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Steel strucure lec # (16)
1. Prof. Dr. Zahid Ahmad Siddiqi
PROPERTIES OF TRUSSES
Truss is a frame structure in which all the
members have axial forces due to the
following facts:
a. Members are arranged in triangles for
stability.
b. All the joints of a truss are actually
semi-rigid or fully rigid. However,
theoretically, these joints may be
considered as pin joints.
2. Prof. Dr. Zahid Ahmad Siddiqi
PANEL LOADS
Concentrated load applied at the interior panel point
of the truss in kN is called Panel Load (P).
It is calculated by multiplying the roof load (load per
unit area) by the horizontal area of roof contributing
load to interior panel point of the truss, as described
in Figures 7.6 and 7.7.
It is separately calculated for dead, live and wind
loads.
3. Prof. Dr. Zahid Ahmad Siddiqi
The truss is analyzed for unit gravity loads, unit
wind on left side of truss and unit wind force on the
right side of the truss.
Principle of superposition is then used to calculate
member forces due to actual loads.
Load at interior panel point
P = load intensity over horizontal plan
area (w)
´ area supported by the panel point
(p ´ s)
= w ´ p ´ s
4. Prof. Dr. Zahid Ahmad Siddiqi
Load at exterior panel point = P / 2
s
Area Contributing
Load At One
Interior Panel
Point = p × s
pp
p/2p/2
s/2
s/2
a) Elevation of Truss
b) Part – Plan of
Truss Roof
Figure 7.6. Area Exerting Load on an Interior Panel Point.
5. Prof. Dr. Zahid Ahmad Siddiqi
p = panel length in a horizontal plane
s = center-to center spacing of trusses
S Truss T-1
Truss T-2
p
Figure 7.2. Isometric View of a Truss Roof.
6. Prof. Dr. Zahid Ahmad Siddiqi
Example 7.1: Find panel loads for the given
truss data.
Data:
Angle of top chord, q = 30°
Dead load of roofing = 160 N/m2
Insulation boards = 50 N/m2
Self weight of purlins = 100 N/m2
Self weight of bracing elements = 30 N/m2
Miscellaneous = 50 N/m2
7. Prof. Dr. Zahid Ahmad Siddiqi
Panel length, p = 2.5 m
Span length of truss, l = 20 m
Spacing of trusses, center-to-center, S = 5.5 m
Solution:
Total dead load except truss self weight
= sum of given dead loads
= 390 N/m2
Live load, from Reference-1, for q = 30o
= 600 N/m2
8. Prof. Dr. Zahid Ahmad Siddiqi
Total gravity load, w = 390 + 600
= 990 N/m2
Using Thayer’s formula,
self weight of truss = (0.37´20+1.7)
@ 122 N/m2
5.5
990
Total dead load = 390 + 122
= 512 N/m2
Leeward wind pressure = 1250(– 0.7)
= –875 N/m2
9. Prof. Dr. Zahid Ahmad Siddiqi
Windward wind pressure = 1250(– 0.9)
= –1125 N/m2
and 1250(0.3) = 375 N/m2
Panel dead load, PD = w ´ p ´ S
= 512 ´ 2.5 ´ 5.5 / 1000
= 7.04 kN
Panel live load, PL = 600 ´ 2.5 ´ 5.5 / 1000
= 8.25 kN
10. Prof. Dr. Zahid Ahmad Siddiqi
The wind load is acting perpendicular to the
inclined roof surface and hence actual inclined
roof area is to be used to calculate the panel
loads.
This can be done by using the inclined panel
length (p / cosq) in the expression for calculation
of the panel loads.
