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Second Moment of Area and Bending Stress

5ME517 Structural Analysis

and Materials

Dr Amal Oraifige a.oraifige@derby.ac.uk

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Bending Stress • The bending stress within the cross

section is a resistant distribution across the neutral axes to the acting bending moment.

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Bending Formula Therefore, the complete formula which describes all aspects of bending is:

y M = σ I = E

R

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Bending Formula

y M = σ I = E

R σ = Bending Stress (N/m2) (Pascal) M = Bending Moment (Nm) y = Distance to Neutral Axis (m) I = Second Moment of Area (m4) E = Young’s Modulus (N/m2) (Pascal) R = Radius of Curvature (m)

y

Beam Loadings

Cross Section

Material’s Properties

Arc of Bending

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Bending Stress Formula

I My = σ y

M = σ I

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Maximum Bending Moment “M”

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The bending moments in beams are produced by the action of applying

perpendicular load/force to the beam.

Bending Moment (last week)

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Max. Bending Moment “+M” A Sagging moment produces the following profile with compression stress on the top and tensile stress on the bottom.

Neutral Axis

F

Tension

Compression

POSITIVE SAGGING

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Bending Stress Distribution due to the Bending Moment The stress distribution will be the resistance to the bending moment distribution in order for the beam to stay in equilibrium. Therefore:

Sagging moment will have positive (tension stress

at the top and negative (compression) stress at the bottom.

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Bending Stress Distribution Stress Distribution Diagram - Sagging

σ = Max.

σ

Tension

Compression

σ = 0

= Max.

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Max. Bending Moment “–M” A hogging moment produces the following profile of tensile stress on the top and compression stress on the bottom.

Neutral Axis

F

Tension

Compression

NEGATIVE HOGGING

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Bending Stress Distribution due to the Bending Moment The stress distribution will be the resistance to the bending moment distribution in order for the beam to stay in equilibrium. Therefore:

Hogging moment will have negative stress

(compression) at the top and positive stress (tension) at the bottom.

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Bending Stress Distribution Stress Distribution Diagram - Hogging

Tension

Compression

σ = 0

σ = Max.

σ = Max.

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Distance to the Neutral Axis “y”

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This is calculated from the beam Cross Section.

“y” will be half of the depth for double symmetrical sections, therefore both stresses are the same value but opposite directions.

Distance to the Neutral Axis “y”

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If the cross section is not double symmetrical, then “y” needs to be calculated according to the relevant axis.

i.e. we have to find the new centroid point which will result in two “y” values.

This will result with two values for the bending stress.

Distance to the Neutral Axis “y”

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Second Moment of Area

“I”

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Second Moment of Area “I” The second moment of area “I” of a beam’s cross

section area indicates the level of resistance to bending.

It is a geometrical property of a beam and depends on the position of the Neutral Axis (symmetrical or non symmetrical sections).

The larger the “I” value the less bending stress produced.

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“I” is calculated according the neutral axis position:

– If the cross section is double symmetrical (neutral axis at the centre – areas are equal) then it can be calculated using the appropriate formula for the cross sections.

Second Moment of Area (I)

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Double Symmetrical Sections

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If the cross section is not double symmetrical, then it needs to be calculated according to the relevant axis.

Single Symmetrical Sections

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Unsymmetrical Sections The calculations of “I” for the following cross sections require a more advanced treatment, unsymmetrical – will be covered next level of study.

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Find the values for “I” and “y” for the cross sections shown below:

Example 1

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Example 2 – Bending Stress Calculate the maximum bending stresses of the cross

section shown below when the max. BM given as 5kNm. Draw the stress distribution diagram. Taking the maximum allowable stress for the carbon steel

used to be 360MPa, Calculate the factor of safety for this beam.

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Example 2 – Stress Diagram

261.1MPa

334.167MPa

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F.O.S = Max. Calculated Stress

Max. Material Stress

F.O.S = 360MPa

334.167MPa

F.O.S = 1.07 is this acceptable??

Example 2 – Factor Of Safety