Loads get transmitted to floor beams after slabs.

Concept:

Influence line diagram:

An influence line for a given function, such as a reaction, axial force, shear force, or bending moment, is a graph that shows variation of that function at any given point on a structure due to application of a unit load at any point on structure. ILD can be drawn for statically determinate as well as indeterminate structures.

Advantages of drawing ILD are as follows:

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

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

Explanation:

Rolling load = 40 kN

Maximum positive shear force:

When rolling load will be acting at point A n shear force will be maximum.

Maximum shear force = Magnitude of load × ordinate of ILD under load

= 40 × 1 = 40 kN (+)

Maximum negative shear force:

When rolling load will be acting at point B n shear force will be maximum.

Maximum shear force = Magnitude of load × ordinate of ILD under load

= 40 × 1 = 40 kN (-).

On drawing curve based on above equation, we will find that maximum point will lie at C.

It would occur at same point as max. Shear, but loads should be applied at far or end of beam so as to maximize it.

Concept:

For simply supported beam of length l, 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/?

Calculation:

$\mathrm{h}=\frac{\frac{l}{4}\times \frac{3\mathrm{l}}{4}}{\mathrm{l}}=\frac{3\mathrm{l}}{16}$

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

$\therefore \mathrm{M}=\mathrm{W}\times \frac{3l}{16}$

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

$\therefore \mathrm{f}=\frac{\mathrm{W}\times \frac{3l}{16}}{\frac{\mathrm{B}{\mathrm{D}}^2}{6}}=\frac{9\mathrm{W}\mathrm{L}}{8\mathrm{B}{\mathrm{D}}^2}$

On applying Muller-Breslau principle, we will see that part right to point can?t move as point C is right next to it. So, ILD will have to remain parallel to x axis.

Portals are eir fixed, pin or partial supported.

Since, it will lie somewhere between end point and L/2, generally it is assumed to lie at a distance of L/3 from base.

For drawing ILD of shear force at section, any section or than support, Assume roller at that section and apply downward push on left arm of section and upward push to right section of arm. Hence assuming roller at E, Downward and Upward push to left and right of section respectively.

On comparing with more exact indeterminate analysis, portal method is more suitable for lower elevation.