Structural Analysis Test 1

In the virtual work method of plastic analysis of steel structure, the virtual quantity is displacement.

Virtual work arises in the application of the principle of least action to the study of forces and movement of a mechanical system. The work of a force acting on a particle as it moves along a displacement will be different for different displacements. Among all the possible displacements that a particle may follow, called virtual displacements, one will minimize the action. This displacement is therefore the displacement followed by the particle according to the principle of least action. The work of a force on a particle along a virtual displacement is known as the virtual work.

Principle of virtual work: (unit-load Method)

Developed by Bernoulli:

To find Δ at point A du to loads P₁, P₂, P₃. Remove all loads, apply virtual load P’ on point A.

For simplicity P’ = 1

It creates internal load u on representative element.

Now remove this load, apply P₁, P₂, P₃ due to which pt. A will be displaced by Δ

∴ External virtual work = 1.Δ

Internal virtual work = u.dL

P^' = 1 = external virtual unit load in direction of Δ.

u = internal virtual load acting on element in direction of a dL.

Δ = external displacement caused by real loads.

dL = internal deformation caused by real loads.

The Maxwell reciprocal theorem, states that the deflection at a point A in the direction of  1 due to load at point B in the direction of 2 is equal in the magnitude to the deflection of point B in the direction of 2 produced by a load applied at A in direction 1.

Hence,

δ₁₂ = δ₂₁

The beam has internal hinges at C & D, so the beam can be break down as shown

In AC section moment applied = M

So the reaction will be = M/L/2 = 2/ML

The carry – over moment at B = 2M/L × L/4 = M/2

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.

Deflection, δ = ∂U/∂F

In which beam deflection is higher, strain energy stored will be maximum.

In cantilever, the deflection will be highest.

The deflection of a simply supported beam is higher than propped cantilever as one of the support is fixed.

In a fixed beam, the deflection will be minimum.

So,

δ_cantilever >  δ_{simple-support} > δ_{propped cantilever} > δ_fixed

∴ U_fixed < U_{proped-cantilever} < U_{simple-support} < U_cantliver

I^st condition:

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

\({\rm{y}}{{\rm{c}}_1} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{2{\rm{L}}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{2{\rm{L}}}}{3}} \right)^2} \times \frac{{\rm{L}}}{3} = {\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}^3}\)

II^nd condition:

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

\({\rm{y}}{{\rm{c}}_2} = \frac{{\rm{W}}}{{3{\rm{EI}}}} \times {\left( {\frac{{\rm{L}}}{3}} \right)^3} + \frac{{\rm{W}}}{{2{\rm{EI}}}}{\left( {\frac{{\rm{L}}}{3}} \right)^2} \times \frac{{2{\rm{L}}}}{3} = \frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}\)

\({\rm{Ratio\;}}\left( {\rm{r}} \right) = \frac{{{\rm{y}}{{\rm{c}}_1}}}{{{\rm{y}}{{\rm{c}}_2}}} = \frac{{{{\frac{{28{\rm{WL}}}}{{162{\rm{EI}}}}}^3}}}{{\frac{{8{\rm{W}}{{\rm{L}}^3}}}{{162{\rm{EI}}}}}} = \frac{{28}}{8} = \frac{7}{2}\)

\(\therefore \frac{{{\rm{y}}{{\rm{c}}_2}}}{{{\rm{y}}{{\rm{c}}_1}}} = \frac{2}{7}\)

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

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

Concept :-

Stiffness value, K when far end is fixed = 4EI/L

Stiffness value, K when far end is roller = 3EI/L

Stiffness value, K when far end is guided roller = EI/L

For stiffness coefficient when far end is guided roller support refer :-

Advanced Structural Analysis, Prof. Devdas MenonDepartment of Civil Engineering, Indian Institute of Technology, Madras

Module - 5.3 ,Lecture - 29Matrix Analysis of Beams and GridsPage No. - 9

D_k = 3J – r_e + r_r

J = No. of joints

r_e = External support Reactions

r_r = Released Reactions

r_e = 7

J = 6

r_r = m – 1

m = No. of members meeting at internal hinge

At B:

r_r = 2 – 1 = 1

At D:

r_r = 2 – 1 = 1

Therefore,

D_k = 3 × 6 – 7 + 2

D_k = 13

D_si = m – (2j – 3)

m = 20

j = 10

D_si = 20 – (2 × 10 – 3) = 3

D_se = r_e – 3 = 4 – 3 = 1

Total indeterminacy = 3 + 1 = 4

As loading is UDL and the shape of the arch is parabolic so there will be no moment in the arch rib.

But as the temperature increases in a two hinged arch (degree of indeterminacy one), the horizontal thrust will increase.

Moment due to horizontal thrust is – Py.

So maximum bending moment will be at crown as crown has highest value of y. 

Stiffness (k) is defined as the force per unit deflection.

K = P/δ 

⇒ δ = P/K

If stiffness matrix is doubled

i.e. K’ = 2K

then,

δ = P/K'

δ = P/2K

δ = (1/2)P/K

δ’ = δ/2.  

Therefore, if the stiffness matrix is doubled with respect to the existing stiffness matrix, the deflection of the resulting frame will be half the existing value.

For k₃₃ Apply unit displacement along 3 and restraining the displacement in 2 and 1

To find the vertical deflection at joint E, a unit vertical load is applied at joint E and the forces in member AB and CD are to be found out.

F_AB sin 60° = 0.5  

F_AB = 0.577 (compressive)

Similarly, F_CD = 0.577 (compressive)

due to increase in temperature, the increase in length of AB & CD = L ∝ t

= 3 × 12 × 10^-6 × 40

= 0.00144 m

So the deflection of joint E,

δ = Σ kδ’

where, k = Force produced in each member when a unit downward load is applied at joint E.

           δ’ = Deflection produced in the concerned member (AB & CD)  due to temparature rise.

= (- 0.577) × (0.00144) + (- 0.577) × (0.00144)

= - 0.00166 m

= 1.66 mm (upward)

[‘–’ means our assumed downward deflection is wrong, it will deflect upward]