ERT352
FARM STRUCTURES
It doesn’t matter what the subject is; once you’ve learnt how to
study, you can do anything you want.
• A truss is essentially a triangulated system of straight
interconnected structural elements.
• The most common use of trusses is in buildings, where
support to roofs, the floors and internal loading such as
services and suspended ceilings, are readily provided.
• The individual elements are connected at nodes; the
connections are often assumed to be nominally pinned.
• The external forces applied to the system and the reactions
at the supports are generally applied at the nodes.
• The principal force in each element in a truss is axial
tension or compression.
• Trusses comprise assemblies of tension and compression
elements.
• Under gravity loads, the top and bottom chords of the truss
provide the compression and tension resistance to overall
bending, and the bracing resists the shear forces.
• A wide range of truss forms can be created. Each can vary
in overall geometry and in the choice of the individual
elements.
• Some of the commonly used types are shown below.
1.
Pratt truss ('N' truss)
•
2.
Pratt trusses are commonly used in long span buildings ranging from
20 to 100 m in span.
Warren truss
•
The Warren truss has equal length compression and tension web
members, and fewer members than a Pratt truss.
3.
North light truss
•
4.
Saw-tooth truss
•
5.
North light trusses are traditionally used for short spans in industrial
workshop-type buildings.
The saw-tooth truss is used in multi-bay buildings
Fink truss
•
This type of truss is commonly used to construct roofs in houses.
1. POINT LOADS AT THE NODES OF THE TRUSS
i.
ii.
This condition occurs when purlins are located at the nodes of the
truss
Purlin spacing is equal to node spacing; Sp = Sn
1. POINT LOADS AT THE NODES OF THE TRUSS
(cont.)
o Given Imposed load on plan, Qk and Dead load on plan,
Gk
o To obtain the point load, P;
• Area of load transferred to each node,
• Design load,
• Point load,
A = Sp x St
q = 1.35Gk + 1.5Qk
P=qxA
o Calculate the tension and compression forces in each
truss members.
• There are possible
combinations of steel sections
that can be used for tension or
compression truss members.
i.
ii.
iii.
Single angle connected through
on leg
Double angle back-to-back,
connected to both sides of
gusset plate
Double angle back-to-back,
connected to one side of gusset
plate
• The loads from the roof sheets, transferred to the purlins, are as
followed:
o Dead load on slope:
Corrugated steel roofing
Lighting and insulation
Self weight of purlins
Self weight of trusses
Total dead load on slope, Gk
= 0.1 kN/m2
= 0.15 kN/m2
= 0.05 kN/m2
= 0.1 kN/m2
= 0.4 kN/m2
= 0.81 kN/m2
o Imposed load on slope, Qk
o Spacing between trusses, St = 5 m
o Purlin Spacing , Sp
=2m
• The purlins are arranged and located at the nodes of the truss as
shown in figure. Determine the point loads that are imposed by
the purlins at all the nodes.
• Design load or factored load, q
q = 1.35Gk + 1.5 Qk
= 1.35 (___) + 1.5 (___)
= _______ kN/m2
• Area of load transferredto any intermediate nodes, A
A = Sp x St = ____ x ____ = _____ m2
• Point load at node, P
P = q x A = ____ x ____ = _____ kN
• The following truss carries a uniform factored load of 26
kN/m. Determine the member forces.
• Point load at each node, P
P
= w * purlin spacing (Sp)
P
= 26 kN/m x 3 m
= 78kN
~ for external nodes, point load is equal to P/2 = 78/2 = 39kN
• By using method of joint or method of section, member
forces can be determined.
• The design value of tension force, NEd at each cross section
shall satisfy as specified in clause 6.2.3 EN 1993-1-1:2005
;
N Ed
 1.0
Nt , Rd
• Where, NEd is design force ; Nt,Rd is design values of
resistance to tension forces
• For sections with holes, the design tension resistance, Nt,Rd
should be taken as the smaller of
a)
The design plastic resistance of the gross cross-section, Npl,Rd
N pl , Rd 
Af y
 m0
Where, A is area of section; fy is yield strength; γm0 is resistance of crosssections (= 1.0);
b)
The design ultimate resistance of the net cross-section at holes for
fasteners, Nu,Rd
N u , Rd 
0.9 Anet f u
 m2
Where, Anet is net area of cross-section; fu is ultimate strength (Table 3.1 of En
1993-1-1:2005); γm2 is resistance of cross-sections in tension fracture (=
1.25);
• The maximum design tension forces in a truss member is
230kN. The member consists of an unequal angle of
100 x 75 x 8 L of steel grade S275. Check the capacity of
the tension member if the ends of the member are:
a)
b)
Connected by two bolts of 20mm nominal diameter
Welded with grade 42 electrode.
a) End members connected with 2 bolts
Tension force, NEd = 230 kN
o Unequal angle; 100 x 75 x 8 L
o
•
From table of properties; Ag = 1350mm, iy = 40mm,
iz = 20.9mm, iu = 42.1mm, iv = 16.3mm
•
From Table 3.1, t = 8mm < 40mm, steel grade S275
 fy = 275N/mm2, fu = 430N/mm2
•
Diameter of bolt, d = 20mm
N Ed
 1.0
Nt , Rd
• For sections with holes, the design tension resistance, Nt,Rd
should be taken as the smaller of
a)
The design plastic resistance of the gross cross-section, Npl,Rd
N pl , Rd 
Af y
 m0
(1350mm2 )( 275 N / mm2 ) x10 3

