Unit 2- Stresses in Beams

  Lecture -1 – Review of shear force and bending moment diagram

  Lecture -2 – Bending stresses in beams

  Lecture -3 – Shear stresses in beams

  Lecture -4- Deflection in beams

  Lecture -5 – Torsion in solid and hollow shafts.

Topics Covered

Theory of simple bending (assumptions)

  Material of beam is homogenous and isotropic => constant E in all direction

  Young’s modulus is constant in compression and tension => to simplify analysis

  Transverse section which are plane before bending before bending remain plain after bending. => Eliminate effects of strains in other direction (next slide)

  Beam is initially straight and all longitudinal filaments bend in circular arcs => simplify calculations

  Radius of curvature is large compared with dimension of cross sections => simplify calculations

  Each layer of the beam is free to expand or contract => Otherwise they will generate additional internal stresses.

Key Points:

1.  Internal bending moment causes beam to deform.

2.  For this case, top fibers in compression, bottom in tension.

Bending in beams

Key Points:

1.  Neutral surface – no change in length.

2.  Neutral Axis – Line of intersection of neutral surface with the transverse section.

3.  All cross-sections remain plane and perpendicular to longitudinal axis.

Bending in beams

Key Points:

1.  Bending moment causes beam to deform.

2.  X = longitudinal axis

3.  Y = axis of symmetry

4.  Neutral surface – does not undergo a change in length

Bending in beams

P A B

RB RA M M

Radius of Curvature, R

Deflected Shape

Consider the simply supported beam below:

M M

What stresses are generated within, due to bending?

Bending Stress in beams

Neutral Surface

M=Bending Moment

M M

Beam

σx (Tension)

σx (Compression)

σx=0

(i) Bending Moment, M (ii) Geometry of Cross-section

σx is NOT UNIFORM through the section depth

σx DEPENDS ON:

Axial Stress Due to Bending:

stress generated due to bending:

Neutral Surface

Bending Stress in beams

Bending Stress in beams

Stresses due to bending

N’ N’

R

E F

A’ C’

B’ D’

Strain in layer EF

€

=yR

€

E =Stress_ in _ the_ layer _ EFStrain _ in _ the_ layer_ EF

E =σyR⎛

⎝ ⎜

⎞

⎠ ⎟

€

σy

=ER

€

σy

=ER

€

σ =ERy

Neutral axis

N A

y dy

dA force on the layer=stress on layer*area of layer

€

=σ × dA

=ER× y × dA

Total force on the beam section

€

=ER× y × dA∫

=ER

y × dA∫

For equilibrium forces should be 0

€

y × dA = 0∫Neutral axis coincides with the geometrical axis

Stress diagram

x

σx

M M

σx

Moment of resistance

N A

y dy

dA force on the layer=stress on layer*area of layer

€

=σ × dA

=ER× y × dA

Moment of this force about NA

€

=ER× y × dA × y

=ER× y 2 × dA

Total moment M=

€

ER× y 2 × dA =

ER∫ y 2∫ × dA

Stress diagram

€

y 2 × dA∫ = I

€

M =ERI⇒ M

I=ER

x

σx

M M

σx

Flexure Formula

€

MI

=ER

=σy

Beam subjected to 2 BM

In this case beam is subjected to moments in two directions y and z. The total moment will be a resultant of these 2 moments.

You can apply principle of superposition to calculate stresses. (topic covered in unit 1).

Resultant moments and stresses

Section Modulus

Section modulus is defined as ratio of moment of inertia about the neutral axis to the distance of the outermost layer from the neutral axis

€

Z =Iymax

MI

=σy

MI

=σmaxymax

M =σmaxIymax

M =σmaxZ

Section Modulus of symmetrical sections

Source:- http://en.wikipedia.org/wiki/Section_modulus

Section Modulus of unsymmetrical sections

In case of symmetrical section neutral axis passes through geometrical center of the section. But in case of unsymmetrical section such as L and T neutral axis does not pass through geometrical center.

The value of y for the outermost layer of the section from neutral axis will not be same.

Composite beams Composite beams consisting of layers with fibers, or rods strategically placed to increase stiffness and strength can be “designed” to resist bending.

Composite beams

b

t t

y

d

€

σ1E1

=σ2E2

σ1 =E1E2σ2

= mσ2 m=modular ratio

€

M =σyI

M = M1 + M2

=σ1yI1 +

σ2yI2

=σ2ymI1 + I2[ ]

Equivalent I (moment of inertia)=

€

mI1 + I2