Question 20.8: The Pratt truss shown in Fig. 20.19(a) carries a dead load o...
The Pratt truss shown in Fig. 20.19(a) carries a dead load of 1.0 \mathrm{kN} / \mathrm{m} applied at its upper chord joints. A uniformly distributed live load, which exceeds 9 \mathrm{~m} in length, has an intensity of 1.5 \mathrm{kN} / \mathrm{m} and is also carried at the upper chord joints. If the diagonal members are designed to resist tension only, find which panels require counterbracing.
A family of influence lines may be drawn as shown in Fig. 20.19(b) for the shear force in each of the 10 panels. We begin the analysis at the centre of the truss where the DLS force has its least effect; initially, therefore, we consider panel 5 . The shear force, S_{5}, in panel 5 with the head of the live load at \mathrm{n}_{5} is given by
S_{5}=1.0\left(\operatorname{area} \mathrm{n}_{5} \mathrm{qa}-\operatorname{area} \mathrm{n}_{5} \mathrm{gb}\right)+1.5\left(\text { area } \mathrm{n}_{5} \mathrm{qa}\right)
i.e.
S_{5}=-1.0 \times \text { area } \mathrm{n}_{5} \mathrm{gb}+2.5 \times \text { area } \mathrm{n}_{5} \mathrm{qa} (i)
The ordinates in the S_{5} influence line at \mathrm{g} and \mathrm{q} are found from similar triangles and are 0.5 and 0.4 , respectively. Also, from similar triangles, \mathrm{n}_{5} divides the horizontal distance between \mathrm{q} and \mathrm{g} in the ratio 0.4: 0.5. Therefore, from Eq. (i)
S_{5}=-1.0 \times \frac{1}{2} \times 5.0 \times 0.5+2.5 \times \frac{1}{2} \times 4.0 \times 0.4
which gives
S_{5}=0.75 \mathrm{kN}
Therefore, since S_{5} is positive, the diagonal in panel 5 will be in compression so that panel 5 , and from symmetry panel 6 , requires counterbracing.
Now with the head of the live load at \mathrm{n}_{4}, S_{4}=1.0 (area \mathrm{n}_{4} \mathrm{ra}-area \mathrm{n}_{4} \mathrm{fb} ) + 1.5 (area \left.\mathrm{n}_{4} \mathrm{ra}\right).
The ordinates and base lengths in the triangles \mathrm{n}_{4} \mathrm{fb} and \mathrm{n}_{4} \mathrm{ra} are determined as before. Then
S_{4}=-1.0 \times \frac{1}{2} \times 6.0 \times 0.6+2.5 \times \frac{1}{2} \times 3.0 \times 0.3
from which
S_{4}=-0.675 \mathrm{kN}
Therefore, since S_{4} is negative, panel 4, and therefore panel 7, do not require counterbracing.
Clearly the remaining panels will not require counterbracing.
Note that for a Pratt truss having an odd number of panels the net value of the dead load shear force in the central panel is zero, so that this panel will always require counterbracing.