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DESIGN OF TRUSS ROOF
Chapter 7
University of Engineering & Technology, Taxila
1
Prof Dr Z. A. 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 the roof
contributing load to interior panel point of the truss.
It is separately calculated for dead, live and wind
loads.
2
Prof Dr Z. A. Siddiqi
The truss is analyzed for unit gravity loads, unit wind
force 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.
Suppose that the deflection at any point of the truss,
called point A, due to unit load acting at some other
point B of the truss is Δ.
3
Prof Dr Z. A. Siddiqi
According to the principle of superposition, for
structures within elastic range, the deflection of the
structure due to combined action of two sets of loadings
is equal to the sum of deflections due to individual
loads acting separately.
If another unit load is applied at A, deflection at point B
will become ∆ +∆ or 2∆.
Similarly for P number of unit point loads, the
deflection will be P x ∆.
4
Prof Dr Z. A. Siddiqi
The same principle is also valid for member forces.
Panel load multiplied with the unit load member forces
gives the magnitudes of member forces for the actual
loads.
This explains the use of calculating the panel loads for
dead, live and wind loads.
5
Prof Dr Z. A. Siddiqi
Load at interior
panel point = P = load intensity over
horizontal plan area (w)
x area supported by the
panel point (p x s)
= w x p x s
Load at exterior panel point = P/ 2
6
Prof Dr Z. A. Siddiqi
However, the exterior panel load needs not to be
calculated as the truss may be analyzed for unit load on
interior panel points and half load on exterior panel
points.
7
Prof Dr Z. A. Siddiqi
p
s
Truss T-1
Truss T-2
p p
p/2 p/2
s/2
s/2
Area contributing load
at one interior panel
Point = p x s
a) Elevation of Truss
s
b) Part-Plan of
Truss Roof
P= panel length in a horizontal plane
S= center-to-center spacing of trusses
s
p
p
s
How much area is contributing load
here as point load
p
s
p
s
s
p
s
p
UDL (w) acting over the panel area
UDL (w) N/m2 is converted
Into Panel concentrated load
P= w x s x p
UDL (wD) N/m2 is converted
Into Panel concentrated load
PD= wD x s x p
UDL (wL) N/m2 is converted
Into Panel concentrated load
PL= wL x s x p
p
s
s
p
How much area is contributing load
here as point load
s
p/2
UDL (w) acting over the half panel area
s
p/2
UDL (w) N/m2 is converted
into Panel concentrated load
P= w x s x p/2
s
p
θ
UDL (ww) N/m2 is converted
Into Panel concentrated load
Pww= ww x s x p/cosθ
θ
p
θ
Example 7.1:
Find panel loads for the given truss data.
Data:
Angle of top chord = 30 0
Dead load of roofing = 17 kg/m2
Insulation boards = 5 kg/m2
Self weight of purlins = 10 kg/m2
Self weight of bracing elements = 3 kg/m2
Miscellaneous = 5 kg/m2
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
= 40 kg/m2
Live load, from Design Aids, for θ = 300 = 60 kg/m2
Total gravity load, w = 40 + 60
= 100 kg/m2
Using Thayer’s formula
Self weight of truss =
12.2 kg/m2
Total dead load = 40 + 12.2
= 52.2 kg/m2
 
5
.
0
5
.
8
20
5
.
5
100


Leeward wind pressure = 1250(-0.7) = -875 N/m2
Windward wind pressure = 1250(-0.9) = -1125 N/m2
and 1250 (0.3) = 375 N/m2
Penal dead load, PD = w x p x S
= 52.22 x 2.5 x 5.5 x 9.81/ 1000
= 7.04 kN
Panel live load, PL = 60 x 2.5 x 5.5 x 9.81 / 1000
= 8.09 kN
• 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/cosθ) in the expression for calculation
of the panel loads.
Panel wind load on leeward side, Pwl
=
= - 13.89 kN (upward)
Panel wind load on windward side, Pww
=
= -17.86 kN
    1000
/
5
.
5
30
cos
5
.
2
875 0 