1.0
 371.3kN
b) The design ultimate resistance of the net cross-section at holes for
fasteners, Nu,Rd
Ah  t (d  2)  8(20  2)  176mm2
Anet  Ag  Ah  1350 176  1174mm2
0.9 Anet f u
N u , Rd 
 m2
0.9(1174mm2 )( 430 N / mm2 ) x10 3

1.25
 363.5kN
 therefore, Nt,Rd = 363.5kN
N Ed
230kN

 0.633  1.0
Nt , Rd 363.5kN
• Section 100 x 75 x 8 L is adequate to carry the force.
b) End member connected with welded
o
o
Tension force, NEd = 230 kN
In this case, there is no hole for the bolts, hence no reduction in
cross sectional of the angle section
o The design tension resistance, Nt,Rd should be taken as the smaller
of
• The design plastic resistance of the gross cross-section, Npl,Rd
N pl , Rd 
Af y
 m0
(1350mm2 )( 275 N / mm2 ) x10 3

1.0
 371.3kN
b) The design ultimate resistance of the net cross-section , Nu,Rd
• Since there is no reduction in the cross section of the angle,
the
Anet  Ag  1350mm2
N u , Rd 
0.9 Anet f u
 m2
0.9(1350mm2 )( 430 N / mm2 ) x10 3

1.25
 418kN
 therefore, Nt,Rd = 371.3kN
N Ed
230kN

 0.619  1.0
Nt , Rd 371.3kN
• Section 100 x 75 x 8 L is adequate to carry the force.
1. FULL SECTION RESISTANCE OF COMPRESSION MEMBER
o The full design resistance of a member that fails by yielding and not
susceptible to buckling is given as Nc,Rd as specified in clause 6.2.4
EN 1993-1-1:2005
N Ed
 1.0
N c , Rd
o Where NEd is the design compression force and Nc,Rd is the full
design resistance for section in classes 1,2, and 3.
N c , Rd 
Af y
 m0
2. BUCKLING RESISTANCEOF COMPRESSION MEMBER
o
The design of compression member susceptible to buckling shall
satisfy as specified in clause 6.3.1 EN 1993-1-1:2005 :
N Ed
 1.0
N b, Rd
o
The buckling resistance, Nb,Rd of a member carrying axially loaded
compression force is given by:
N b, Rd
Af y

 m1
3. BUCKLING RESISTANCE ABOUT Y-Y AXIS, Ny,b,Rd
N y ,b, Rd
 y Af y

 m1
• Where; the capacity of reduction factor is given as
y 
1
y  

2
y
 y
 1.0
 y  0.5 1   ( y  0.2)   y
y 
Af y
N cr
Lcr 1

x
i y 1
And λ1 = 93.9ε = 93.9 x 0.92 = 86.39
2

4. BUCKLING RESISTANCE ABOUT Z-Z AXIS, Nz,b,Rd
N z ,b, Rd
 z Af y

 m1
• Where; the capacity of reduction factor is given as
z 
1
 z   z  z
2
 1.0

 z  0.5 1   (z  0.2)  z
z 
Af y
N cr
Lcr 1

x
iz 1
And λ1 = 93.9ε = 93.9 x 0.92 = 86.39
2

• Buckling Length, Lcr
o Lcr of the top chord equals to the distance between nodes
• The following figure shows a truss with one of the member
carries compression force of 50kN. The member consists of
a single angle section with both ends are connected to long
leg. Design the compression member using steel grade
S275 with end members are connected with two bolts.
• Axial compression load, NEd = 50 kN
• Try unequal angle 80 x 60 x 7 L
o Ag = 938 mm2; T = 7 mm;
o iy = 25.1mm; iz = 17.4mm; iu = 27.7mm; iv = 12.8mm
• From Table 3.1: Steel grade S275, t = 7mm < 40mm
•
fy = 275 N/mm2
• From Table 5.2(sheet 3 of 3) (page 44)
• ε = (235/fy)0.5 = (235/275)0.5 = 0.92
• For angle; h/t = 80/7 = 11.4 ≤ 15 ε = 15(0.92) = 13.8
(b + h)2t = (60+80)/(2x7) = 10.0 ≤ 11.5ε = 11.5(0.92) = 11.5
 Section is class 3 and not susceptible to local buckling
• The smallest radius of gyration, iv = 12.8mm will contribute
to the largest slenderness = Lcr/iv, hence v-v is the weakest
axis.
*Note:
the larger the slenderness, the
weaker is the axis.
• Buckling resistance about
axis, v-v axis, Nv,b,Rd
 weakest
Af
N v ,b , Rd 
v
 m1
y
v 
1
 v   v  v
2
 1 .0

 v  0.5 1   (v  0.2)  v
v 
Af y
N cr
Lcr 1

x
iv 1
• And λ1 = 93.9ε = 93.9 x 0.92 = 86.39
• Lcr = 1.0L = 1.0 (2154) = 2154 mm
2

Lcr 1 2154mm
1
v 
x 
x
 1.95
iv 1 12.8mm 86.39
• From Table 6.2: L – sectionshave buckling curve type ’b’
• From Table 6.1:for type ‘b’ , imperfection factor, α = 0.34

v  0.5 1   (v  0.2)  v

2


 v  0.5 1  0.34(1.95  0.2)  (1.95) 2  2.7
v 

1
 v   2 v  v
1
2.7 
2.7 2  1.95
 0 .2  1.0
N v ,b , Rd
 v Af y

 m1
0.2(938)(275) x10 3

1 .0
 51.6kN
• Check force equilibrium
NEd =50kN < Nv,b,Rd = 51.6kN  OK!
• Angle section 80 x 60 x 7 L is adequate.