Steel strucure lec # (16)

Steel strucure lec # (16)

1. Prof. Dr. Zahid Ahmad Siddiqi

PROPERTIES OF TRUSSES

Members are arranged in triangles for
stability.

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

A load intensity over horizontal plan
area (w)
x area supported by the panel point
(px)

wxpxs

4. Prof. Dr. Zahid Ahmad Siddiqi

Load at exterior panel point = BI?
ge

a) Elevation of Truss
pS T
Se T

Area Contributing

ad At One
f Panel

Interior
Point =p x s

b) Part - Plan of
Truss Roof

Figure 7.6. Area Exerting Load on an Interior Panel Point.

5. Prof. Dr. Zahid Ahmad Siddiqi

panel length in a horizontal plane
center-to center spacing of trusses

Truss T-1

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, @ 30°
Dead load of roofing 160 N/m?

Insulation boards 50 N/m2
Self weight of purlins 100 N/m?
Self weight of bracing elements 30 N/m?

Miscellaneous 50 N/m2

7. Prof. Dr. Zahid Ahmad Siddiqi

Panel length, p = 25m

Span length of truss, / = 20m

Spacing of trusses, center-to-center,S = 5.5m
Solution:
Total dead load except truss self weight
= sum of given dead loads
= 390 N/m?
Live load, from Reference-1, for 9 = 30°
= 600 N/m?

8. Prof. Dr. Zahid Ahmad Siddiqi

Total gravity load, w = 390 + 600
= 990 N/m?

Using Thayer’s formula, 990
self weight of truss = 35 (0.37x20+1.7)

= 122 N/m?
Total dead load = 390 + 122
= 512 N/m2
Leeward wind pressure = 1250(- 0.7)
= -875 N/m?

9. Prof. Dr. Zahid Ahmad Siddiqi

Windward wind pressure = 1250(- 0.9)
= -1125 N/m?
and 1250(0.3) = 375 N/m?

Panel dead load, P, = wxpxS

= 52215555000
= 7.04 kN

Panel live load, P, 600 x 2.5 x 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 / cos6) in the expression for calculation
of the panel loads.

Panel wind load on leeward side, P,,
= ces) 2 (5.5) / 1000

5
30°

cos

= -13.89 KN (upward)

11. Prof. Dr. Zahid Ahmad Siddiqi

Panel wind load on windward side,
25

cos30°

-17.86 kN

= (375) (es) 000

Pay 1129) Js 51000

= 5.95 kN
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

1.2D + 1.6L, + 0.65W (Wind effect is small
and may be ignored, especially if suction is
present throughout)

1.2D + 0.5 L,+ 1.3W
a) Wind towards the Right

b) Wind towards the Left
0.9D + 1.3W

a) Wind towards the Right
b) Wind towards the Left

13. Member
No.

Unit Member
gravity force due
load to unit
member wind load

force on hinge

side

Member
force due
to unit
wind load
on roller
side

Prof. Dr. Zahid Ahmad Siddiqi

Table 7.2. Sample Table of Forces.

(1.2P+0.5P,) — (12P,+0.5P,)
xCol.2

xCol.2
11328

sr +L3P,,, X
Col.3

Col.4
+1.3P yp x

+13P,,x
Col.4 Col3

14. Prof. Dr. Zahid Ahmad Siddiqi

0.9P,xCol2 0.9P,xCol.2 Maximum Maximum Remarks
+1.3P,,, x Col.3 +1.3P,,, x Col.4 factored factored

+1.3P,,, x Col.4 +1.3P,,, x Col.3 tension (7,,) Compression
(2)

15. Prof. Dr. Zahid Ahmad Siddiqi

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 Sidi
Allowable bending strength, M, = F,2,/Q,
= Fyx 1.10 S,/1.67
= 0.66F, S,
Allowable bending stress, F, = 0.66 F,

Allowable tensile stress, F, = F,/9Q,
= 0.60F,

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/,,'"
of the purlin span to control deflections.

Ss
ar 2 Lao

18. Prof. Dr. Zahid Ahmad Siddiqi

E. Order of preference for member selection may
generally be as under:

Single angle section with no sag rod
Single angle section with one sag rod
Single angle section with two sag rod
C-section with no sag rod

C-section with one sag rod

C-section with two sag rod

Wor S-section with no sag rod

Wor S-section with one sag rod

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 230x102 mm, 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 x
purlin spacing

Note:lf 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 No
Present 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
(S/2 or Z,/2) is reserved for torsion and only the
other half (S/2 or Z,/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,

M, M,
= —=+
fy Ss, 5S,
For an ordinary beam (where only M, is present),
the section is selected on the basis of section
modulus and not cross-sectional area as in tension

and compression members.
S, = M,/F,

+ stresses due to torque

However, in case of a purlin, two unknowns (S, and
Sy) occur in a single equation.