    1000
/
5
.
5
30
cos
5
.
2
1125 0 






And
=
= 5.95 kN
    1000
/
5
.
5
30
cos
5
.
2
375 0 





Assignment:
Find the panel loads for your
truss.
Time Allowed: 1 week

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steel_pr_3.pptx

  • 1. DESIGN OF TRUSS ROOF Chapter 7 University of Engineering & Technology, Taxila 1 Prof Dr Z. A. Siddiqi
  • 2. 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 the roof contributing load to interior panel point of the truss. It is separately calculated for dead, live and wind loads. 2 Prof Dr Z. A. Siddiqi
  • 3. The truss is analyzed for unit gravity loads, unit wind force 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. Suppose that the deflection at any point of the truss, called point A, due to unit load acting at some other point B of the truss is Δ. 3 Prof Dr Z. A. Siddiqi
  • 4. According to the principle of superposition, for structures within elastic range, the deflection of the structure due to combined action of two sets of loadings is equal to the sum of deflections due to individual loads acting separately. If another unit load is applied at A, deflection at point B will become ∆ +∆ or 2∆. Similarly for P number of unit point loads, the deflection will be P x ∆. 4 Prof Dr Z. A. Siddiqi
  • 5. The same principle is also valid for member forces. Panel load multiplied with the unit load member forces gives the magnitudes of member forces for the actual loads. This explains the use of calculating the panel loads for dead, live and wind loads. 5 Prof Dr Z. A. Siddiqi
  • 6. Load at interior panel point = P = load intensity over horizontal plan area (w) x area supported by the panel point (p x s) = w x p x s Load at exterior panel point = P/ 2 6 Prof Dr Z. A. Siddiqi
  • 7. However, the exterior panel load needs not to be calculated as the truss may be analyzed for unit load on interior panel points and half load on exterior panel points. 7 Prof Dr Z. A. Siddiqi
  • 9. p p p/2 p/2 s/2 s/2 Area contributing load at one interior panel Point = p x s a) Elevation of Truss s b) Part-Plan of Truss Roof P= panel length in a horizontal plane S= center-to-center spacing of trusses
  • 10.
  • 11.
  • 12.
  • 13.
  • 14.
  • 15.
  • 16. s p
  • 17. p s How much area is contributing load here as point load
  • 18. p s
  • 20. s p
  • 21. UDL (w) acting over the panel area
  • 22. UDL (w) N/m2 is converted Into Panel concentrated load P= w x s x p
  • 23. UDL (wD) N/m2 is converted Into Panel concentrated load PD= wD x s x p
  • 24. UDL (wL) N/m2 is converted Into Panel concentrated load PL= wL x s x p
  • 25. p s s p How much area is contributing load here as point load
  • 26. s p/2
  • 27. UDL (w) acting over the half panel area s p/2
  • 28. UDL (w) N/m2 is converted into Panel concentrated load P= w x s x p/2
  • 30. UDL (ww) N/m2 is converted Into Panel concentrated load Pww= ww x s x p/cosθ θ p θ
  • 31.
  • 32.
  • 33.
  • 34. Example 7.1: Find panel loads for the given truss data. Data: Angle of top chord = 30 0 Dead load of roofing = 17 kg/m2 Insulation boards = 5 kg/m2 Self weight of purlins = 10 kg/m2 Self weight of bracing elements = 3 kg/m2 Miscellaneous = 5 kg/m2 Panel Length, p = 2.5 m Span length of truss, L = 20 m Spacing of trusses, center-to-center, S = 5.5 m
  • 35. Solution: Total dead load except truss self weight = sum of given dead loads = 40 kg/m2 Live load, from Design Aids, for θ = 300 = 60 kg/m2 Total gravity load, w = 40 + 60 = 100 kg/m2
  • 36. Using Thayer’s formula Self weight of truss = 12.2 kg/m2 Total dead load = 40 + 12.2 = 52.2 kg/m2   5 . 0 5 . 8 20 5 . 5 100  
  • 37. Leeward wind pressure = 1250(-0.7) = -875 N/m2 Windward wind pressure = 1250(-0.9) = -1125 N/m2 and 1250 (0.3) = 375 N/m2 Penal dead load, PD = w x p x S = 52.22 x 2.5 x 5.5 x 9.81/ 1000 = 7.04 kN Panel live load, PL = 60 x 2.5 x 5.5 x 9.81 / 1000 = 8.09 kN
  • 38. • 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/cosθ) in the expression for calculation of the panel loads.
  • 39. Panel wind load on leeward side, Pwl = = - 13.89 kN (upward) Panel wind load on windward side, Pww = = -17.86 kN     1000 / 5 . 5 30 cos 5 . 2 875 0            1000 / 5 . 5 30 cos 5 . 2 1125 0       
  • 40. And = = 5.95 kN     1000 / 5 . 5 30 cos 5 . 2 375 0      
  • 41.
  • 42. Assignment: Find the panel loads for your truss. Time Allowed: 1 week