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Use of Flexure Formula:
Illustrative Problems:
An I - section girder, 200mm wide by 300 mm depth flange and web of thickness is 20 mm is used as simply supported beam for a span of 7 m. The girder carries a distributed load of 5 KN /m and a concentrated load of 20 KN at mid-span.
Determine the
(i). The second moment of area of the cross-section of the girder
(ii). The maximum stress set up.
Solution:
The second moment of area of the cross-section can be determained as follows :
For sections with symmetry about the neutral axis, use can be made of standard I value for a rectangle about an axis through centroid i.e. (bd 3 )/12. The section can thus be divided into convenient rectangles for each of which the neutral axis passes through the centroid. Example in the case enclosing the girder by a rectangle
Computation of Bending Moment:
In this case the loading of the beam is of two types
(a) Uniformly distributed load
(b) Concentrated Load
In order to obtain the maximum bending moment the technique will be to consider each loading on the beam separately and get the bending moment due to it as if no other forces acting on the structure and then superimpose the two results.
Hence
Shearing Stresses in Beams
All the theory which has been discussed earlier, while we discussed the bending stresses in beams was for the case of pure bending i.e. constant bending moment acts along the entire length of the beam.
Concept of Shear Stresses in Beams :
By the earlier discussion we have seen that the bending moment represents the resultant of certain linear distribution of normal stresses x over the cross-section. Similarly, the shear force Fx over any cross-section must be the resultant of a certain distribution of shear stresses.
Derivation of equation for shearing stress :
Assumptions :
- Stress is uniform across the width (i.e. parallel to the neutral axis)
- The presence of the shear stress does not affect the distribution of normal bending stresses.
It may be noted that the assumption no.2 cannot be rigidly true as the existence of shear stress will cause a distortion of transverse planes, which will no longer remain plane.
In the above figure let us consider the two transverse sections which are at a distance ‘ x' apart. The shearing forces and bending moments being F, F + F and M, M + M respectively. Now due to the shear stress on transverse planes there will be a complementary shear stress on longitudinal planes parallel to the neutral axis.
Let be the value of the complementary shear stress (and hence the transverse shear stress) at a distance
‘ Y' 0 from the neutral axis. Z is the width of the x-section at this position
A is area of cross-section cut-off by a line parallel to the neutral axis.
= distance of the centroid of Area from the neutral axis.
Let + d are the normal stresses on an element of area A at the two transverse sections, then there is a difference of longitudinal forces equal to ( d . A) , and this quantity summed over the area A is in equilibrium with
the transverse shear stress on the longitudinal plane of area z x.
The figure shown below indicates the pictorial representation of the part.
