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Samantha Ramirez, MSE

Stress

• The intensity of the internal force acting on a specific plane (area) passing through a point.

• ΔA is an infinitesimal size area with a uniform force acting on it.

ΔFΔFz

ΔFx ΔFy

ΔAF1

F2

ΔF

ΔA

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Stress Equations

• Normal Stress (σ) is the intensity of force, or force per unit area, acting normal to ΔA.

𝜎𝑧 = lim∆𝐴→0

∆𝐹𝑧∆𝐴

• Shear Stress (τ) is the intensity of force, or force per unit area, acting parallel to ΔA.

𝜏𝑧𝑥 = lim∆𝐴→0

∆𝐹𝑥∆𝐴

𝜏𝑧𝑦 = lim∆𝐴→0

∆𝐹𝑦

∆𝐴

• Subscripts• First letter is the orientation of

force normal to ΔA• Second letter is the orientation of

force causing shear

Stress Statesσx , σy , σz , τxy , τxz , τyx , τyz , τzx , τzy

τxy = τyx , τxz = τzx , τyz = τzy

σx , σy , σz , τxy , τxz , τyz

Volume Element Units

N/m2 (Pa)

1000 N/m2 (kPa)

lbf/in2 (psi)

1000 lbf/in2 (ksi)

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Saint-Venant’s Principle• The stress and strain at points in a body sufficiently away from the

region of load application will be the same as the stress and strainproduced by any applied loadings that have the same staticallyequivalent resultant.

Deformation

• Deformation is defined as the changes to a body’s shape and size when a load is applied.

• Deformation is specified by normal and shear strains and occurs because of loads and temperature.

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Average Normal Strain (ε)

• The elongation or contraction of a line segment per unit length

휀𝑎𝑣𝑔 =∆𝑠′ − ∆𝑠

∆𝑠=∆𝐿

∆𝑠

• Strain depends on geometry of formation

• Stress and strain are independent of each other

Average Shear Strain (γ)

• The change in angle between two lines that were originally perpendicular

𝛾𝑛𝑡 =𝜋

2− lim

𝐵→𝐴 𝑎𝑙𝑜𝑛𝑔 𝑛𝐶→𝐴 𝑎𝑙𝑜𝑛𝑔 𝑡

𝜃′

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Tensile Test

•P is measured using a load cell

• 𝜎𝑎𝑣𝑔 =𝑃

𝐴

•δ is elongation measured using an extensometer (optical or mechanical)

• 휀𝑎𝑣𝑔 =𝛿

𝐿𝑜

• A strain gauge can be used to measure strain directly

Stress-Strain Diagram

• The most important result from the tension test is the stress-strain diagram (σ-ε diagram).

• Nominal or engineering stress and strain are used to create a σ-ε diagram.

• 𝜎 =𝐹

𝐴

• 휀 =∆𝐿

𝐿𝑜

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Stress-Strain Diagram Properties

• Proportional Limit• The upper stress limit in the linear elastic region of a σ-ε diagram

• Elastic Limit• If the load is removed before reaching the elastic limit, the specimen will

return to its original shape• If the load passes the elastic limit, the specimen will permanently deform

• Yield Strength• The stress that causes a material to breakdown and deform permanently• When yield strengths is not well defined, the offset yield strength can be

calculated• A line parallel to the initial straight-line portion of the stress-strain curve at

0.2% strain (0.002 in/in).

• Ultimate Tensile Strength• The maximum stress the material will reach through testing• Necking occurs (visibly see permanent deformation)

• Fracture Stress• The stress when the material breaks

Hooke’s Law

• The proportional limit was observed by Robert Hooke in 1676

𝐸 =𝜎

휀• E: Young’s Modulus, the constant of proportionality

• Young’s Modulus or the modulus of elasticity is a mechanical property of a material that indicates stiffness.

•𝜎

: Slope of the straight-line portion of the stress-strain curve

• Theoretical Moduli• Steel: E=29 Msi or 200 GPa• Aluminum: E=10 Msi or 69 Gpa

• The modulus of elasticity can only be used if a materials has linear elastic behavior and is being subjected to stresses below the proportional limit.• If a material has yielded, the modulus of elasticity cannot be used.

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Poisson’s Ratio

• A deformable body subjected to a normal tensile load not only elongates but also contracts laterally.

휀𝑙𝑜𝑛𝑔 =𝛿

𝐿

휀𝑙𝑎𝑡 =𝛿′

𝑟

𝜈 = −휀𝑙𝑎𝑡휀𝑙𝑜𝑛𝑔

Shear Stress- Shear Strain Diagram

• Hooke’s Law for Shear• 𝐺 =

𝜏

𝛾

• G: Shear modulus of elasticity or modulus of rigidity

• τ: Shear stress

• γ: Shear strain

• Relation of modulus of elasticity and rigidity

• 𝐺 =𝐸

2(1+𝜈)

• For steel,• E=29000 ksi• G=11000 ksi• ν=0.32

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Uniaxial Tensile Test

• Axially Loaded Bar (Uniaxial Tensile Test)• Assumptions

• Homogeneous material

• Isotropic material

• Bar remains straight and cross-section flat

• P is applied along the centroid axis

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Average Normal Stress

න𝑑𝐹 = න𝜎𝑑𝐴

𝑃 = 𝜎𝐴𝑣𝑔𝐴

𝝈𝑨𝒗𝒈 =𝑷

𝑨

Example 1• The built-up shaft consists of a pipe AB and solid rod BC.

