Question

The joints C, D and E of a bridge truss carry equal vertical loads P = 10,000 kgf. The inclined web members form angles of 45° with the horizontal. Calculate the forces in the web members 1, 2, 3, 4, 5, 6, 7, 8 and 9, caused by the mentioned loads.

PLEASE SOLVED AND EXPLAIN

Answer 1

THE PROBLEM IS VERY SIMPLE.

Read the solution carefully and if have any problem then comment .I will resolve the problem immediately.

WE USE THE METHOD OF SECTION TO SOLVE THIS PROBLEM.

IN SOLUTION THE VALUE OF

Sin(45)=Cos (45)=0.707

$\sum$Fx= Summation of forces in X direction.

$\sum$Fy= Summation of forces in Y direction.

$\sum$M=Moment aabout the point

Above equation are used to solve the problem and All above equation are equated to zero to obtain the value of the force in member.

$\sum$Fx= 0 means Equationg the forces in X direction to zero

Same is for $\sum$ Fy= 0 means Equationg the forces in Y direction to zero

,$\sum$M=0 means Equationg the MOMENT to zero

Clockwise moment is Negative and Anticlockwise moment is negative.

- H I THI THI FDI Fot DE FDE method of section]

Fy=0 RATRB = 30,000kgf [EMA=o 100009 +20000 780000K = RB 60,000 X= 4 RB X RB = 15000 KgF RA= 15000 kg F - RB (4x)

www.dar Applying method of & FFG 0 - A) FFC section Efx=0 FF Gt FFC C0345+ FAC= [ST - ο RA - Ffc Bin45=0 RA= FFC Bin 45 FC = 21016.41 FAC RA RA= FAF = 15,000N

foom Equation (6),B) Retting values cell get ITACE - 15000kgf using another section Follt FOD с to, ooo Еср RA (m) & Fx=0, Efy = 0 FGH + FAD COB 45 + FD = 0 PA-1000 o-FGogin 450 Boleing we -TGD: 7072. 13 kg F

7 FGH Efy=0 + FGP 10,00 of F G C EfY=0 - 10000 - FGC - FGD Sin 45 - 0 we have FGD = 7072.13 KgF I FG C= - 20605 KgF]

TEMA=0 10ooox + FG.Oasinus' + FGD & COS45 +FGHX=0 Petting values & Cancelling as we get, FGH = - 20000kg F - Petting values in equation. ale get, I FCD = 15,000 kgf Forces 1 = FAF 2 = FAC 3 = FFC 4 - FFG 5 - FGC 6 = FED 7 - FGD 8 - FGH 9 = FHD in memben = 15,000 = - 15000 = 212 16.6. = 0 -20605 kg = 15,000 = 7072.13 = -20,000 =