1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
|
//Finding unknown forces
//Refer fig. 4.8
//Let us assume joint E as origin,EC as x-axis,EA as y-direction
//accordingly the co-ordinates are
//A(0,3),B(3,3),C(6,0),D(3,0),E(0,0)
YD=-40 //kN
YC=-40 //kN
//Using co-ordinates lengths are found out to be
LAB=3 //m
LBC=3*sqrt(2) //m
LCD=3 //m
LDE=3 //m
LBD=3 //m
LBE=3*sqrt(2) //m
//Consider joint C
//applying equilibrium conditions
tCB=40/3
tCD=-40/3
FCB=tCB*LBC //kN
FCD=-13.333*LCD //kN
//Consider joint D
//applying equilibrium conditions
tDE=tCD
FDE=tCD*LCD //kN
tDB=40/3
FDB=tDB*LBD
//Consider joint B
//applying equilibrium conditions
tBE=-(13.333+13.333)
FBE=tBE*LBE //kN
tBA=40
FBA=tBA*LAB
printf("The required forces are:-\nForce in member AB=%.2d kN\nForce in member BC=%.2d kN\nForce in member CD=%.2d kN\nForce in member DE=%.2d kN\nForce in member EB=%.2d kN\nForce in member BD=%.2d kN",FBA,FCB,FCD,FDE,FBE,FDB)
|