24. Prof. Dr. Zahid Ahmad Siddiqi

We cannot calculate S, and S, 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: Wp (Nm) = (load of roofing + insulation
+part of miscellaneous loads) x purlin pacing
+ (purlin self weight) x purlin spacing

25. Prof. Dr. Zahid Ahmad Siddiqi

The two terms are kept separate as one is
producing torque while the other is not.

yw
Me)
w, x
\
Y

>

Figure 7.10. Components of
Load Acting On a Purlin.

26. Prof. Dr. Zahid Ahmad Siddiqi

w, (Nm) = live load (N/m?) x purlin
spacing

Again, self weight of the purlin is kept as a
separate entity.

Calculate w, and w, by referring to Figure 7.10.

Calculate maximum values of M, and M, by using
bending moment diagrams for the given sag rod
case.

Further, calculate M, 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.

Mass
(M,)

x? ass

(M,)

ass

(M,)

ass

0

M,+ 4 M, for single unequal leg
angle purlins

M,+2M, for single equal leg
angle purlins

M, + 15 M, 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.
is) = Ue Ud,
om F, 0.66F,
Select the section such that S, + (S,).., d > S/30
and the preference of section is satisfied.

8. Actual bending stress is then evaluated by
using the following expression:

M,
5, Su (with torsion) + — (no torsion)

Sy

29. Prof. Dr. Zahid Ahmad Siddiqi

Always consider magnitudes of M, and M, without
their signs because each combination gives
addition of stresses at some points within the
section.

9. If the stress due to M, is more than two
times the stress due to M,, revise the section by

a) increasing the sag rods

b) selecting section with bigger S,/S, ratio

However, if sag rods are limited due to
construction difficulties, the first option is not
employed.

30. Prof. Dr. Zahid Ahmad Siddiqi

otherwise, revise the section.

Check bit for angles, b;/ t, for channels and
b,/ 2t, for W-sections (called A-value).

A < 4 OK

otherwise, revise the section.

For single angles, only shorter leg is in
compression throughout and hence is to be used to
check À value.

31. Prof. Dr. Zahid Ahmad Siddiqi

The value of Ao 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) x 9.81 x number of purlin / span of
the truss

Provided self-weight

< 1.20 x assumed purlin weight

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

x (no. of purlins on one side — 1)

33. Prof. Dr. Zahid Ahmad Siddiqi

Component of tie rod force in the direction of
sag rod direction should provide the required
force F (Figure 7.11).

Rcos@=F

Force in tie rod = R = F/cos0

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/m?
Insulation = 50 N/m?

Assumed self weight of purlin = 100 N/m?

(approximately 15% of the applied load)
Live load 590 N/m?

30°

2.5m

5.5m

35. Prof. Dr. Zahid Ahmad Siddiqi

No. of truss panels

Solution:
210 x 2.5 + 100 x 2.5
525 + 250 N/m
590x2.5 = 1475 N/m

2000 + 250 N/m
w cos@ $2
8

225000830" ¿52 = 7368 N-m

36. Prof. Dr. Zahid Ahmad Siddiqi

250 sin30°
— %

2000 sin 30° x
32
945.3 + 118.2 N-m

5.5? + 55

(Mass = M, + 15M, = 23,320 N-m
23,320x1000
Soros Den

S/30 = as = 183mm

= 141.3 x 103 mm?

Trial Section No. 1: C 230 x 19.9
S, = 174 x 10% mmé : S,= 15.8 x 10% mm?

37. Prof. Dr. Zahid Ahmad Siddiqi

d>d,,, oK
_ 7368x1000 945.3x1000 118.2x1000
174x10° (15.8/2)x10° 15.8x10°

= 42.34 + 127.14 = 169.5MPa>F,
(revise)

f,

Note:The stress due to M, is more than two times
that due to M,.

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 M, divided by very less value of
S,/2 may give higher answer for the stresses.

Trial Section No. 2: MC150x22.5
S, = 136 x 10% mm? S, = 28.7 x 103 mm?

The depth of this section is less than the required
minimum depth and hence it must be revised.