Panel wind load on leeward side, Pwl
= (–875) (5.5) / 1000
= –13.89 kN (upward)
÷
ø
ö
ç
è
æ
o
30cos
5.2
11. Prof. Dr. Zahid Ahmad Siddiqi
Panel wind load on windward side,
Pww = (–1125) (5.5)/1000
= –17.86 kN
÷
ø
ö
ç
è
æ
o
30cos
5.2
and = (375) (5.5)/1000
= 5.95 kN
÷
ø
ö
ç
è
æ
o
30cos
5.2
TABLE OF FORCES
In case only dead, live and wind loads are acting
on a truss, following combinations may be
investigated:
12. Prof. Dr. Zahid Ahmad Siddiqi
1. 1.2D + 1.6Lr + 0.65W (Wind effect is small
and may be ignored, especially if suction is
present throughout)
2. 1.2D + 0.5 Lr + 1.3W
a) Wind towards the Right
b) Wind towards the Left
3. 0.9D + 1.3W
a) Wind towards the Right
b) Wind towards the Left
13. Prof. Dr. Zahid Ahmad Siddiqi
Member
No.
Unit
gravity
load
member
force
Member
force due
to unit
wind load
on hinge
side
Member
force due
to unit
wind load
on roller
side
(1.2PD+
1.6PL)
´Col.2
(1.2PD+0.5PL)
´Col.2
+1.3Pww ´
Col.3
+1.3Pwl ´
Col.4
(1.2PD+0.5PL)
´Col.2
+1.3Pww ´
Col.4
+1.3Pwl ´
Col.3
(1) (2) (3) (4) (5) (6) (7)
Table 7.2. Sample Table of Forces.
14. Prof. Dr. Zahid Ahmad Siddiqi
0.9PD´Col.2
+1.3Pww ´ Col.3
+1.3Pwl ´ Col.4
0.9PD´Col.2
+1.3Pww ´ Col.4
+1.3Pwl ´ Col.3
Maximum
factored
tension (Tu)
Maximum
factored
Compression
(Pu)
Remarks
(8) (9) (10) (11) (12)
15. Prof. Dr. Zahid Ahmad Siddiqi
PURLIN DESIGN
General Notes
A. Allowable stress design (ASD) or load and
resistance factor design (LRFD) may be used for
the design of a purlin.
However, only ASD method is explained here in
detail. Service loads and reduced material
strengths are involved in allowable stress design.
It is assumed that the roof sheathing provides the
necessary lateral support to the purlin through J-
bolts and the purlin behaves as a continuously
braced beam.
16. Prof. Dr. Zahid Ahmad Siddiqi
Allowable bending strength, Mb = Fy Zx / Wb
» Fy ´ 1.10 Sx / 1.67
= 0.66 Fy Sx
Allowable bending stress, Fb = 0.66 Fy
Allowable tensile stress, Ft = Fy / Wt
= 0.60 Fy
B. The dead plus live load (D + L) combination is
used because it is proved to be critical for purlin
and roof sheet design.
17. Prof. Dr. Zahid Ahmad Siddiqi
C. Dead load on purlin acts due to roofing,
insulation and self-weight of the purlin.
Insulation load is considered if it is directly attached
or hanged from the sheet or the purlin.
Approximately one-third or half of the miscellaneous
load may also be included.
D. Depth of section should not be lesser than 1/30
th
of the purlin span to control deflections.
dmin ³ S/30
18. Prof. Dr. Zahid Ahmad Siddiqi
E. Order of preference for member selection may
generally be as under:
i) Single angle section with no sag rod
ii) Single angle section with one sag rod
iii) Single angle section with two sag rod
iv) C-section with no sag rod
v) C-section with one sag rod
vi) C-section with two sag rod
vii) W or S-section with no sag rod
viii) W or S-section with one sag rod
ix) W or S-section with two sag rod
19. Prof. Dr. Zahid Ahmad Siddiqi
Z-section is behavior-wise the best section for a
purlin. However, as it is not a hot-rolled section
and is to be made by cold bending, it may not be
readily available.
In case the section modulus required for the first
option is much greater than 230´103 mm3, the
option of channel section may be selected directly.
F. The width of angle section may not commonly
exceed 102 mm.
G. The roof load is converted into beam load per
unit length by the formula given below:
20. Prof. Dr. Zahid Ahmad Siddiqi
Load per unit length = load per unit area of roof ´
purlin spacing
Note:If the panel length is excessive and it is
difficult to design the roofing, purlins are also
placed in between the panel points reducing the
purlin spacing and span for the roof sheet.
This induces bending moment in the top chord of
the truss, which must be checked as a beam
column for such cases.
