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: