Concept:

The No. of compatibility equation required if the frame is to be analyzed using Force method is equal to its degree of static indeterminacy by definition.

Degree of Static Indeterminacy (D_s) can be determined as below:

D_s = D_se + D_si

D_se = Degree of external static indeterminacy

D_si = Degree of internal static indeterminacy

D_se = Support reactions – No of Equilibrium equations

D_si = 3 × C – R

C = No of closed loops

R= No of releases = m - 1

Calculations:

No of reactions = 3 × 3 + 2 = 11 (3 fixed supports and one is pinned support)

No of equilibrium reactions = 3

D_se = 11 - 3 = 8

No of closed loops, C = 1

Releases

Moment releases at joint, A = 3 - 1 = 2

Moment releases at joint, B = 2 - 1 = 1

D_si = 3  C – R

= 3 × 1 – (2 + 1)

= 0

D_s = D_se + D_si

= 8 + 0

= 8

∴ Static Indeterminacy, D­_s = 8

Concept

We know that a cable takes shape of a Bending moment diagram (BMD).

Now, If only concentrated load is acting on a mid-span of cable, then BMD of equivalent simply supported beam subjected to load at mid span is triangular shape, so shape of cable will be triangular as shown in above figure.

Due to symmetry tension in AC and BC is same.

$\sin \theta =\frac{\sqrt{17}}{9}$

$\sum F_y=0$

FBD at C.

⇒ 2T sin θ = 120

$\Rightarrow T=\frac{60}{\sin \theta }=\frac{60\times 9}{\sqrt{17}}$

∴ T = 130.95 kN ≈ 131 kN

Concept:

For two hinged semi-circular arches:

Horizontal thrust is given by:

${\mathrm{H}}_{\mathrm{A}}={\mathrm{H}}_{\mathrm{B}}=\frac{\int \frac{{\mathrm{M}}_{\mathrm{x}}\mathrm{y}\mathrm{d}\mathrm{x}}{\mathrm{E}{\mathrm{I}}_{\mathrm{c}}}}{\int \frac{{\mathrm{y}}^2\mathrm{d}\mathrm{x}}{\mathrm{E}{\mathrm{I}}_{\mathrm{c}}}}$

Where,

M_x = Moment at A – A’, y = vertical distance,

E = Modulus of elasticity, and

 I = Moment of inertia of c/s of the arch.

When  W load acts at the crown, then

Horizontal thrust (H) = w/π

∴ H is independent of the radius of the arch.

So, the ratio will be 1: 1: 1.

Note

Important Points:

 For a Two Hinged Semicircular arch subjected to concentrated load (W) at any other point which makes an angle α with the horizontal, then the horizontal thrust,

$H=\left(\frac{W}{\pi }\right){\sin }^2\left(\alpha \right)$

Concept:

Flexural stiffness factor for rigid joint O is defined as the moment required to produce one-unit rotation at that joint.

Stiffness factor for various end conditions are specified below in tabulated form:

Calculation:

Let ‘M’ concentrated Moment is acting at joint O.

Since ‘O’ is rigid joint, so, rotation of members OA, OB, OC, OD are equal under moment ‘M’ and at ‘O’ let this rotation is ‘∝’.
For OA: For end is fixed.

Moment required to causes ‘∝’ rotation at $\text{O},{\text{M}}_{\text{OA}}=\frac{4\text{EI}}{\text{L}}\propto$

For OC: For end is free.

Moment required to cause ‘∝’ rotation at O, M_OC = O

For OB: For end is pinned

Moment required to cause ‘∝’ rotation at $\text{O},{\text{M}}_{\text{OB}}=\frac{3\text{EI}}{\text{L}}\propto$

For OD: Far end is roller, but rotation develop at ‘D’ is along the length of member i.e. axial reaction and hence does not produce any bending.

