Concept:

Statically Determinate Structures: Those structures in which reactions and internal member forces can be obtained from equations of static equilibrium only the called statically determinate structures. e.g. simply supported Beam, 3 hinged arch.

Statically Indeterminate Structure: Those structures where all the member forces and support reactions cannot be evaluated using equations of static equilibrium only and need additional compatibility equations are called statically indeterminate structures. e.g. fixed Beam, 2-hinged arch.

Degree of static Indeterminacy for a beam is given as

D_S = (Number of support reactions removed) – (Number of restraints added)

The above result is for making a given beam to a cantilever.

Calculation:

It is a case of vertical loading, and in a hinge support if all the supports are at same level the number of restraints is only 1 as we do not need to bother about horizontal displacements.

There is one restraint applied at the slider and 1 at the hinge A in order to make it a cantilever.

D_S = Number of support reactions removed – R’ = 3 – 2 = 1 

$\begin{array}{l}1.\mathrm{S}\mathrm{t}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{A}\mathrm{B}\mathrm{a}\mathrm{t}\mathrm{B}=\frac{3\mathrm{E}\mathrm{I}}{3\mathrm{L}}=\frac{\mathrm{E}\mathrm{I}}{\mathrm{L}}\\ \mathrm{S}\mathrm{t}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{C}\mathrm{D}\mathrm{a}\mathrm{t}\mathrm{C}=\frac{4\mathrm{E}\mathrm{I}}{4\mathrm{L}}=\frac{\mathrm{E}\mathrm{I}}{\mathrm{L}}\end{array}$

K_CD = K_BA

2. Loading is symmetrical

So, it is observed than neither the loading is unsymmetrical nor the stiffness is different. So the frame will not sway in either direction.

Kinematic indeterminacy for the 2D rigid frame is given by

D_k = 3_j­ - r_e + r_r - h_r

J → no. of joints

r_e­ → Reactions components at joints

r_r­ → Released Reactions due to hybrid joints

r_r = m - 1

m → no. of members meeting at a joint

h_r­ → Number of axially rigid members due to which axial displacements are prevented

J = 12

r_e­ = 8

r_r = m - 1

Hybrid joints are available at j, H and at joint C.

At J : r_r = 3 - 1 = 2

At H : r_r = 3 - 1 = 2

At C : r_r = 4 - 1 = 3

Total released reaction = 2 + 2 + 3 = 7r_r = 7

D_k = 3 × 12 - 8 + 7 - h_r

h_r = 14

D_k = 36 - 8 + 7 - 14

D_k = 21

Since point B is hinged, so summation of moments about pt.B = 0

$M_A=\left(\frac{1}{2}\times 1200\times 4\right)\times \left(\frac{2}{3}\times 4\right)$

An influence line represents the variation of either the reaction, shear, moment, or deflection at a specific point in a member as a concentrated force moves over the member.

Advantages of drawing ILD are as follows:

i) To determine the value of quantity (shear force, bending moment, deflection, etc.) for a given system of loads on the span of structure.

ii) To determine the position of a live load for the quantity to have the maximum value and hence to compute the maximum value of the quantity.

ILD can be drawn for statically determinate as well as indeterminate structures. ILD for statically determinate and indeterminate structures are linear and nonlinear respectively.

Flexibility matrix method and stiffness matrix method are two matrix methods.

Following are the assumptions made for matrix analysis of the structure:

a) Material of all the structural member obeys Hooke’s law.

b) Members are subjected to small deflections.

c) Change in length under a deflection perpendicular to member length is 0.

d) Principal of superposition is applicable.

