STATICS AND STRENGTH OF MATERIALS REVIEW: MODULE...
Transcript of STATICS AND STRENGTH OF MATERIALS REVIEW: MODULE...
STATICS AND STRENGTH OF MATERIALS REVIEW:
MODULE #1
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FINDING INTERNAL REACTIONS TO LOADING
• In statically determinate problems, the internal
reaction will be found by:1. Cutting through a body with a “section”.
2. Isolating the free body on either side of the section.
3. Applying the equilibrium equations.
• Forces shown at the cut are the internal reactions.
• If we consider those forces to be distributed over
the cross section:
– Then we have force per unit of area and call it stress.
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INTERNAL REACTIONS: THE FBD
Consider the free-body diagram of a member
involving loads and reactions in a plane
Reactions:
–Ax, Ay, and B are found from equations
of equilibrium and FBD.
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INTERNAL REACTIONS: THE FBD
• Internal reactions at the cut A–A:1.we cut the member at A-A.
2.separate the member at that cut
3.draw FBD of one side
• The axial force F, the shear force V, and the bending
moment M can be found by equations of equilibrium.
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INTERNAL REACTIONS
• Here is the right side of the same beam.
• The equal and oppositely directed forces and moment
acting on the right-hand part are shown.
• If the two parts are put back together, the internal
reactions add to zero.
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STRESS IN AN AXIALLY LOADED MEMBER
Consider a straight two-force
member of uniform cross section.
The line of action of the loads passes through
the centroid of the cross section, as shown.
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AXIAL STRESS
• The intensity of the force normal to the area is called the
normal stress.• Represented by Greek lower case letter σ (sigma).
The average value of the normal stress over the area is defined by:
Stress is called tensile stress when it stretches the material on which it acts.
It is called compressive stress when it shortens the material on which it acts.
…where P is the axial force and A is the cross-sectional area.
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AVERAGE SHEAR STRESS
• Many problems feature applied forces transmitted from one body to
another by developing internal reactions on planes parallel to the
applied force.
The pin is in “double shear”.
Two I-bars are connected by a shear pin
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AVERAGE SHEAR STRESS
• Cutting the shear pin along the planes at “A”
results in the following:
The average shear stress in this case (double shear) is:
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AVERAGE SHEAR STRESS
• In this example, two plates are joined by a rivet.
• The rivet is said to be in single shear
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BEARING STRESS
• In certain structural and mechanical problems,
one body is supported by another, as shown:
Bearing stress between two
bodies in contact can be
calculated by:
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STRESS: UNITS
• Usual U.S. customary system units for stress
are pounds per square inch, abbreviated psi. Or kip per square inch, abbreviated ksi.
• In metric, or SI units, stress is in newtons per
square meter, abbreviated as N/m2, or also
designated as a pascal (Pa).The pascal is a small unit of stress and it may be
more convenient to use the kilopascal (103 Pa) or
megapascal (106 Pa)
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STRESS
• Several other points concerning stress:
– Stress developed in an object is independent of
the material from which that object is made.
– Stress is a computed quantity related only to
internal reactions and area of an object.
– Strength is defined as the maximum allowable
stress a material can sustain.
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ALLOWABLE STRESS
• Allowable stress:
– maximum stress that is considered safe for a material
to support under certain loading conditions.
– Values are determined by tests, and experience gained
from performance of previous designs under service
conditions.
– Also sometimes called the design stress
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ALLOWABLE STRESS
• With the allowable normal stress
known, we can solve for the required
area in a compression or tension member:
σa
• With the allowable shear stress Ƭ
known, we can solve for the required
area in a shear member:
a
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FACTOR OF SAFETY
• The factor of safety is defined as: