Influence Line Diagram and Rolling Loads MCQ Quiz in मराठी - Objective Question with Answer for Influence Line Diagram and Rolling Loads - मोफत PDF डाउनलोड करा

Explanation:

Muller-Breslau Principle:

1. It states that "If an internal stress component or a reaction component is considered to act through some small distance and thereby to deflect or displace a structure. The curve of deflected structure will influence line stress or reaction component".
2. It is used to draw influence line diagrams for determinate and indeterminate structures.
3.  ILD for determinate structure is linear while that for the indeterminate structure will be curvilinear.
4. This method is based on the concept that Influence lines are deflection curves.
5. It is a straight application of Maxwell's reciprocal theorem.
6. ILD is a graphical representation of variation in the reactions, Shear force, and bending moment at each and every section when unit load moves from one to another end of the structure. They can be drawn for any type of structure – beams, arches, trusses, etc.

For example, to draw the influence line diagram for a vertical reaction at A in a simply supported beam as shown below.

Remove the ability to resist movement in the vertical direction at A by using the guided roller

Concept:

Suppose 4 loads P1, P2, P3, P4 move in a series. Maximum bending moment will occur under P2 when the centre of span is midway between the centre of gravity of the load system R and the wheel load P2.

A maximum bending moment under the wheel load is always below some point load. To find maximum bending moment below a point load, the point load should be placed so that the resultant load and the point load are equidistant from the mid-span.

To find absolute bending moment following steps are followed:

Step 1: Find the resultant of all the loads.

Step 2: Place the individual load as per the condition of absolute maximum bending moment and compute the bending moment under the concentrated load.

Step 3: The maximum of all the possible cases will give the absolute maximum bending moment.

Here the maximum bending moment below the load is obtained.

Important point:

The influence line diagram represents the variation of the stress function at a specified point in a member as a concentrated load moves over the member.

To find the value of stress function, the value of the ordinate of ILD at the location of point load is multiplied by the value of point load.

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 Piecewise linear and nonlinear respectively.

Explanation:

ILD for the shear force at fixed end A of the cantilever beam

⇒ from Muller Principle: To draw ILD for the shear force at R. H. S. of  A without releasing fixed end moment and the support at A, cut the beam at A, and give unit displacement.

Then the deflected shape of the beam itself is the ILD for the shear force at R. H. S. of A.

Concept:

For simply supported beam of length l, the maximum value of ILD for bending moment at a distance ‘a’ from left support and at a distance ‘b’ from right support, is given by ab/ℓ

Calculation:

\({\rm{h}} = \frac{{\frac{l}{4} \times \frac{{3{\rm{l}}}}{4}}}{{\rm{l}}} = \frac{{3{\rm{l}}}}{{16}}\)

Now, maximum bending moment at ℓ/4 is W × h

\(\therefore {\rm{M}} = {\rm{W}} \times \frac{{3l}}{{16}}\)

Given: Breadth = B, Depth = D, Section Modulus (z) = BD²/6, and Flexural stress (f) = M/Z

\(\therefore {\rm{f}} = \frac{{{\rm{W}} \times \frac{{3l}}{{16}}}}{{\frac{{{\rm{B}}{{\rm{D}}^2}}}{6}}} = \frac{{9{\rm{WL}}}}{{8{\rm{B}}{{\rm{D}}^2}}}\)

Concepts:

Case 1: When UDL Longer than Span:

When a uniformly distributed load longer than the span of the simply supported beam moves from left to right, the maximum bending moment at mid-span exists if the load occupies the whole span.

Case 2: When UDL is shorter than Span

When a uniformly distributed load shorter than the span of the simply supported beam moves from left to right, the maximum bending moment at any point (say C) occurs when the load is kept such that when average loading on the left side of C is equal to the average loading on right side of C. It means section C will divide load in the same ratio as it divides the span.

Let us take a beam of length L in which a uniformly distributed load of length L' (L' < L) moves from left to right.

To get Maximum BM at 'c', UDL is placed such that.

\(\rm \frac{wx}{a}=\frac{w(L'-x)}{b}\) a + b = L

Explanation:

At any section C, influence line ordinate for negative shear is (z/L) and positive shear is \(\left( {\frac{{L - z}}{L}} \right)\). Hence, when z = 0 i.e., at support A, ILD ordinate for positive support B ILD ordinate for negative shear force force is maximum (= 1) and when z = L i.e. at support B, ILD ordinate is maximum (= 1) for shear force at support section A and B are as shown in figure.

Obviously, maximum shear force occurs when the load is on support.

Explanation

The influence line diagram for the reaction at B is shown below:-

To get the maximum reaction at B, avg. load on AD = avg. load on DC

\(\begin{array}{l} \frac{x}{{12}} = \frac{{5 - x}}{4}\\ x = 3.75\ m \end{array}\)

For right side: Ordinate of ILD at end of the udl load → \(\rm{=(4-1.25)\times \frac{1.5}{4}=\frac{1.5\times 2.75}{4}=1.03}\)

For right side: Ordinate of ILD at end of the udl load → \(\rm{=(12-3.75)\times \frac{1.5}{12}=\frac{1.5\times 8.25}{12}=1.03}\)

So the maximum reaction at B will be

\(= \frac{1}{2} \times \left( {3.75} \right) \times \left( {1.03 + 1.5} \right) \times 3 + \frac{1}{2} \times \left( {1.25} \right) \times \left( {1.03 + 1.5} \right) \times 3\)

= 18.975 KN

Concept:

Above is the simply supported beam AB. A section x-x is taken at x distance from left side A.

A unit load is applied and unit rotation is provided at x-x section. Then, BMs due to rolling action of load at different section is given by:

\({{\rm{M}}_{\rm{x}}} = {{\rm{R}}_{\rm{a}}} \times {\rm{x}} - 1 \times {\rm{x}} \Rightarrow {{\rm{M}}_{\rm{x}}} = \left( {{{\rm{R}}_{\rm{A}}} - 1} \right){\rm{x}}\)

\({{\rm{R}}_{\rm{A}}} = \frac{{{\rm{L}} - {\rm{x}}}}{{\rm{L}}}\)

\({{\rm{M}}_{\rm{x}}} = \frac{{{\rm{x}}\left( {{\rm{L}} - {\rm{a}}} \right)}}{{\rm{L}}}\)

X = L, Mc = 0 and at x = 0, Mc = 0

The above equation is a two degree equation in terms of x and hence and shape ILD for BM will be parabolic

An influence line for a given function, such as a reaction, axial force, shear force, or bending moment, is a graph that shows the variation of that function at any given point on a structure due to the application of a unit load at any point on the structure.

The ordinates of influence line diagram for bending moment always have the dimension of length.