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DESIGN OF TRUSS ROOF
Chapter 7
University of Engineering & Technology, Taxila
Prof Dr Z. A. Siddiqi
1
TABLE OF FORCES
There are three important points to be considered while
calculating the member forces:
1. Panel load multiplied with unit load forces gives the
member forces by the principle of superposition.
According to this principle, which is applicable for
elastic structures, if a unit load is applied on a truss
and the force in any member is calculated as F; then if
we apply another unit load simultaneously at the same
point, the force in the member will be 2F.
Prof Dr Z. A. Siddiqi
2
Similarly, this member force will be 3F for three unit
loads or P x F if P times unit loads are applied.
2. Effects of various types of loads are to be added while
calculating member forces.
Vertical and inclined loads on the truss cannot be
added directly because of different lines of action of
each.
However, separate member forces due to vertical and
inclined loads have the same direction (along the
member longitudinal axis) and hence can be
algebraically added together.
Prof Dr Z. A. Siddiqi
3
3. Probability of occurrence of various loads in a load
combination and corresponding load factors (factor of
safety) are also applied during these calculations.
In case only dead, live and wind loads are acting on a
truss, following combinations may be investigated:
1.
1.2D +
1.6Lr + 0.65W
(Wind effect is small and may be ignored especially
suction is present throughout)
Prof Dr Z. A. Siddiqi
4
2.
1.2D + 0.5Lr + 1.3W
Wind towards the Right
Wind towards the left
3.
0.9D + 1.3W
Wind towards the Right
Wind towards the Left
For the roof design, the first (gravity) or second (wind)
load combination is critical. It is to be noted that the
downward wind load is to be considered in the second
combination.
Prof Dr Z. A. Siddiqi
5
The third combination is critical for reversal of forces
and hence is evaluated for upward wind pressures.
In case the wind load has two values, one downward
and one upward, the downward value should be used
for the second combination and the upward value
should be used for the third combination.
It would be unreasonable to include full wind and full
snow (or live load) stresses together in a single
combination.
Prof Dr Z. A. Siddiqi
6
Similarly, when the wind is blowing at its full strength,
the live load intensity may be reduced.
The second combination reflects the condition when
most severe windstorm is blowing and hence the live
load intensity may be reduced to 0.5/ 1.6 or 0.31 times
its maximum intensity, showing less probability of
occurrence of full live load together with wind.
The third combination represents an unoccupied
building subjected to the heaviest wind.
Prof Dr Z. A. Siddiqi
7
The negative sign with the wind forces only indicates
that the combination is critical when the wind is
producing member forces opposite in sense to that
produced by the dead loads.
The design forces may be calculated by entering the
values into a Table of Forces as in Table 7.2.
The first four columns of this table are directly taken
from the unit load analysis of the truss while columns 5
to 9 are calculated using the first four columns and the
algebraic values of the panel loads, already determined.
Prof Dr Z. A. Siddiqi
8
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Prof Dr Z. A. Siddiqi
(9)
(10)
(11)
Remarks
Maximum factored
Compression (Pu)
Maximum factored tension
(Tu)
(0.9 PD) x Col.2
+1.3 P ww x Col.4
+1.3 PwL x Col.3
(0.9 PD) x Col.2
+ 1.3 P ww x Col.3
+ 1.3 PwL x Col.4
(1.2 PD + 0.5 PL) x Col.2
+ 1.3 P ww x Col.4
+ 1.3 PwL x Col.3
(1.2 PD + 0.5 PL) x Col.2
+ 1.3 Pww x Col.3
+ 1.3 PwL x Col.4
(1.2PD+1.6PL) x Col.2
Member force due to unit wind load on roller side
Member force due to unit Wind load on hinge
side
Unit gravity load member force
Member No.
Sample Table of Forces
(12)
9
After getting the values in these columns, the maximum
factored tension and compression may be found out and
entered in the next two columns.
Usually tension is represented by positive sign and
compression by a negative sign in columns 2 to 4 and
hence maximum tension is defined as the maximum
positive value and maximum compression as the
minimum value (maximum negative answer) in
columns 5 to 9.
Prof Dr Z. A. Siddiqi
10
The remarks column is used to describe the type of the
member for design such as pure tension member, pure
compression member, member under reversal of
stresses and zero force member.
A computer spreadsheet may conveniently be used to
construct such a table.
The truss members may now be selected by using the
procedure given in earlier chapters and connections
may be designed by the methods outlined in the coming
chapters.
