Som 3

CHAPTER 3STRESSES IN

BEAMS

SURYAVANSHI PARTH Dept. MECH LJ Polytechnic

Stresses in beamsBasic definitions

Faculty Name Dept. LJ Polytechnic

Shear stresses in beams of rectangular cross-section

Vertical and horizontal shear stresses.

We can isolate a small element mn of the beam. There are horizontal shear stresses acting between horizontal layers of the beam as well as vertical shear stresses acting on the cross-sections

At any point in the beam, these complementary shear stresses are equal in magnitude

τ = 0 where y = ±h/2

Stresses in beams Basic definitions


Copyright 2005 by Nelson, a division of Thomson Canada Limited

FIG. 5-26

Shear stresses in a beam of rectangular cross section



A formula for the shear stress τ in a rectangular beam can be derived

Where V is the shear force, I is the moment of inertia and b is the width of the beam. Q is the first moment of the cross-sectional area above the level at which the shear stress τ is being evaluated. The shear formula can be used to determine the shear stress τ at any point in the cross-section of a rectangular beam



Copyright 2005 by Nelson, a division of Thomson Canada Limited

FIG. 5-28 Shear stresses in a beam of rectangular cross section

Stresses in beams Shear stress in beams


We can determine the distribution of the shear forces in a beam of rectangular cross-section

The distribution of shear stresses over the height of the beam is parabolic. Note that τ = 0 where y = ±d/2

The maximum value of shear stress occurs at the neutral axis (y1 = 0) where the first moment Q has its maximum value.

Where A = bd is the cross-sectional area

Stresses in beams Shear stress in beams


Stresses in beams Bending moment in beams


Beams are subjected to bending moment and shearing forces which vary from section to section. To resist the bending moment and shearing force, the beam section develops stresses.

Bending is usually associated with shear. However, for simplicity we neglect effect of shear and consider moment alone ( this is true when the maximum bending moment is considered---- shear is ZERO) to find the stresses due to bending. Such a theory wherein stresses due to bending alone is considered is known as PURE BENDING or SIMPLE BENDING theory.

Stresses in beams Bending moment in beams


BENDING ACTION Sagging : Fibers below the neutral axis (NA)

get stretched -> Fibers are under tension

Fibers above the NA get compressed : Fibers are in compressionHogging : Vice-versa

In between there is a fiber or layer which neither undergoes tension nor compression. This layer is called Neutral Layer (stresses are zero).The trace of this layer on the c/s is called the Neutral Axis.

Stresses in beams Assumptions in bending theory


Assumptions made in Pure bending theory

1) The beam is initially straight and every layer is free to expand or contract.

2) The material is homogenous and isotropic.3) Young’s modulus (E) is same in both tension

and compression.4) Stresses are within the elastic limit.5) The radius of curvature of the beam is very

large in comparison to the depth of the beam.

6) A transverse section of the beam which is plane before bending will remain plane even after bending. 7) Stress is purely longitudinal.

Stresses in beams Bending stress in beams


Consider the beam section of length “dx” subjected to pure bending. After bending the fiber AB is shortened in length, whereas the fiber CD is increased in length. In b/w there is a fiber (EF) which is neither shortened in length nor increased in length (Neutral Layer).Let the radius of the fiber E'F′ be R . Let us select one more fibre GH at a distance of ‘y’ from the fiber EF as shown in the fig. EF= E'F′ = dx = R dθThe initial length of fiber GH equals R dθAfter bending the new length of GH equals



G'H′= (R + y) dθ= R dθ + y dθ



Change in length of fiber GH = (R dθ + y dθ) - R dθ = y dθ Therefore the strain in fiber GH Є= change in length / original length= y dθ/ R dθ Є = y/RIf σ ь is the bending stress and E is the Young’s modulus of the material, then strain Є = σ ь/E

σ ь /E = y/R => σ ь = (E/R) y---------(1)σ ь = (E/R) y => i.e. bending stress in any fiber is proportional to the distance of the fiber (y) from the neutral axis and hence maximum bending stress occurs at the farthest fiber from the neutral axis.



Consider an elemental area ‘da’ at a distance ‘y’ from the neutral axis. The force on this elemental area = σ ь × da

= (E/R) y × da {from (1)}

The moment of this resisting force about neutral axis =

(E/R) y da × y = (E/R) y² da



Total moment of resistance offered by the beam section,

M'= (E/R) y² da= E/R y² da

y² da =second moment of the area =moment of inertia about the neutral axis.

\ M'= (E/R) INA

For equilibrium moment of resistance (M') should be equal to applied moment Mi.e. M' = MHence. We get M = (E/R) INA



(E/R) = (M/INA)--------(2)From equation 1 & 2, (M/INA)= (E/R) = (σ ь /y) ---- BENDING EQUATION.(Bernoulli-Euler bending equation)

Where E= Young’s modulus, R= Radius of curvature, M= Bending moment at the section, INA= Moment of inertia about neutral axis, σ ь= Bending stressy = distance of the fiber from the neutral axis

Stresses in beams Section modulus


(M/I)=(σ ь /y) or σ ь = (M/I) yIts shows maximum bending stress occurs at the greatest distance from the neutral axis. Let ymax = distance of the extreme fiber from the N.A. σ ь(max) = maximum bending stress at distance ymax

σ ь(max) = (M/I) y max where M is the maximum moment carrying capacity of the section, M = σ ь(max) (I /y max) M = σ ь(max) (I/ymax) = σ ь(max) ZWhere Z= I/ymax= section modulus (property of the section) Unit ----- mm3 , m3

Som 3

Engineering