ME302: Materials MechanicsFall 2015Lecture 6

Chap. 3 Mechanical Properties of Materials (3) Chap. 4 Axial Load (1)

Jaeyun Moon, Ph.D.Assistant Professor

Department of Mechanical Engineering

Creep and Fatigue (3.8)

(1) Creep : a time-dependent permanent deformation at elevated temperature and constant stress

Æ the rate of creep is determined by temperature and constant stress

• How to determine Creep Strength� the maximum stress that the material can

withstand during a specified time without exceeding an allowable creep strain.

� Affected by temperature, duration of loading…

� Allowable creep strain- 0.1%/year for steel in bolts and piping

• σ-t diagram� A curve of stress vs. time (up to 1000hrs)

Creep strength σc=20ksi

If a metal bar is loaded to just below its yield strength at room temperature.. - no plastic deformation

If a metal bar is loaded to just below its yield strength at 1500oF.. - a small amount of deformation will occur at first (Stage I)- the metal bar start to stretch very slowly (Stage II)- it will finally break into two after certain time period (Stage III)

� Creep Failure & Collapse of World Trade Center

Bazant, Z. P. & Zhou, Y. Why did the World Trade Center collapse? - Simple analysis. J Eng Mech-Asce 128, 2-6 (2002).http://en.wikipedia.org/wiki/Collapse_of_the_World_Trade_Center

A destructive fire caused by the aircraft fuel spilled into the structure

→ the steel of the columns to be exposed to sustained temperature >800oC

→ A loss of the protective thermal insulation of steel during the initial blast

→ A decrease of yield strength of structural steel and significant deformation(creep)

→ Creep buckling of columns

What happened for 60-100mins after planes hit the buildings??

(2) Fatigue : Repeated cycles of stress or strain cause structure’s fracture.

• Fatigue failure• With cyclic loading, fracture will occur at a stress that is less

the material’s yield stress.• Brittle behavior

Bridge

EngineCrankshaftFatigue fracture surface of the broken

stub of the secondary shaft

• Minute cracks Æ stress concentration at tip region Æ crack propagation Æ failure

� S-N diagram (stress-cycle diagram)

• Endurance or Fatigue Limit: a limiting stress below which no failure(or cracks) occurs after applying a load for a specified number of cycles.

• Steel and titanium alloy have a distinct limitex) for steel, (Sel)st = 27ksi

• Aluminum and copper do not have a distinct limit

Î Choose a strength at 500 million (x106) cyclesEx) for aluminum, (Sel)al = 19ksi

Average Properties of Common Engineering Materials

Example 3.5A specimen of titanium alloy is tested in torsion and the shear stress-strain diagram is shown in Fig. 3-25a. Determine the shear modulus G, the proportional limit, and the ultimate shear stress. Also, determine the maximum distance d that the top of a block of this materials, shown in Fig. 3-25b, could be displaced horizontally if the materials behaves elastically when acted upon by a shear force V. What is the magnitude of V necessary to cause this displacement?

Example 3.6An aluminum specimen shown in Fig. 3-26 has a diameter of 𝑑0=25mm and a gauge length of 𝐿0=250mm. If a force of 165kN elongates the gauge length 1.20mm, determine the modulus of elasticity. Also, determine by how much the force causes the diameter of the specimen to contract. Take 𝐺𝑎𝑙=26GPa and 𝜎𝑌=440MPa.

Summary : Chap 1 & Chap 2

(1) Stress: Normal stress, Shear stressσ : Normal stress

τ : Shear stress• Normal stress • Shear stress

(2) Strain: Normal strain, Shear strain• Normal strain • Shear strain

l’ : the new lengthl : the original length

θ : the new angle

• Small angle approximation

Summary : Chap 3

• Stress-Strain Diagram for ductile materials

• σpl: the proportional limit stress• σY: the yield stress or yield point• σu: the ultimate stress

= the maximum stress • σf: the fracture stress

𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 = 𝐿𝑓 − 𝐿0𝐿0(100%)

𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎𝑟𝑒𝑎 = 𝐴0 − 𝐴𝑓𝐴0(100%)

• Ductility

𝜖𝑓 =0.38 38%For a mild steel,

60%For a mild steel,

For a mild steel,

• Brittle Materials

� Little or no yielding before failure.

• Hooke’s Law

Linear relationshipbetween 𝜎 − 𝜖

E: the modulus of elasticity Young’s modulus𝜖: strain𝜎: stress

(1) (2)

(3)

(4)

• Stain Hardening

After strain hardening • Higher yield point AÆA’• A greater elastic region (1)Æ(2)• A smaller plastic region (3)Æ(4)

Larger percent of cold work causes that…• strength increases• ductility decreases

• Strain Energy : the external work is stored in the material as internal energy

Strain-energy density

Strain-energy density(linear elastic materials)

𝜎 = 𝑬𝜖E: Young’s modulusσ: stress

• Modulus of Resilience: the strain-energy density, when the stress reaches the proportional limit.

