4 m 15 m PI 15 kN 3 m 3 m L 2 m 2 m P 15 kN P2 5 kN 3 m Pr 1

4 m 1.5 m PI 15 kN 3 m 3 m L 2 m 2 m. P 15 kN P2 5 kN 3 m Pr 10 kN

Solution

solution:

here first problem of simplysupported truss is ABCD is solved by method by joints

1)this method works on principle of static equllibrium

take moment about A

Rd*9-15*3+5*4=0

Rd=2.777 KN

Fy=0

Vy+Rd-P1=0

Vy=12.222 KN

Fx=0

Vx-P2=0

Vx=5 KN

here moment about D is

Md=0

15*6+5*4-12.22*9=0

hence correct

2)here method of joint for joint A

Fx=0

Racos67.75+Fae-Fabcos53.13=0

Fy=0

Rasin67.75-Fabsin53.13=0

fab=15.27 KN

Fae=4.164 KN

for joint B

Fy=0

-P1-Fbe-Fabcos69.86=0

-Fabsin36.86+Fbc=0

Fbc=9.159 KN

Fbe=-27.2 KN

for joint E

Fy=0

Fbs+Fcesin53.13=0

Fae+Fef+Fcfcos53.13=0

Fce=34.01 KN

Fef=-24.57 KN

for joint C

Fbc-P2-Fcesin36.86+Fcdsin36.86=0

-Fcecos36.86-Fcf-Fcdcos36.86=0

Fcd=3.471 KN

Fcf=48.87 KN

for joint D

Ffd-Fcdcos53.13=0

Ffd=2.08 KN

in this way this truss is solved by method of joits for calculating forces in each member