TAM 2xx References

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Pure Bending

Transverse loads applied to a beam causes deflections, primarily up or down is referred to as bending. Bending stresses depend on the beams cross-section, length, and material properties.

Sign Conventions

Internal
Fig: internalSignConvention
from reference pages
External
Fig: externalSignConvention
from reference pages

Boundary Conditions

Statically Determinate Beams
Fig: determinateBeams
from reference pages
Statically Indeterminate Beams
Fig: indeterminateBeams
from reference pages
Loadings
Fig: loadings1
from reference pages

Relations Among Load, Shear, and Mending Moments - does this go with Shear/Moment Diagrams?

Pure Bending

Fig: PureBending

Taken from TAM251 Lecture Notes - L6S11

Take a flexible strip, such as a thin ruler, and apply equal forces with your fingers as shown. Each hand applies a couple or moment (equal and opposite forces a distance apart). The couples of the two hands must be equal and opposite. Between the thumbs, the strip has deformed into a circular arc. For the loading shown here, just as the deformation is uniform, so the internal bending moment is uniform, equal to the moment applied by each hand.

Stress-Strain Variations

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Geometry of Deformation

Fig: Geometry

Taken from TAM251 Lecture Notes - L6S12

Assumptions:
  • Plane sections remain plane \( \rightarrow \) no shear stress/strains.
    $$ \gamma_{xy} = \gamma_{xz} = 0\ $$
    Therefore:
    $$ \tau_{xy} = \tau_{xz} = 0\ $$
    Also, traction free boundary conditions yields...
    $$ \sigma_y = \sigma_z = \tau_{yz} = 0\ $$
  • There is a Neutral axis between the top and the bottom where the length does not change.
    $$ \sigma_x = 0 \rm\ and \rm\ \epsilon_x = 0\ $$
  • The beam deforms into a circular arc where the top surface (\( AB \)) is in compression (\( \sigma_x, \varepsilon_x <0 \)), and the bottom surface (\( A \)) is in tension(\( \sigma_x, \varepsilon_x >0 \)).
  • Any point in the beam is in a state of uniaxial normal stress.
  • Finding stresses is a statically indeterminate problem.

Material Behavior: linear elastic beams

Elastic range: bending moment is such that the normal stresses remain below the yield strength. Hooke’s law combined with equilibrium gives: Defines longitudinal and neutral axes:
$$ \int_A y dA =0\ $$
Moment-curvature equation:
$$ M(x) = \frac{E(x)I_z(x)}{\rho(x)}\ $$

Expandable Derivation

Constitutive Relationship:

$$ L_{y_i} = L_{NAi} = L_{NAf} = \rho\theta\ $$
$$ L_{y_f} = (\rho - y)\theta\ $$
$$ \varepsilon_x = \frac{(\rho-y)\theta - \rho\theta}{\rho\theta}\ $$
$$ \varepsilon_x = \frac{-y\theta}{\rho\theta} = \frac{-y}{\rho}\ $$
$$ \sigma_x = E\varepsilon_x = -\frac{Ey}{\rho}\ $$

Force Equilibrium:

$$ \Sigma F_x = 0\ $$
$$ \Sigma F_x = \int dF\ $$
$$ \Sigma F_x = \int \sigma_x dA\ $$
$$ \Sigma F_x = \int\frac{-Ey}{\rho}dA\ $$
$$ \Sigma F_x = \frac{-E}{\rho}\int y dA = 0\ $$

Moment Equilibrium:

$$ \Sigma M_z = -M - \int y dF = 0\ $$
$$ \Sigma M_z = -M - \int y\sigma_x dA\ $$
$$ \Sigma M_z = -M - \int y (\frac{-Ey}{\rho}) dA\ $$
$$ \Sigma M_z = -M - + \frac{E}{\rho} \int y^2 dA = 0\ $$

**End Derivation**

First Moment of Area: Centroid of an Area

The first moment of the area A with respect to the z-axis is given by \( Q_z = \int_A y dA = \Sigma yA \) . The first moment of the area A with respect to the y-axis is given by \( Q_y = \int_A z dA = \Sigma zA \).

The centroid of an area is at the coordinates \( (\overline{x}, \overline{y}) \).

Fig: Centroid

Taken from TAM251 Lecture Notes - L6S19

$$ \bar{y} = \frac{1}{A}\int_A y dA\ $$
$$ \bar{z} = \frac{1}{A}\int_A z dA\ $$

Complex (or composite) areas can be divided into smaller, easier parts.

Fig: CompositeCentroid

Taken from TAM251 Lecture Notes - L6S19

$$ \bar{Y} = \frac{1}{A_{tot}}\sum_i (A_i\bar{y}_i)\] \[\bar{z} = \frac{1}{A_{tot}}\sum_i (A_i\bar{z}_i)\ $$

Second Moment or Area Moment of Inertia

\ The moment of inertia of the area A with respect to the z-axis is given by \( I_z = \int_A y^2 dA \). The moment of inertia of the area A with respect to the y-axis is given by \( I_y = \int_A z^2 dA \). Note: polar moment of inertia in this plane
$$ J = \int_A \rho^2 dA = \int_A (y^2 + z^2)dA = I_y + I_z\ $$
Fig: CommonShapes

From the formula sheet

Parallel-axis theorem: the moment of inertia about an axis through C parallel to the axis through the centroid C is related to \( I_C \) by
$$ I_C = I_{C'} +A'd_{CC'}^2\ $$

Maximum Normal Stress

From equilibrium: the centroid is located at \( \bar{y}=0 \), i.e., the neutral axis passes through the centroid of the section. Elastic Flextural Formula
$$ \sigma_x (x,y) = - \frac{M(x)y}{I_z(x)}\ $$

**Expandable Derivation**

$$ M_z = \frac{EI_z}{\rho}\ $$
$$ M_z = \frac{-\sigma_x}{y}I_z\ $$

**End Derivation**

To evaluate the maximum absolute normal stress, denoting “c” the largest distance from the neutral surface, we use:
Fig: MaxStress

Taken from TAM251 Lecture Notes - L6S17

$$ \sigma_{max} = \frac{|M|c}{I_z}\ $$
Note that the ratio I/c depends only upon the geometry of the cross section. This ratio is called the ELASTIC SECTION MODULUS and is denoted by S:
$$ \sigma_{max} = \frac{|M|}{S}\ $$
Where
$$ S = \frac{I}{c}\ $$

Composite Beams /!\ BSM: no longer covered in TAM 251, not sure if/when it is covered in ME curriculum.

Recall that \( \epsilon_x = -\frac{y}{\rho} \) does not depend on the material properties of the beam, and is based only on the assumptions of geometry done so far.

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In non-homogeneous beams, we can no longer assume that the neutral axis passes through the centroid of the composite section. We should now determine that location… After obtaining the TRANSFORMED CROSS SECTION, we get
$$ \int_{A_t}y d A_t = 0\ $$
Therefore, the neutral axis passes through the centroid of the transformed cross section. Note that the widening (\( n > 1 \)) or narrowing (\( n < 1 \)) must be done in a direction parallel to the neutral axis of the section, since we want y-distances to be the same in the original and transformed section, so that the distance y in the flexural formula is unaltered.
$$ \sigma_1 = -\frac{My}{I_t}\ $$
$$ \sigma_2 = -\frac{nMy}{I_t}\ $$

Eccentric Axial Loading in a Plane of Symmetry

Equilibrium gives:
$$ F=P \ \ \text{and} \ \ M=Pd\ $$

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Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due to a pure bending moment.

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Validity requires stresses below proportional limit (elastic region), deformations have negligible effect on geometry, and stresses not evaluated near points of load application.