mechanics of materials chp7
TRANSCRIPT
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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.
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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
12
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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