Concept:

For a two hinged circular arch carrying a load P at a section, the radius vector corresponding to which makes an angle θ with the horizontal the horizontal reaction is given as

$H=\frac{P}{\pi }{\sin }^2\theta$

Calculation:

R = 12/2 = 6m, P = 20 kN,

$\sin \theta =\frac{4}{6}=\frac{2}{3}$

$\therefore H=\frac{P}{\pi }{\sin }^2\theta =\frac{20}{\pi }\times {\left(\frac{2}{3}\right)}^2=2.83kN$

I^st condition:

For a cantilever beam subjected to load W at distance of L/3 from free end, the deflection is given by:

$\mathrm{y}{\mathrm{c}}_1=\frac{\mathrm{W}}{3\mathrm{E}\mathrm{I}}\times {\left(\frac{2\mathrm{L}}{3}\right)}^3+\frac{\mathrm{W}}{2\mathrm{E}\mathrm{I}}{\left(\frac{2\mathrm{L}}{3}\right)}^2\times \frac{\mathrm{L}}{3}={\frac{28\mathrm{W}\mathrm{L}}{162\mathrm{E}\mathrm{I}}}^3$

II^nd condition:

For a cantilever beam subjected to load W at distance of 2L/3 from free end:

$\mathrm{y}{\mathrm{c}}_2=\frac{\mathrm{W}}{3\mathrm{E}\mathrm{I}}\times {\left(\frac{\mathrm{L}}{3}\right)}^3+\frac{\mathrm{W}}{2\mathrm{E}\mathrm{I}}{\left(\frac{\mathrm{L}}{3}\right)}^2\times \frac{2\mathrm{L}}{3}=\frac{8\mathrm{W}{\mathrm{L}}^3}{162\mathrm{E}\mathrm{I}}$

$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\left(\mathrm{r}\right)=\frac{\mathrm{y}{\mathrm{c}}_1}{\mathrm{y}{\mathrm{c}}_2}=\frac{{\frac{28\mathrm{W}\mathrm{L}}{162\mathrm{E}\mathrm{I}}}^3}{\frac{8\mathrm{W}{\mathrm{L}}^3}{162\mathrm{E}\mathrm{I}}}=\frac{28}{8}=\frac{7}{2}$

$\therefore \frac{\mathrm{y}{\mathrm{c}}_2}{\mathrm{y}{\mathrm{c}}_1}=\frac{2}{7}$

Concept:

If the number of external reactions is less than the number of static equilibrium equations, then the truss is said to be externally unstable.

For Internal stability: For Plane truss, the number of members required for internal stability is:

m ≥ 2j - r

Where,

m = Number of members, j = Number of joints, and r = No. of static equilibrium equations

Calculation:

External stability:

R = Number of external reactions = 2

r = Conditions of static equilibrium= 3 i.e. (ΣF_x = 0, ΣF_y = 0, and ΣM_z = 0)

R < r, this implies truss is externally unstable.

Internal stability:

m = 10, and j = 7

∴ 2j – r = 2 x 7 - 3 = 11

Hence, m < 2j-r, Truss is internally unstable too.

Concept:

The flexibility matrix has the following two main properties:

1. The flexibility matrix is square symmetric matrix due to Maxwell’s reciprocal theorem which me Ans: that its diagonal elements are always equal.

2. The diagonal elements of flexibility matrix are always positive due to Castigilanostheorem–II.

Calculation:

a = d (using 1st property)

b > 0 and c > 0 (using 2nd property)

∴ b = c, b > 0, c > 0

Slope deflection equation:

$M_{BC}=M_{BC}^f+\frac{2EI}{8}\left[2{\theta }_B+{\theta }_C-\frac{3\delta }{L}\right]$

Since there is no loading on the portion BC, therefore there will not be any fixed end moments. Also, there is no settlement or deflection, therefore, δ = 0.

Therefore, the slope deflection equation becomes

$M_{BC}=\frac{2EI}{L}\left[2{\theta }_B+{\theta }_C\right]$

$M_{BC}=\frac{2\times 2EI}{8}\left[2{\theta }_B+{\theta }_C\right]$

$M_{BC}=\frac{4EI}{8}\left[2{\theta }_B+{\theta }_C\right]$

If at a joint 3 members meet out of which 2 are collinear then force in 3^rd member will be zero provided there is no load/reaction at that joint.

∴ F_GC = 0, F_FB = 0 and F_HD = 0

i.e. forces in the member GC, FB, HD is equal to zero

Now, considering joint E:

∵ F_EH = R_E

There is no horizontal force.