Moment required to cause ‘∝’ rotation at O, M_OD = o

Consider equilibrium of Joint O

∑M_o = 0

$\text{M}=\frac{4\text{EI}}{\text{L}}+\frac{3\text{EI}}{\text{L}}\propto =\frac{7\text{EI}}{\text{L}}\propto$

Now, Moment required to cause a unit rotation (∝ = 1) at rigid Joint O, is called stiffness factor, (k) $\therefore \text{k}=\frac{\text{M}}{\propto }=\frac{7\text{EI}}{\text{L}}$

Calculation

ΣF_y = 0

 i.e. R_A + R_B = 5t

ΣM_A = 0

 4t × L + 5t × L = R_B × 2L

R_B = 9t/2 = 4.5t

Taking joint B,

Force direction is towards joint B ∴ Compressive

∴ F_B.F = 4.5 t (compression)

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.

Approach: To get the maximum compressive force place the UDL only up to the region of compressive force

Concept:

ILD (Influence lines are important in designing beams and trusses used in bridges, etc. Where loads will move along their span.

The influence lines show where a load will create the maximum effect for any of the functions studied.)

Calculations:

Given, ILD of the truss member

The intensity of UDL = 30 kN/m.

From the ILD figure, we know that the upper triangle is in the tension part and the bottom triangle is in the compression part.

The length of the compression span needs to be determined

From the similarity of triangles:

$\frac{0.29}{x}=\frac{0.58}{4-x}$

$x=\frac{4}{3}$

In the above figure,

In order to get the maximum compressive force, UDL is placed till the compressive zone.   

 Force = UDL × (Area of the Compression triangle of ILD)

 Maximum compressive force = 30 × ($\frac{1}{2}\times 0.29\times \left(4+\frac{4}{3}\right)$ ) kN

 = 30 × 0.7733

 = 23.2 kN. 

∴ Maximum compressive force = 23.2 kN.

Concept:

The kinematic indeterminacy of the given 2 D frame can be evaluated b = using the following formula:

D_k = 3 × J + R – S - m

Where,

J is the total no of joints

R is the releases and it is given as: R = ∑ (m’-1), m’ is the total no of members meeting at hybrid joint.

S is the support reactions

m is the total no of axial rigid members

Calculation

No of Joints, J = 12

S = 3 + 1 + 2 = 6 (one is fixed support, one is roller support and last one is guided roller support)

Releases:

For joint A, R = 3 - 1 = 2

For joint N, R = 2 - 1 = 1

Now,

D_k = 3 × 12 + (2 + 1) – 6 – m = 33 – m

Case 1:

If all columns are axially rigid, then m = total no. of columns i.e. m = 9

∴ D_K = 33 - 9 = 24

Case 2:

If all beams are axially rigid, then m = total no. of beams i.e. m = 6

∴ D_K = 33 - 6 = 27

Concept:

Slope Deflection Equation:

The slope deflection equation at the end A for member AB can be written as:

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

Considering downward settlement to be negative we will have the equation as

${\mathrm{M}}_{\mathrm{A}\mathrm{B}}=\overline{{\mathrm{M}}_{\mathrm{A}\mathrm{B}}}+\frac{2\mathrm{E}\mathrm{I}}{\mathrm{l}}\left(2{\theta }_{\mathrm{A}}+{\theta }_{\mathrm{B}}+\frac{3\mathrm{\Delta }}{\mathrm{l}}\right)$

Calculation:

Fixed end moments:

${\mathrm{M}}_{\mathrm{F}\mathrm{A}\mathrm{B}}=\frac{-\mathrm{P}\mathrm{L}}{\mathrm{S}}=\frac{-60\times 4}{8}=-30\mathrm{k}\mathrm{N}\mathrm{m}$

M_FBA = 30 kNm

${\mathrm{M}}_{\mathrm{F}\mathrm{B}\mathrm{C}}=\frac{-\mathrm{W}{\mathrm{L}}^2}{12}=\frac{-30\times {\left(4\right)}^2}{12}=-40\mathrm{k}\mathrm{N}\mathrm{m}$