H. Lateral component of loads at the top flange
producing torsion should be considered separate
from the self-weight of purlin not producing torsion.
21. Prof. Dr. Zahid Ahmad Siddiqi
Torque is
Present
No
Torque
Figure 7.8. Purlin Loads With And Without Torque.
I. In place of using complicated formulas for
torsion design, half strength in lateral direction
(Sy/2 or Zy/2) is reserved for torsion and only the
other half (Sy/2 or Zy/2) is used for lateral bending.
No calculations for torsion are required afterwards.
22. Prof. Dr. Zahid Ahmad Siddiqi
J. Purlin is assumed to be simply supported on
trusses, both for x and y direction bending. The
bending moments may be calculated by using the
typical bending moment diagrams given in
Reference-1.
K. Sag rod is considered as a lateral roller support
for purlin with no effect on major axis bending
(Figure 7.9).
a) Major Axis Bending
b) Minor Axis Bending
Figure 7.9.Major Axis And Lateral Bending of a Purlin With
Mid-Point Sag Rod.
23. Prof. Dr. Zahid Ahmad Siddiqi
L. Applied stress,
fb = + stresses due to torque
y
y
x
x
S
M
S
M
+
For an ordinary beam (where only Mx is present),
the section is selected on the basis of section
modulus and not cross-sectional area as in tension
and compression members.
Sx = Mx / Fb
However, in case of a purlin, two unknowns (Sx and
Sy) occur in a single equation.
24. Prof. Dr. Zahid Ahmad Siddiqi
We cannot calculate Sx and Sy as such, making
it necessary to use some simplifying
assumption for the selection of the trial section.
Once the trial section is selected, its stresses
may easily be back checked to verify that they
remain within the permissible range.
Procedure For Purlin Design
1. wD (N/m) = (load of roofing + insulation
+part of miscellaneous loads) ´ purlin pacing
+ (purlin self weight) ´ purlin spacing
25. Prof. Dr. Zahid Ahmad Siddiqi
wx
wy
q
w
Figure 7.10. Components of
Load Acting On a Purlin.
The two terms are kept separate as one is
producing torque while the other is not.
26. Prof. Dr. Zahid Ahmad Siddiqi
2. wL (N/m) = live load (N/m2) ´ purlin
spacing
Again, self weight of the purlin is kept as a
separate entity.
Calculate wx and wy by referring to Figure 7.10.
Calculate maximum values of Mx and My by using
bending moment diagrams for the given sag rod
case.
Further, calculate My for loads producing torsion
and loads not producing torsion separately.
27. Prof. Dr. Zahid Ahmad Siddiqi
6. For the selection of trial section, make the
following approximation applicable only for this
step.
(My)ass = 0
(Mx)ass = Mx + 4 My for single unequal leg
angle purlins
(Mx)ass = Mx + 2 My for single equal leg
angle purlins
(Mx)ass = Mx + 15 My for C and W sections
purlins
28. Prof. Dr. Zahid Ahmad Siddiqi
7. Calculate the required elastic section
modulus about the major axis according to the
assumption of step number 6.
(Sx)req =
( ) ( )
y
assx
b
assx
F
M
F
M
66.0
=
Select the section such that Sx » (Sx)req, d ³ S/30
and the preference of section is satisfied.
8. Actual bending stress is then evaluated by
using the following expression:
fb = (with torsion) + (no torsion)
2/y
y
x
x
S
M
S
M
+
y
y
S
M
29. Prof. Dr. Zahid Ahmad Siddiqi
Always consider magnitudes of Mx and My without
their signs because each combination gives
addition of stresses at some points within the
section.
9. If the stress due to My is more than two
times the stress due to Mx, revise the section by
a) increasing the sag rods
b) selecting section with bigger Sy / Sx ratio
However, if sag rods are limited due to
construction difficulties, the first option is not
employed.
30. Prof. Dr. Zahid Ahmad Siddiqi
10. If fb £ Fb OK
otherwise, revise the section.
11. Check b/t for angles, bf / tf for channels and
bf / 2tf for W-sections (called l-value).
l £ lp OK
otherwise, revise the section.
For single angles, only shorter leg is in
compression throughout and hence is to be used to
check l value.