∴ F_DE = 0

Also considering joint D, we can say that force in member CD = 0

For the given truss:

At point D, point C, and point F

No external force is acting

∴ Member AD, PE, CB & CE are zero force member.

Similarly, member EF is also a zero force member.

By symmetricity, RA = RB = 35 kN

ΣH = 0

∴ HA = 0

At joint B

ΣV_B = 0

∴ F_BE sin θ = 35

$\therefore {\mathrm{F}}_{\mathrm{B}\mathrm{E}}=\frac{35}{0.8}=43.75\mathrm{k}\mathrm{N}\left(\mathrm{C}\right)$

ΣH_B = 0

F_BF = F_BE × cos θ,

${\mathrm{F}}_{\mathrm{B}\mathrm{F}}=\frac{35}{0.8}\times 0.6=26.25\mathrm{k}\mathrm{N}$

∴ F_BF = 26.25 kN (C)

From symmetricity

F_BF = F_AF and F_BE = F_AE

Now, applying the unit vertical load at point F

At point D, point C, and point F

No external force is acting

∴ Member AD, PE, CB & CE are zero force member.

From symmetricity

R_A = R_B = 0.5 kN

At joint A

ΣV = 0

F_AE × sin θ = 0.5 kN

${\mathrm{F}}_{\mathrm{A}\mathrm{E}}=\frac{0.5}{0.8}\therefore 0.625\mathrm{k}\mathrm{N}$

∴ F_AE = 0.625 kN (C)

ΣH = 0

F_AE cos θ = F_AF

$\therefore {\mathrm{F}}_{\mathrm{A}\mathrm{E}}=\frac{0.5}{0.8}\times 0.6=0.375\mathrm{k}\mathrm{N}$

F_AF = 0.375 kN (C)

At point F

ΣV = 0

∴ F_FE = 1 kN (T)

ΣH = 0

∴ F_FB = 0.375 kN (C)

$\therefore \delta =\frac{136.72\times 2+29.53\times 2}{\mathrm{A}\mathrm{E}}$

$\therefore \delta =\frac{332.5}{\mathrm{A}\mathrm{E}}\mathrm{m}$

Concept:

Standard fixed end moments:

Calculation:

Consider a infinitesimal strip of width dx at a distance x from the fixed end A,

Using the standard result,

$M_{BA}=\frac{Pb{a}^2}{{\ell }^2}=\underset{0}{\overset{4}{\int }}\frac{\left(2dx\right)\cdot \left(8-x\right)\cdot x^2}{8^2}$ = 3.33 kN-m

Concept:

The truss shown above is a determinate truss and all member forces can be calculated using the equilibrium equations only.

Calculation:

Calculating the reactions first,

∵ ∑ F_y = 0 ⇒ R_A + R_I = 20 kN

∵ ∑ F_x = 0 ⇒ H_A + 10 = 0 ⇒ H_A = -10 kN

∵ ∑ M_A   = 0 ⇒ 10 × 6 + 20 × 8 - R_I × 16 = 0

$\Rightarrow R_I=\frac{55}{4}kN$

$\Rightarrow R_A=\frac{25}{4}kN$

Using method of sections,

For moment equilibrium of the section shown, ∑ M_D = 0

$\Rightarrow F_{CE}\times 3+\frac{25}{4}\times 4+10\times 3=0$

$\Rightarrow F_{CE}=\frac{-55}{3}kN=-18.33kN$

negative sign indicates compressive force in CE.

Fixed end moments:

${\mathrm{F}}_{\overline{\mathrm{A}\mathrm{B}}}=\frac{4\mathrm{E}\mathrm{I}{\theta }_{\mathrm{A}}}{\mathrm{L}}=\frac{4\times 2\times {10}^7\times 0.0015}{4000}=30\mathrm{k}\mathrm{N}\mathrm{m}$

${\mathrm{F}}_{\overline{\mathrm{B}\mathrm{A}}}=\frac{2\mathrm{E}\mathrm{I}{\theta }_{\mathrm{A}}}{\mathrm{L}}=\frac{2\times 2\times {10}^7\times 0.0015}{4000}=15\mathrm{k}\mathrm{N}\mathrm{m}$