M_FCB = 40 kNm

We have slope deflection equation

${\mathrm{M}}_{\mathrm{B}\mathrm{C}}={\mathrm{M}}_{\mathrm{B}\mathrm{C}}^{\pi }+\frac{4\mathrm{E}\mathrm{I}}{\mathrm{L}}{\theta }_{\mathrm{B}\mathrm{C}}+\frac{2\mathrm{E}\mathrm{I}}{\mathrm{L}}{\theta }_{\mathrm{C}\mathrm{B}}$

${\mathrm{M}}_{\mathrm{B}\mathrm{C}}=-40+\frac{4\mathrm{E}\mathrm{I}}{4}{\theta }_{\mathrm{B}}+\frac{2\mathrm{E}\mathrm{I}}{4}{\theta }_{\mathrm{C}}$

$\therefore {\mathrm{M}}_{\mathrm{B}\mathrm{C}}=-40+\mathrm{E}\mathrm{I}{\theta }_{\mathrm{B}}+0.5\mathrm{E}\mathrm{I}{\theta }_{\mathrm{C}}$

Concept

Flexibility: It is deflection produced due to unit force.

Axial flexibility (δ ₁₁) = $\frac{1\times L}{A\times E}$

 subscript 11 denotes ( Deflection along (1) caused by Unit force in direction (1) )

similarly, subscript 12 denotes ( Deflection along (1) caused by Unit force in direction (2) )

Here , (1) is in the axial direction , (2) is in rotation 

Calculation

${\mathrm{f}}_{11}=\frac{\left(1\right)\times \mathrm{L}}{\mathrm{A}\mathrm{E}}=\frac{\mathrm{L}}{\mathrm{A}\mathrm{E}}$

Due to axial load there is no deformation in direction (2)

∴ f₂₁ = 0

 Due to unit rotation in direction (2) there is no deflection in direction (1)

∴ F₁₂ = 0

 ${\mathrm{f}}_{22}=\frac{\mathrm{L}}{\mathrm{E}\mathrm{I}}$

 $\left[\mathrm{f}\right]=\left[\begin{array}{cc}\frac{\mathrm{L}}{\mathrm{A}\mathrm{E}}& 0\\ 0& \frac{\mathrm{L}}{\mathrm{E}\mathrm{I}}\end{array}\right]$

Concept:

The given beam is indeterminate as we have 6 unknown reactions and only 3 equilibrium equation.

The moment developed at the fixed end due to applied forces is called a fixed end moment.

The fixed moment here is obtained using the standard equation and the superposition principle.

Let the end moments be M_AB and M_BA respectively at ends A and B.

The standard formula used is as shown below:

Calculating fixed end moments for each load individually:

${\mathrm{M}}_{\mathrm{A}\mathrm{B}1}=\frac{\mathrm{W}\times {\left(\frac{2\mathrm{L}}{3}\right)}^2\times \frac{\mathrm{L}}{3}}{{\mathrm{L}}^2}=\frac{4\mathrm{W}\mathrm{L}}{27}$

${\mathrm{M}}_{\mathrm{B}\mathrm{A}1}=\frac{\mathrm{W}\times \frac{2\mathrm{L}}{3}\times {\left(\frac{\mathrm{L}}{3}\right)}^2}{{\mathrm{L}}^2}=\frac{2\mathrm{W}}{27\mathrm{L}}$

Similarly for load 2,

${\mathrm{M}}_{\mathrm{A}\mathrm{B}2}=\frac{\mathrm{W}\times \frac{2\mathrm{L}}{3}\times {\left(\frac{\mathrm{L}}{3}\right)}^2}{{\mathrm{L}}^2}=\frac{2\mathrm{W}}{27\mathrm{L}}$

