The portion of truss shown represents the upper part of a power transmission lin

Question

The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression. Fig. P6.24

Explanation / Answer

Use section method first. Cut EJ, EH and CH members. Sum of the forces in x-direction gives: EH=0 Taking the moments about the points E and C gives EJ=CH= 1.2 kN (C) (C stands for compression and T for tension) Now, sum of the forces at joint F in x-direction gives DF+EF=0 Sum of the forces at joint F in y-direction DF sin (theta) - EF sin (theta) -1.2 = 0 (theta is half of the angle DFE) 2DF=1.2/sin (theta) = 1.2 / 0.262 => DF= 2.29 kN (T), EF= 2.29 kN (C) similarly for joint A we have: AB= 2.29 kN (T), AC= 2.29 kN (C) Sum of the forces at joint F in y-direction -DF sin (theta) - DE = 0 => DE = 0.6 kN (C) Sum of the forces at joint F in x-direction DF cos (theta) - BD = 0 => BD = 2.21 kN (T) Sum of the forces at joint B in x-direction -AB cos (theta) + BD + BE (4/5)= 0 => BE=0 Sum of the forces at joint B in y-direction -AB sin (theta) - BC= 0 => BC= 0.6 kN (C) Finally, sum of the forces at joint C in x-direction: -AC cos (theta) + CE= 0 => CE= 2.21 kN (C)

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