4.7 If the maximum tensile force in any of the truss members must be limited to 22 kN, and the maximum compressive force must be limited to 20 kN, determine the largest permissible mass m which may be supported by the truss.

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

Summation of forces in the Y direction

∑Fy = 0

= BC sin30° - W = 0.   hence BC = 2WT

summation of forces in the X direction

∑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

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