The pipe has an inner diameter of 20 mm and an outerdiameter of 28 mm. The rod has a diameter of 12 mm.Determine the average normal stress at D and E andrepresent the stress on a volume element located at eachof these points.

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Example 2

• If the turnbuckle is subjected to an axial force of P=900 lb, determine the average normal stress developed in section a-a and in each bolt shank at B and C. Each bolt shank has a diameter of 0.5 in.

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Average Shear Stress in Connections

• Direct Shear

• Shear Stress is the force per unit area that acts in a plane of the cross-sectional area.

𝝉𝑨𝒗𝒈 =𝑽

𝑨

Single ShearF

F

VM

N

F

F

FV

FF

F

Pin F

F

F

Situation #2

Situation #3

Situation #1

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Double Shear

Situation #5

FF

𝐹

2

Pin

F

𝐹

2

𝐹

2

𝐹

2𝐹

2

Situation #4

Example 3

• If the joint is subjected to an axial force of P=9 kN, determine theaverage shear stress developed in each of the 6-mm diameter boltsbetween the plates and the members and along each of the fourshaded shear planes.

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Factor of Safety

• Factor of Safety (FS) is a ratio of the failure load, Ffail, divided by the allowable load, Fallow. (FS>1)

• If the load applied to the member is linearly related to the stress developed within the member, then

𝐹𝑆 =𝐹𝐹𝑎𝑖𝑙𝐹𝐴𝑙𝑙𝑜𝑤

𝐹𝑆 =𝑁𝐹𝑎𝑖𝑙𝑁𝐴𝑙𝑙𝑜𝑤

=𝜎𝐹𝑎𝑖𝑙𝜎𝐴𝑙𝑙𝑜𝑤

𝐹𝑆 =𝑉𝐹𝑎𝑖𝑙𝑉𝐴𝑙𝑙𝑜𝑤

=𝜏𝐹𝑎𝑖𝑙𝜏𝐴𝑙𝑙𝑜𝑤

Example 4

• The pin is made of a material having a failure shear stress of 100 MPa.Determine the minimum required diameter of the pin to the nearestmm. Apply a factor of safety of 2.5 against shear failure.

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Example 5

• Determine the maximum vertical force P that can be applied to thebell crank so that the average normal stress developed in the 10 mmdiameter rod, CD, and the average shear stress developed in the 6mm diameter double sheared pin B not exceed 175 MPa and 75 MPa,respectively.

Example 6

• To the nearest 1/16”, determine the required thickness of member BC and the diameter of the pins at A and B if the allowable normal stress for member BC is σAllow=29 ksi and the allowable shear stress for the pins is τAllow=10 ksi.

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Normal Stress Concentrations

• Holes and sharp transitions at a cross section will create stress concentrations.

• In engineering practice, only the maximum stress at these sections must be known.

• The maximum normal stress occurs at the smallest cross-sectional area.

Stress Concentration Factor

• A stress concentration factor “K” is used to determine the maximum stress at sections where the cross-sectional area changes.

𝐾 =𝜎𝑚𝑎𝑥

𝜎𝑎𝑣𝑔

• K is found from graphs in handbooks of stress analysis

• σavg is the average normal stress at the smallest cross section

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Stress Concentration Factor Tables

Example 7

• Determine the maximum normal stress developed in the bar when it is subjected to a tension of P=2 kip.

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Thin-Walled Pressure Vessels• Cylindrical or spherical vessels are commonly used in industry to

serve boilers or tanks.

• A “thin wall” refers to a vessel having an inner radius to wall

thickness ratio of 10 or more𝑟

𝑡≥ 10 .

• We will assume a uniform or constant stress distributionthroughout the thickness because it is thin.

• Pressure vessels are subjected to loadings in all directions.

Cylindrical Vessels

Circumferential or Hoop Stress (𝜎1) Longitudinal or Axial Stress (𝜎2)

𝜎1 =𝑝𝑟

𝑡𝜎2 =

𝑝𝑟

2𝑡

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Spherical Vessels

• σ1,σ2: the normal stress in the hoopand longitudinal directions,respectively.• Each is assumed to be constant

throughout the wall of the cylinder, andeach subjects the material to tension.

• p: the internal gauge pressuredeveloped by the contained gas

• r: the inner radius of the cylinder

• t: the wall thickness (r/t≥10)

𝜎2 =𝑝𝑟

2𝑡

Example 8

• The tank of the air compressor is subjected to an internalpressure of 90 psi. If the internal diameter of the tank is 22 in,and the wall thickness is 0.25 in, determine the stresscomponents acting at point A. Draw a volume element of thematerial at this point, and show the results on the element.

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Example 9

• The A-36 steel band is 2 in wide and is secured around the smoothrigid cylinder. If the bolts are tightened so that the tension in themis 400 lb, determine the normal stress in the band, the pressureexerted on the cylinder, and the distance half the band stretches.

Example 10

• A pressure-vessel head is fabricated by gluing the circular plate tothe end of the vessel as shown. If the vessel sustains an internalpressure of 450 kPa, determine the average shear stress in theglue and the state of stress in the wall of the vessel.