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7/24/2019 1.1 - Introduction - Simple Stresses
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MEC 103MECHANICS OF DEFORMABLE BODIES
INTRODUCTION
ENGR. ROGELIO FRETTEN C. DELA CRUZ, CEINSTRUCTOR
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ENGINEERING MECHANICS:
1. Statics
2. Dynamics
We considered only the external effect of forces acting
on a body.
The bodies are assumed perfectly rigid
(no deformation).
STRENGTH OF MATERIALS:
Internal effects of the forces on the body will beconsidered.
Deformations will be of great importance.
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The difference between rigid-body mechanics
and mechanics of materials can be appreciated if we
consider the bar shown in Fig. 1.1.
In mechanics of materials, the statics solution is
extended to include analysis of the forces acting
inside the bar to be certain that the bar will neither
break nor deform excessively
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ANALYSIS OF INTERNAL FORCES
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It is convenient to represent both R and CR in terms of two
components: one perpendicular to the cross-section and the
other lying in the cross-section. These components are given
physically meaningful names.
P - the component of the resultant
force that is perpendicular to the
cross-section, tending to elongateor shorten the bar. It is called the
normal force or axial force.
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V - the component of the
resultant force lying in theplane of the cross-section,
tending to shear (slide) one
segment of the bar relative to
the other segment. It is calledthe shear force.
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T - the component of theresultant couple that tends
to twist (rotate) the bar. It is
called the twistingmoment
ortorque.
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M - the component of theresultant couple that
tends to bend the bar. It
is called the bending
moment.
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SIMPLE STRESSES
Stress is known as the intensity ofload per unit area.
Stress is also a measure of the unit
strength of a material.
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Three types of simple stress:
1. Normal Stress
2. Shear ing Stress3. Bearing Stress.
SIMPLE STRESSES
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NORMAL STRESS
The resisting area is perpendicular
to the applied force, thus normal.
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NORMAL STRESS
Two types of normal stress:
1. Tensile stress2. Compressive stress
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NORMAL STRESS
The normal stress acting at any point
on a cross-section is given by theformula:
Where:
= Normal Stress
P = Axial force
A = Cross-sectional Area
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NORMAL STRESS
The normal stress acting at any point
on a cross-section is given by theformula:
Units of stress:
22
22
in
kipksi;
in
lbpsi
MPa1mm
N1;Pa1
m
N1
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For water at 4C
= 62.4 lb/ft3
= 9.81 kN/m3
= 1000kg/m3
Acceleration due to influence of gravity:
g= 32.2 ft/sec2
g = 9.81 m/sec2
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Illustrative Problems
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The compound bar ABCD consis ts of three segments,
each of a dif ferent mater ia l w ith dif ferent d imension s.
Compu te the stress in each segment when the axia lloads are app lied.
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Determine the largest weight W that can be
supported by the two wires AB and AC. The
working stresses are 100 MPa forABand 150MPa for AC. The cross-sectional areas of AB
and AC are 400 mm2 and 200 mm2,
respectively.
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The 1000-kg uniform barABis suspended from two cables
ACand BDeach with cross-sectional area 400 mm2. Find
the magnitude P and location x of the largest additional
vertical force that can be applied to the bar. The stresses in
AC and BD are limited to 100 MPa and 50 MPa,
respectively.
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Determine the largest weight W that can be
supported safely by the structure shown in the
figure. The working stresses are 16,000 psi for thesteel cable AB and 720 psi for the wood strut BC.
Neglect the weight of the structure.
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The homogeneous 120-N sign is suspended from
a ball and socket joint at Oand cablesADand
BC
. Determine the tensile stresses in the cables ifeach cable has a cross-sectional area of 10 mm2.
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The wood pole is supported by two cables of 1/4-in.
diameter. The turn buckles in the cables are tightened
until the stress in the cables reaches 60,000 psi. If the
working compressive stress for wood is 200 psi,
determine the smallest permissible diameter of the
pole.
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END
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