${\mathrm{F}}_{\overline{\mathrm{B}\mathrm{C}}}=-\frac{6\mathrm{E}\mathrm{I}\left(\delta \right)}{{\mathrm{L}}^2}=\frac{-6\times 2\times {10}^7\times 5}{{\left(6000\right)}^2}=-16.67\mathrm{k}\mathrm{N}\mathrm{m}$

${\mathrm{F}}_{\overline{\mathrm{C}\mathrm{B}}}=-\frac{6\mathrm{E}\mathrm{I}\left(\delta \right)}{{\mathrm{L}}^2}=\frac{-6\times 2\times {10}^7\times 5}{{\left(6000\right)}^2}=-16.67\mathrm{k}\mathrm{N}\mathrm{m}$

${\mathrm{F}}_{\overline{\mathrm{B}\mathrm{D}}}=0$

Distribution factor:

Moment distribution table:

Concept:

Cable is a structure member which can resists only tensile forces. The bending moment and shear force are zero everywhere on cable.

Calculation:

Cables carry only tensile forces

At joint B:

∵ ∑ F_y = 0

⇒  T_AB sin θ = P₁

$\Rightarrow T_{AB}\times \frac{4}{5}=P_1$          ---(1)

∑ F_x = 0 ⇒ T_AB cos θ = T_BC

$\Rightarrow \frac{3}{5}{T}_{AB}=T_{BC}$         ---(2)

Again at joint C,

∑ F_y = 0

⇒ T_CD sin θ = P₂

$\Rightarrow P_2=\frac{T_{CD}}{\sqrt{2}}$          ---(3)

And, $T_{CD}\cos \theta =T_{BC}\Rightarrow T_{BC}=\frac{T_{CD}}{\sqrt{2}}$         ---(4)

From (3) and (4)

$P_2=\frac{T_{CD}}{\sqrt{2}}=\frac{T_{BC}\sqrt{2}}{\sqrt{2}}=T_{BC}$            ---(5)

From (1) and (2)

$P_1=\frac{4}{5}{T}_{AB}and{T}_{AB}=\frac{5}{3}{T}_{BC}$

$\Rightarrow P_1=\frac{4}{5}\times \frac{5}{3}{T}_{BC}=\frac{4}{3}{T}_{BC}$         ---(6)

∴ using (5) and (6),

$P_1=\frac{4}{3}{P}_2$

Concept:

An influence line for a member is a graph representing the variation of internal force( reaction, shear, moment, or deflection) in a fixed section of the member, due to a unit load traversing a structure.

The maximum bending moment is always below a point load when there are only point loads moving on a girder. Hence to identify which load if placed on section B gives the maximum B.M at section B, allow each of the load one by one to cross over section B, and calculate average load to the left and right of section B.

The load crossing over of which leads to heavier left side portion (in terms of average weight), changing to lighter, is the load which if placed on section B leads to maximum B.M at section B.

Calculation:

Hence due to the passing of 15 kN load from section B, the average load to the left becomes lesser than that on the right side of the girder.

Hence when 15 kN load is placed at the section B, the bending moment is maximum at the section.

Concept:

Muller-Breslau principle for Qualitative ILD:

For determinate structures, the influence line for any stress function may be obtained by removing the restraint offered by the stress function and introducing a directly related unit displacement at the location of the stress function. The deflected shape represents Influence Line Diagram for the stress function.

While giving the above displacement if one part of the beam has a tendency to take curved shape ILD will not exist for that part. Thus all displacement has to be given on the other side. Hence ILD for determinate structures is always composed of straight lines.

To calculate the value of stress function, the load acting at the point is multiplied with the ordinate of the ILD.

Calculation:

ILD for the shear force at a section 8 m from A,

α + β = 1

Also, $\frac{\alpha }{\beta }=\frac{\beta }{12}\Rightarrow \frac{\alpha }{\beta }=\frac{2}{3}$

$\Rightarrow \frac{2}{3}\beta +\beta =1\Rightarrow \beta =\frac{3}{5}$ unit

$\therefore \alpha =\frac{2}{5}$ units

For maximum positive shear force, the load should be placed such that the tail of the UDL is just to the right of C and span CB is loaded.

