For the loaded truss shown in Fig. 2 Calculate the internal axial force in the following members CF, DE and DF. 6 m 5 m 10 KN A 8.25 KN Oa. FDF=0 KN Ob. FCF= -20.2 KN Oc. FCF=-23.75 KN Od. FDF= 10 KN Oe. FDE44.375 KN- Ot. FDE 40.425 KN - O& FCF -18.2 KN Oh. FDF=2.5 KN OL FCF=-22.25 KN O FDE 42.375 KN Ok FDE 47.375 KN- OL FDF- 12.5 KN H B 5 m 5m 15 kN G Figure 2 15 KN F 5 m D E 21.75 KN 10 KN

Explain how agency benefits of a transportation project is estimated? Use a toll highway project to outline your proposed method.

АО 3 KN VB C -1.26 m--1.26 m- 6 KN VD E 1.26 m- 1 kN /F T 1.2 m G 2. Determine the force in each member of the half hip truss shown. State whether each member is in tension of compression. Reaction at G is a roller support. (Ans: FAB = 5.80 KN (C), FAC = 4.20 KN (T), FBC = 0 KN, FCE = 4.20 KN (T), FBE = 1.450 kN (T), FBD = 5.25 KN (C), FDE = 6 KN (C), FDF = 5.25 kN (C), FEF = 7.25 KN (T), FEG = 0 KN, FFG = 6 KN (C)) SHOW ALL THE FBD AND FORCES GIVEN. DONT FORGET THE SYMBOLS COMPLETE WORK FOR UPVOTE

Determine the axial load at point B for the beam below. 12k 60° 8 k SASn sn sn -5 ft 5 ft- -5 ft- Select the correct response 6 k -2.39 k -6 k 2.39k