Determinacy & Indeterminacy - 3 - Free MCQ Practice Test with solutions,

Kinematic indeterminacy means degree of freedom of structure at various joints.
No rotation or translation is possible at A so degree of freedom at A is zero. There is a possibility of rotation at C but no translation so degree of freedom is one. At G both rotation and translation is possible so. degree of freedom is 2. At J no rotation but translation so d.o.f. is 1 . At B, D, H and K there are 4 rotations and 1 translation so d.o.f. is 5. At E, F and I three rotations and two translations so d.o.f., is 5.
So kinematic indeterminacy,
= 0 + 1 + 2 + 1 + 5 + 5 = 14 
Alternate:
From direct formula
External reactions r_e = 3 + 2 + 1 + 2 = 8 
Number of members (m) = 11
Number of rigid joints (j) = 9
Number of hinged joints (j') = 2
There are no internal hinges so number of releases is zero.
r_r = 0
Degree of kinematic indeterminacy,
D_k = 3(j + j') - r_e + r_r - m
= 3 x (9 + 2) - 8 - 11
= 33 - 19 = 14

Static indeterminacy D_s = 3 m + r_e - r_r - 3(j + j') Where,
m = total members = 12
r_e = total external reactions = 12
j = total number of rigid joints = 11
j' = total number of hybrid joints = 1
r_r = total number of released reactions

∴ D_s = (3 x 12) + 12 - 1 - 3 (11 + 1)
= 36 + 11 - 36 = 11

D_s = D_se + D_si
D_se = r_E - 6
r_E = 6 + 3 + 3 + 6 = 18
D_se = 12
D_si = 6_c - r_R = 6 x 1 - 3 
= 6 - 3 [(2 - 1) + (3 - 1)]
= 6 - 9 = -3 
∴ D_s = 12 - 3 = 9

Static indeterminacyDs=3m+re−rr−3(j+j′)
where m = total members = 12
re = total external reactions = 12
j = total number of rigid joints = 11
j' = total number of hybrid joints = 1
rr = total number of released reactions
=∑(mj−1)=2−1=1
∴DS=(3×12)+12−1−3(11+1)
=36+11−36=11

• A stable frame is one where all members are correctly restrained and form a rigid structure.

• Frame 2: Stable because all joints are connected in a triangular geometry, ensuring rigidity.

• Frame 3: Stable due to sufficient members and proper constraints preventing movement.

• Frame 4: Stable, as it includes additional bracing (diagonal members) ensuring no instability.

• Frame 1: Unstable as it lacks sufficient members or bracing for overall rigidity.

Correct choice: c (2, 3, and 4).

The kinematic indeterminacy of the beam is given by,
D_k = 3j - r_e + r_r
Now, j = 5 
r_e = 7
r_r = 1
∴ D_k = 3 x 5 - 7 + 1 = 9

D_s = D_se + D_si
D_se = r_e - 3 = 14 - 3 = 11
D_si = 3c - r_r

= 18 - ( 2 + 2 + 3 + 2 + 3 + 2 + 2) 
= 18 - (16) = 2 
∴ D_s = 11 + 2 = 13

D_k = [3(j + j) - r_e] + r_r - m
where,
j = total number of rigid joints = 12
j' = total number of hybrid joints = 2
r_e = total number of external reactions
= 3+ 2 + 2 = 7
r_r =  = (2 - 1) + (2 - 1) = 2
m = total number of axially rigid member = 9
∴ D_k = [3(12 + 2) - 7] + 2 - 9 = 35 - 7 = 28