Answer:
588.55 kg
attached below is a sketch of the members of the truss
Explanation:
Given data:
Maximum tensile force = 22 KN
maximum compressive force = 20 KN
To determine the largest permissible mass m which may be supported by the truss
First we have have to find the summation of forces in the Y and X direction along Joint C
<u>Summation of forces in the Y direction</u>
∑Fy = 0
= BC sin30° - W = 0. hence BC = 2WT
<u>summation of forces in the X direction </u>
∑Fx = 0
= -CD - BC cos30°
therefore CD = -2W cos30°
hence ; CD = - 1.732W
ignoring the compression on CD. therefore CD = 1.732W C
Next we have to analyze Joint B
summation of forces in the X direction
∑Fx = 0
= - AB + BC cos30° = 0
therefore AB = BC cos30°
AB = 1.732W T
∑Fy = 0
- BD - BC sin30° = 0
therefore BD = -W = W C
Analyzing Joint D
∑Fy = 0
AD sin30° + BD = 0
hence ;AD = 2W T
∑Fx = 0
-DE - ADcos30° + CD = 0
hence DE = -3.464W = 3.464W C
At Joint E
∑Fy = 0
i.e. AE = 0
from the above analysis The tensile force is greatest on members AD and BC
AD = BC = 2W
from the question the maximum tensile force in any of the truss members = 22KN
Hence ;
2W = 22KN
2mg = 22000N
m = 22000 / ( 2 * 9.81 )
= 1121.30 kg
from the above analysis the greatest compressive force is found in DE which is the critical part of the Truss hence the maximum mass it can carry is the largest permissible mass which may be supported by the Truss
DE = 3.464 W
from the question the maximum compressive force = 20 KN
hence ;
3.464 W = 20KN
3.464 mg = 20000N
m = 20000 / ( 3.464 * 9.81 )
m = 588.55 kg