${\mathrm{M}}_{\mathrm{B}\mathrm{A}2}=\frac{\mathrm{W}\times {\left(\frac{2\mathrm{L}}{3}\right)}^2\times \frac{\mathrm{L}}{3}}{{\mathrm{L}}^2}=\frac{4\mathrm{W}\mathrm{L}}{27}$

Applying the superposition principle to add the effect of two loads individually,

${\mathrm{M}}_{\mathrm{A}\mathrm{B}}=\frac{4\mathrm{W}\mathrm{L}}{27}+\frac{2\mathrm{W}\mathrm{L}}{27}=\frac{2\mathrm{W}\mathrm{L}}{9}$

${\mathrm{M}}_{\mathrm{B}\mathrm{A}}=\frac{2\mathrm{W}\mathrm{L}}{27}+\frac{4\mathrm{W}\mathrm{L}}{27}=\frac{2\mathrm{W}\mathrm{L}}{9}$

Important Point:

Superposition principle:

As per the principle of superposition, each of the loadings produces its effect independent of other and total effect is the summation of effects due to individual loads.

For the principle of superposition to be valid:-

a) Stress should be proportional to strain.

b) Deformation should be small.

Concept

The flexural stiffness factor for rigid joint O is defined as the moment required to produce one-unit rotation at that joint.

Stiffness factor for various end conditions are specified below in the tabulated form:

Distribution Factor (D.F):

$\mathrm{D}\mathrm{F}=\frac{\mathrm{S}\mathrm{t}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}}{\mathrm{S}\mathrm{u}\mathrm{m}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{m}\mathrm{e}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\mathrm{j}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}}$

Calculation

Stiffness of member BA is given by (Far end fixed)

 $\frac{4E\left(2I\right)}{L}=\frac{8EI}{L}$

Stiffness of member BC is given by (Far end hinged)

  $\frac{3E\left(3I\right)}{2L}=\frac{9EI}{2L}$

Distribution factor of BA at $B=\frac{K_{BA}}{K_{BA}+K_{BC}}$

${\left(D.F\right)}_{BA}=\frac{\frac{8EI}{L}}{\frac{8EI}{L}+\frac{9EI}{2L}}=\frac{16}{25}$

∴ DF = 16/25

Case I: when members are considered extensible

At A : Three reaction components are available [F_x , F_y , F_z]

At C : Three reaction components are available [F_x , F_y , F_z]

At E : Six reaction components are available [F_x , F_y , F_z , M_x , M_y , M_z]

At H : Four reaction components are available [F_y , M_x , M_y , M_z]

D_k = 6j - r_e - h_r

r_e = 3 + 3 + 6 + 4 = 16

j = 8

h_r = 0 (because members are extensible)

D_k = 6 × 8 - 16 - 0 = 32

Case II: When members are inextensible

D_k = 6j - r_e - h_r

j → 8

r_e = 3 + 3 + 6 + 4 = 16

h_r = 4 [4 horizontal members]

D_k = 6 × 8 - 16 - 4

D_k = 28

$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}=\frac{32}{28}=1.14$

For the given truss:

Only the forces in half the members have to be determined since the truss is symmetric with respect to both loading and geometry.

Considering Joint A:

From equilibrium equation:

ΣF_y = 0

4 - F_AG sin 30° = 0

∴ F_AG = 8 kN (C)

ΣF_x = 0

F_AB - 8 cos 30° = 0

∴ F_AB = 6.928 kN (T)

Considering Joint G:

ΣF_y = 0

F_GB sin 60° - 3 cos 30° = 0

∴ F_GB = 3.00 kN (C)

ΣF_X = 0

8 - 3 sin 30° - 3 cos 60° - F_GF = 0

∴ F_GF = 5.00 kN (C)

Considering Joint B:

ΣF_y = 0

F_BF sin 60° - 3 sin 30° = 0

∴ F_BF = 1.73 kN (T)