Trial Section No. 3: C230x22

S, = 185 x 10? mm? S, = 16.6 x 10% mm?
f, = 39.83 + 121.0 = 160.84 MPa
< F, OK

39. Prof. Dr. Zahid Ahmad Siddiqi

b,/t, = 63/10.5 =6 <108 OK

Final Selection: C230 x 22

Check For Self Weight
: . 22x 9.81x10
Actual self weight of purlin = a —

= 108 N/m? < 1.20 x 100 N/m? OK

Design Of Sag Rod
A 5/8 w sin0 x Sx 4
5/8 x 2250 x sin 30° x 5.5 x 4

15,469 N

40. Prof. Dr. Zahid Ahmad Siddiqi

FI cos8
15,469 / cos30° = 17,862 N

R
Ara 7 06H,
T2 _ _17862
4 0.6 x 250

d = 12.31 mm

req

Use 15 mm diameter steel bar as sag rods

41. 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. |. Corrugated Sheet.

The nomenclature for various dimensions is shown
in Figures 7.13 and 7.14.

42. Prof. Dr. Zahid Ahmad Siddiqi

Figure 7.13. View of G. L Corrugated Sheets Along the Corrugations.

Figure 7.14. View of G. I. Corrugated Sheets Perpendicular to Corrugations.

43. Prof. Dr. Zahid Ahmad Siddiqi

Sheet Designation
65x13 75x20

Nominal pitch, mm 65 75

Actual pitch, mm 66 73

Nominal depth, mm 13 20

Actual depth, mm 13 19

Total width of one sheet, mm

Side laps, mm

L/, corrugations with fasteners placed at a

maximum spacing of 300 mm in
perpendicular direction.

Cover (Effective width covered by one
sheet), mm

Number of corrugations in cover

Minimum end lap, mm
Fasteners are to be provided not less than at
every third corrugation over each purlin.

44. Thickness of sheet, mm

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
inclination (0) plus end lap (E).

Allowable working stress, MPa

Maximum allowable deflection.

roof

Prof. Dr. Zahid Ahmad Siddiqi

Varies according
to gage.

(n p/eos0 + E)

0.60 F,

span/90

Varies
according to
gage.

(n p/cosO +

0.60 Ey

span/90

45. Prof. Dr. Zahid Ahmad Siddiqi

us Weight Thickness Weight Properties per meter of Corrugated width
Gage t without laps

(im?) I s
(x 10° mm‘) (x 10° mm’)

46. Weight Thickness
t

(Oz. per
Sft)

70
50

40

Prof. Dr. Zahid Ahmad Siddiqi

Weight Properties per meter of Corrugated width
without laps

(Nim?) 4 I s
(«104 mm‘) (10? mm’)

250.6 14.45 13.28

181.6 10.31 9.84

147.1 8.01

1193

91.5

776

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/m? 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:

Az 0.013 x w, p*/ El

For sheet with one end continuous considering some
constraint at ends:

Amax = 0.001 x w, p4/ El

For sheet with one end continuous:

Amax = 0.0054 x w, p*/ El
Sheets per 100 m? of inclined roof area are:
10° à 10°
Nioo = y1-(SL+EC) CUE)

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

Rain Water

Incorrect

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.

Correct Al

(a) End Lap Over Purlins (b) End Lap Within Purlins

Figure 7.13.Correct Position of End Lap.

53. Prof. Dr. Zahid Ahmad Siddiqi

Total load on the sheet = dead load of
roofing + insulation + live load (N/m?).

Consider unit width of slab (1 m) and design
this strip as a beam.

Load per unit length of roof strip, w (N/m), is

calculated as follows:

w = load per unit roof area
x 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.

2
wx p

(N — m)

M ax * 1000

max

req = 067,

Select gage of sheet from 15 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
LOS

< 1.2 x 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 (A,,,,,)
and check against the allowable value (A,).

Amax < À

max

56. Prof. Dr. Zahid Ahmad Siddiqi

10. Calculate the number of sheet panels required for
100 m? of roof area (Nj 99) 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 N, x S, where N, is the number of

spaces between the trusses.

The inclined roof area on one side, A

_ (@/2+Sheet Projection )
cosO

x Nx S

57. Prof. Dr. Zahid Ahmad Siddiqi
Number of sheets on one side, N,

=A x N 00

1 (round to higher whole number)

Total number of sheets =2xN,

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

. Standard designation

. Sheet panel size
. Bolt spacing in the two directions

. Number of sheets required