Question

The truss shown below is supported by a pinned support at A and roller support at C. The length of the members is shown in the Figure below. Set the forces acting on the truss as P1 = 9 kN, P2 = 15 kN. PF 3 m - 5. State whether the members CE and CB are in tension or compression. 6. Determine the cross-sectional areas of the members CE and CB. The stresses are not to exceed 20.2 MPa in tension and 8.45 MPa in compression. A reduced stress in compression is specified to reduce the danger of buckling.

Answer 1

P = 9KN RAHT RAV 2 - ISKNR Given. PI= GKN and P2 = 15 KN. RAV + Rev= 15kN -O RAMK6 = -9 KN Copp diretion as indicated in the figure) Taking moment A LHA 4x9+15*3- Rev X6 =0 Rev 6=81 Rev = 81, KN I Rev = 13.5 kw ] - | Rev = 1.5 kN{ substituting @ ino) Scanned with Lam S.

Resolving forces 6 membes by using section method Fee tono and = 3 . FCB T 0 - 36686° 3m V. Be-15 RAV Erzo - RAV + 15KN - FCE CO$136-86) co - 105 kW + 15 KW - Fre (0.80) -0 I FCE = 16.875 KN ( compressive) Ź H=0 9 KN + F€D - FCE SPY (3686) +FCB FED + FCB = 10.122.2 - © Paking momento , A qr4 + F€D Y4 - FCE 8P1 (36-86) ry - FCE COA (3686) X3 + 15 x3 = 0 36+ feoxy - Fee (2.399)- Fee (24)+45 20 Scanned with CamScanner co

14F€D - Yo8 fçe +8150 -0 : Solving sariation 0.80 I fes = 10:122XN (tenpron) Given that The Allowable Stresses - 20.2 Hpa potensson = 8.5 tpa in Compression * Since the member TB es in tension Allowable = Force in member Stress in Area tensloy Area Requised = In member Aneon of LB = crospechon 10.129 X 103 PO-2 51.08 mm & The momber ce 99 Pn compression CS Scanned with CamScanner

Allowable stress force in ce member in comporssion 1. Area = 16.825 x 103 of Cross sechron of member 8.45 I Areace = 1997. mm in 2000 mm W / Pls. do contact incase of any doubt. Don't forget to like it. Thank in advance. Scanned with CamScanner