∴ Maximum positive shear force at section C

= (Area under ILD from B to C) × (UDL) $=\left(\frac{1}{2}\times 12\times \frac{3}{5}\times 20\right)=72kN$

Concept:

A three-hinged arch is a statically determinate structure with two supports hinged and an internal hinge provided in the body of the arch generally at the crown. Such arches are analyzed using equations of static equilibrium only.

Calculation:

Let the span be L and rise be R.

For vertical equilibrium,

$R_A+R_B=\frac{4L}{2}$

Due to an internal hinge at C, the Bending moment at C should be zero.

Vertical reaction $R_A=R_B=\frac{4\times L}{2}kN$       (Due to symmetry)

Horizontal reaction at support (H):

B.M at C due to forces on the left $=4\times \frac{L}{2}\times \frac{L}{4}-R_A\times \frac{L}{2}+H\times R=0$

$\Rightarrow \frac{L^2}{2}-\frac{4L}{2}\times \frac{L}{2}+HR=0$

$\Rightarrow HR=\frac{L^2}{2}\Rightarrow H=\frac{L^2}{2R}$

For, Horizontal reaction = Vertical reaction

$\Rightarrow \frac{L^2}{2R}=\frac{4L}{2}\Rightarrow \frac{L}{R}=4$

Fixed end moments:

${\mathrm{F}}_{\overline{\mathrm{A}\mathrm{B}}}=-\frac{\mathrm{w}{\mathrm{L}}^2}{12}-\frac{6\mathrm{E}\mathrm{I}\left(\delta \right)}{{\mathrm{L}}^2}=-\frac{15\times 4^2}{12}-\frac{6\mathrm{E}\mathrm{I}\times \left(\frac{96}{\mathrm{E}\mathrm{I}}\right)}{{12}^2}=-24\mathrm{k}\mathrm{N}\mathrm{m}$

${\mathrm{F}}_{\overline{\mathrm{B}\mathrm{A}}}=+\frac{\mathrm{w}{\mathrm{L}}^2}{12}-\frac{6\mathrm{E}\mathrm{I}\left(\delta \right)}{{\mathrm{L}}^2}=\frac{15\times 4^2}{12}-\frac{6\mathrm{E}\mathrm{I}\times \left(\frac{96}{\mathrm{E}\mathrm{I}}\right)}{{12}^2}=16\mathrm{k}\mathrm{N}\mathrm{m}$

${\mathrm{F}}_{\overline{\mathrm{B}\mathrm{A}}}=+\frac{\mathrm{w}{\mathrm{L}}^2}{12}-\frac{6\mathrm{E}\mathrm{I}\left(\delta \right)}{{\mathrm{L}}^2}=\frac{15\times 4^2}{12}-\frac{6\mathrm{E}\mathrm{I}\times \left(\frac{96}{\mathrm{E}\mathrm{I}}\right)}{{12}^2}=16\mathrm{k}\mathrm{N}\mathrm{m}$

${\mathrm{F}}_{\overline{\mathrm{C}\mathrm{B}}}=+\frac{\mathrm{w}\mathrm{L}}{8}=\frac{24\times 3}{8}=9\mathrm{k}\mathrm{N}\mathrm{m}$

Distribution factor:

Moment distribution table:

Concept:

Stiffness Matrix is a displacement method of analysis in which Degree of Kinematic Indeterminacy is found and the number of coordinate direction are chosen accordingly. The structure is analyzed by developing a stiffness matrix.

Development of stiffness matrix:

To develop j^th column of a stiffness matrix - unit displacements will be given in j^th coordinate direction without giving displacement in any other coordinate directions and forces developed in all the coordinate direction are found out.