ΣF_x = 0

F_BC + 1.73 cos 60° + 3 cos 30° - 6.928 = 0

Concept:

Slope deflection equation:

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

Calculation:

$\begin{array}{l}{M}_{FBC}=-\frac{w{l}^2}{12}=-\frac{15\times 8^2}{12}=-80kN-m\\ M_{BC}=M_{FBC}+\frac{2EI}{I}\left(2{\theta }_B+{\theta }_C-\frac{3\delta }{l}\right)\\ \Rightarrow M_{BC}=-80+\frac{2EI}{8}\left(2{\theta }_B+{\theta }_C\right)=0.25EI\left(2{\theta }_B+{\theta }_C\right)-80\end{array}$

Concept:

A 3 hinged arch is a statically determinate structure due to internal hinge at its crown.

The equation of trajectory of 3 hinged parabolic arches by considering support A as the origin is given by:

$\text{y }=\frac{4\text{h}}{{\text{L}}^2}\left(\text{x}\right)(\text{L}-\text{x})$

Where h = crown height, and L = horizontal length of the arch.

Calculation:

Equation of Parabolic circle:

$\text{y }=\frac{4\text{h}}{{\text{L}}^2}\left(\text{x}\right)(\text{L}-\text{x})$

For crown height (h) = 10 m and span (L) = 40 m

$\text{y}=\frac{4\times 10}{{\left(40\right)}^2}\times \text{x}\times (40-\text{x})=\frac{\text{x}\times (40-\text{x})}{40}$

Location of maximum bending moment (BM):

The maximum sagging B.M. will occur in the region where udl is applied i.e. region AC.

Cut the arch at (x, y) and draw its FBD of left portion:

∑M_x = 0 ⇒ M =750 x – 500 y – 25x²

$\text{M}=750\text{x}-500\left[\frac{\text{x}(40-\text{x})}{40}\right]-25{\text{x}}^2$

M = 12.5 x² + 950x – 25x²

∴ M = - 12.5 x² + 250 x

For BM to be maximum at minimum, $\frac{\text{dM}}{\text{dx}}=0$

$\frac{\text{dM}}{\text{dx}}=-25\text{x}+250=0\Rightarrow \text{x}=10\text{ m}$

$\frac{{\text{d}}^2\text{M}}{\text{d}{\text{x}}^2}=-25<0\Rightarrow \text{At }\text{ x}=10\text{ m }\text{ BM }\text{ will }\text{ be }\text{ maximum}$

Concept:

The given problem can be solved by using force method. The given beam has 1 degree of static indeterminacy, so we need only 1 more equation called compatibility equation in addition to equilibrium equations.

The overhang can be removed and replaced by its equivalent Moment and reaction at support B and hence, beam will behave like a cantilever beam with one end is fixed and other end is free and subjected to loads.

Consider reaction B is redundant and hence, it is removed and replaced by its reaction.

By compatibility equation, net vertical displacement of joint B is Zero.

Further, when a cantilever beam is subjected to load at its free end, the displacement occurred at its free is shown below:

Calculation

By compatibility

(Def. due to 40 kN and 80 kN-m) is equal to vertical deflection due to R­_B at B

$\frac{40\times 4^3}{3EI}+\frac{80\times 4^2}{2EI}=\frac{R_B\times 4^3}{3EI}$ 

⇒ R_B = 70 kN (↑) 

∑ M_A = 0 ⇒ M_A + 70 × 4 = 40 × 6

M_A = -40 kN-m

-ve sign implies M_A will be in clockwise dir.

∑ F_y = 0

R_A + 40 = 70

R_A = 30 kN

M_BC = 40 × 2

= 80 kN-m (↓)

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

Let the resultant of wheel load be at distance X̅ from the leading load (70 kN)

Taking a moment of loads along the leading load 70 kN.

