Theory of Simple Bending

Pure Bending: Only B.M. but no S.F.

Neutral Layer : The layer which does not under go any change in length (n.a).

Neutral Axis : Line of intersection of Neutral Layer with plane of cross section.

• It passes through C.G. of cross section.
• The axis where the strain changes its sign.
  

Equation of Pure Bending:

EI = Flexural rigidity

M/I = f/y= E/R .

Curvature= (I/R) = (M/EI)

ftop = (M/I).yt,

fbottom = (M / I). Yb

Assumptions:

• The transverse sections which are plane and normal before bending remain plane and normal to the longitudinal fibres after bending (Bernoulli’s Assumption).
• Material is homogeneous, isotropic and obeys Hook’s Law and limits of eccentricity are not exceeded.
• Every layer is free to expand or contract.
• Modulus of elasticity has same value for tension and compression
• The beam is subjected to pure bending and therefore bends in an arc of a circle.
• Radius of curvature is large compared to the dimensions of the cross section.

Other Important Points:

• The strain of a fibre is proportional to its distance from neutral axis e = y / R.
• The sum of all the compressive forces above neutral layer must be equal to the sum of tensile forces below neutral layer.

 Section Modulus:

• M = f. (1/Y) = f x moment of inertia/ Distance of extreme fibre = fz
• Z = (I / Y) = Section Modulus.

Significance: It represents the strength the of section. Greater the value of ‘z’, stronger will be the section.

Illustration: Solid circular section of equal weight when compared to other section such as rectangle, I - section and hollow circular sections uneconomical in bending as its section modulus ill be small and consequently strength in bending.

Beam of Uniform Strength: If extreme fibre stresses (or maximum bending stress) is same at every section along its longitudinal axis, it is called a beam of uniform strength.

Necessary Condition: Z a M at every section.

S.S.Beam with a point load ‘W’ at centre: To have uniform strength, by

• Keeping width constant  and varying the  depth,
  

Parabolic variation for ‘d’.

• Depth constant, width varying:
  

NOTE: Please workout the above values for u.d.l.

Beams of Composite Sections:

Section made of different materials: For determining moment of inertia of composite section, the section is to be transformed into equivalent section using modular ratio.

Ex:          Let (EB / EA) = m

The equivalent width = (b + 2 m. t)

Flitched Beam: A timber beam strengthened by steel strips.

M.R = fw.(Zeq)

        = fw.[(b + 2mt) d2/6]

where, fs =  m.fw

(strain will be same in both the materials due to bending.)

Two beams one over another behaving, independently (No rigid connection): Each bar bends about its own axis. ‘R’ for both the sections is assumed approximately equal.

R = (EbIb / Mb) = (EsIs / Ms)

Ms/Mb = (EsIs / EbIb)

If both are of same material: (f1/f2) = (t1/t2)

NOTE: If there is a rigid connection, they will behave like one piece and bending take place about combined axis.

Important Practical Observations:

• The ratio of moment of resistance of a beam of square section placed with two sides horizontal to a diagonal horizontal is (2)1/2
• A beam of rectangular section is to be cut from a circular log of dia ‘d’. The ratio of depth to width for maximum strength in pure bending is (2)1/2 :1 

• The M.R. of a beam of square section with the diagonal in the plane of bending is by flattening the top and bottom corners as shown in figure.

                                        Y1 = 1/9Y

For this % increase in section modulus Z = 5.35%.

Bending stress would be reduced by 5.35%.

• Circular Section: Depth of cutting for improving

             Z=0.011 d.,

              ΔZ = 0.7%

• Three beams have the same length, same allowable bending stress and are subjected to same maximum bending moment. The cross section of the beams are a circle, a square and a rectangle with depth twice the width.

Ratio of weights of (rectangle .. square .. circle) = (0.8 : 1 : 1.12)