# An introduction to Mechanics of solids, Lecture notes for Mechanics. Indian Institute of Technology (IIT)

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ESO204: Mechanics of Soilds

ESO204: Mechanics of Solids

Stresses due to bending

Strain energy

 The strain energy per unit volume in presence of axial stress σx is ½ σxεx

Strain energy

 The strain energy per unit volume in presence of shear stress τxy is ½ τxyγxy

Strain Energy in Pure Bending  The only non-zero stress in pure bending is

the stress σx. Since the strain energy per unit volume in presence of axial stress σx is ½ σxεx, the strain energy over the entire beam length can be written as:

 U = ∫V ½ σxεxdV = ∫A ∫L ½ M2/EI2 y2dAdx

 U = ∫L ½ M2/EI dx

Strain Energy for beams subjected to bending moment and shear

 For beams carrying shear loads, the non- zero stresses are the stress σx, and the shear stresses τxy and τxz . The total strain over the volume is:

 U = ∫L ½ M2/EI dx + ∫V ½ τxyγxy dV

+ ∫V ½ τxzγxz dV Usually the energy contribution due to shear

is small and is often neglected.

Deformation of Elastic Beams  We have already

obtained:  dΦ/ds= 1/ρ = Mb/E Izz  Substituting for radius

of curvature in terms of deformation:

 d2v/dx2= Mb/E Izz  Where v is the

transverse displacement of the neutral axis.

Example: Beam Deflection using Castigliano's Theorem

 Find the deflection of a cantilever beam subjected to a tip load P and a distributed loading w per unit length

Example: Beam Deflection using Castigliano's Theorem

Example: Beam Deflection using integration of moment curvature relation

 Find the deflection of a cantilever beam subjected to a tip load P and a tip bending moment M0.

Cantilever beam subjected to tip load and bending moment

EI d2v/dx2 = M = -M0- P(L-x)

EI dv/dx = -M0x -P(Lx-x2/2)+c1 EIv = -M0x2/2- P(Lx2/2-x3/6)+c1x+c2 Boundary Conditions At x=0, v=0 and dv/dx=0 Substituting, we get: c1=c2=0

Cantilever beam subjected to tip load and bending moment

 The general expressions for deflection and slope are:

EIv = -M0x2/2- P(Lx2/2-x3/6)

EI dv/dx = -M0x-P(Lx-x2/2)  At x=L , the values are: (v)x=L = -[M0L2/2+ PL3/3]/EI

(dv/dx)x=L = -[M0L+ PL2/2]/EI

Negative sign of v denotes that the deflection is in the negative y direction.

Cantilever beam subjected to tip load and bending moment

Example: Beam Deflection using integration of moment curvature relation

 Find the deflection of a simply supported beam of length L subjected to a load W applied at a distance a from the left end.

Singularity Functions  The singularity function <x-a>n are

defined as: <x-a>n = 0 if x<=a = (x-a)n if x>=a Differentiation of singularity

functions is similar to regular differentiation i.e.

d/dx <x-a>n = n <x-a>(n-1)

Singularity function <x-a>0

 The singularity function <x-a>0 (also called unit step function u(x-a)) is defined as:

<x-a>0 = 0 if x<=a = (x-a)0 i.e. 1 if x>=a

Singularity function <x-a>1

 The singularity function <x-a>1 is defined as: <x-a>1 = 0 if x<=a = (x-a)1 i.e. (x-a) if x>=a

Singularity function <x-a>2

 The singularity function <x-a>2 is defined as: <x-a>2 = 0 if x<=a = (x-a)2 if x>=a

Unit Impulse function  The analytical expression for the unit

impulse function is denoted as δ(t) where δ(t) = unbounded when t=o δ(t) =0 when t is not equal to zero  The impulse function is abstraction of the

pulse with an infinitely large amplitude and infinitesimally small pulse width. The unit impulse can be visualized as a pulse with amplitude of 1/ε and width of ε , or as a triangle.

Unit Impulse function  The unit

impulse can be visualized as a pulse with amplitude of 1/ε and width of ε , or as a triangle. Note that δ(t) is unbounded when t=o

Relation between unit impulse function and unit step function

Summary of singularity functions

Deflection of a simply supported beam

 EId2v/dx2= Mb= Wbx/L – W<x-a>1

 EIdv/dx=Wbx2/(2L)- W<x-1>2/2+c1  EIv=Wbx3/(6L)- W<x-1>3/6+c1x+c2 Where c1 and c2 are constants of integration.  Geometric Boundary Conditions: V=0 at x=0 and x=L  0=c2  0=WbL3/(6L)-Wb3/6 + c1L  V=-W/(6EI) [bx(L2-b2-x2)/L +<x-a>2]

Superposition  If solutions to a number of deflection

problems involving simple conditions of load and support are available, then a convenient method is the method of superposition.

 We have linearity between applied load and bending moment; also linearity between deflection and bending moment.

 We can superpose deflections due to a combination of loads and moments.

Standard Results for deflection

Standard Results for deflection