Calculation:

I^st Column:

$K_{11}=\frac{12EI}{L^3}$

$K_{21}=\frac{6EI}{L^2}$

K₃₁ = 0

II^nd Column:

$K_{12}=\frac{6EI}{L^2}$

$K_{22}=\frac{4EI}{L}+\frac{EI}{L}=\frac{5EI}{L}$

$K_{32}=\frac{-EI}{L}$  {negative sign due to opposite moment to the direction shown in fig}

III^rd column:

K₁₃ = 0

$K_{23}=\frac{-EI}{L}$

$K_{33}=\frac{EI}{L}$

The stiffness matrix obtained is $\left[\begin{array}{ccc}\frac{12EI}{L^3}& \frac{6EI}{L^2}& 0\\ \frac{6EI}{L^2}& \frac{5EI}{L}& \frac{-EI}{L}\\ 0& \frac{-EI}{L}& \frac{EI}{L}\end{array}\right]$

From the stiffness matrix obtained α = 12, β = 5 and γ = 1

∴ α + β +γ = 12+5+1 = 18

Important Point:

Flexibility Matrix is a force method of analysis in which Degree of Static Indeterminacy is found and the number of coordinate direction are chosen accordingly. The structure is analyzed by developing a flexibility matrix.

Calculation:

FBD of the beam considering reaction at B as R_B

Deflection at B considering R_B as redurdant

$\therefore {\delta }_B=A_1{\overline{x}}_1+A_2{x}_2=\frac{M_0}{EI}\times \frac{L}{2}\times \frac{3L}{4}+\frac{M_0}{2EI}\times \frac{L}{2}\times \frac{L}{4}$

$=\frac{7}{16}\frac{M_0{L}^2}{EI}$

Deflection at B due to R_B

$\therefore {\delta }_B^{\prime }=\left[\frac{R_B{\left(\frac{L}{2}\right)}^3}{3EI}+\frac{\left(R_B\frac{L}{2}\right){\left(\frac{L}{2}\right)}^2}{2EI}\right]+\left[\frac{R_B{\left(\frac{L}{2}\right)}^2}{2EI}+\frac{(R_B\frac{L}{2}}{2EI}\right]\times \frac{L}{2}+\frac{R_B{\left(\frac{L}{2}\right)}^3}{3E\left(2I\right)}$

$=\left[\frac{R_B{L}^3}{24EI}+\frac{R_B{L}^3}{16EI}\right]+\left[\frac{R_B{L}^3}{16EI}\right]+\left[\frac{R_B{L}^3}{16EI}+\frac{R_B{L}^3}{24EI}\right]+\left[\frac{R_B{L}^3}{48EI}\right]$

$\therefore {\delta }_B^{\prime }=\frac{11R_B{L}^3}{48EI}$

${\delta }_B^{\prime }-{\delta }_B=0$   (Due to support at B)

$\Rightarrow \frac{11R_B{L}^3}{48EI}=\frac{7}{16}\frac{M_0{L}^2}{EI}$

$\Rightarrow R_B=\frac{21}{11}\frac{M_0}{L}$

Thus the FBD of the beam is,

For rotational equilibrium, ∑ M_A = 0

$\Rightarrow M^{\prime }+M_0-\frac{21}{11}\frac{M_0}{L}\times L=0$

$\Rightarrow M^{\prime }=\frac{10}{11}{M}_0$

Concept:

For the given structure, degree of kinematic indeterminacy (D_k) = 1 (Ignoring the axial deformation)

There is only one degree of freedom i.e. rotation at joint B. Since Joint B is rigid, the rotation in span BA and BC will be same. Let the rotation at joint B is ‘θ_B’.

This rotation can be determined using slope deflection equations as follow:

${\mathrm{M}}_{\mathrm{B}\mathrm{A}}={\left(\mathrm{F}\mathrm{E}\mathrm{M}\right)}_{\mathrm{B}\mathrm{A}}+\frac{2\mathrm{E}\mathrm{I}}{\mathrm{L}}\left(2\times {\theta }_{\mathrm{B}}+{\theta }_{\mathrm{A}}-\frac{3\mathrm{\Delta }}{\mathrm{L}}\right)$

${\mathrm{M}}_{\mathrm{A}\mathrm{B}}={\left(\mathrm{F}\mathrm{E}\mathrm{M}\right)}_{\mathrm{A}\mathrm{B}}+\frac{2\mathrm{E}\mathrm{I}}{\mathrm{L}}\left(2\times {\theta }_{\mathrm{A}}+{\theta }_{\mathrm{B}}-\frac{3\mathrm{\Delta }}{\mathrm{L}}\right)$

Where, (FEM)_BA and (FEM)_AB are fixed end moments for span BA and AB respectively and ∆ is the sway.

