1) Degree of static indeterminacy :-
$D_{se}$ = r-3 = 3-3 =0
$D_{si}$ = M-[2j-3] = 6-[2(4)-3] = 6-5 = 1
$D_{s}$ = 0+1 = 1
2) Let AC be the redundant :-
Remove Member AD
3) P-Analysis :-
Σ$M_D =0 =20*4-V_c$*4 =0
$V_c$ = 20KN
Σ$F_y =0 =V_D$-30+20 =0
$V_D$ = 10KN
Joint D:-
Σ$F_y =0$
10-30+$P_{DB}$sin45 =0
$P_{DB}$ = 28.28KN
4) K-Analysis:-
5) Table:-
Member |
P |
K |
L |
PKL |
$k^2$L |
R |
PK |
$P_f$=P+PK |
AB |
-20 |
-0.7 |
4 |
56 |
1.96 |
-18.58 |
13 |
-7 |
BC |
-20 |
-0.7 |
4 |
56 |
1.96 |
-18.58 |
13 |
-7 |
CD |
0 |
-0.7 |
4 |
0 |
1.96 |
-18.58 |
13 |
13 |
DA |
-30 |
-0.7 |
4 |
84 |
1.96 |
-18.58 |
13 |
17 |
AC |
0 |
1 |
5.65 |
0 |
5.66 |
-18.58 |
-18.58 |
-18.58 |
DB |
28.28 |
1 |
5.65 |
160.06 |
5.66 |
-18.58 |
-18.56 |
9.7 |
|
|
|
|
ΣPKL =356.06 |
Σ$K^2$L = 19.16 |
-18.58 |
|
|
R= -[$\frac{ΣPKL}{ΣK^2L}$] = -[$\frac{356.06}{19.16}$] = -18.58KN(c)