Prof Dr Z. A. Siddiqi
11
Unit Gravity Loads
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Prof Dr Z. A. Siddiqi
(9)
(10)
(11)
Remarks
Maximum factored
Compression (Pu)
Maximum factored tension
(Tu)
(1.2PD+1.6PL) x Col.2
Member force due to unit wind load on roller side
Member force due to unit Wind load on hinge
side
Unit gravity load member force
Member No.
Sample Table of Forces
(12)
13
Unit Wind Load on Hinge Side
Unit Wind Load on Roller Side
If Panel wind load is negative = Suction
Pww
Pww= +ve
(1.2PD+0.5PL) X COL. 2
+ 1.3 PWW X COL. 3
Pwl
Pww
Pww= +ve
Pwl = -ve
(1.2PD+0.5PL) X COL. 2
+ 1.3 PWW X COL. 3
Load Combination in Column No 6
Pwl
Pww
Pww= +ve
Pwl = -ve
(1.2PD+0.5PL) X COL. 2
+ 1.3 PWW X COL. 3
+ 1.3 PWL X COL. 4
Pwl
Pww
Pww= +ve
Pwl = -ve
(1.2PD+0.5PL) X COL. 2
+ 1.3 PWW X COL. 3
+ 1.3 PWL X COL. 4
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Prof Dr Z. A. Siddiqi
(9)
(10)
(11)
Remarks
Maximum factored
Compression (Pu)
Maximum factored tension
(Tu)
(1.2 PD + 0.5 PL) x Col.2
+ 1.3 Pww x Col.3
+ 1.3 PwL x Col.4
(1.2PD+1.6PL) x Col.2
Member force due to unit wind load on roller side
Member force due to unit Wind load on hinge
side
Unit gravity load member force
Member No.
Sample Table of Forces
(12)
21
Load Combination in Column No 7
Pwl
Pww= +ve
Pwl = -ve
Pww
(1.2PD+0.5PL) X COL. 2
+ 1.3 PWW X COL. 4
+ 1.3 PWL X COL. 3
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Prof Dr Z. A. Siddiqi
(9)
(10)
(11)
Remarks
Maximum factored
Compression (Pu)
Maximum factored tension
(Tu)
(1.2 PD + 0.5 PL) x Col.2
+ 1.3 P ww x Col.4
+ 1.3 PwL x Col.3
(1.2 PD + 0.5 PL) x Col.2
+ 1.3 Pww x Col.3
+ 1.3 PwL x Col.4
(1.2PD+1.6PL) x Col.2
Member force due to unit wind load on roller side
Member force due to unit Wind load on hinge
side
Unit gravity load member force
Member No.
Sample Table of Forces
(12)
23
Load Combination in Column No 8
Pwl
Pww
Pww= -ve
Pwl = -ve
(0.9PD) X COL. 2
+ 1.3 PWW X COL. 3
+ 1.3 PWL X COL. 4
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Prof Dr Z. A. Siddiqi
(9)
(10)
(11)
Remarks
Maximum factored
Compression (Pu)
Maximum factored tension
(Tu)
(0.9 PD) x Col.2
+ 1.3 P ww x Col.3
+ 1.3 PwL x Col.4
(1.2 PD + 0.5 PL) x Col.2
+ 1.3 P ww x Col.4
+ 1.3 PwL x Col.3
(1.2 PD + 0.5 PL) x Col.2
+ 1.3 Pww x Col.3
+ 1.3 PwL x Col.4
(1.2PD+1.6PL) x Col.2
Member force due to unit wind load on roller side
Member force due to unit Wind load on hinge
side
Unit gravity load member force
Member No.
Sample Table of Forces
(12)
25
Load Combination in Column No 9
Pwl
Pww= -ve
Pwl = -ve
Pww
(0.9PD) X COL. 2
+ 1.3 PWW X COL. 4
+ 1.3 PWL X COL. 3
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Prof Dr Z. A. Siddiqi
(9)
(10)
(11)
Remarks
Maximum factored
Compression (Pu)
Maximum factored tension
(Tu)
(0.9 PD) x Col.2
+1.3 P ww x Col.4
+1.3 PwL x Col.3
(0.9 PD) x Col.2
+ 1.3 P ww x Col.3
+ 1.3 PwL x Col.4
(1.2 PD + 0.5 PL) x Col.2
+ 1.3 P ww x Col.4
+ 1.3 PwL x Col.3
(1.2 PD + 0.5 PL) x Col.2
+ 1.3 Pww x Col.3
+ 1.3 PwL x Col.4
(1.2PD+1.6PL) x Col.2
Member force due to unit wind load on roller side
Member force due to unit Wind load on hinge
side
Unit gravity load member force
Member No.
Sample Table of Forces
(12)
27
Assignment:
Draw the Table of Forces for
Your Data & find the Truss
Member Forces.
Time Allowed: 1 week

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