31. Prof. Dr. Zahid Ahmad Siddiqi
The value of lp for unstiffened elements is 10.8
and for stiffened elements is 31.6 for A36 steel.
Any section meeting these requirements and
continuously braced in lateral direction is called
compact section.
12. Check self-weight of the purlin:
Actual self-weight of purlin = weight of purlin
section (kg/m) ´ 9.81 ´ number of purlin / span of
the truss
Provided self-weight
£ 1.20 ´ assumed purlin weight OK
32. Prof. Dr. Zahid Ahmad Siddiqi
otherwise, revise purlin self-weight and all the
calculations.
Write the final selection using standard
designation.
Design the sag rod, if required.
Design Of Sag Rod
1. Force in sag rod, F =
force due to one purlin from Reference-1
´ (no. of purlins on one side – 1)
33. Prof. Dr. Zahid Ahmad Siddiqi
2. Component of tie rod force in the direction of
sag rod direction should provide the required
force F (Figure 7.11).
R cos q = F
Force in tie rod = R = F / cosq
3. Calculate required area of the sag and tie
rods and select section.
34. Prof. Dr. Zahid Ahmad Siddiqi
Example 7.2: Design a channel section purlin
with midpoint sag rod for the following data:
Dead load of roofing = 160 N/m2
Insulation = 50 N/m2
Assumed self weight of purlin = 100 N/m2
(approximately 15% of the applied load)
Live load = 590 N/m2
q = 30°
p = 2.5 m
S = 5.5 m
35. Prof. Dr. Zahid Ahmad Siddiqi
No. of truss panels = 8
Solution:
wD = 210 ´ 2.5 + 100 ´ 2.5
= 525 + 250 N/m
wL = 590 ´ 2.5 = 1475 N/m
w = 2000 + 250 N/m
Mx = S2
= = 7368 N-m
8
cosqw
2
5.5
8
30cos2250
´
o
36. Prof. Dr. Zahid Ahmad Siddiqi
My = +
= 945.3 + 118.2 N-m
2
5.5
32
30sin2000
´
o
2
5.5
32
30sin250
´
o
(Mx)ass = Mx + 15 My = 23,320 N-m
(Sx)req = = 141.3 ´ 103 mm3
25066.0
1000320,23
´
´
dmin = S / 30 = = 183 mm
30
10005.5 ´
Trial Section No. 1: C 230 ´ 19.9
Sx = 174 ´ 103 mm3 : Sy = 15.8 ´ 103 mm3
37. Prof. Dr. Zahid Ahmad Siddiqi
d > dmin OK
fb =
= 42.34 + 127.14 = 169.5 MPa > Fb
(revise)
333
108.15
10002.118
10)2/8.15(
10003.945
10174
10007368
´
´
+
´
´
+
´
´
Note:The stress due to My is more than two times
that due to Mx.
The numerical values of stresses due to bending in
the two directions also explain the importance of
lateral bending compared with the major axis
bending.
38. Prof. Dr. Zahid Ahmad Siddiqi
Smaller value of My divided by very less value of
Sy/2 may give higher answer for the stresses.
Trial Section No. 2: MC150´22.5
Sx = 136 ´ 103 mm3 Sy = 28.7 ´ 103 mm3
The depth of this section is less than the required
minimum depth and hence it must be revised.
Trial Section No. 3: C230´22
Sx = 185 ´ 103 mm3 Sy = 16.6 ´ 103 mm3
fb = 39.83 + 121.0 = 160.84 MPa
< Fb OK
39. Prof. Dr. Zahid Ahmad Siddiqi
bf / tf = 63/10.5 = 6 < 10.8 OK
Final Selection: C230 ´ 22
Check For Self Weight
Actual self weight of purlin =
@ 108 N/m2 < 1.20 ´ 100 N/m2 OK
20
1081.922 ´´
Design Of Sag Rod
F = 5/8 w sinq ´ S ´ 4
= 5/8 ´ 2250 ´ sin 30° ´ 5.5 ´ 4
= 15,469 N
40. Prof. Dr. Zahid Ahmad Siddiqi
R = F / cosq
= 15,469 / cos30° = 17,862 N
Areq =
=
dreq = 12.31 mm
Use 15 mm diameter steel bar as sag rods
yF
R
6.0
2
4
reqd
p
2506.0
17862
´
41. Prof. Dr. Zahid Ahmad Siddiqi
GALVANIZED IRON (G. I.)