$\overline{X}=\frac{70\times 0+150\times 1+60\times 1.5+120\times 2}{70+150+60+120}=1.2m$

Hence we get the loading diagram with the resultant as,

For getting more bending moment under a chosen wheel load, the resultant of wheel load and the chosen load should be at equal distance from the center.

In this question 150 KN load is maximum and nearer to resultant, hence maximum bending moment will occur under this load.

Maximum bending moment will be,

M$=\left(\frac{4.18}{5.1}\times 2.5\times 120\right)+\left(\frac{4.6}{5.1}\times 2.5\times 60\right)+\left(2.5\times 150\right)+\left(\frac{3.9}{4.9}\times 2.5\times 70\right)$

∴ M = 890 kNm

Concept:

There are two unknown i.e. rotation at joint B and C. Since Joint B is rigid, the rotation in span BA and BC will be same. Let the rotation at joint B and joint C is θB and θC.

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, consider the rotational equilibrium of joint B.

MBA + MBC = 0

Further, Slope at A is Zero because it is fixed support.

There is no sway, so ∆ = 0.

Calculation

$M{F}_{AB}=-\frac{W{L}^2}{12}=\frac{-30\times 2^2}{12}=-10kN-m$

$M{F}_{BA}=+\frac{W{L}^2}{12}=\frac{+30\times {\left(2\right)}^2}{12}=+10kN-m$

There is no load on span BC, so FEMS for BC and CB are zero.

Slope Deflections:

$M_{AB}=-10+\frac{2EI}{2}\left(+{\theta }_B\right)$

$M_{BA}=10+\frac{2EI}{2}\left(2{\theta }_B\right)$

$M_{BC}=0+\frac{2E\left(2I\right)}{4}\left(2{\theta }_B+{\theta }_C\right)$

$M_{CB}=0+\frac{2E\left(2I\right)}{4}\left(2{\theta }_C+{\theta }_B\right)$

Joint equilibrium of support B:

∑ MB = 0

MBA + MBC = 0

⇒ 10 + EI (20B) = EI (20B + θC) = 0

$4{\theta }_B+{\theta }_C=\frac{-10}{EI}$      ...*1)

MCB = Applied external moment

EI (2θC + θB) = 10

$\Rightarrow 2{\theta }_C+{\theta }_B=\frac{10}{EI}$     ...2)

On solving equation (1) and (2), we get

${\theta }_C=\frac{50}{7EI}(\leftarrow )$ Clockwise Direction

Concept:

Let movement of Joint C is _c and it is calculated using Castigliano’s theorem -II as shown below:

${\mathrm{\Delta }}_{\mathrm{C}}=\sum \frac{\mathrm{P}\mathrm{K}\mathrm{L}}{\mathrm{A}\mathrm{E}}$

Where,

P = Forces in all the truss members due to applied load,

K = Forces in all truss members due to unit load at C applied parallel to roller support

(∵ Only displacement possible is parallel to roller the support i.e. along direction AB)

L = Length of each truss member, and AE = Axial rigidity for each truss member.

The ‘P’ and ‘K’ values are evaluated using joint equilibrium method and tensile and compressive forces are assumed to be positive and negative respectively.

Calculations:

Calculation as ‘P’ Values

Let reaction developed at C is R_C and it will be perpendicular to roller.

$\sum {\mathrm{F}}_{\mathrm{x}}=\mathrm{O}\Rightarrow {\mathrm{H}}_{\mathrm{A}}-\frac{{\mathrm{R}}_{\mathrm{c}}}{\sqrt{2}}=10$

$\sum {\mathrm{F}}_{\mathrm{y}}=\mathrm{O}\Rightarrow {\mathrm{V}}_{\mathrm{A}}-\frac{{\mathrm{R}}_{\mathrm{c}}}{\sqrt{2}}=10$

$\sum {\mathrm{M}}_{\mathrm{A}}=\mathrm{O}\Rightarrow \frac{{\mathrm{R}}_{\mathrm{c}}}{\sqrt{2}}\times \sqrt{2}=\frac{10\times 1}{\sqrt{2}}+\frac{10\times 1}{\sqrt{2}}$

