I. Member LK, LA and LB are meeting at joint ‘L’ and AL and LK re co-linear so force in the member LB is zero.

F_LB = 0

II. Member FG, FE, and FD are meeting at joint F, So force in the member FD = 0

F_FD = 0

III. Removing Member LB, DF, thus member BK, BC and BA meets at point B out of which BA and BC are co-linear and vice versa for member DG, DC and DE.

So, F_BK = 0

F_DG = 0

IV. Similarly, Member HM, HI and HG are meeting at joint H

F_HM = 0

The degree of indeterminacy,

Number of external reactions = r_e
= 3 + 3 + 3 + 3 = 12
Number of rigid joints,
j= 10
Number of joints at which releases are located, 
j= 1
Number of members, 
m = 12
As the hinge is located at a point where 4 members meet. Hence it is equivalent to three hinges.
Therefore number of releases, r_r = 3.

Concept

According to the principle of superposition

For a linearly elastic structure, the load effects caused by two or more loadings are the sum of the load effects caused by each loading separately.

Note that the principle is limited to:

• Linear material behaviour only;

• Structures undergoing small deformations only (linear geometry).

It is not applicable when:

1. The material does not obey Hooke’s law.

2. The effect of temperature changes are taken into consideration.

Ds for a pin jointed frame = m+r-2j

Ds for a rigid jointed frame = 3m+r -3j

m- No. of members, r- no. of reactions, j- no. of joints

Dk = 2j –r for a pin jointed frame

Dk = 3j –r for a rigid jointed frame

j number of joints and r number of reactions.

- Kinematic indeterminacy refers to the number of possible independent movements of a structure.
- For a pin-jointed space frame, each joint has 3 degrees of freedom (translations in x, y, and z directions).
- Formula: (3j - r), where (j) is the number of joints and (r) is the number of reaction components or constraints.
- This formula accounts for the three-dimensional nature of space frames, unlike planar frames, which have 2 degrees of freedom per joint.

m=3; r= 3+3 =6 ; j=4

Ds = 3m+r-3j = 3(3) + 6 -3(4) = 3

A perfect plane frame means a determinate structure, so

j = 11; r=8; m=11

Dk = 3j-r-no. of inextensible members = 3(11) – 8 -11 = 14

4 reactions mean that the system is definitely indeterminate. But stability would depend upon the nature of forces acting on the planar structure.

For plane truss degree of indeterminacy,

D_s = m + r_e - 2j

r_e = 4; m= 10; j = 5

D_s = 4 + 10 - 2 x 5 = 4

A structure will be statically determinate if the external reactions can be determined from force- equilibrium equations. A structure is stable when the whole or part of the structure is prevented from large displacements on account of loading.

The structure in figure (b) is stable but statically indeterminate to the second degree.

The structure shown in figure (c) has both reaction components coinciding with each other, so the moment equilibrium condition wilt never be satisfied and the structure will not be under equilibrium. In figure'(a), the structure is stable and there are three reaction components which can be determined by two force equilibrium conditions and one moment equilibrium condition.

D_s=D_se+D_si
D_ss= r_E- 3
= 4 - 3 = 1
D_si = 3C - r_R = 3C - ∑(m¹- 1)
= 3 x 2 -(2-1) +(2-1)
= 6 - 2 = 4 
∴ D_s = D_ss + D_si = 5

Degree of indeterminacy,

n = (m + r_E) - 2j

= (9 + 3 ) - 2 x 6 = 0

Since the degree of indeterminacy is zero and the frame is stable so it is a perfect frame.

The centre of span is midway between the centre of gravity of the load system and the wheel load under consideration

The maximum bending moment at section-Xwill occur when UDL occupies the whole span.

Maximum bending = Area under the influence line diagram

SF will be equal to-the area under the influence line diagram,

Maximum bending moment at a section occurs when a particular load is on the section which changes the ratio

Given:

Distribution Factor:

Fixed End Moments:

M_AB = M_BA = M_PE = M_EP = M_CD = M_DC = 0

\({{\rm{M}}_{{\rm{BP}}}} = - \frac{{{\rm{W}}{{\rm{L}}^2}}}{{12}} = - \frac{{24 \times {8^2}}}{{12}} = - 128{\rm{\;kN}}.{\rm{m}}\)

\({{\rm{M}}_{{\rm{PB}}}} = \frac{{{\rm{W}}{{\rm{L}}^2}}}{{12}} = \frac{{24 \times {8^2}}}{{12}} = 128{\rm{\;kN}}.{\rm{m}}\)

\({{\rm{M}}_{{\rm{PC}}}} = - \frac{{{\rm{W}}{{\rm{L}}^2}}}{{12}} = - \frac{{24 \times {8^2}}}{{12}} = - 128{\rm{\;kN}}.{\rm{m}}\)

\({{\rm{M}}_{{\rm{CP}}}} = \frac{{{\rm{W}}{{\rm{L}}^2}}}{{12}} = \frac{{24 \times {8^2}}}{{12}} = 128{\rm{\;kN}}.{\rm{m}}\)

Distribution Table:

Hence the magnitude of bending moment at P is 176 kNm

Maximum Horizontal Thrust occurs when the load is at center and is given by
H = (25WL)/(128h)
W = Concentrated Load = 100kN
Span= L = 32m; Rise= h=5m
Substituting the values we get H=125kN

Increase in temperature in a two hinged arch (degree of indeterminacy one ) will cause horizontal thrust only.

Moment due to horizontal thrust is -Hy

So, max B.M will be at crown as crown has height value of y.