Past Year Questions: Determinacy and Indeterminacy | Structural Analysis - Civil Engineering (CE) PDF Download

Q1: The plane frame shown in the figure has fixed support at joint A, hinge support at joint F, and roller support at joint 1 . In the figure, A to I indicate joints of the frame.    [2024, Set-1]

If the axial deformations are neglected, the degree of kinematic indeterminacy is (integer).
Ans: 9
Sol: From inextensibility of BF and BA, 
ΔB_x = 0 ΔB_y = 0 
From inextensibility of BE and EF, 
ΔE_x = 0 ΔE_y = 0
From inextensibility of CE and BC
ΔC_x = 0 ΔC_y = 0
From inextensibility of BD and CD,
ΔD_x = 0 ΔD_y = 0
From inextensibility of DG,
ΔG_x = 0 
From inextensibility of EH and HF,
ΔH_x = 0 ΔH_y = 0
From inextensibility FI,
Δl_x = 0
⇒ If at joint (F),FB, FE, FH, FI are rigidly connected then possible displacements at F = θ_F. 
If at joint I, FI and IH are rigidly connected, possible displacements at I = θ_I.
Hence unknown joint displacements are
θ_B, θ_C, θ_D, θ_E, θ_F, θ_G, Δ_Gy, θ_H, θ_I
⇒ D_k = 9
Note: However, if somebody assumes all members at joint (F) to be connected with pin then at (F) we have unknown joint displacements as
θ_FB, θ_FE, θ_FH, θ_FI.
Similarly, if at joint I if somebody assumes FI and HI to be pin connected then unknown joint displacements at I are θ_IF, θ_IH.
Hence, D_k will increase by 4.
⇒ D_k = 9 + 4  = 13

Q1: Consider the following three structures:

Structure I: Beam with hinge support at A, roller at C, guided roller at E, and internal hinges at B and D.

Structure II: Pin-jointed truss, with hinge support at A, and rollers at B and D.

Structure III: Pin-jointed truss, with hinge support at A and roller at C.
Which of the following statements is/are TRUE?    [2023, Set-1]
(a) Structure I is unstable
(b) Structure II is unstable
(c) Structure III is unstable
(d) All three structures are stable
Ans: (a, b and c)
Sol: Unstability in beam can be checked
(i) If support reactions are not enough.
(ii) If reactions are concurrent.
(iii) If reactions are parallel.
(iv) If there is mechanism.
where as for truss also along with the above given points the triangular panels are generally stable.
But, a virtual inspection must be conducted to understand the stability

(I) It is unstable as it shows mechanisms.
Also to understand if we apply a vertical force at slider side, the deflected shape will be as follows.

(II) The frame is internally stable but all the reactions are concurrent and meeting at hinge A, and the frame can rotate about A. Hence, it is unstable.

(III) The frame is unstable because it cannot resist shear in DE and AB since DE and AB are slender members.

Q1: Consider a beam PQ fixed at P, hinged at Q, and subjected to a load F as shown in figure (not drawn to scale). The static and kinematic degrees of indeterminacy, respectively, are    [2022, Set-2]
(a) 2 and 1
(b) 2 and 0
(c) 1 and 2
(d) 2 and 2
Ans: (a)
Sol:

Static indeterminacy, SI = r - 3 = (3 + 2) - 3 = 2
Kinematic indeterminacy = 0 + 1 = 1

Q1: The degree of static indeterminacy of the plane frame as shown in the figure is______    [2019, Set-2]

Ans: 15
Sol:
Static indeterminacy D_s = D_se + D_si - Force releases
External indeterminacy, D_se = r - s
No. of support reactions, r = 7
Number of equilibrium equations, s = 3
D_se = 7 - 3 = 4 
Internal indeterminacy D_si = 3 x No of Closed boxes
= 3 x 4 = 12
Force releases = 1 [At the internal hinge]
D_s  = 4 + 12 - 1 = 15

Q1: Consider the frame shown in the figure

If the axial and shear deformations in different members of the frame are assumed to be negligible, the reduction in the degree of kinematic indeterminacy would be equal to    [2017, Set-2]
(a) 5
(b) 6
(c) 7
(d) 8
Ans: (b)
Sol: 
DOF of Joints:

When all members are extensible,

D_k (when extensible) = 14
D_k(when inextensible) = D_k (when extensible) - No. of axially rigid member
= 14 - 6 = 8
So, reduction in

Note : Shear deformation is not considered in calculation of D_K.

Q2: A planar truss tower structure is shown in the figure.

Consider the following statements about the external and internal determinacies of the truss.
P. Externally Determinate
Q. External Static indeterminacy = 1
R. External Static Indeterminacy = 2
S. Internally Determinate
T. Internal Static Indeterminacy = 1
U. Internal Static Indeterminacy = 2
Which one of the following options is correct?    [2017: 2 Marks, Set-I]
(a) P-False; Q-True; R-False; S-False; T-False; U-True
(b) P-False; Q-True; R-False; S-False; T-True; U-False
(c) P-False; Q-False; R-True; S-False; T-False; U-True
(d) P-True; Q-True; R-False; S-True; T-False; U-True
Ans. (a)
Solution. D_se= r_e - 3 - s = 4 - 3 = 1
D_si = m -(2j - 3) = 15 - (2 x 8 - 3)
= 2
Trick : For a truss formed due to combination of simple triangles.
D_si = No. of double diagonal panels = 2
D_se = R - 3 = 4 - 3 = 1

Question for Past Year Questions: Determinacy and Indeterminacy

Try yourself:The kinematic indeterminacy of the plane truss shown in the figure is    
[2016: 1 Mark, Set-II]

• A.

  11
• B.

  8
• C.

  3
• D.

  0

Explanation

Kinematic indeterminacy,
D_k = 2j - r_e
= 2 x 7 - 3 = 11

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Question for Past Year Questions: Determinacy and Indeterminacy

Try yourself:A guided support, as shown in the figure below, is represented by three springs (horizontal, vertical and rotational) with stiffness k_x, k_y and ke respectively. The limiting values of k_x, k_y and k_e are   [2015: 1 Mark, Set-II]

• A.

  ∞, 0, ∞
• B.

  ∞,∞,∞
• C.

  0,∞,∞
• D.

  ∞,∞,0

Explanation

Guided roller support has 2 reactions:

For given support, kx = ∞, k_y = 0, k_{θ = ∞}

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Question for Past Year Questions: Determinacy and Indeterminacy

Try yourself:The static in determinacy of the two-span continuous beam with an internal hinge, shown below, is ____________.    [2014: 1 Mark, Set-II]

• A.

  0
• B.

  1
• C.

  2
• D.

  3

Explanation

Number of members, m = 4
► Number of external reaction, r_e = 4
► Number of joint, j = 5
► Number of reaction released, r_r = 1
► Degree of static indeterminacy, D_s = 3 m + r_e - 3j - r_r
= 3 x 4 + 4 - 3 x 5 - 1
= 0

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Question for Past Year Questions: Determinacy and Indeterminacy

Try yourself:The degree of static indeterminacy of a rigid jointed frame PQR supported as shown in figure is    [2014: 1 Mark, Set-I]

• A.

  zero
• B.

  one
• C.

  two
• D.

  unstable

Explanation

D_s = D_se + D_si
= (r_e -3 ) + 3 C - r_r,
= (4-3)+ 3 x 0- 1
= 0
Hence, the structure is stable.

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Note: Stable for vertical loading unstable for horizontal loading.