Question

USE METHOD OF JOINTS. I am stuck so please provide detailed explination as well. nothing crazy though. Thanks

Use the method of Joints to Fnd Tcf, IDE and . The truss shown in the figure is made up of five horizontal and six inclined rods. All inclined rods are 1 m long and at right angles to each other. The truss carries two vertical loads, Fi = 4kN and F2 = 1 kN as shown. Find the tensions in rods CE, DE, and DF. Filetaraflg5-trussjonta-3

Answer 1

In this problem, first of all find the all angles and length of members from geometry then Calculate the reactions at support by using three equilibrium equation after that use Method of Joints and Calculate the force force in required members as below:

solution: - ts. HT 4 kn > AC - AN + NC = AB. Cel 45° + BC. cos 45° AC 1. Cod 45° + 1. Co 45' = 0.7071 + 0.7071 AC = 1.414 m 1 (AC-CE - EG = 3D - DF) Calculation of Reactions :- c) EFx=0 → Ax =o Ü) EFyzo → Ay - A-1 + Gay co Ay + Gy = 5 KN. (ii) EMA = 0 = -(4x 1.414) - 14(2x1.414) +(GyX 3x1.414) - 5.656 – 2.828 + 4.242 Gry zo O 4.242 = 8.484 Cny = 2 kN. . Ay = 5-2= 3 kN. Now, use " Method of Joints" for calculation of force in members CE,

DE A DF :- olt joint G:- (1) Efyzo Fea' bin 45 ° +2=0 45° a a EC 12 kn - 2 = - 2.828 KN. ni Jin 45° *) E FR20 > - Fec - Frge cas 45' = 0 FEC = -(-2.898) 6845° = 2 kN. At joint F:- (i) EFY=0 - Ffe. Din A5 - Feri bindsso - FEF = - FFG =-(-2.828) PEF Fer = .818 kv. Ü) EF x =0 » FEG. Los 45°- FDF - FEE Cas45°=0 FDF = (FFG - FEF) cos45° FDF - (-2.898 – 2.828) Cod 45° FDF = -4 KN. FDE FEF A joint E- (i) e Fy=0 > FDE bih 45 + FEE: Sinds -1=0 0.707. FDE +(2.898 x 0707) -1=0 T AS" AS ee E FEG

0.707 FDE + 2-1=0 FDE = -0,03 = -1.414 kW. Ü EFx=0 · Cod 45 - 0 Ce E . > FEG- FLE + FER ( 45 - Fee = Fea + (Fer - FDE) cos 45° Fee = 2 + (2.898 + 1.414). Cos 45° FCE = 2+3 = 5 KN. Inwer' Fee = 5 kN (Tensile) FDE = 1.414 KN ( compression) F DF = 4 KN ( compression)