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BENDING STRESSES IN BEAMS
THEORY OF PURE BENDING
Following results may be drawn form the observation of pure bending:
However they remain straight after the deformation.
This means plane section before bending remains plane after bending.
Let s be the longitudinal stress produced by bending then we have,
Assumptions made in the theory of simple bending: (i) The beam is initially straight. (ii) The beam has constant cross sectional area, with an axis of symmetry. the axis of symmetry is vertical. (iii) The material is homogeneous & isotropic (iv) Hooke's law is valid. (v) Plane transverse section remains plane after bending. (vi) Every layer of material is free to expand or contract longitudinally and laterally under stress and do not exert pressure upon each other. Thus the poison's effect and the interface of the adjoining differently stressed fibre are ignored. (vii)The value of Young's modulus (E) for the material is the same in tension and in compression
Standard result to be noted:
Hence arrangement (I) is 41.4% more strong than (II).
out of a circular log of diameter 'D' is b = b= d/√2
For maximum strength, Z should be maximum
Hence
b= d/√2
• Three beams have the same length, same allowable bending stress and are subjected to the same maximum bending moment. The cross section of the beams are a circle, a square and rectangle with depth twice the width. Find the ratio of the weights of the circular and rectangular beams with respect to the square beams.
Let the circular section be of diameter d
Let the square section be of side x
Let the rectangular section be of width b & depth 2b.
Since M = sZ
Hence, for same bending moments under same bending stresses, the section modulus
should be equal for all three beams.
Þ d = 1.193 x
b = 0.6299 x
W _{rectangular }< W _{square} < W _{circular}
Following points may be noted with reference to above problem.
(i) For maximum strength (Z_{max}), The portion EB to be cutoff is given by
(ii) Increase in the section modulus
(i) For a circular section it is possible to increase the section modulus by 0.7% by cutting the segmental portions shaded.
(ii) The depth of segment d should be 0.011 times the diameter of the section.
FLITCHED BEAMS
Flitched beam is a composite section consisting of a wooden beam strengthened by metal plates. The arrangement is so connected that the components act together as one beam.
The modular ratio m =
MOMENT OF RESISTANCE OF THE SECTION:
Case– 1 : Flitched are attached symmetrically at sides:
Flitched beam section Equivalent wooden section
Let M_{r} be the moment of resistance of the section.
Let M_{w} & M_{s} be the moments of resistance of wood & steel sections respectively.
Let wooden joint be b unit wide & d unit deep.
Let each steel plate be t units thick and d units deep.
Let s_{w} & s_{s} be the extreme stresses in wood & steel respectively.
M_{r} = M_{w} + M_{s}
Hence the moment of resistance of the section is the same as that of a wooden member of breadth b + m (2t) and depth d. The rectangular section b + m (2t) units wide and d units deep is called the equivalent wooden section.
Case  2 : Flitches are attached symmetrically at the top and bottom.
In this case the moment carrying capacity is greater than the moment carrying capacity of side flitches.
BEAMS OF UNIFORM STRENGTH
For economical design, the section of the beam may be reduced towards the support a since bending moment decrease towards the supports. The beam may be designed such that at every section, the extreme fibre stress reaches the permissible stress.
A beam so designed is called a beam of uniform strength. There are three possible cases for such design.
(1) Varying width by keeping constant depth.
(2) Varying depth by keeping constant width.
(3) Varying width & depth both.
Case1 : Beam of uniform strength  Constant depth
Let the permissible stress be s
• The width of the beam at x is given by
• The width of the beam at mid span is given by
Case 2: Beam of Uniform strength Constant width.
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