• Modulus of Toughness: the entire area under the stress-strain diagram = the maximum amount of strain energy that the material can absorb before it fractures

1) The modulus of elasticity E is a measure of the linear relationship between stress and strain. The common unit is:

a) kN/mm2

b) MPa

c) GPa

d) All of them

Pa=N/m2

𝑘𝑁𝑚𝑚2 =

1000𝑁10−6𝑚2 = 1𝐺 𝑁/𝑚

2 = 1𝐺𝑃𝑎(a)

2) An aluminum rod, shown in (a), has a circular cross section and is subjected to an axial load of 10kN. If a portion of the stress-strain diagram is shown in (b), determine the elastic behavior or plastic behavior for section AB and section BC.

Section AB Section BC(a) plastic plastic(b) elastic elastic(c) elastic plastic(d) plastic elastic

𝐴𝐴𝐵 = 0.25 × 𝜋 × 20𝑚𝑚 2 = 314.6𝑚𝑚2𝐴𝐵𝐶 = 0.25 × 𝜋 × 15𝑚𝑚 2 = 176.71𝑚𝑚2

𝜎𝐴𝐵 =𝑃𝐴𝐴𝐵= 10𝑘𝑁314.6𝑚𝑚2 = 31.83𝑀𝑃𝑎

Cross sectional area

Stress

𝜎𝐵𝐶 =𝑃𝐴𝐵𝐶= 10𝑘𝑁176.71𝑚𝑚2 = 56.59𝑀𝑃𝑎

< 40MPa, elastic region > 40MPa, plastic region

Chap4. Axial Load

9 Determine the elastic deformation of axially loaded member9 Apply the principle of superposition for total effect of different loading cases9 Deal with compatibility conditions9 Use ‘force method’ of analysis.

• Chapter Objectives

Saint-Venant’s Principle (4.1)• Saint-Venant’s Principle

The stresses and strains in a body at points that are sufficiently remote from points of application of load depends only on the static resultant of the loads and not on the distribution of loads. .

[Image courtesy] http://coefs.uncc.edu/mwhelan3/files/2010/10/ICD_Saint_Venant1.pdf

(1) Stress concentration at point load

(2) Stress concentration around hole

(3) Stress concentration around abrupt change in cross-section

section a-a

section b-b

section c-c

• Consider the rectangular shaped bar which deforms elastically.

• Localized deformation that occurs at each end tends to even out and become uniformthroughout the midsection of the bar.

• The minimum distance from the bar’s end to sufficiently remove the localized deformation should be at least be equal to the largest dimension of the loaded cross section. (the width, not the thickness)

Î Saint-Venant’s Principle� sections a-a, b-b : the localized deformation� section c-c: the uniform deformation

Elastic Deformation of an Axially Loaded Member(4.2)

• How to determine the elastic displacement(deformation or elongation) of a member subjected to axial loads.

𝛿=displacement of barL=original length of barP=internal axial forceA=constant cross-sectional area of the barE=modulus of elasticity for the material

𝜹 = 𝑷𝑳𝑨𝑬

𝛿 = 𝑃𝐿𝐴𝐸

� For a general case,

𝜎 = 𝑃(𝑥)𝐴(𝑥) 𝜖 = 𝑑𝛿𝑑𝑥the elongation

the original length

𝜎 = 𝐸(𝑥)𝜖

the stress-strain is in the elastic region: the Hooke’s law

𝑃(𝑥)𝐴(𝑥) = 𝐸(𝑥)

𝑑𝛿𝑑𝑥

∙ ∙ ∙ ∙ ∙ ① ∙ ∙ ∙ ∙ ∙ ②

Let’s plug ① ② into ③∙ ∙ ∙ ∙ ∙ ③

𝒅𝜹 = 𝑷(𝒙)𝒅𝒙𝑨(𝒙)𝑬(𝒙)

� Derivation 𝜹 = 𝑷𝑳𝑨𝑬

𝛿 = 0

𝐿 𝑃(𝑥)𝑑𝑥𝐴(𝑥)𝐸(𝑥)

𝛿=displacement of one point on the bar relative to the other pointL=original length of barP(x)=internal axial force at the section, located a distance x from one endA(x)=cross-sectional area of the bar expressed as a function of xE(x)=modulus of elasticity for the material expressed as a function of x

• For a Constant Load and Cross-sectional Area case,

𝛿 = 𝑃𝐿𝐴𝐸𝛿=displacement of barL=original length of barP=internal axial forceA=constant cross-sectional area of the barE=modulus of elasticity for the material

𝛿 = 𝑃𝐿𝐴𝐸

• For a bar which is subjected to several different axial forces,

• Sign Convention� Tension/elongation: + (positive)� Compression/contraction: -(negative)

• Example

Normal force diagram

𝛿𝐴/𝐷 = 𝑃𝐿𝐴𝐸 =(5𝑘𝑁)𝐿𝐴𝐵𝐴𝐸 + (−3𝑘𝑁)𝐿𝐵𝐶𝐴𝐸 + (−7𝑘𝑁)𝐿𝐶𝐷𝐴𝐸𝛿 = 𝑃𝐿𝐴𝐸

: the summation of the relative displacements of the ends of each segment

Example 4.1 The assembly shown in Fig. 4-6a consists of an aluminum tube AB having a cross-sectional area of 400mm2. A steel rod having a diameter of 10mm is attached to a rigid collar and passes through the tube. If a tensile load of 80kN is applied to the rod, determine the displacement of the end C of the rod. Take Est=200GPa, Eal=70GPa.