CORRUGATED ROOFING SHEETS
Standard Designation:
Nominal pitch in mm x Nominal depth in mm.
Following two sheets are commonly used.
65 x 13 G. I. Corrugated Sheet.
75 x 20 G. I. Corrugated Sheet.
The nomenclature for various dimensions is shown
in Figures 7.13 and 7.14.
42. Prof. Dr. Zahid Ahmad Siddiqi
C
S
P
D t
W
Figure 7.13. View of G. I. Corrugated Sheets Along the Corrugations.
E
E
L
Figure 7.14. View of G. I. Corrugated Sheets Perpendicular to Corrugations.
43. Prof. Dr. Zahid Ahmad Siddiqi
Symbol Description Sheet Designation
65 x 13 75 x 20
Pn Nominal pitch, mm 65 75
P Actual pitch, mm 66 73
Dn Nominal depth, mm 13 20
D Actual depth, mm 13 19
W Total width of one sheet, mm 700 700
S Side laps, mm
11/2 corrugations with fasteners placed at a
maximum spacing of 300 mm in
perpendicular direction.
105 115
C Cover (Effective width covered by one
sheet), mm
595 585
Nc Number of corrugations in cover 9 8
E Minimum end lap, mm
Fasteners are to be provided not less than at
every third corrugation over each purlin.
200 200
44. Prof. Dr. Zahid Ahmad Siddiqi
t Thickness of sheet, mm Varies according
to gage.
Varies
according to
gage.
L Length of one sheet, m
1.5m to 4.0m, 0.25m increments.
Preferably should be close to
multiples of horizontal panel length
(p) divided by cosine of roof
inclination (q) plus end lap (E).
( n p/cosq + E) ( n p/cosq +
E)
Fb Allowable working stress, MPa 0.60 Fy 0.60 Fy
Da Maximum allowable deflection. span/90 span/90
45. Prof. Dr. Zahid Ahmad Siddiqi
US
Gage
Weight Thickness
t
Weight
without laps
Properties per meter of Corrugated width
(Oz. per
Sft.)
(mm) (N/m2) A
mm2
I
(x 104 mm4)
S
(x 103 mm3)
12 70 2.753 236.7 2919 5.69 7.42
14 50 1.994 171.5 2098 4.03 5.48
16 40 1.613 139.0 1687 3.22 4.51
18 32 1.311 112.6 1361 2.58 3.70
20 24 1.006 86.7 1031 1.95 2.86
22 20 0.853 73.3 868 1.64 2.42
46. Prof. Dr. Zahid Ahmad Siddiqi
US
Gage
Weight Thickness
t
Weight
without laps
Properties per meter of Corrugated width
(Oz. per
Sft.)
(mm) (N/m2) A
mm2
I
(x 104 mm4)
S
(x 103 mm3)
12 70 2.753 250.6 3107 14.45 13.28
14 50 1.994 181.6 2235 10.31 9.84
16 40 1.613 147.1 1797 8.26 8.01
18 32 1.311 119.3 1448 6.65 6.56
20 24 1.006 91.5 1099 5.04 5.03
22 20 0.853 77.6 923 4.23 4.26
24 16 0.701 63.7 747 3.43 3.47
26 12 0.551 50.3 576 2.64 2.69
28 10 0.475 43.1 487 2.23 2.29
29 9 0.437 39.8 445 2.03 2.09
47. Prof. Dr. Zahid Ahmad Siddiqi
Notes:
U.S. Standard Gage is officially a weight gage, in
Ounces per Sft. of flat sheet.
The approximate thickness is calculated by using
the steel density 7846 Kgs/m3 plus 2.5 percent
allowance for average over-run in area and
thickness.
Smaller gage always means more sheet
thickness.
