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MECH 5312 – Solid Mechanics II Dr. Calvin M. Stewart Department of Mechanical Engineering The University of Texas at El Paso

MECH 5312 Solid Mechanics II - me.utep.edume.utep.edu/cmstewart/documents/ME5312/Lecture 1 - Review of...Department of Mechanical Engineering ... •Fracture Mechanics •Fatigue •Creep

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MECH 5312 – Solid Mechanics IIDr. Calvin M. Stewart

Department of Mechanical Engineering

The University of Texas at El Paso

Course Topics

• Review of Elementary Concepts

• Stress, Stress Components, and Equilibrium

• Strain-Displacement Equations, Compatibility

• Linear Stress-Strain Temperature Relations

• Inelastic Material Behavior

• Stress Concentrations

• Fracture Mechanics

• Fatigue

• Creep

Course Objectives

• To educate students on the three-dimensional mechanics of materials. By the end of the course, students will have developed the mathematical skills necessary to solve advanced mechanics of materials problems beyond the scope of typical undergraduate knowledge.

Definition of Mechanics

• Mechanics – the study of equilibrium of matter and forces that cause such motion or equilibrium.

• Variables include ….

Time – Space – Force – Energy – Matter

Definition of MechanicsMotion and Equilibrium of Matter

Important Equations in Mechanics

Conservation of Mass

Conservation of Linear and Angular

Moment (Equilibrium)

Compatibility

Conservation of Energy

(Thermodynamics)

Constitutive Equations

(Mechanical Behavior of Material)

Gas Turbine Blades

Review of Elementary Mechanics of Materials

Axially Loaded Members

• Axial Stress

• Elongation

• Axial strain

2PF L

A

PL

e LAE

e P

L AE E

Torsionally Loaded Members

• Shear Stress

• Rotation

• Shear Strain

2TF L

J

TL

radGJ

L G

Bending of Beams

• Normal Bending Stress

• Displacement v in the y direction

• Transverse Shear Stress

2M x y

F LI

2

2

M xv

x EI

2V x Q

F LIb

1 1

2 2

2 2 3

1

1 4

8

y a a

y y y

Q y dA yb dy b a y L

Examples

Example 1

• The uniform A-36 Steel bar has a diameter of 50 mm and is subjected to the loading shown. Determine the displacement at D, and the displacement of point B relative to C.

Example 2

• The 1.5in diameter shaft shown is supported by two bearings and is subject to three torques. Determine the shear stress developed at points A and B located at section a-a of the shaft.

• Point A: r=0.75in

• Point B: r= 0.15in

Example 3

• The simply support beam has the cross-sectional area shown. Determine the absolute maximum bending stress in the beam and draw the stress distribution over the cross section at this location. Also, what is the stress at point B?

Example 4

• The beam shown is made of two boards. Determine the maximum shear stress in the glue necessary to hold the boards together along the seam where they are joined.

Stress-Strain Relations

Engineering Stress and Strain

0

P

A

L e

L L

Engineering Stress-Strain Curve

True Stress and Strain

t

t

P

A

tL L e

tt

t

dLd

L

ln ln 1tL

tt t

L

Ld

L

Material Properties

• Elastic Limit

• Proportional Limit

• Yield Strength

• Ultimate Tensile Strength

• Modulus of Elasticity

• Percent Elongation

Material Properties

21

2R

YU

E

Ignore Upper Yield Point, YU

Use Lower Yield Point, YL Y

Total Strain EnergyTU

• Yield Point for Structural Steel

• Modulus of Resilience

• Modulus of Toughness

Material Properties

• Poisson’s Ratio

• Necking of a Mild Steel Tension Specimen

1 0.5l

a

Material Properties

• True-Stress Strain Curve

Limits on Design

Limits on Design

• Historically, limits on the design of a system have been established using Factor of Safety

where Rn is the nominal resistance (the critical parameter associated with failure, usually a material property)

where Rw is the safe working magnitude of the same parameter (usually calculated from the design)

n

w

RSF

R w

f

n

NSF

Nw

c

n

tSF

t

Static Creep Fatigue

Safety Factors in Application

• Industrial Applications SF range from 1.0 to 3.0

• In aircraft and space vehicle design, where it is critical to reduce the weight of the vehicle as much as possible, the SF may be nearly 1.0

• In the nuclear reactor industry, where safety is of prime importance in the face of many unpredictable effects, SFmay be as high as 5.0

Design Inequality

• When several different load might be active in a design at the same time, a design inequality is necessary

where each Qi represents the effect of a particular working (or service-level) load, such as internal pressure or temperature change, N denotes the number of load types considered.

Nn

i

i

RQ

SF

Static

Limit-States Design

• It has been recognized that a single SF is inadequate to account for all the unknowns in design.

• As an alternative, we can introduce the Limit-States Design inequality

where γi are the load factors for load effects Qi and φ is the resistance factor for the nominal resistance Rn.

N

i i n

i

Q R

Static

Example

Example 5

• A steel rod is used as a tension brace in a structure. The structure is subject to dead load (the load from the structure itself), live load ( the load from the structure’s content), and wind load.

• The effect of each of the individual loads on the tension brace is D=25kN, L=60kN, W=30kN.

• Select a circular rod of appropriate size to carry these loads safely. Use steel with a yield strength of 250MPa.

• Make the selection using (a) factor-of-safety design and (b) limit-states design

Modes of Failure

Modes of Failure

• Failure by excessive deflection• Elastic deflection

• Deflection caused by creep

• Failure by general yielding

• Failure by fracture• Sudden failure of brittle materials

• Fracture of cracked or flawed members

• Progressive fracture (fatigue)

• Failure by instability The DeHavilland Comet Crash – Progressive Fracture

Calvin M. StewartAssistant ProfessorDepartment of Mechanical EngineeringThe University of Texas at El Paso500 W. University Blvd.Suite A126El Paso, Texas 79968-0717

Email: [email protected]: http://me.utep.edu/cmstewart/Phone: 915-747-6179Fax: 915-747-5019

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