Further, considering the rotational equilibrium of joint B.

M_BA + M_BC = 0

Slope at A and C is Zero because they are fixed supports and ∆ = 0 (no sway).

Fixed End Moments:

Fixed end moments for a span subjected to concentrated load is shown in the following figure:

Calculation:

Calculation of Fixed End Moments:

$\mathrm{F}\mathrm{E}{\mathrm{M}}_{\mathrm{B}\mathrm{C}}=-\frac{10\times 4}{8}=-5\mathrm{k}\mathrm{N}\mathrm{m},\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{F}\mathrm{E}{\mathrm{M}}_{\mathrm{C}\mathrm{B}}=\frac{10\times 4}{8}=5\mathrm{k}\mathrm{N}\mathrm{m}$

$\mathrm{F}\mathrm{E}{\mathrm{M}}_{\mathrm{B}\mathrm{A}}=\frac{90\times 1^2\times 2}{3^2}=20\mathrm{k}\mathrm{N}\mathrm{m},\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{F}\mathrm{E}{\mathrm{M}}_{\mathrm{A}\mathrm{B}}=-\frac{90\times 2^2\times 1}{3^2}=-40\mathrm{k}\mathrm{N}\mathrm{m}$

Slope deflection equation:

Span AB:

${\mathrm{M}}_{\mathrm{A}\mathrm{B}}=-40+\frac{2\mathrm{E}\mathrm{I}}{3}\times \left({\theta }_{\mathrm{B}}\right),\mathrm{a}\mathrm{n}\mathrm{d}{\mathrm{M}}_{\mathrm{B}\mathrm{A}}=20+\frac{2\mathrm{E}\mathrm{I}}{3}\times \left(2{\theta }_{\mathrm{B}}\right)$

Span BC:

${\mathrm{M}}_{\mathrm{B}\mathrm{C}}=-5+\frac{2\mathrm{E}\mathrm{I}}{4}\times \left(2{\theta }_{\mathrm{B}}\right),\mathrm{a}\mathrm{n}\mathrm{d}{\mathrm{M}}_{\mathrm{C}\mathrm{B}}=5+\frac{2\mathrm{E}\mathrm{I}}{4}\times \left({\theta }_{\mathrm{B}}\right)$

Consider the rotational equilibrium of Joint B:

M_BA + M_BC = 0

⇒ θ_B = - 6.42/EI (-ve sign indicates that θ_B is in anticlockwise)

Substitute the value of θ_B in above equations:

We get, M_BC = -11.42 kNm, M_CB = 1.79 kNm, M_AB = -44.29 kNm and M_BA = 11.42kNm

Consider the FBD of Member AB and let horizontal reaction developed at A is H_A.

Consider moment equilibrium at B.

∑M_B = 0

⇒ 90 × 2 - 44.20 + 11.42 - H_A × 3 = 0

⇒ H_A = 49.07 kN

Concept:

The moment at any joint is distributed as per the distribution factor of any member meeting at that joint.

Distribution factor $=\frac{Memberstiffness}{Jointstiffness}$

Joint stiffness = Σ  Members stiffness meeting at the joint.

Calculation:

Joint stiffness at O = ∑ Joint stiffness of members

$=\frac{2EI}{L}+\frac{3EI}{L}+\frac{4EI}{L}=\frac{9EI}{L}$

Stiffness of member $OD=\frac{4EI}{L}$

∴ Distribution factor $=\frac{Jointstiffness}{Memberstiffness}=\frac{\left(9EI/L\right)}{\left(4EI/L\right)}=\frac{9}{4}$

∴ Moment carried by $OD=Distributionfactor\times Momentat0=\frac{9}{4}\times PL$

∴ Moment developed at $D=\frac{9}{4}PL$           (∵ Carryover factor at a fixed end is -1)