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Chapter 7-Bending

Members that are slender and support loadings that are applied perpendicular to their longitudinal axis are called beams.

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V: Shear M: Moment

Beam sign convention

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;0yF

V-w(x)Δx-(V+ΔV) = 0

ΔV=-w(x)Δx

;0oM+

0<k<1 -VΔx-M+w(x)Δx[k(Δx)]+(M+ΔM)=0

ΔM=VΔx-w(x)k(Δx)2.

Dividing by Δx.

Graphical method for constructing shear and moment diagrams

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xwdx

dV

Slope of

the shear

diagram at

each point.

=

- Distributed

load intensity

at each point.

Vdx

dM

Slope of

moment

diagram at

each point.

=

Shear

at each

point.

If V(x) is linear, M(x) is parabolic.

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Solution

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Bending deformation of a straight member:

Flexural formula:

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Where:

σmax = maximum normal stress in the member, which occurs at a point on the cross

sectional area farthest away from the neutral axis.

M = resultant internal moment, determined from the method of sections and the

equations of equilibrium, and computed about the neutral axis of the cross section.

I = moment of inertia of the cross-sectional area computed about the neutral axis.

c = the perpendicular distance from the neutral axis to a point farthest away from

the neutral axis, where σmax acts.

I

yM In general:

I

Mcmax dAy

cM 2max

I: Moment of Inertia

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Solution I

Mcmax

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Unsymmetric bending:

Where the moment is applied along the

principle axis of a non-symmetrical beam

we need to find out where the PRINCIPLE

AXES OF INERTIA are so that the following three conditions are met:

Moment applied along principle axis

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max

c

y

Product of inertia:

This “product of inertia” will be zero for the correct selection of the Principle Axes of

Inertia. These can be found using a few techniques that we will look into later.

However, it is important to note that if a shape has an axis of symmetry then the

Principle axes will always be located on the axis of symmetry and perpendicular to it.

This may arise as a result of poor design or complex design circumstances. A

bending moment is applied about an axis that is not a principle axes of the cross-

section.

Moment Arbitrarily Applied

A bending moment is applied at Theta about an axis that is not a principle axes of the

cross-section.

To get the internal resultant normal stress one must:

1- resolve the applied moment into components about the

principle axes.

2- Then use the flexure formula to get normal stresses.

3- Then use superposition to get resultant.

+ =

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The resultant internal

stress at any point on

the cross section is

thus found as:

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Orientation of Neutral Axis For unsymmetrical bending the angle theta (the direction of the applied moment) is

not equal to the angle alpha (the inclination of the neutral axis)

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Composite beams Beams made from more than one type of material are called composite beams.

Stiff=rigid

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So we define the Transformation Factor as the ratio of the two stiffnesses

So for the analysis of a composite beam one must transform one of the materials

into an “equivalent width” of the other material. Less stiff materials will be

transformed into thinner sections of the stiffer and vice versa

The stress in the transformed beam is same for the original beam, however, the

stress in the transformed material has to be multiplied by the n factor to get the true

stresses in this area,

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Reinforced Concrete Beams Concrete is very susceptible to cracking when in tension and therefore is not

suitable for use as a beam, UNLESS, steel is placed where tensile resistance is

required.

Because tensile resistance of concrete is unpredictable and often very low it is

assumed to be ZERO for the sake of design and steel alone resists tension.

Typical Practice: • Reinforcement is placed far away from the neutral axis so that a larger moment can be resisted. • Reinforcement must be covered by some concrete for fire protection, erosion coverage and other.

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So, stress analysis requires locating the neutral axis by transforming the steel into

an equivalent concrete area using:

We can obtain h’ (distance from the top of

the beam to the neutral axis) using the fact

that the centroid C of the cross-sectional

area of the transformed section lies on the

neutral axis..

With reference to the neutral axis, therefore, the moment of the two areas, must

be zero: Ay~

0'2

''

hdnA

hbh st

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Solution:

Cos

Sin

Internal moment components:

mNM z .48052013

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mNM y .20052013

5

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Section properties:

400mm

20mm

180mm

20mm

yB

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400mm

20mm

180mm

20mm

Using the parallel-axis theorem of Appendix A, I = I + A.d2

A d : distance from C to G.C

90mm

90mm C

B

200mm -57.4mm-90mm= 52.6mm

? ?

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Coordinate of point A from point C

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θ

From the figure : θ’ = tan-1 (5/12) = 22.62 degree

Angle θ : See slide 19.

θ = 180-θ’

θ = 180-22.62=157.38 degree

θ = -22.62 degree

θ’

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