Question

2. Determine the forces in members BC, CD, and DE of the truss shown below by the method of sections. State if the members are in tension or compression. 10 kN 15 kN

Answer 1

3m- TokN TOKM to Iska Since soller *30 *3m → A is hinge support and fis support reaction are as shown above.

Efn=o RAx-o f Efyzo 14 Ray & Rp - 10-15=0. Ray + Ray = 25 - 0 Taking moment at point f. SEME=0 (10x6) + (15×3) - (Ray x 9) =0. I 60+45 = hayxq Ray - 11.67 (kn) i Rey= 25-11.67 - Rey- 13.33 (kN) Now using method of section 0 - 1 as shown . Considering the right section we get fed as Rey=13.33 (kN) 15 KN *3m first finding angle o.

O = 45 Taking moment of forces at (& Emo (Rex 3) - (Foex 3) =0. FDE - Ray I foE = 13.33 (kN) Positive means tensile force. +P8fyzo Pfy - 15 - foc sind = o. 13.33-15 = for sin (45) -1.67= focx sin (45) = foc -2.36 (kN) Negative sign means compressive force. efnzo -Fec- fpe coso - FDE = 0 -FBc - (-2.36 x cos45) - 13.33 20. fac = -11.66 (KN) Negative sign means Compressive force. Therefore, DE 13.33 (kN) (Tension) DC = 2.36 (kN) (Compression) BC = 11.66 (kN) (compression)