On solving above, we get, R_C = 10√2 kN, H_A = 20 kN, and V_A = 0 KN

Consider the Equilibrium of Joint (A)

∑F_y = 0, F_AB sin 45° = 0

⇒ F_AB = 0

∑F_x = 0 ⇒ F_AC = -20 KN (compressive)

Consider the equilibrium of joint (B)

∑F_x = 0 ⇒ F_BC = 10√2 KN (compressive))

Calculation of ‘K’ values:

∑ H_C = 0 ⇒ V_A = 0

$\sum {\mathrm{F}}_{\mathrm{y}}=0\Rightarrow \frac{{\mathrm{R}}_{\mathrm{C}}}{\sqrt{2}}=\frac{1}{\sqrt{2}}={\mathrm{H}}_{\mathrm{A}}$

H­_A = √2 kN

Consider equilibrium of Joint (A)

∑ F_y = 0, ⇒ F_AB = 0

∑ F_x = 0, F_AC = -√2 K_N (Compressive)

Consider equilibrium of Joint (B)

F_AB = 0, ⇒ F_BC = 0

Deflection at joint C $({\mathrm{\Delta }}_{\mathrm{c}})=\sum \frac{\mathrm{P}\mathrm{K}\mathrm{L}}{\mathrm{A}\mathrm{E}}=0+\frac{40}{\mathrm{A}\mathrm{E}}+0=\frac{40}{\mathrm{A}\mathrm{E}}$

AE = 4000 kN (given)

$\therefore {\mathrm{\Delta }}_{\mathrm{c}}=\frac{40}{4000}\times {10}^3=10\mathrm{m}\mathrm{m}$

Generation of second column of the stiffness matrix

Restrain the direction 1 and allow unit rotation at 2.

K₁₂ = force at 1 due to unit rotation at 2

$=\frac{4EI}{L}$

K₂₂ = Force at 2 due to unit rotation at 2

$=\frac{8EI}{L}$

Generation of first column of the stiffness matrix Restrain the joint 2 and apply unit rotation at 1

K21 = force at 2 due to unit rotation at 1.

$=\frac{4EI}{L}$

K11 = force at 1 due to unit rotation at 1

$=\frac{4EI}{L}+\frac{8EI}{L}=\frac{12EI}{L}$

∴ The stiffness matrix $\equiv \left[\begin{array}{cc}{K}_{11}& K_{12}\\ K_{21}& K_{22}\end{array}\right]$

$\begin{array}{l}=\left[\begin{array}{cc}\frac{12EI}{L}& \frac{4EI}{L}\\ \frac{4EI}{L}& \frac{8EI}{L}\end{array}\right]\\ =\frac{4EI}{L}\left[\begin{array}{cc}3& 1\\ 1& 2\end{array}\right]\end{array}$

Distribution Factor:

Fixed End Moments:

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

${\mathrm{M}}_{\mathrm{B}\mathrm{P}}=-\frac{\mathrm{W}{\mathrm{L}}^2}{12}=-\frac{24\times 8^2}{12}=-128\mathrm{k}\mathrm{N}.\mathrm{m}$

${\mathrm{M}}_{\mathrm{P}\mathrm{B}}=\frac{\mathrm{W}{\mathrm{L}}^2}{12}=\frac{24\times 8^2}{12}=128\mathrm{k}\mathrm{N}.\mathrm{m}$

${\mathrm{M}}_{\mathrm{P}\mathrm{C}}=-\frac{\mathrm{W}{\mathrm{L}}^2}{12}=-\frac{24\times 8^2}{12}=-128\mathrm{k}\mathrm{N}.\mathrm{m}$

${\mathrm{M}}_{\mathrm{C}\mathrm{P}}=\frac{\mathrm{W}{\mathrm{L}}^2}{12}=\frac{24\times 8^2}{12}=128\mathrm{k}\mathrm{N}.\mathrm{m}$

Distribution Table:

Hence the magnitude of bending moment at P is 176 kNm

Taking the left side column of the frame. i.e. member AB separately.

