Mechanics of Materials

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14 January 2011
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Mechanics of Materials
CIVL 3322 / MECH 3322

Statics Review
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A Quiz
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Stress and Internal Forces
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¢

In Statics, we spent most of our time
looking at reactions at supports
¢

Two variations from this were when
we considered the forces in trusses
and
¢

The shear and moment at a point in a
beam (you may or may not have
looked at this)
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¢

In Mechanics of Materials, it is these
internal forces and how they affect the
behavior of the system that will be
critical to our analysis

¢

Depending on the loading pattern,
there can vary greatly through the
beam
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Stress and Internal Forces
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¢

At any point in a continuous system,
we can look at the internal forces
acting on the system by modeling the
point we are looking at in the system
as a fixed end support.
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¢

This means that at any point in a
continuous member, we will have a
force parallel to the member, a force
perpendicular to the member and a
resistive moment.
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Stress and Internal Forces
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¢

The force that acts parallel to the cut
we make to separate the member is
known as the axial force or the normal
force
¢

The force that acts perpendicular to
the cut is known as the shear force
¢

The moment is known as the bending
moment
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Stress and Internal Forces
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¢

The dark blue arrows along the axis of
the beam are the axial forces
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¢

The axial force is considered as a
positive axial force if it would cause
the section under consideration to be
in tension.
¢

The axial force is considered as a
negative axial force if it would cause
the section under consideration to be
in compression.

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Stress and Internal Forces
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¢

This shear force is always normal to
the axis of the cut face of the beam
and is typically labeled as V.
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¢

The force acting on the plane of the
cut generates a stress on that plane
¢

If the force is normal to the plane, the
stress is a normal stress or an axial
stress
¢

If the force is parallel to the plane, the
stress is a shear stress
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Stress and Internal Forces
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¢

The magnitude of the stress is the
ratio of the magnitude of the applied
force to the area of the cross section
¢

For normal stress, the expression is
σ
av
g

F
A
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¢

This expression assumes that the
force is equally distributed across the
cut face
¢

To determine the axial stress at a
point required an ability to divide the
force according to some function
σ
av
g

F
A
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Stress and Internal Forces
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¢

Units of stress are similar to units of
pressure
l

Lb
f
/ft
2
l

N/m
2


A N/m
2
is given the unit of Pa (Pascal)
σ
av
g

F
A
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¢

Pressure is a scalar, stress is a vector
so they are not exactly the same
¢

Usually you don’t see Pa by itself, the
units are
KPa
,
MPa
,
GPa

σ
av
g

F
A
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¢

Pressure is a scalar, stress is a vector so
they are not exactly the same
¢

Usually you don’t see Pa by itself, the units
are
KPa
,
MPa
,
GPa

E01
A solid 0.5 in diameter steel hanger
rod is used to hold up one end of a
walkway support beam. The force
carried by the rod is 5,000 lb.
Determine the normal stress in the
rod. (Disregard the weight of the rod).
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Rigid bar ABC is supported by a pin at A and axial member (1), which has a cross
section of 540 mm2. The weight of the rigid bad ABC can be neglected.
(a)

Determine the normal stress in member (1) is a load P=8kN is applied at C.
(b)

If the maximum normal stress in member (1) must be limited to 50
Mpa
, what is
the maximum load magnitude P that may be applied to the rigid bar at C?
A 50-mm-wide steel bar has axial loads applied at points B, C, and D. If the normal
stress magnitude in the bar must not exceed 60
MPa
, determine the minimum
thickness that can be used for the bar.