, u = Ju1 (2L 60x40x5) = 93,29 cm4
Jb, v = Jv1 (2L 60x40x5) = 24,44 cm4
Jw (2L 60x40x5) = 59,02 cm6 (calculations according to Mechanics of Materials, thin-walled
theory)
JT ≈ 2 JT (L 60x40x5) = 0,750 cm4
NRd = 2 A fy = 225,130 kN
mu = mv = mT = 1,00
L0u = L0v = L0T = 4,000 m
Ncr, u = p2 EJu1 / (mu L0u)2 = 120,847 kN
Ncr, v = p2 EJv1 / (mv L0v)2 = 31,659 kN
zs = 2,73 cm
i0 = √ (iu2 + iv
2) = 2,20 cm
is = √ (i02 + zs
2) = 3,50 cm
Ncr, T = [p2 EJw / (mT l0T)2 + GJT] / is2 = 495,740 kN
m = min[√ (mv / mT) ; √ (mT / mv)] = 1,000
x = 1 - (m zs2 / is
2) = 0,392
Ncr, zT = {Ncr, v + Ncr, T - √ [(Ncr, v + Ncr, T)2 - 4 Ncr, v Ncr, T x] } / (2 x) = 30,448 kN
lu = √(A fy / Ncr, u) = 1,365
lv = √(A fy / Ncr, u) = 2,667
lT = √(A fy / Ncr, T) = 0,674
lvT = √(A fy / Ncr, vT) = 2,706
L 60x40x5 → tab. 6.1, 6.2, 1993-1-1 → au = av = aT = avT = 0,49
Fu = [1 + au (lu - 0,2) + lu2] / 2 = 1,717
Fv = [1 + av (lv - 0,2) + lv2] / 2 = 4,661
FT = [1 + aT (lT - 0,2) + lT2] / 2 = 0,843
FvT = [1 + avT (lvT - 0,2) + lvT2] / 2 = 4,776
cu = min{1/[Fu + √ (Fu2 - lu
2)] ; 1,0} = 0,362
cv = min{1/[Fv + √ (Fv2 - lv
2)] ; 1,0} = 0,118
cT = min{1/[FT + √ (FT2 - lT
2)] ; 1,0} = 0,741
cvT = min{1/[FvT + √ (FvT2 - lvT
2)] ; 1,0} = 0,114
c = min(cu ; cz ; cT ; cvT) = 0,114
A fy = 225,130 kN
c A fy = 25,844 kN
NEd / A fy = 0,089
OK.
NEd / c A fy = 0,774 < 1,000
OK.
Summary:
No batten plates: effort 2,212 > 1,000
Battens in one plane (big distance): effort 0,941 > 0,800 ≈ 1,176 > 1,000
Battens in two planes (small distance): effort 0,774 < 1,000
Conclusions:
Batten plates in one planes is more effective than no batten plates, but the best are two
planes of batten plates
Results of calculations - verification
Fch-up
Fch-down
Fwm
Sx = 0 ; Fwm / Fch-down ≈ 0 →
|Fch-down| ≈ |Fch-up|
Fi
Photo: Author
q
There is possible to estimate force in top and bottom chords; we use beam-analogy
q = S Fi / L
Mmax = q L2 / 8 or q L2 / (8 cos2 a)
|Fch-down| ≈ |Fch-up| ≈ Mmax / H
Photo: Author
Similar shape of axial forces in chord and bending moment for single-span beam;
the same for axial forces in web members and shear force in single-span beam.
Photo: Author
But this estimation is not true for symmetrical supports - there are additional axial forces in chords.
Photo: Author
There is big difference between symmetrical supports and unsymmetrical supports (approximation)
Photo: Author
Ntop:
Nc, Ed, sym ≈ 1,5 Nc, Ed, unsym
Nbottom:
Nt, Ed, sym ≈ 0,5 Nt, Ed, unsym
Photo: Author
There is big horizontal reaction for symmetrical support. Because of this reaction, truss will
behave as for unsymmetrical supports after short time of exploatation (deformations of
columns, local destruction of masonry or concrete structure around anchor bolts).
Calculation as for symmetrical supports means, that forces taken under consideration in
design project are completely different than real forces in structure: for bottom chord real are
much more bigger than theoretical. This means big probability destruction of bottom chord
and collapse total structure.
Better way is to model the truss with unsymmetrical supports - this is closer its real
behaviour.
Photo: Author
There is difference between two directions of web members
Photo: Author
Deformations: elongation (tensile force) and abridgement (compressive force):
Transmission of forces form chords to supports and zero force members:
Photo: Author
Stifness of I-beam:
JI = Jtop flange + Jweb + Jbottom flange
For symmetrical cross-section:
Jtop flange = Jbottom flange ≈ 2 (h1, I / 2)2 Aflange
JI ≈ (h1, I2 / 2) Aflange + Jweb
Photo: Autor
Stifness of truss (Konstrukcje metalowe, M. Łubiński, A Filipiak, W. Żółtowski,
Arkady 2000):
Jtruss ≈ 0,7 [Atop chord Abottom chord / (Atop chord + Abottom chord)] (h1, truss / 2)2
For symmetrical cross-section:
Atop chord = Abottom chord = Achord
Jtruss ≈ 0,7 [Achord2 / (2 Achord)] (h1, truss / 2)2 = 0,7 (h1, truss
2 / 2) Achord
Stifness of I-beam:
JI ≈ (h1, I2 / 2) Aflange + Jweb
For different types of I-beams, Jweb = 7% - 25% of Jflange
Stifness of truss:
JI ≈ 0,7 (h1, truss2 / 2) Achord
For Jtruss = JI , we need Achord >> Aflange and h1, truss >> h1, I
h1, I = L / 20 - L / 25 ; h1, truss = L / 10 - L / 15 → h1, truss > h1, I
Achord - CHS, RHS, I-beam ; Aflange - plate → Achord >> Aflange
Death weight of truss
Ist proposal (PN B 02001):
gT
= [ 2 / a + 0,12 (g + q)] L / 100
gT, g (roofing + purlins), q (snow + wind) → [kN/m2], characteristic values
a (distance between trusses), L → [m]
IInd proposal:
|Fch-down| ≈ |Fch-up| ≈ Mmax / H →
Ach-down ≈ Ach-up = A = Mmax / (H fy) →
gch-down ≈ gch-up ≈ gweb members = A dsteel L →
gT
= 3 dsteel L Mmax / (H fy)
Types of truss structures
Algorithm of calculation for compressed closely spaced build-up compression members
Examination issues
Laced members - słup wielogałęziowy skratowany
Battened members - słup wielogałęziowy z przewiązkami
Closely spaced build-up members - pręt wielogałęziowy