From Moment equilibrium equation: ∑M_B = 0

${\mathrm{M}}_{\mathrm{B}\mathrm{A}}-\frac{\mathrm{W}\times {\mathrm{L}}^2}{\mathrm{L}}-{\mathrm{M}}_{\mathrm{A}\mathrm{B}}-{\mathrm{H}}_{\mathrm{A}}\times \mathrm{L}=0$

${\mathrm{H}}_{\mathrm{A}}\times \mathrm{L}=-{\mathrm{M}}_{\mathrm{B}\mathrm{A}}+\frac{\mathrm{W}{\mathrm{L}}^2}{\mathrm{L}}+{\mathrm{M}}_{\mathrm{A}\mathrm{B}}$

${\mathrm{H}}_{\mathrm{A}}=\frac{{-\mathrm{M}}_{\mathrm{B}\mathrm{A}}}{\mathrm{L}}+\frac{\mathrm{W}\mathrm{L}}{2}+\frac{{\mathrm{M}}_{\mathrm{A}\mathrm{B}}}{\mathrm{L}}$

Taking the right-side column of the frame. i.e. member CD separately.

From Moment equilibrium equation: ∑Mc = 0

M_CD – M_DC – H_D × L = 0

${\mathrm{H}}_{\mathrm{D}}=\frac{{\mathrm{M}}_{\mathrm{C}\mathrm{D}}-{\mathrm{M}}_{\mathrm{D}\mathrm{C}}}{\mathrm{L}}$

Column shear equation: ${\mathrm{H}}_{\mathrm{A}}+{\mathrm{H}}_{\mathrm{D}}=\frac{{\mathrm{M}}_{\mathrm{A}\mathrm{B}}-{\mathrm{M}}_{\mathrm{B}\mathrm{A}}}{\mathrm{L}}+\frac{{\mathrm{M}}_{\mathrm{C}\mathrm{D}}-{\mathrm{M}}_{\mathrm{D}\mathrm{C}}}{\mathrm{L}}+\frac{\mathrm{W}\mathrm{L}}{2}$

Concept:

The maximum reaction at supports can be evaluated by drawing the Influence Line Diagrams (ILDs). The ILD can be drawn by using Muller Breslau Principal.

To draw the ILD of a reaction at any support, remove that support and lift the beam by one unit, then corresponding deflected shape of beam is itself the ILD of the beam for the support reaction.

In this case, to draw the ILD of a support reaction A, remove support A (do not remove support B and C) and lift the beam by one unit, then corresponding deflected shape of beam is itself the ILD of the beam for the support reaction at A as shown in calculation part.

In this case, to draw the ILD of a support reaction C, remove support C (do not remove support A and B) and lift the beam by one unit, then corresponding deflected shape of beam is itself the ILD of the beam for the support reaction at A as shown in calculation part.

Maximum Reaction = Intensity of Udl × Area under influence line diagram

Calculation:

 i) For Reaction A

ILD for reaction at A.

It is shown above, that to get maximum reaction at A, entire UDL has to be placed in b/w A and D.

∴ maximum Reaction at A, $V_{A,max}=\frac{1}{2}\left(1\times 4\right)\times 7=14kN$

 ii) For Reaction C

It is dear from above that to get maximum +ve reaction at ‘C’ uDL has to be placed such a way that area of ΔCdB is maximum and this is possible is uDL is placed from D’ to C.

Now, maximum +ve Reaction at C

  $V_{C,max}^{+}=\left[\frac{1}{2}\times \frac{4}{7}\times 4+\frac{3}{7}\times 4\right]\times 7=20kN$

∴ V_{C max} = 20 kN