48. Prof. Dr. Zahid Ahmad Siddiqi
Maximum actual deflection due to live udl:
For simply supported sheet:
Dmax. = 0.013 ´ wL p4 / EI
For sheet with one end continuous considering some
constraint at ends:
Dmax. @ 0.001 ´ wL p4 / EI
For sheet with one end continuous:
Dmax. @ 0.0054 ´ wL p4 / EI
Sheets per 100 m2 of inclined roof area are:
N100 =
( ) ( )ELCECSLWL -
@
+-
88
1010
49. Prof. Dr. Zahid Ahmad Siddiqi
DESIGN OF CORRUGATED SHEET
1. Use Reference-1 for the related definitions
and data.
Allowable stress design (ASD) is used here as
for the purlin design.
Dead plus live load combination seems to be
critical for the sheet design and hence wind
combinations are not considered.
2. End lap should be exactly on the purlin
(Figures 7.13 and 7.14).
50. Prof. Dr. Zahid Ahmad Siddiqi
Incorrect
Correct
Rain Water
Figure 7.14.Correct Placement of Overlap Within End Lap.
51. Prof. Dr. Zahid Ahmad Siddiqi
Accordingly, the length of sheet panel (L) is
corresponding to 1,2 or 3 times the inclined panel
length plus the required end lap.
The resulting dimension may be rounded to upper
0.25 m length.
If the required length of sheet corresponding to
single panel length is more than 4.0 m or if the
available section modulus does not satisfy the
flexural criterion, extra purlins may be placed
between the two panel points.
If one purlin is used at center of a panel length,
the span of the sheet reduces to p/2.
52. Prof. Dr. Zahid Ahmad Siddiqi
However, the top chord of the truss must be
checked for the combined action of compression
and bending moment.
Similarly, purlin design must also be made using
spacing of purlins equal to the modified c/c
distance between the purlins.
(a) End Lap Over Purlins (b) End Lap Within Purlins
Correct
Incorrect
Figure 7.13.Correct Position of End Lap.
53. Prof. Dr. Zahid Ahmad Siddiqi
3. Total load on the sheet = dead load of
roofing + insulation + live load (N/m2).
4. Consider unit width of slab (1 m) and design
this strip as a beam.
5. Load per unit length of roof strip, w (N/m), is
calculated as follows:
w = load per unit roof area
´ 1 m width
= load per unit roof area,
magnitude-wise only, in N/m
units (only applicable for a roof
and not for a beam)
54. Prof. Dr. Zahid Ahmad Siddiqi
6. Assume the sheet to be simply supported
over the purlins. Even if it is continuous, the
maximum bending moment will nearly be the
same.
Mmax = (N – m)
8
2
pw´
7. (Sx)req =
yF
M
6.0
1000max ´
Select gage of sheet from 1st column of the
corresponding table in Reference-1 for properties
of the corrugated sheets.
55. Prof. Dr. Zahid Ahmad Siddiqi
8. Actual self weight of roofing
= value from Col.4 of the properties table
´ 1.35
£ 1.2 ´ assumed weight OK
If self-weight is significantly greater, revise the
sheet as well as the purlin design.
9. Calculate the deflection due to live load (Dmax)
and check against the allowable value (Da).
Dmax < Da OK
56. Prof. Dr. Zahid Ahmad Siddiqi
10. Calculate the number of sheet panels required for
100 m2 of roof area (N100) using the expression given in
Reference – 1. It is better to use the actual length of the
panel before rounding.
11. The total length of building may be
represented by Nt ´ S, where Nt is the number of
spaces between the trusses.
The inclined roof area on one side, A
= ´ Nt ´ S
( )
qcos
ProjectionSheet2+l
57. Prof. Dr. Zahid Ahmad Siddiqi
Number of sheets on one side, N1
= A ´ (round to higher whole number)
100
100N
Total number of sheets = 2 ´ N1
12. Decide spacing of bolts in end and side laps such
that bolts are only applied at the crests.
13. Summarize the design results as under:
Final Results For Corrugated
Roof Sheet Design:
1. Gage of sheet
58. Prof. Dr. Zahid Ahmad Siddiqi
2. Standard designation
3. Sheet panel size
4. Bolt spacing in the two directions
5. Number of sheets required