Question

IT

Can you put the measurements from you ... and explain to me how to calculate the truss in this design ... it's not a bridge it's like the platform ... all the image is the same design but from different sides

please ...dr

Answer 1

To design the truss, you first need to provide the weight which it has to bear. Here, lets assume, you design of the platform is good enough to bear a load- of 1400kg or 14 kN.

(Note; you can make your own assumption for the data. It depends on the purpose for which you are using the platform).

From, your figure; it seems there are 14 vertical rods or stands to withstand the load of 14kN. Thus:

load on each vertical section=14kN/14= 1kN

For load at the corner rods, refer the image...

IKN I KN I KN IKN I KN assuming angle theta as 30 loloso A Fi 30 F//sin150 = Fz/sin180= 1/sin30 This gives; F =1kN and Fz=0 180 ( ) TIN 150 TIKN

Thus, all straight/vertical rods bear a load of 1kN, while the inclined rod has no force to bear. Thus, you see that all the vertical rods are under compression., with compressive force of 1kN.

Now, to determine the size of rod, you first need to decide the material of rod to be used. Here, taking cost as a factor, we are using alloy steel as the material of rod for reducing the costs.

The alloy steel has compressive strength of 200kPa. So, the cross section area of the rod is:

Area=Load/allowable stress

But, assuming factor of safety as 4. So, working stress= 200/4= 50kPa

Area=1/50 =0.02 m^2

So, diameter= 2.55cm

Hence the rod should have a dia of 2.55 cm. (Note: more number of rods would give less diameter size, but if you use less number of rods for support then it will give more diameter as a result.)

Hence in order to bear a load of 14kN you must use alloy steel rods of dia 2.55 cm. where the inclination angle of inclined rods is 30°

(Note: if you use angle other than 30° then the values would change.)

Conclusion: This solution is fully based on assumed data's. Anyone can assume his/her own data and the results will be different. Hence, , if you prefer your own material or any other load then please solve accordingly.

All the best.......