A propped cantilever beam A B of length L carries a concentrated load P acting at the position s

A propped cantilever beam A B of length L carries a...

Key Concepts

Static Equilibrium

Static equilibrium is a fundamental concept in mechanics where a structure remains at rest because the sum of forces and the sum of moments acting on it are zero. This principle is applied to determine unknown reactions at supports by setting up equilibrium equations that account for all applied loads, reaction forces, and moments.

Propped Cantilever Beam

A propped cantilever beam is a structural element that is fixed at one end (cantilevered) and simply supported at the other end (prop). This configuration induces both bending and shear stresses and provides both moment resistance and vertical support, resulting in a statically indeterminate system where additional compatibility equations may be required for analysis.

Concentrated Load

A concentrated load is a force that is applied at a single, specific point along a beam. This load causes a sudden change in the internal shear force diagram and contributes to bending moments along the beam, and its effects must be carefully considered when establishing the equilibrium conditions for reaction calculations.

Support Reactions

Support reactions are the forces and moments developed at the supports of a structure to maintain equilibrium under applied loads. In a propped cantilever beam, these include the vertical reactions at both the fixed and the simply supported ends, as well as an additional moment reaction at the fixed support, all determined using equilibrium of forces and moments.

Shear Force Diagram

The shear force diagram is a graphical representation showing how the shear force varies along the length of a beam. It is constructed by plotting the net vertical force at several points along the beam, with discontinuities typically occurring at points where loads are applied, offering insight into the internal force distribution.

Bending Moment Diagram

The bending moment diagram illustrates the variation of bending moments along a beam, reflecting the beam’s resistance to bending. It is derived by integrating the shear force diagram or applying moment equilibrium, with the locations of maximum and zero moments identified as critical for design and analysis.

Transcript

00:01 So here on this problem we are given a propped cantilever beam of length l and a considered load of magnitude piece acting at a distance of small a from the fixed support and we have to calculate the reactions are a r b and m a at a moment acting at point a for the given beam and we have also we would also draw the s fd and bmd and a bmd and we're to lay the critical ordnates now to solve this problem will be using the super imposition principle and if we consider that this support is not there then the deflection would be like this it would behave like this and now if we consider that this is not there but is acting on this image if we consider this is not acting then the deflection diagram would be like this now let us make more schematic diagram of the deflection so this is a deflection curve when we assume that the reaction force rb is not acting and this deflection delta b1 we can obtain from any of the book this formula we have p a square over 6e i times 3l minus a where l is the total distance of the beam this total distance l and a is the distance at which the point p is the at least the load p is acting similarly if we consider at the point load p is not acting the deflection would be in the upward direction and its formula would be equals to delta b 2 will be equals to r b the reaction here l cube over 3 ei now we can use a compatibility equation to calculate the value of r b and by using the compatibility equation we can say that the deflection at point b should be equals to zero so it will be equals to delta b1 minus delta b2 will be equals to zero so from here delta b1 we have this much so we can plug it here p a square times 3 l minus a divide by this is 6ei minus delta b2 we have this much so i can plug it here rb l cube over 3ei and it is equals to 0 so from here we can easily calculate the expression for r b so r b will be equals to p a square times 3 l minus a divide by 2 l cube so this is a required expression for r b now next i will be calculating the expression for our a and m a for that we'll be using the equilibrium condition of the beam now we'll be equating the forces in the vertical direction that is sigma fy is equals to zero now the forces in the downward direction will be taking it to be positive and in the upward direction will be taking it to be negative so we have minus r…

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